NCERT Solutions Class 12 Mathematics Chapter 11

NCERT Solutions Class 12 Mathematics Chapter 11 Three-Dimensional Geometry

Class 12 Mathematics syllabus has a total of 13 essential chapters. Chapter 11 is on Three-Dimensional Geometry, which is a continuation of the chapter: Vector Algebra. The important topics covered are the distance between two points, section formulae and coordinates of a point in space. Geometry guides you in deciding how to design houses and buildings in different geometric shapes in the real world. NCERT Solutions Class 12 Mathematics Chapter 11 helps students find answers to the questions necessary for final exams. This chapter makes the process of learning Three-Dimensional Geometry simpler. This would be helpful for you in your daily practice sessions and while completing homework assignments. The step-by-step walkthrough of the solution is self-explanatory.

NCERT Solutions Class 12 Mathematics Chapter 11 assists students:

• In studies.
• Increases competency levels.

The solutions provided come in handy for completing the Classwork and preparing for the exams. This focuses on:

• 3D shapes involving three coordinates, namely X, Y, & Z coordinates.
• The shapes that absorb space are called 3D shapes.
• They are rigid shapes having three dimensions: length, width, and height.

Three-Dimensional Geometry plays an essential role in exams as multiple questions are included from this chapter in the assessment. NCERT Solutions Class 12 helps students solve complex Three-Dimensional Theories in more straightforward ways. This gives students a solid introduction to Three-Dimensional Mathematics while studying geometry in three different directions.

Key Topics Covered In NCERT Class 12 Mathematics Chapter 11

This is a practical lesson that provides better suggestions for implementing formulas. In this account, we will explore different formulas, examples, and theories thoroughly to help students build a solid Mathematical foundation. NCERT Solutions Class 12 Mathematics Chapter 11 applies many real-life examples to deliver students a better understanding of the subject.

Studying Geometry provides many basic skills and builds:

• logical thinking.
• Analytical reasoning.
• Problem-Solving Skills.

The main topics covered NCERT Solutions for Class 12 Mathematics Chapter 11 - Three-Dimensional Geometry are

 Serial No Topic 1 Introduction 2 Coordinate Axes and Coordinate Planes in Three-Dimensional Space 3 Coordinates of a Point in Space 4 Distance Between Two Points 5 Section Formula

1. Introduction

The theory covers the notes prepared as per the CBSE syllabus (2022–2023) and the NCERT curriculum. Prime topics covered in NCERT Solutions Class 12 Mathematics Chapter 11 are Direction Cosine and Direction Ratios of a line joining two points. You can also consider the equation of lines and planes in space under different conditions, the angle between line and plane, space between two lines etc. NCERT Solutions Class 12 Mathematics Chapter 11 will help you practice the formulas thoroughly to know the subject well. This topic has been introduced in Mathematics to find different shapes and figures. In the real world, all the objects are in a three-dimensional form. For example, household objects like tables, kitchen utensils, beds etc., have 3D geometry. You would learn the latest advanced version of 3D geometry in the 12th standard.

1. Coordinate Axes and Coordinate Planes in Three-Dimensional Space

The three coordinate axes of a rectangular Cartesian coordinate system are the x-axis, y-axis, and z-axis. These lines are three mutually perpendicular lines. NCERT Solutions Class 12 Mathematics Chapter 11 further explains that the Three-Dimensional Geometry shape has a face, an edge, and a vertex. The size of these forms has been decided by the area they occupy. Space exists in 2D structures, whereas surface space exists in three-dimensional shapes. The opening of all the faces of a three-dimensional form is the surface area, cube, cuboid, cone, and Cylinder shapes that can be seen around us. A coordinate plane is formed by the pair of axes of the three-dimensional coordinate system x, y, and z. NCERT Solutions Class 12 Mathematics Chapter 11 explains this coordination in detail.

1. Coordinates of a Point in Space

NCERT Solutions Class 12 Mathematics Chapter 11 explains a three-dimensional space called 3-space or tri-dimensional space. The point in the space of a geometric setting that contains three values, x, y and z, is required to determine the position of an element. This performs as a three-parameter physical world model where tangible matter exists. These three values are selected from the term's width, height, depth, and length. Just as two-dimensional space consists of many boundless lines, three–dimensional space contains infinitely many planes. These planes are of a particular value. For example, the XY plane consists of the x-axis and y-axis, the YZ plane, which consists of y-axis and z-axis and the XZ plane, which holds the x-axis n z-axis. NCERT Solutions Class 12 Mathematics Chapter 11 provides many examples for a more straightforward understanding of the above-mentioned points.

1. Distance Between Two Points

Distance between two points is the length of the line segment that connects the two given points. NCERT Solutions Class 12 Mathematics Chapter 11 preferably uses a formula in Mathematics to calculate the distance between the two points in the coordinate plane. By replacing those points in the formula, we can get the distance between two points. To find out the position of a point in a plane, we require a pair of the coordinate axis. The distance of the point through the x-axis from the Centre is called the x-coordinate. The distance of the point across the y-axis from the starting point is called y-coordinate. The ordered point x and y represents the coordinate of the point.

1. Section Formula

The section formula has been applied to find out the coordinates of a point. The coordinates of a point in the three–dimensional space are used to locate the point given in the system. This can decide the point that divides a line internally into a specific ratio. Section formula could be divided into internal division formula and external division formula. NCERT Solutions Class 12 Mathematics Chapter 11 also focuses on exceptional cases of section formulas like the midpoint formula and trisection points. Section formulas in three-dimensional can be used to gain more valuable results in

• Coordinates of the centroid of a triangle.
• Collinearity of three points.

NCERT Solutions Class 12 Mathematics Chapter 11 provides detailed clarity of internal and external section points.

The exercise and solutions paper are available on the Extramarks website. These questions will clear all doubts, help students understand the concept better, and quickly solve the Mathematical problems. NCERT Solutions Class 12 Mathematics Chapter 11 includes questions according to the latest CBSE syllabus of 2022-23.

Exercise 11.2 Coordinate Axes and Coordinate Planes in Three-Dimensional Space - Questions & Solutions refer to the below link.

Exercise 11.3 Coordinates of a Point in Space - Questions & Solutions click on the below link.

Exercise 11.4 Distance Between Two Points - Questions & Solutions click here to know more.

Exercise 11.5 Section Formula - Questions & Solutions refer to the link below.

Along with Class 12 Mathematics solutions, you can explore NCERT Solutions on our Extramarks website for all primary and secondary Classes. NCERT Solutions Class 1, NCERT Solutions Class 2, NCERT Solutions Class 3, NCERT Solutions Class 4, NCERT Solutions Class 5, NCERT Solutions Class 6, NCERT Solutions Class 7, NCERT Solutions Class 8, NCERT Solutions Class 9, NCERT Solutions Class 10, NCERT Solutions Class 11, NCERT Solutions Class 12.

NCERT Exemplar for Class 12 Mathematics

NCERT Exemplars are practice books that include additional higher-level questions and are meant to aid meticulous learning. They are mainly used for competitive exams. These topics are up to date, consisting of different topics explained in every chapter of the Class 12 Mathematics book. Students can click on the links provided by Extramarks for these Exemplar questions and answers for Class 12 Mathematics chapter wise notes.

Considering the standard and variety, practising Exemplar problems is essential for school boards and competitive examinations. Students can perfectly evaluate themselves by working on these problems and altering their problem-solving skills for future examinations. NCERT Solutions Class 12 Mathematics provides a series of exemplar solutions for Class 12 for detailed clarification to all questions given in the NCERT exemplar Class 12 books.

Key Features of NCERT Solutions for Class 12 Mathematics Chapter 11

To score well in competitive exams, one should have a strong command in Mathematics. The NCERT books present you with demanding questions that improveyour analytical skills and give you sufficient exposure to all kinds of questions in any of these exams. Our application is highly beneficial for your preparation for the following reasons.

• NCERT Solutions Class 12 Mathematics answers to all numerical problems in the textbook are given with a step-by-step simplification.
• Free unlimited access.
• Highly detailed solutions to all the theoretical as well as logical reasoning questions from the NCERT Mathematics textbook of Class 12 that are verified by respective experts of the subject.
• The reputed teachers' accurate and straightforward answers cover all the complex and specific topics.
• The material provided by NCERT helps students to score remarkably in the exams.

Q.1 If a line makes angles 90°, 135°, 45° with the x, y and z-axes respectively, find its direction cosines.

Ans.

$\begin{array}{l}\mathrm{Let}\mathrm{the}\mathrm{direction}\mathrm{cosines}\mathrm{of}\mathrm{the}\mathrm{line}\mathrm{be} \mathrm{l},\mathrm{m}\mathrm{and}\mathrm{n}.\mathrm{}\\ \mathrm{Then}, \\ \mathrm{l}=\mathrm{}\mathrm{cos}90\mathrm{°}=0, \\ \mathrm{m}=\mathrm{cos}\mathrm{ }135\mathrm{°}=-\frac{1}{\sqrt{2}},\\ \mathrm{n}=\mathrm{cos}\mathrm{ }45\mathrm{°}=\frac{1}{\sqrt{2}}\\ \mathrm{Therefore},\mathrm{the}\mathrm{direction}\mathrm{cosines}\mathrm{of}\mathrm{the}\mathrm{line}\mathrm{are}0,-\frac{1}{\sqrt{2}}\mathrm{and} \frac{1}{\sqrt{2}}.\end{array}$

Q.2 Find the direction cosines of a line which makes equal angles with the coordinate axes.

Ans.

$\begin{array}{l}\mathrm{Let}\mathrm{the}\mathrm{direction}\mathrm{cosines}\mathrm{of}\mathrm{the}\mathrm{line},\mathrm{which}\mathrm{makes}\mathrm{equal}\mathrm{angle}\mathrm{\alpha }\mathrm{}\\ \mathrm{with}\mathrm{each}\mathrm{of}\mathrm{the}\mathrm{coordinate}\mathrm{axes}\mathrm{be} \mathrm{l},\mathrm{m}\mathrm{and}\mathrm{n}.\mathrm{}\\ \mathrm{Then}, \\ \mathrm{l}=\mathrm{}\mathrm{cos\alpha }, \mathrm{m}=\mathrm{cos}\mathrm{ }\mathrm{\alpha },\mathrm{n}=\mathrm{cos}\mathrm{ }\mathrm{\alpha }\\ \mathrm{Since}, {\mathrm{l}}^{2}+{\mathrm{m}}^{2}+{\mathrm{n}}^{2}=1\\ ⇒{\mathrm{cos}}^{2}\mathrm{\alpha }+{\mathrm{cos}}^{2}\mathrm{\alpha }+{\mathrm{cos}}^{2}\mathrm{\alpha }=1\\ ⇒3{\mathrm{cos}}^{2}\mathrm{\alpha }=1\\ ⇒ {\mathrm{cos}}^{2}\mathrm{\alpha }=\frac{1}{3}\\ ⇒ \mathrm{ }\mathrm{cos\alpha }=±\frac{1}{\sqrt{3}}\\ \mathrm{Therefore},\mathrm{the}\mathrm{direction}\mathrm{cosines}\mathrm{of}\mathrm{the}\mathrm{line},\mathrm{which}\mathrm{is}\mathrm{equally}\mathrm{inclined}\mathrm{}\\ \mathrm{to}\mathrm{the}\mathrm{coordinate}\mathrm{axes},\mathrm{are}±\frac{1}{\sqrt{3}},±\frac{1}{\sqrt{3}},±\frac{1}{\sqrt{3}}.\end{array}$

Q.3 If a line has the direction ratios –18, 12, –4 then what are its direction cosines?

Ans.

$\begin{array}{l}\mathrm{If}\mathrm{}\mathrm{a}\mathrm{}\mathrm{line}\mathrm{}\mathrm{has}\mathrm{}\mathrm{the}\mathrm{}\mathrm{direction}\mathrm{}\mathrm{ratios}\mathrm{}–18,\mathrm{}12,\mathrm{}–4,\mathrm{then}\mathrm{direction}\\ \mathrm{cosines}\mathrm{are}:\\ \mathrm{l}=\frac{-18}{\sqrt{{\left(-18\right)}^{2}+{\left(12\right)}^{2}+{\left(-4\right)}^{2}}}\\ =\frac{-18}{\sqrt{324+144+16}}\\ =\frac{-18}{22}\\ =\frac{-9}{11}\\ \mathrm{m}=\frac{12}{\sqrt{{\left(-18\right)}^{2}+{\left(12\right)}^{2}+{\left(-4\right)}^{2}}}\\ =\frac{12}{\sqrt{324+144+16}}\\ =\frac{12}{22}\\ =\frac{6}{11}\\ \mathrm{n}=\frac{-4}{\sqrt{{\left(-18\right)}^{2}+{\left(12\right)}^{2}+{\left(-4\right)}^{2}}}\\ =\frac{-4}{\sqrt{324+144+16}}\\ =\frac{-4}{22}\\ =\frac{-2}{11}\\ \mathrm{Thus},\mathrm{the}\mathrm{direction}\mathrm{cosines}\mathrm{are}\frac{-9}{11},\frac{6}{11},\frac{-2}{11}.\end{array}$

Q.4 Show that the points (2, 3, 4), (–1, –2, 1), (5, 8, 7) are collinear.

Ans.

The given points are: A(2, 3, 4), B(–1, –2, 1) and C(5, 8, 7).
Direction ratios of line joining A (2, 3, 4) and B (–1, –2, 1) are:
–1–2, –2 – 3, 1 – 4 i.e., –3, –5, –3.
Direction ratios of line joining B (–1, –2, 1) and C (5, 8, 7) are:
5–(–1), 8 –(–2), 7 –1 i.e., 6, 10, 6.
It can be seen that the direction ratios of BC are –2 times of that of AB.
Therefore, AB is parallel to BC. Since, B is common to both AB and BC, so, points A, B and C are collinear.

Q.5 Find the direction cosines of the sides of the triangle whose vertices are (3, 5, –4), (–1, 1, 2) and (–5, –5, –2).

Ans.

$\begin{array}{l}\mathrm{Let}\mathrm{ }\mathrm{the}\mathrm{ }\mathrm{vertices}\mathrm{ }\mathrm{of}\mathrm{ }\mathrm{triangle}\mathrm{ }\mathrm{ABC}\mathrm{ }\mathrm{be}\mathrm{ }\mathrm{A}\mathrm{ }\left(3,5,-4\right),\\ \mathrm{B}\left(-1,1,2\right)\mathrm{ }\mathrm{and}\mathrm{ }\mathrm{C}\left(-5,-5,-2\right).\\ \mathrm{The}\mathrm{ }\mathrm{direction}\mathrm{ }\mathrm{ratios}\mathrm{ }\mathrm{of}\mathrm{ }\mathrm{side}\mathrm{ }\mathrm{AB} \mathrm{are}:\\ -1-3, 1-5, 2-\left(-4\right) \mathrm{i}.\mathrm{e}.,-4,-4,6.\\ \mathrm{So},\mathrm{ }\mathrm{the}\mathrm{ }\mathrm{direction}\mathrm{ }\mathrm{cosines}\mathrm{ }\mathrm{of}\mathrm{ }\mathrm{AB}\mathrm{ }\mathrm{are}:\\ \frac{-4}{\sqrt{{\left(-4\right)}^{2}+{\left(-4\right)}^{2}+{6}^{2}}},\frac{-4}{\sqrt{{\left(-4\right)}^{2}+{\left(-4\right)}^{2}+{6}^{2}}},\frac{6}{\sqrt{{\left(-4\right)}^{2}+{\left(-4\right)}^{2}+{6}^{2}}}\\ \mathrm{i}.\mathrm{e}., \frac{-4}{2\sqrt{17}},\frac{-4}{2\sqrt{17}},\frac{6}{2\sqrt{17}}\\ ⇒\frac{-2}{\sqrt{17}},\frac{-2}{\sqrt{17}},\frac{3}{\sqrt{17}}\\ \mathrm{The}\mathrm{ }\mathrm{direction}\mathrm{ }\mathrm{ratios}\mathrm{ }\mathrm{of}\mathrm{ }\mathrm{side}\mathrm{ }\mathrm{BC}\mathrm{ }\mathrm{are}:\\ -5-\left(-1\right), -5-1,\mathrm{ }-2-2 \mathrm{i}.\mathrm{e}.,-4,-\mathrm{ }6,-\mathrm{ }4.\\ \mathrm{So},\mathrm{ }\mathrm{the}\mathrm{ }\mathrm{direction}\mathrm{ }\mathrm{cosines}\mathrm{ }\mathrm{of}\mathrm{ }\mathrm{AB}\mathrm{ }\mathrm{are}:\\ \frac{-\mathrm{ }4}{\sqrt{{\left(-4\right)}^{2}+{\left(-4\right)}^{2}+{6}^{2}}},\frac{-\mathrm{ }6}{\sqrt{{\left(-4\right)}^{2}+{\left(-4\right)}^{2}+{6}^{2}}},\frac{-\mathrm{ }4}{\sqrt{{\left(-4\right)}^{2}+{\left(-4\right)}^{2}+{6}^{2}}}\\ \mathrm{i}.\mathrm{e}., \frac{-\mathrm{ }4}{2\sqrt{17}},\frac{-\mathrm{ }6}{2\sqrt{17}},\frac{-\mathrm{ }4}{2\sqrt{17}}\\ ⇒\frac{-2}{\sqrt{17}},\frac{-3}{\sqrt{17}},\frac{-2}{\sqrt{17}}\\ \mathrm{The}\mathrm{ }\mathrm{direction}\mathrm{ }\mathrm{ratios}\mathrm{ }\mathrm{of}\mathrm{ }\mathrm{side}\mathrm{ }\mathrm{AC}\mathrm{ }\mathrm{are}:\\ -5-3, -5-5,\mathrm{ }-\mathrm{ }2-\left(-4\right) \mathrm{i}.\mathrm{e}.,-8, -10, 2.\\ \mathrm{So},\mathrm{ }\mathrm{the}\mathrm{ }\mathrm{direction}\mathrm{ }\mathrm{cosines}\mathrm{ }\mathrm{of}\mathrm{ }\mathrm{AB}\mathrm{ }\mathrm{are}:\\ \frac{-8}{\sqrt{{\left(-8\right)}^{2}+{\left(-10\right)}^{2}+{2}^{2}}},\frac{-10}{\sqrt{{\left(-8\right)}^{2}+{\left(-10\right)}^{2}+{2}^{2}}},\frac{2}{\sqrt{{\left(-8\right)}^{2}+{\left(-10\right)}^{2}+{2}^{2}}}\\ \mathrm{i}.\mathrm{e}., \frac{-\mathrm{ }8}{2\sqrt{42}},\frac{-10}{2\sqrt{42}},\frac{2}{2\sqrt{42}}\\ ⇒\mathrm{ }\frac{-\mathrm{ }4}{\sqrt{42}},\frac{-5}{\sqrt{42}},\frac{1}{\sqrt{42}}\end{array}$

Q.6

$\begin{array}{l}\mathrm{Show}\mathrm{that}\mathrm{the}\mathrm{three}\mathrm{lines}\mathrm{with}\mathrm{direction}\mathrm{cosines}\\ \frac{\mathrm{12}}{\mathrm{13}}\mathrm{,}\frac{–3}{\mathrm{13}}\mathrm{,}\frac{–4}{\mathrm{13}};\frac{\mathrm{4}}{\mathrm{13}}\mathrm{,}\frac{\mathrm{12}}{\mathrm{13}}\mathrm{,}\frac{\mathrm{3}}{\mathrm{13}};\frac{\mathrm{3}}{\mathrm{13}}\mathrm{,}\frac{–4}{\mathrm{13}}\mathrm{,}\frac{\mathrm{12}}{\mathrm{13}}\mathrm{are}\mathrm{mutually}\mathrm{perpendicular}\mathrm{.}\end{array}$

Ans.

$\begin{array}{l}Two\text{lines with direction cosines}{l}_{\text{1}}{\text{,m}}_{\text{1}}{\text{,n}}_{\text{1}}\text{and}{l}_{\text{2}}{\text{,m}}_{\text{2}}{\text{,n}}_{\text{2}}\text{are}\\ \text{perpendicular to each other if}{l}_{\text{1}}{l}_{\text{2}}+{\text{m}}_{\text{1}}{\text{m}}_{\text{2}}+{\text{n}}_{\text{1}}{\text{n}}_{\text{2}}=0\\ \left(i\right)\text{For the lines with direction cosines,}\frac{12}{13},\frac{-3}{13},\frac{-4}{13}\text{and}\frac{4}{13},\frac{12}{13},\frac{3}{13},\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}we\text{have}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{l}_{\text{1}}{l}_{\text{2}}+{\text{m}}_{\text{1}}{\text{m}}_{\text{2}}+{\text{n}}_{\text{1}}{\text{n}}_{\text{2}}=\frac{12}{13}.\frac{4}{13}+\frac{-3}{13}.\frac{12}{13}+\frac{-4}{13}.\frac{3}{13}\\ \text{}\text{}\text{}\text{}=\frac{48}{169}-\frac{36}{169}-\frac{12}{169}=0\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}Thus,\text{lines are perpendicular}\text{.}\\ \left(ii\right)\text{For the lines with direction cosines}\frac{4}{13},\frac{12}{13},\frac{3}{13}\text{and}\frac{3}{13},\frac{-4}{13},\frac{12}{13},\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}we\text{have}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{l}_{\text{1}}{l}_{\text{2}}+{\text{m}}_{\text{1}}{\text{m}}_{\text{2}}+{\text{n}}_{\text{1}}{\text{n}}_{\text{2}}=\frac{4}{13}.\frac{3}{13}+\frac{12}{13}.\left(\frac{-4}{13}\right)+\frac{3}{13}.\frac{12}{13}\\ \text{}\text{}\text{}\text{}=\frac{12}{169}-\frac{48}{169}+\frac{36}{169}=0\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}Thus,\text{lines are perpendicular}\text{.}\\ \left(iii\right)\text{For the lines with direction cosines}\frac{3}{13},\frac{-4}{13},\frac{12}{13}\text{and}\frac{12}{13},\frac{-3}{13},\frac{-4}{13},\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}we\text{have}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{l}_{\text{1}}{l}_{\text{2}}+{\text{m}}_{\text{1}}{\text{m}}_{\text{2}}+{\text{n}}_{\text{1}}{\text{n}}_{\text{2}}=\frac{3}{13}.\frac{12}{13}+\frac{-4}{13}.\left(\frac{-3}{13}\right)+\frac{12}{13}.\left(\frac{-4}{13}\right)\\ \text{}\text{}\text{}\text{}=\frac{36}{169}+\frac{12}{169}-\frac{48}{169}=0\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}Thus,\text{lines are perpendicular}\text{.}\\ \text{Therefore, all lines are mutually perpendicular}\text{.}\end{array}$

Q.7 Show that the line through the points (1,–1, 2), (3, 4, –2) is perpendicular to the line through the points (0, 3, 2) and (3, 5, 6).

Ans.

Let line AB has end points A(1,–1, 2) and B(3, 4, –2). And line CD has end points C (0, 3, 2) and D(3, 5, 6).

The direction ratios a1, a2 and a3 of AB are (3 – 1), {4 –(–1)} and (–2–2) i.e., 2, 5, – 4.The direction ratios b1, b2 and b3 of CD are (3 – 0), (5 –3) and (6 – 2) i.e., 3, 2, 4.AB and CD will be perpendicular to each other, if a1b1+ a2b2 + a3b3 = 0.Then,
a1b1+ a2b2 + a3b3 = (2)(3) + (5)(2) + (– 4)(4)

= 6 + 10 – 16
= 0

Therefore, AB and CD are perpendicular to each other.

Q.8 Show that the line through the points (4, 7, 8), (2, 3, 4) is parallel to the line through the points (–1, –2, 1), (1, 2, 5).

Ans.

Let line AB has end points A (4, 7, 8) and B (2, 3, 4). And line CD has end points C (–1, –2, 1) and D (1, 2, 5).

The direction ratios a1, a2 and a3 of AB are (2 – 4), (3 –7) and (4 – 8) i.e., –2, – 4, – 4.The direction ratios b1, b2 and b3 of CD are {1–(–1)}, {2 –(–2)} and (5 – 1) i.e., 2, 4, 4.AB will be parallel to CD, if(a1/b1) = (a2/b2) = (a3/b3). Then,
(a1 / b1) = (–2/2) = –1

(a2 / b2) = (– 4/4) = – 1

(a3 / b3) = (– 4/4) = – 1

So, (a1/b1) = (a2/b2) = (a3/b3)

Therefore, AB is parallel to CD.

Q.9 Find the equation of the line which passes through the point ( 1 , 2 , 3 ) and is parallel to the vector 3 i ^ + 2 j ^ 2 k ^ .

Ans.

$\begin{array}{l}\mathrm{The}\mathrm{position}\mathrm{vector}\mathrm{of}\mathrm{point}\mathrm{A}\left(1,2,3\right)\mathrm{is}\stackrel{\to }{\mathrm{a}}\mathrm{=}\stackrel{^}{\mathrm{i}}+2\stackrel{^}{\mathrm{j}}+3\stackrel{^}{\mathrm{k}}\mathrm{and}\mathrm{it}\mathrm{is}\\ \mathrm{parallelt}\mathrm{to}\mathrm{vector}\stackrel{\to }{\mathrm{b}}=3\stackrel{^}{\mathrm{i}}+2\stackrel{^}{\mathrm{j}}–2\stackrel{^}{\mathrm{k}}\mathrm{.}\\ \mathrm{Since},\mathrm{the}\mathrm{line}\mathrm{passes}\mathrm{through}\mathrm{the}\mathrm{point}\mathrm{A}\left(1,2,3\right)\mathrm{and}\mathrm{parallel}\mathrm{to}\\ \mathrm{vector}\stackrel{\to }{\mathrm{b}}\mathrm{is}\mathrm{given}\mathrm{by}\\ \stackrel{\to }{\mathrm{r}}\mathrm{=}\stackrel{\to }{\mathrm{a}}+\mathrm{\lambda }\stackrel{\to }{\mathrm{b}},\mathrm{where}\mathrm{\lambda }\mathrm{is}\mathrm{a}\mathrm{constant}\mathrm{.}\\ ⇒ \stackrel{\to }{\mathrm{r}}\mathrm{=}\left(\stackrel{^}{\mathrm{i}}+2\stackrel{^}{\mathrm{j}}+3\stackrel{^}{\mathrm{k}}\right)+\mathrm{\lambda }\left(\mathrm{3}\stackrel{^}{\mathrm{i}}+2\stackrel{^}{\mathrm{j}}–2\stackrel{^}{\mathrm{k}}\right)\\ =\left(1+3\mathrm{\lambda }\right)\stackrel{^}{\mathrm{i}}\mathrm{+}\left(2+2\mathrm{\lambda }\right)\stackrel{^}{\mathrm{j}}\mathrm{+}\left(3-2\mathrm{\lambda }\right)\stackrel{^}{\mathrm{k}}\\ \mathrm{This}\mathrm{is}\mathrm{the}\mathrm{required}\mathrm{equation}\mathrm{of}\mathrm{the}\mathrm{line}\mathrm{.}\end{array}$

Q.10 Find the equation of the line in vector and in cartesian form that passes through the point with position vector 2 i ^ j ^ + 4 k ^ and is in the direction i ^ + 2 j ^ k ^ .

Ans.

$\begin{array}{l}\text{Let}\text{\hspace{0.17em}}\text{}\stackrel{\to }{\text{a}}\text{=2}\stackrel{^}{\text{i}}-\stackrel{^}{\text{j}}\text{+4}\stackrel{^}{\text{k}}\text{}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{}\stackrel{\to }{\text{b}}\text{=}\stackrel{^}{\text{i}}\text{+2}\stackrel{^}{\text{j}}\text{–}\stackrel{^}{\text{k}}\\ \text{Since, the line passes through the point with position vector}\stackrel{\to }{\text{a}}\text{}\\ \text{and parallel to vector}\stackrel{\to }{\text{b}}\text{is given by}\\ \text{}\stackrel{\to }{\text{r}}\text{=}\stackrel{\to }{\text{a}}\text{+λ}\stackrel{\to }{\text{b}}\text{, where λ is a constant}\text{.}\\ ⇒\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\stackrel{\to }{\text{r}}\text{=}\left(\text{2}\stackrel{^}{\text{i}}-\stackrel{^}{\text{j}}\text{+4}\stackrel{^}{\text{k}}\right)\text{+λ}\left(\stackrel{^}{\text{i}}\text{+2}\stackrel{^}{\text{j}}-\stackrel{^}{\text{k}}\right)\text{\hspace{0.17em}}\text{\hspace{0.17em}}\dots \left(\text{i}\right)\\ \text{This is the required equation of the line in vector form}\text{.}\\ \text{Putting}\stackrel{\to }{\text{r}}\text{=x}\text{\hspace{0.17em}}\stackrel{^}{\text{i}}\text{+y}\stackrel{^}{\text{j}}\text{+z}\text{\hspace{0.17em}}\stackrel{^}{\text{k}}\text{in equation}\left(\text{i}\right)\text{, we get}\\ \text{x}\text{\hspace{0.17em}}\stackrel{^}{\text{i}}\text{+y}\stackrel{^}{\text{j}}\text{+z}\text{\hspace{0.17em}}\stackrel{^}{\text{k}}\text{=}\left(\text{2}\stackrel{^}{\text{i}}-\stackrel{^}{\text{j}}\text{+4}\stackrel{^}{\text{k}}\right)\text{+λ}\left(\stackrel{^}{\text{i}}\text{+2}\stackrel{^}{\text{j}}-\stackrel{^}{\text{k}}\right)\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{=}\left(\text{2+λ}\right)\stackrel{^}{\text{i}}-\left(\text{1}-\text{2λ}\right)\stackrel{^}{\text{j}}\text{+}\left(\text{4}-\text{λ}\right)\stackrel{^}{\text{k}}\\ ⇒\text{x=2+λ,}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{y=2λ-1 and z=4-λ}\\ ⇒\text{λ=x-2,}\text{​}\text{\hspace{0.17em}}\text{λ=}\frac{\text{y+1}}{\text{2}}\text{and λ=4-z}\\ ⇒\frac{\text{x}-\text{2}}{\text{1}}\text{=}\frac{\text{y+1}}{\text{2}}\text{=}\frac{\text{z}-\text{4}}{-\text{1}}\\ \text{This is the required equation of the given line in cartesian form}\text{.}\end{array}$

Q.11 Findthecartesianequation of the line which passes through the point 2, 4, 5 and parallel to the line given by x+3 3 = y4 5 = z+8 6 .

Ans.

$\begin{array}{l}\text{The direction ratios of}\frac{\text{x+3}}{\text{3}}\text{=}\frac{\text{y}-\text{4}}{\text{5}}\text{=}\frac{\text{z+8}}{\text{6}}\text{are 3,5 and 6}\text{.}\\ \text{Let the direction ratios of required line parallel to}\text{\hspace{0.17em}}\\ \text{\hspace{0.17em}}\frac{\text{x+3}}{\text{3}}\text{=}\frac{\text{y}-\text{4}}{\text{5}}\text{=}\frac{\text{z+8}}{\text{6}}\text{are 3k, 5k and 6k}\text{.}\\ \text{Since, equation of the line through the point}\left({\text{x}}_{\text{1}}\text{,}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{\text{y}}_{\text{1}}\text{,}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{\text{z}}_{\text{1}}\right)\text{and with}\\ \text{the direction ratios a, b and c is}\\ \text{}\frac{\text{x}-{\text{x}}_{\text{1}}}{\text{a}}\text{=}\frac{\text{y}-{\text{y}}_{\text{1}}}{\text{b}}\text{=}\frac{\text{z}-{\text{z}}_{\text{1}}}{\text{c}}\\ \text{Therefore, the equation of required line}\text{\hspace{0.17em}}\text{through the point}\\ \left(-\text{2,}\text{\hspace{0.17em}}\text{4,}\text{\hspace{0.17em}}-\text{5}\right)\text{is}\\ \text{}\frac{\text{x+2}}{\text{3k}}\text{=}\frac{\text{y}-\text{4}}{\text{5k}}\text{=}\frac{\text{z+5}}{\text{6k}}\\ \text{or}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\frac{\text{x+2}}{\text{3}}\text{=}\frac{\text{y}-\text{4}}{\text{5}}\text{=}\frac{\text{z+5}}{\text{6}}\\ \text{which is the required equation of line}\text{.}\end{array}$

Q.12

$\begin{array}{l}\mathrm{The}\mathrm{ }\mathrm{cartesian}\mathrm{ }\mathrm{equation}\mathrm{ }\mathrm{of}\mathrm{ }\mathrm{a}\mathrm{ }\mathrm{line}\mathrm{ }\mathrm{is}\mathrm{ }\frac{\mathrm{x}-5}{3}=\frac{\mathrm{y}+4}{7}=\frac{\mathrm{z}-6}{2}.\\ \mathrm{Write}\mathrm{ }\mathrm{its}\mathrm{ }\mathrm{vector}\mathrm{ }\mathrm{form}.\end{array}$

Ans.

$\begin{array}{l}\mathrm{The}\mathrm{ }\mathrm{cartesian}\mathrm{ }\mathrm{equation}\mathrm{of}\mathrm{the}\mathrm{line}\mathrm{is}\\ \frac{\mathrm{x}-\mathrm{5}}{\mathrm{3}}\mathrm{=}\frac{\mathrm{y}+4}{\mathrm{7}}\mathrm{=}\frac{\mathrm{z}-\mathrm{6}}{\mathrm{2}}...\left(\mathrm{i}\right)\\ \mathrm{The}\mathrm{given}\mathrm{lines}\mathrm{passes}\mathrm{through}\mathrm{the}\mathrm{point}\left(5,-\mathrm{ }4,6\right).\mathrm{The}\mathrm{position}\\ \mathrm{vector}\mathrm{of}\mathrm{this}\mathrm{point}\mathrm{is}\stackrel{\to }{\mathrm{a}} = 5\mathrm{ }\stackrel{^}{\mathrm{i}}-\mathrm{4}\stackrel{^}{\mathrm{j}}+6\stackrel{^}{\mathrm{k}}\mathrm{.}\\ \mathrm{The}\mathrm{direction}\mathrm{ratios}\mathrm{of}\mathrm{the}\mathrm{given}\mathrm{line}\mathrm{are}3, 7\mathrm{and}2\mathrm{.}\\ \mathrm{The}\mathrm{direction}\mathrm{vector}\mathrm{of}\mathrm{given}\mathrm{line}\mathrm{is}\stackrel{\to }{\mathrm{b}} = 3\mathrm{ }\stackrel{^}{\mathrm{i}}+7\stackrel{^}{\mathrm{j}}+2\mathrm{ }\stackrel{^}{\mathrm{k}}\mathrm{.}\\ \mathrm{So},\mathrm{the}\mathrm{equation}\mathrm{of}\mathrm{line}\mathrm{through}\mathrm{the}\mathrm{point}\stackrel{\to }{\mathrm{a}}\mathrm{and}\mathrm{in}\mathrm{the}\mathrm{direction}\\ \mathrm{of}\mathrm{the}\mathrm{vector}\stackrel{\to }{\mathrm{b}}\mathrm{is}\mathrm{given}\mathrm{by}\mathrm{the}\mathrm{equation},\\ \stackrel{\to }{\mathrm{r}} =\mathrm{ }\stackrel{\to }{\mathrm{a}}+\mathrm{\lambda }\stackrel{\to }{\mathrm{b}},\mathrm{where}\mathrm{\lambda }\mathrm{is}\mathrm{a}\mathrm{contant}\mathrm{.}\\ \mathrm{So},\stackrel{\to }{\mathrm{r}} =\mathrm{ }\left(5 \stackrel{^}{\mathrm{i}}–4\stackrel{^}{\mathrm{j}}+6\stackrel{^}{\mathrm{k}}\right)+\mathrm{\lambda }\left(3 \stackrel{^}{\mathrm{i}}+7\stackrel{^}{\mathrm{j}}+2\mathrm{ }\stackrel{^}{\mathrm{k}}\right)\\ \mathrm{This}\mathrm{is}\mathrm{the}\mathrm{required}\mathrm{equation}\mathrm{of}\mathrm{the}\mathrm{given}\mathrm{line}\mathrm{in}\mathrm{the}\mathrm{vector}\mathrm{form}\mathrm{.}\end{array}$

Q.13 Find the vector and the cartesian equations of the lines that passes through the origin and (5, –2 , 3).

Ans.

$\begin{array}{l}\text{The required line passes through origin}\text{.Therefore its position}\\ \text{vector is}\stackrel{\to }{\text{a}}\text{ }\text{=}\text{ }\text{0}\text{.}\\ \text{The direction ratios of line through origin and the point}\left(\text{5,}-\text{2,3}\right)\\ \text{are:}\text{\hspace{0.17em}}\left(\text{5-0}\right)\text{,}\left(-\text{2}-\text{0}\right)\text{,}\left(\text{3}-\text{0}\right)\text{i}\text{.e}\text{., 5,-2,3}\\ \text{The line is parallel to the position vector}\stackrel{\to }{\text{b}}\text{=5}\text{\hspace{0.17em}}\stackrel{^}{\text{i}}\text{-2}\stackrel{^}{\text{j}}\text{+3}\text{\hspace{0.17em}}\stackrel{^}{\text{k}}\text{.}\\ \text{So, the equation of line passes through origin and parallel to}\stackrel{\to }{\text{b}}\text{is}\\ \text{}\stackrel{\to }{\text{r}}\text{=}\stackrel{\to }{\text{a}}\text{+λ}\stackrel{\to }{\text{b}}\text{,}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{λ}\in \text{R}\\ ⇒\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\stackrel{\to }{\text{r}}\text{=0+λ}\left(\text{5}\text{\hspace{0.17em}}\stackrel{^}{\text{i}}-\text{2}\stackrel{^}{\text{j}}\text{+3}\text{\hspace{0.17em}}\stackrel{^}{\text{k}}\right)\\ ⇒\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\stackrel{\to }{\text{r}}\text{=λ}\left(\text{5}\text{\hspace{0.17em}}\stackrel{^}{\text{i}}-\text{2}\stackrel{^}{\text{j}}\text{+3}\text{\hspace{0.17em}}\stackrel{^}{\text{k}}\right)\\ \text{The equation of line through the point}\left({\text{x}}_{\text{1}}{\text{,y}}_{\text{1}}{\text{,z}}_{\text{1}}\right)\text{and direction}\\ \text{ratios a,b,c is given by}\text{}\frac{\text{x}-{\text{x}}_{\text{1}}}{\text{a}}\text{=}\frac{\text{y}-{\text{y}}_{\text{1}}}{\text{b}}\text{=}\frac{\text{z}-{\text{z}}_{\text{1}}}{\text{c}}\\ \text{Therefore, equation of the required line in the cartesian form is}\\ \text{}\text{}\text{}\text{}\text{}\frac{\text{x-0}}{\text{5}}\text{=}\frac{\text{y-0}}{-\text{2}}\text{=}\frac{\text{z-0}}{\text{3}}\\ ⇒\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\frac{\text{x}}{\text{5}}\text{=}\frac{\text{y}}{-\text{2}}\text{=}\frac{\text{z}}{\text{3}}\end{array}$

Q.14 Find the vector and the cartesian equations of the line that passes through the points (3, –2, –5), (3, –2, 6).

Ans.

$\begin{array}{l}\mathrm{Let}\mathrm{the}\mathrm{given}\mathrm{points}\mathrm{are}\mathrm{A}\left(3,–2,-\mathrm{5}\right)\mathrm{and}\mathrm{B}\left(3,-2, 6\right)\mathrm{through}\\ \mathrm{which}\mathrm{line}\mathrm{AB}\mathrm{passes}\mathrm{.}\\ \mathrm{So},\mathrm{position}\mathrm{vector}\mathrm{of}\mathrm{point}\left(3,–2,-\mathrm{5}\right)\left(\stackrel{\to }{\mathrm{a}}\right)=3\mathrm{ }\stackrel{^}{\mathrm{i}}-2\stackrel{^}{\mathrm{j}}-5\mathrm{ }\stackrel{^}{\mathrm{k}}\\ \mathrm{The}\mathrm{direction}\mathrm{ratios}\mathrm{of}\mathrm{AB}\mathrm{are}\mathrm{given}\mathrm{as}\\ \left(3-3\right),\left(-2+2\right),\left(6+5\right)\mathrm{i}.\mathrm{e}., 0,0,11\\ \mathrm{The}\mathrm{equation}\mathrm{of}\mathrm{vector}\mathrm{in}\mathrm{the}\mathrm{direction}\mathrm{of}\mathrm{AB}\mathrm{is}\\ \stackrel{\to }{\mathrm{b}}=0\mathrm{ }\stackrel{^}{\mathrm{i}}+0\mathrm{ }\stackrel{^}{\mathrm{j}}+11\stackrel{^}{\mathrm{k}}=11\stackrel{^}{\mathrm{k}}\\ \mathrm{The}\mathrm{equation}\mathrm{of}\mathrm{AB}\mathrm{in}\mathrm{the}\mathrm{vector}\mathrm{form}\mathrm{is}\\ \stackrel{\to }{\mathrm{r}}=\left(3\mathrm{ }\stackrel{^}{\mathrm{i}}-2\stackrel{^}{\mathrm{j}}-5\mathrm{ }\stackrel{^}{\mathrm{k}}\right)+\mathrm{\lambda }\left(11\stackrel{^}{\mathrm{k}}\right)\\ \mathrm{The}\mathrm{equation}\mathrm{of}\mathrm{AB}\mathrm{in}\mathrm{the}\mathrm{cartesian}\mathrm{form}\mathrm{is}\\ \frac{\mathrm{x}-3}{0}=\frac{\mathrm{y}+2}{0}=\frac{\mathrm{z}+5}{11}\end{array}$

Q.15

$\begin{array}{l}\mathrm{Find}\mathrm{ }\mathrm{the}\mathrm{ }\mathrm{angle}\mathrm{ }\mathrm{between}\mathrm{ }\mathrm{the}\mathrm{ }\mathrm{following}\mathrm{ }\mathrm{pairs}\mathrm{ }\mathrm{of}\mathrm{ }\mathrm{lines}:\\ \left(\mathrm{i}\right)\stackrel{\to }{\mathrm{r}}=2\mathrm{ }\stackrel{^}{\mathrm{i}}-5\stackrel{^}{\mathrm{j}}+\stackrel{^}{\mathrm{k}}+\mathrm{\lambda }\left(3\mathrm{ }\stackrel{^}{\mathrm{i}}+2\stackrel{^}{\mathrm{j}}+6\mathrm{ }\stackrel{^}{\mathrm{k}}\right)\mathrm{ }\mathrm{and}\\ \stackrel{\to }{\mathrm{r}}=7\mathrm{ }\stackrel{^}{\mathrm{i}}-6\stackrel{^}{\mathrm{k}}+\mathrm{\mu }\left(\mathrm{ }\stackrel{^}{\mathrm{i}}+2\stackrel{^}{\mathrm{j}}+2\stackrel{^}{\mathrm{k}}\right)\\ \left(\mathrm{ii}\right)\stackrel{\to }{\mathrm{r}}=3\mathrm{ }\stackrel{^}{\mathrm{i}}+\stackrel{^}{\mathrm{j}}-2\mathrm{ }\stackrel{^}{\mathrm{k}}+\mathrm{\lambda }\left(\mathrm{ }\stackrel{^}{\mathrm{i}}-\stackrel{^}{\mathrm{j}}-2\mathrm{ }\stackrel{^}{\mathrm{k}}\right) \mathrm{and}\\ \stackrel{\to }{\mathrm{r}}=2\mathrm{ }\stackrel{^}{\mathrm{i}}-\stackrel{^}{\mathrm{j}}-56\mathrm{ }\stackrel{^}{\mathrm{k}}+\mathrm{\mu }\left(3\mathrm{ }\stackrel{^}{\mathrm{i}}-5\stackrel{^}{\mathrm{j}}-4\mathrm{ }\stackrel{^}{\mathrm{k}}\right)\end{array}$

Ans.

$\begin{array}{l}\left(\mathrm{i}\right)\mathrm{ }\mathrm{Here},\mathrm{ }\stackrel{\to }{{\mathrm{b}}_{1}}=3\mathrm{ }\stackrel{^}{\mathrm{i}}+2\stackrel{^}{\mathrm{j}}+6\mathrm{ }\stackrel{^}{\mathrm{k}}\mathrm{ }\mathrm{and}\mathrm{ }\stackrel{\to }{{\mathrm{b}}_{2}}=\mathrm{ }\stackrel{^}{\mathrm{i}}+2\stackrel{^}{\mathrm{j}}+2\stackrel{^}{\mathrm{k}}\\ \mathrm{Then}\mathrm{ }\mathrm{angle}\mathrm{ }\mathrm{\theta }\mathrm{ }\mathrm{between}\mathrm{ }\mathrm{the}\mathrm{ }\mathrm{two}\mathrm{ }\mathrm{lines}\mathrm{ }\mathrm{is}\mathrm{ }\mathrm{given}\mathrm{ }\mathrm{by}\\ \mathrm{cos\theta }=\left|\frac{\stackrel{\to }{{\mathrm{b}}_{1}}.\stackrel{\to }{{\mathrm{b}}_{2}}}{\left|\stackrel{\to }{{\mathrm{b}}_{1}}\right|\left|\stackrel{\to }{{\mathrm{b}}_{1}}\right|}\right|\\ =\left|\frac{\left(3\mathrm{ }\stackrel{^}{\mathrm{i}}+2\stackrel{^}{\mathrm{j}}+6\mathrm{ }\stackrel{^}{\mathrm{k}}\right).\left(\mathrm{ }\stackrel{^}{\mathrm{i}}+2\stackrel{^}{\mathrm{j}}+2\stackrel{^}{\mathrm{k}}\right)}{\left|3\mathrm{ }\stackrel{^}{\mathrm{i}}+2\stackrel{^}{\mathrm{j}}+6\mathrm{ }\stackrel{^}{\mathrm{k}}\right|\left|\mathrm{ }\stackrel{^}{\mathrm{i}}+2\stackrel{^}{\mathrm{j}}+2\stackrel{^}{\mathrm{k}}\right|}\right|\\ =\left|\frac{3+4+12}{\sqrt{{3}^{2}+{2}^{2}+{6}^{2}}\sqrt{{1}^{2}+{2}^{2}+{2}^{2}}}\right|\\ =\left|\frac{19}{\sqrt{9+4+36}\sqrt{1+4+4}}\right|\\ =\frac{19}{7×3}\\ =\frac{19}{21}\\ \therefore \mathrm{ }\mathrm{\theta }={\mathrm{cos}}^{–1}\left(\frac{19}{21}\right)\\ \left(\mathrm{ii}\right)\mathrm{ }\mathrm{Here},\mathrm{}\stackrel{\to }{{\mathrm{b}}_{1}}=\mathrm{ }\stackrel{^}{\mathrm{i}}-\stackrel{^}{\mathrm{j}}-2\mathrm{ }\stackrel{^}{\mathrm{k}}\mathrm{ }\mathrm{and}\mathrm{ }\stackrel{\to }{{\mathrm{b}}_{2}}=3\mathrm{ }\stackrel{^}{\mathrm{i}}-5\mathrm{ }\stackrel{^}{\mathrm{j}}-4\mathrm{ }\stackrel{^}{\mathrm{k}}\\ \mathrm{Then}\mathrm{angle}\mathrm{\theta }\mathrm{between}\mathrm{the}\mathrm{two}\mathrm{lines}\mathrm{is}\mathrm{given}\mathrm{by}\\ \mathrm{cos\theta }=\left|\frac{\stackrel{\to }{{\mathrm{b}}_{1}}.\stackrel{\to }{{\mathrm{b}}_{2}}}{\left|\stackrel{\to }{{\mathrm{b}}_{1}}\right|\left|\stackrel{\to }{{\mathrm{b}}_{1}}\right|}\right|\\ =\left|\frac{\left(\mathrm{ }\stackrel{^}{\mathrm{i}}-\stackrel{^}{\mathrm{j}}-2\mathrm{ }\stackrel{^}{\mathrm{k}}\right).\left(3\mathrm{ }\stackrel{^}{\mathrm{i}}-5\mathrm{ }\stackrel{^}{\mathrm{j}}-4\mathrm{ }\stackrel{^}{\mathrm{k}}\right)}{\left|\mathrm{ }\stackrel{^}{\mathrm{i}}-\stackrel{^}{\mathrm{j}}-2\mathrm{ }\stackrel{^}{\mathrm{k}}\right|\left|3\mathrm{ }\stackrel{^}{\mathrm{i}}-5\mathrm{ }\stackrel{^}{\mathrm{j}}-4\mathrm{ }\stackrel{^}{\mathrm{k}}\right|}\right|\\ =\left|\frac{3+5+8}{\sqrt{{1}^{2}+{\left(-1\right)}^{2}+{\left(-2\right)}^{2}}\sqrt{{3}^{2}+{\left(-5\right)}^{2}+{\left(-4\right)}^{2}}}\right|\\ =\left|\frac{16}{\sqrt{\mathrm{ }1+1+4}\sqrt{9+25+16}}\right|\\ =\frac{16}{\sqrt{6}×\sqrt{50}}\\ =\frac{16}{10\sqrt{3}}\\ \therefore \mathrm{ }\mathrm{\theta }={\mathrm{cos}}^{-1}\left(\frac{8}{5\sqrt{3}}\right)\end{array}$

Q.16

$\begin{array}{l}\mathrm{Find}\mathrm{ }\mathrm{the}\mathrm{ }\mathrm{angle}\mathrm{ }\mathrm{between}\mathrm{ }\mathrm{the}\mathrm{ }\mathrm{following}\mathrm{ }\mathrm{pair}\mathrm{ }\mathrm{of}\mathrm{ }\mathrm{lines}:\\ \left(\mathrm{i}\right) \frac{\mathrm{x}-2}{2}=\frac{\mathrm{y}-1}{5}=\frac{\mathrm{z}+3}{-3}\mathrm{ }\mathrm{and}\mathrm{ }\frac{\mathrm{x}+2}{-1}=\frac{\mathrm{y}-4}{8}=\frac{\mathrm{z}-5}{4}\\ \left(\mathrm{ii}\right) \frac{\mathrm{x}}{2}=\frac{\mathrm{y}}{2}=\frac{\mathrm{z}}{1}\mathrm{ }\mathrm{and}\mathrm{ }\frac{\mathrm{x}-5}{4}=\frac{\mathrm{y}-2}{1}=\frac{\mathrm{z}-3}{8}\end{array}$

Ans.

$\begin{array}{l}\left(\mathrm{i}\right) \frac{\mathrm{x}-2}{2}=\frac{\mathrm{y}-1}{5}=\frac{\mathrm{z}+3}{-3}\mathrm{and}\frac{\mathrm{x}+2}{-1}=\frac{\mathrm{y}-4}{8}=\frac{\mathrm{z}-5}{4}\\ \mathrm{The}\mathrm{direction}\mathrm{ratios}\mathrm{of}\mathrm{the}\mathrm{first}\mathrm{line}\mathrm{are}2,5,-3\mathrm{and}\mathrm{the}\mathrm{direction}\\ \mathrm{ratios}\mathrm{of}\mathrm{the}\mathrm{second}\mathrm{line}\mathrm{are}-1,8,4.\mathrm{If}\mathrm{\theta }\mathrm{is}\mathrm{the}\mathrm{angle}\mathrm{between}\\ \mathrm{them},\mathrm{then}\\ \mathrm{cos\theta }=|\frac{{\mathrm{a}}_{1}{\mathrm{a}}_{2}+{\mathrm{b}}_{1}{\mathrm{b}}_{2}+{\mathrm{c}}_{1}{\mathrm{c}}_{2}}{\sqrt{{{\mathrm{a}}_{1}}^{2}+{{\mathrm{b}}_{1}}^{2}+{{\mathrm{c}}_{1}}^{2}}.\sqrt{{{\mathrm{a}}_{2}}^{2}+{{\mathrm{b}}_{2}}^{2}+{{\mathrm{c}}_{2}}^{2}}}|\\ \mathrm{ }=|\frac{2×-1+5×8+\left(-3\right)×4}{\sqrt{{\left(2\right)}^{2}+{\left(5\right)}^{2}+{\left(-3\right)}^{2}}.\sqrt{{\left(-1\right)}^{2}+{8}^{2}+{4}^{2}}}|\\ \mathrm{ }=|\frac{-2+40-12}{\sqrt{4+25+9}.\sqrt{1+64+16}}|\\ \mathrm{ }=|\frac{-2+40-12}{\sqrt{38}.\sqrt{81}}|\\ \mathrm{ }\mathrm{\theta }={\mathrm{cos}}^{-1}\left(\frac{26}{9\sqrt{38}}\right)\\ \left(\mathrm{ii}\right) \frac{\mathrm{x}}{2}=\frac{\mathrm{y}}{2}=\frac{\mathrm{z}}{1}\mathrm{and}\frac{\mathrm{x}-5}{4}=\frac{\mathrm{y}-2}{1}=\frac{\mathrm{z}-3}{8}\\ \mathrm{The}\mathrm{direction}\mathrm{ratios}\mathrm{of}\mathrm{the}\mathrm{first}\mathrm{line}\mathrm{are}2,\mathrm{ }2,\mathrm{ }1\mathrm{and}\mathrm{the}\mathrm{direction}\\ \mathrm{ratios}\mathrm{of}\mathrm{the}\mathrm{second}\mathrm{line}\mathrm{are}4, 1, 8.\mathrm{If}\mathrm{\theta }\mathrm{is}\mathrm{the}\mathrm{angle}\mathrm{between}\\ \mathrm{them},\mathrm{then}\\ \mathrm{cos\theta }=|\frac{{\mathrm{a}}_{1}{\mathrm{a}}_{2}+{\mathrm{b}}_{1}{\mathrm{b}}_{2}+{\mathrm{c}}_{1}{\mathrm{c}}_{2}}{\sqrt{{{\mathrm{a}}_{1}}^{2}+{{\mathrm{b}}_{1}}^{2}+{{\mathrm{c}}_{1}}^{2}}.\sqrt{{{\mathrm{a}}_{2}}^{2}+{{\mathrm{b}}_{2}}^{2}+{{\mathrm{c}}_{2}}^{2}}}|\\ \mathrm{ }=|\frac{2×4+2×1+1×8}{\sqrt{{\left(2\right)}^{2}+{\left(2\right)}^{2}+{\left(1\right)}^{2}}.\sqrt{{\left(4\right)}^{2}+{1}^{2}+{8}^{2}}}|\\ \mathrm{ }=|\frac{8+2+8}{\sqrt{4+4+1}.\sqrt{16+1+64}}|\\ \mathrm{ }=|\frac{18}{\sqrt{9}.\sqrt{81}}|=|\frac{18}{3×9}|\\ \mathrm{ }\mathrm{\theta }={\mathrm{cos}}^{-1}\left(\frac{2}{3}\right)\end{array}$

Q.17

$\begin{array}{l}\mathrm{Find}\mathrm{ }\mathrm{the}\mathrm{ }\mathrm{values}\mathrm{ }\mathrm{of}\mathrm{ }\mathrm{p}\mathrm{ }\mathrm{so}\mathrm{ }\mathrm{that}\mathrm{ }\mathrm{the}\mathrm{ }\mathrm{lines}\mathrm{ }\frac{1-\mathrm{x}}{3}=\frac{7\mathrm{y}-14}{2\mathrm{p}}=\frac{\mathrm{z}-3}{2}\\ \mathrm{and} \frac{7-7\mathrm{x}}{3\mathrm{p}}=\frac{\mathrm{y}-5}{1}=\frac{6-\mathrm{z}}{5}\mathrm{ }\mathrm{are}\mathrm{ }\mathrm{at}\mathrm{ }\mathrm{right}\mathrm{ }\mathrm{angles}.\end{array}$

Ans.

$\begin{array}{l}\mathrm{The}\mathrm{eqution}\mathrm{of}\mathrm{given}\mathrm{lines}\mathrm{are}\\ \frac{1-\mathrm{x}}{3}=\frac{7\mathrm{y}-14}{2\mathrm{p}}=\frac{\mathrm{z}-3}{2}\mathrm{and}\frac{7-7\mathrm{x}}{3\mathrm{p}}=\frac{\mathrm{y}-5}{1}=\frac{6-\mathrm{z}}{5}\\ ⇒\frac{\mathrm{x}-1}{-3}=\frac{\mathrm{y}-2}{\left(\frac{2\mathrm{p}}{7}\right)}=\frac{\mathrm{z}-3}{2}\mathrm{and}\frac{\mathrm{x}-1}{\left(-\frac{3\mathrm{p}}{7}\right)}=\frac{\mathrm{y}-5}{1}=\frac{\mathrm{z}-6}{-5}\\ \mathrm{The}\mathrm{direction}\mathrm{ratios}\mathrm{of}\mathrm{the}\mathrm{first}\mathrm{line}\mathrm{are}-3,\mathrm{ }\frac{2\mathrm{p}}{7}, 2\mathrm{and}\mathrm{the}\mathrm{direction}\\ \mathrm{ratios}\mathrm{of}\mathrm{the}\mathrm{second}\mathrm{line}\mathrm{are}-\frac{3\mathrm{p}}{7}, 1,\mathrm{ }-\mathrm{5}.\mathrm{Since},\mathrm{both}\mathrm{lines}\mathrm{are}\\ \mathrm{perpendicular},\mathrm{so}\\ \mathrm{ }{\mathrm{a}}_{1}{\mathrm{a}}_{2}+{\mathrm{b}}_{1}{\mathrm{b}}_{2}+{\mathrm{c}}_{1}{\mathrm{c}}_{2}=0\\ ⇒\left(-3\right).\left(-\frac{3\mathrm{p}}{7}\right)+\frac{2\mathrm{p}}{7}.1+2.\left(-5\right)=0\\ ⇒ \frac{9\mathrm{p}}{7}+\frac{2\mathrm{p}}{7}-10=0\\ ⇒ 11\mathrm{p}-70=0\\ ⇒ \mathrm{p}=\frac{70}{11}\\ \mathrm{Thus},\mathrm{the}\mathrm{value}\mathrm{of}\mathrm{p}\mathrm{is}\frac{70}{11}.\end{array}$

Q.18

$\begin{array}{l}\mathrm{Show}\mathrm{ }\mathrm{that}\mathrm{ }\mathrm{the}\mathrm{ }\mathrm{lines}\mathrm{ }\frac{\mathrm{x}-5}{7}=\frac{\mathrm{x}+2}{-5}=\frac{\mathrm{z}}{1}\mathrm{ }\mathrm{and}\mathrm{ }\frac{\mathrm{x}}{1}=\frac{\mathrm{y}}{2}=\frac{\mathrm{z}}{3}\mathrm{ }\mathrm{are}\\ \mathrm{perpendicular}\mathrm{ }\mathrm{to}\mathrm{ }\mathrm{each}\mathrm{ }\mathrm{other}.\end{array}$

Ans.

$\begin{array}{l}\mathrm{The}\mathrm{eqution}\mathrm{of}\mathrm{given}\mathrm{lines}\mathrm{are}\\ \frac{\mathrm{x}-5}{7}=\frac{\mathrm{x}+2}{-5}=\frac{\mathrm{z}}{1}\mathrm{ }\mathrm{and}\mathrm{ }\frac{\mathrm{x}}{1}=\frac{\mathrm{y}}{2}=\frac{\mathrm{z}}{3}\\ \mathrm{The}\mathrm{direction}\mathrm{ratios}\mathrm{of}\mathrm{the}\mathrm{first}\mathrm{line}\mathrm{are}7, -5, 1\mathrm{and}\mathrm{the}\mathrm{direction}\\ \mathrm{ratios}\mathrm{of}\mathrm{the}\mathrm{second}\mathrm{line}\mathrm{are}1, 2,\mathrm{ }3.\mathrm{}\mathrm{So},\\ {\mathrm{a}}_{1}{\mathrm{a}}_{2}+{\mathrm{b}}_{1}{\mathrm{b}}_{2}+{\mathrm{c}}_{1}{\mathrm{c}}_{2}=7.1+\left(-5\right).2+1.3\\ \mathrm{ }=7-10+3\\ \mathrm{ }=0\\ \mathrm{Thus},\mathrm{both}\mathrm{lines}\mathrm{are}\mathrm{perpendicular}\mathrm{to}\mathrm{each}\mathrm{other}\mathrm{.}\end{array}$

Q.19

$\begin{array}{l}\mathrm{Find}\mathrm{ }\mathrm{the}\mathrm{ }\mathrm{shortest}\mathrm{ }\mathrm{distance}\mathrm{ }\mathrm{between}\mathrm{ }\mathrm{the}\mathrm{ }\mathrm{lines}\\ \stackrel{\to }{\mathrm{r}}=\left(\stackrel{^}{\mathrm{i}}+2\stackrel{^}{\mathrm{j}}+\stackrel{^}{\mathrm{k}}\right)+\mathrm{\lambda }\left(\stackrel{^}{\mathrm{i}}-\stackrel{^}{\mathrm{j}}+\stackrel{^}{\mathrm{k}}\right)\mathrm{ }\mathrm{and}\\ \stackrel{\to }{\mathrm{r}}=\left(2\mathrm{ }\stackrel{^}{\mathrm{i}}-\stackrel{^}{\mathrm{j}}+\stackrel{^}{\mathrm{k}}\right)+\mathrm{\mu }\left(2\mathrm{ }\stackrel{^}{\mathrm{i}}+\stackrel{^}{\mathrm{j}}+2\mathrm{ }\stackrel{^}{\mathrm{k}}\right)\end{array}$

Ans.

$\begin{array}{l}\mathrm{The}\mathrm{ }\mathrm{equations}\mathrm{ }\mathrm{of}\mathrm{ }\mathrm{the}\mathrm{ }\mathrm{given}\mathrm{ }\mathrm{lines}\mathrm{ }\mathrm{are}:\\ \mathrm{ }\stackrel{\to }{\mathrm{r}}=\left(\stackrel{^}{\mathrm{i}}+2\stackrel{^}{\mathrm{j}}+\stackrel{^}{\mathrm{k}}\right)+\mathrm{\lambda }\left(\stackrel{^}{\mathrm{i}}-\stackrel{^}{\mathrm{j}}+\stackrel{^}{\mathrm{k}}\right) \mathrm{ }...\left(\mathrm{i}\right)\\ \mathrm{ }\stackrel{\to }{\mathrm{r}}=\left(2\mathrm{ }\stackrel{^}{\mathrm{i}}-\stackrel{^}{\mathrm{j}}-\stackrel{^}{\mathrm{k}}\right)+\mathrm{\mu }\left(2\mathrm{ }\stackrel{^}{\mathrm{i}}+\stackrel{^}{\mathrm{j}}+2\mathrm{ }\stackrel{^}{\mathrm{k}}\right)...\left(\mathrm{ii}\right)\\ \mathrm{Comparing}\mathrm{equatin}\mathrm{ }\left(\mathrm{i}\right)\mathrm{with}\stackrel{\to }{\mathrm{r}}=\stackrel{\to }{{\mathrm{a}}_{1}}+\mathrm{\lambda }\stackrel{\to }{{\mathrm{b}}_{1}}\mathrm{and}\mathrm{equation}\mathrm{ }\left(\mathrm{ii}\right)\\ \mathrm{with} \stackrel{\to }{\mathrm{r}}=\stackrel{\to }{{\mathrm{a}}_{2}}+\mathrm{\mu }\stackrel{\to }{{\mathrm{b}}_{2}},\mathrm{we}\mathrm{get}\\ \stackrel{\to }{{\mathrm{a}}_{1}}=\stackrel{^}{\mathrm{i}}+2\stackrel{^}{\mathrm{j}}+\stackrel{^}{\mathrm{k}},\stackrel{\to }{{\mathrm{b}}_{1}}=\stackrel{^}{\mathrm{i}}-\stackrel{^}{\mathrm{j}}+\stackrel{^}{\mathrm{k}}\\ \mathrm{and} \mathrm{ }\stackrel{\to }{{\mathrm{a}}_{2}}=2\mathrm{ }\stackrel{^}{\mathrm{i}}-\stackrel{^}{\mathrm{j}}-\stackrel{^}{\mathrm{k}},\stackrel{\to }{{\mathrm{b}}_{2}}=2\mathrm{ }\stackrel{^}{\mathrm{i}}+\stackrel{^}{\mathrm{j}}+2\mathrm{ }\stackrel{^}{\mathrm{k}}\\ \mathrm{The}\mathrm{shortest}\mathrm{distance}\mathrm{between}\mathrm{the}\mathrm{lines}\\ \mathrm{d}=\left|\frac{\left(\stackrel{\to }{{\mathrm{b}}_{1}}×\stackrel{\to }{{\mathrm{b}}_{2}}\right).\left(\stackrel{\to }{{\mathrm{a}}_{2}}-\stackrel{\to }{{\mathrm{a}}_{1}}\right)}{\left|\stackrel{\to }{{\mathrm{b}}_{1}}×\stackrel{\to }{{\mathrm{b}}_{2}}\right|}\right|...\left(\mathrm{iii}\right)\\ \mathrm{So},\stackrel{\to }{{\mathrm{b}}_{1}}×\stackrel{\to }{{\mathrm{b}}_{2}}=\left|\begin{array}{ccc}\stackrel{^}{\mathrm{i}}& \stackrel{^}{\mathrm{j}}& \stackrel{^}{\mathrm{k}}\\ 1& -1& 1\\ 2& 1& 2\end{array}\right|\\ \mathrm{ }=\left(-2-1\right)\stackrel{^}{\mathrm{i}}-\left(2-2\right)\stackrel{^}{\mathrm{j}}+\left(1+2\right)\stackrel{^}{\mathrm{k}}\\ \mathrm{ }=-3\mathrm{ }\stackrel{^}{\mathrm{i}}-0.\mathrm{ }\stackrel{^}{\mathrm{j}}+3\mathrm{ }\stackrel{^}{\mathrm{k}}\\ \left|\stackrel{\to }{{\mathrm{b}}_{1}}×\stackrel{\to }{{\mathrm{b}}_{2}}\right|=\left|-3\mathrm{ }\stackrel{^}{\mathrm{i}}-0.\mathrm{ }\stackrel{^}{\mathrm{j}}+3\mathrm{ }\stackrel{^}{\mathrm{k}}\right|\\ \mathrm{ }=\sqrt{{\left(-3\right)}^{2}+{3}^{2}}\\ \mathrm{ }=\sqrt{9+9}\\ \mathrm{ }=3\sqrt{2}\\ \mathrm{ }\stackrel{\to }{{\mathrm{a}}_{2}}-\stackrel{\to }{{\mathrm{a}}_{1}}=\left(2\mathrm{ }\stackrel{^}{\mathrm{i}}-\stackrel{^}{\mathrm{j}}-\stackrel{^}{\mathrm{k}}\right)-\left(\stackrel{^}{\mathrm{i}}+2\stackrel{^}{\mathrm{j}}+\stackrel{^}{\mathrm{k}}\right)\\ =\stackrel{^}{\mathrm{i}}-3\stackrel{^}{\mathrm{j}}-2\stackrel{^}{\mathrm{k}}\\ \mathrm{From}\mathrm{equation}\left(\mathrm{i}\right),\mathrm{we}\mathrm{get}\\ \mathrm{ }\mathrm{d}=\left|\frac{\left(-3\mathrm{ }\stackrel{^}{\mathrm{i}}+3\mathrm{ }\stackrel{^}{\mathrm{k}}\right).\left(\stackrel{^}{\mathrm{i}}-3\stackrel{^}{\mathrm{j}}-2\stackrel{^}{\mathrm{k}}\right)}{3\sqrt{2}}\right|\\ \mathrm{ }=\left|\frac{-3-6}{3\sqrt{2}}\right|\\ \mathrm{ }=\frac{3}{\sqrt{2}}×\frac{\sqrt{2}}{\sqrt{2}}\\ \mathrm{ }=\frac{3\sqrt{2}}{2}\\ \mathrm{Thus},\mathrm{the}\mathrm{distance}\mathrm{between}\mathrm{the}\mathrm{given}\mathrm{two}\mathrm{lines}\mathrm{is}\frac{3\sqrt{2}}{2}\mathrm{units}\mathrm{.}\end{array}$

Q.20

$\begin{array}{l}\mathrm{Find}\mathrm{ }\mathrm{the}\mathrm{ }\mathrm{shortest}\mathrm{ }\mathrm{distance}\mathrm{ }\mathrm{between}\mathrm{ }\mathrm{the}\mathrm{ }\mathrm{lines}\\ \frac{\mathrm{x}+1}{7}=\frac{\mathrm{y}+1}{-\mathrm{ }6}=\frac{\mathrm{z}-1}{1}\mathrm{ }\mathrm{and}\mathrm{ }\frac{\mathrm{x}-3}{1}=\frac{\mathrm{y}-5}{-2}=\frac{\mathrm{z}-7}{1}\end{array}$

Ans.

$\begin{array}{l}\mathrm{The}\mathrm{ }\mathrm{equations}\mathrm{ }\mathrm{of}\mathrm{ }\mathrm{the}\mathrm{ }\mathrm{given}\mathrm{ }\mathrm{lines}\mathrm{ }\mathrm{are}:\\ \frac{\mathrm{X}+1}{7}=\frac{\mathrm{Y}+1}{-6}=\frac{\mathrm{Z}-1}{1}\mathrm{ }\mathrm{and}\mathrm{ }\frac{\mathrm{X}-3}{1}=\frac{\mathrm{Y}-5}{-2}=\frac{\mathrm{Z}-7}{1}\\ \mathrm{Comparing}\mathrm{ }\mathrm{both} \mathrm{the} \mathrm{equations} \mathrm{with}\\ \frac{\mathrm{X}-{\mathrm{X}}_{1}}{{\mathrm{a}}_{1}}=\frac{\mathrm{Y}-{\mathrm{Y}}_{1}}{{\mathrm{b}}_{1}}=\frac{\mathrm{Z}-{\mathrm{Z}}_{1}}{{\mathrm{C}}_{1}} \mathrm{and} \frac{\mathrm{X}-{\mathrm{X}}_{2}}{{\mathrm{a}}_{2}}=\frac{\mathrm{Y}-5}{{\mathrm{b}}_{2}}=\frac{\mathrm{Z}-7}{{\mathrm{C}}_{2}} \mathrm{respectively}.\\ \\ {\mathrm{X}}_{1}=-1, \mathrm{ }{\mathrm{X}}_{2}=3\\ {\mathrm{Y}}_{1}=-1, \mathrm{ }{\mathrm{Y}}_{2}=5\\ {\mathrm{Z}}_{1}=1, \mathrm{ }{\mathrm{Z}}_{2}=7\\ {\mathrm{a}}_{1}=7, \mathrm{ }{\mathrm{a}}_{2}=1\\ {\mathrm{b}}_{1}=-6, \mathrm{ }{\mathrm{b}}_{2}=-2\\ {\mathrm{c}}_{1}=1, \mathrm{ }{\mathrm{c}}_{2}=1\\ \mathrm{Distance} \mathrm{between} \mathrm{two} \mathrm{lines} \mathrm{is} \mathrm{given} \mathrm{by}\\ \mathrm{ }\mathrm{d}=\left|\frac{\left|\begin{array}{ccc}{\mathrm{x}}_{2}-{\mathrm{x}}_{1}& {\mathrm{y}}_{2}-{\mathrm{y}}_{1}& {\mathrm{z}}_{2}-{\mathrm{z}}_{1}\\ {\mathrm{a}}_{1}& {\mathrm{b}}_{1}& {\mathrm{c}}_{1}\\ {\mathrm{a}}_{2}& {\mathrm{b}}_{2}& {\mathrm{c}}_{2}\end{array}\right|}{\sqrt{{\left({\mathrm{b}}_{1}{\mathrm{c}}_{2}-{\mathrm{b}}_{2}{\mathrm{c}}_{1}\right)}^{2}+{\left({\mathrm{c}}_{1}{\mathrm{a}}_{2}-{\mathrm{c}}_{2}{\mathrm{a}}_{1}\right)}^{2}+{\left({\mathrm{a}}_{1}{\mathrm{b}}_{2}-{\mathrm{a}}_{2}{\mathrm{b}}_{1}\right)}^{2}}}\right|\\ =\left|\frac{\left|\begin{array}{ccc}3-\left(-1\right)& 5-\left(-1\right)& 7-\left(-1\right)\\ 7& -6& 1\\ 1& -2& 1\end{array}\right|}{\sqrt{{\left\{-6×1-\left(-2\right).1\right\}}^{2}+{\left(1×1-1×7\right)}^{2}+{\left(7×-2-1×-6\right)}^{2}}}\right|\\ \mathrm{d}=\left|\frac{\left|\begin{array}{ccc}4& 6& 8\\ 7& -6& 1\\ 1& -2& 1\end{array}\right|}{\sqrt{{\left(-4\right)}^{2}+{\left(-6\right)}^{2}+{\left(-8\right)}^{2}}}\right|\\ =\left|\frac{4\left(-6+2\right)-6\left(7-1\right)+8\left(-14+6\right)}{\sqrt{16+36+64}}\right|\\ =\left|\frac{-16-36-64}{\sqrt{116}}\right|\\ =\left|\frac{-116}{2\sqrt{29}}\right|\\ =\left|-2\sqrt{29}\right|\\ \mathrm{So},\mathrm{ }\mathrm{the}\mathrm{ }\mathrm{distance}\mathrm{ }\mathrm{between}\mathrm{ }\mathrm{two}\mathrm{ }\mathrm{given}\mathrm{ }\mathrm{lines}\mathrm{ }\mathrm{is} 2\sqrt{29}\mathrm{ }\mathrm{units}.\end{array}$

Q.21

$\begin{array}{l}\mathrm{Find}\mathrm{ }\mathrm{the}\mathrm{ }\mathrm{shortest}\mathrm{ }\mathrm{distance}\mathrm{ }\mathrm{between}\mathrm{ }\mathrm{the}\mathrm{ }\mathrm{lines}\mathrm{ }\mathrm{whose}\mathrm{ }\mathrm{vector}\\ \mathrm{equations}\mathrm{ }\mathrm{are}\\ \stackrel{\to }{\mathrm{r}}=\left(\stackrel{^}{\mathrm{i}}+2\stackrel{^}{\mathrm{j}}+3\stackrel{^}{\mathrm{k}}\right)+\mathrm{\lambda }\left(\stackrel{^}{\mathrm{i}}-\mathrm{ }\stackrel{^}{\mathrm{j}}+2\mathrm{ }\stackrel{^}{\mathrm{k}}\right)\mathrm{ }\mathrm{and}\\ \stackrel{\to }{\mathrm{r}}=\left(4\mathrm{ }\stackrel{^}{\mathrm{i}}+\mathrm{ }\stackrel{^}{\mathrm{j}}+6\mathrm{ }\stackrel{^}{\mathrm{k}}\right)+\mathrm{\mu }\left(2\mathrm{ }\stackrel{^}{\mathrm{i}}+3\mathrm{ }\stackrel{^}{\mathrm{j}}+\mathrm{ }\stackrel{^}{\mathrm{k}}\right)\end{array}$

Ans.

$\begin{array}{l}\mathrm{The}\mathrm{ }\mathrm{equations}\mathrm{ }\mathrm{of}\mathrm{ }\mathrm{the}\mathrm{ }\mathrm{given}\mathrm{ }\mathrm{lines}\mathrm{ }\mathrm{are}:\\ \stackrel{\to }{\mathrm{r}}=\left(\stackrel{^}{\mathrm{i}}+2\stackrel{^}{\mathrm{j}}+3\stackrel{^}{\mathrm{k}}\right)+\mathrm{\lambda }\left(\stackrel{^}{\mathrm{i}}-3\mathrm{ }\stackrel{^}{\mathrm{j}}+2\mathrm{ }\stackrel{^}{\mathrm{k}}\right)...\left(\mathrm{i}\right)\\ \stackrel{\to }{\mathrm{r}}=\left(4\mathrm{ }\stackrel{^}{\mathrm{i}}+5\mathrm{ }\stackrel{^}{\mathrm{j}}+6\mathrm{ }\stackrel{^}{\mathrm{k}}\right)+\mathrm{\mu }\left(2\mathrm{ }\stackrel{^}{\mathrm{i}}+3\mathrm{ }\stackrel{^}{\mathrm{j}}+\mathrm{ }\stackrel{^}{\mathrm{k}}\right) \mathrm{ }...\left(\mathrm{ii}\right)\\ \mathrm{Comparing}\mathrm{ }\mathrm{equatin}\mathrm{ }\left(\mathrm{i}\right)\mathrm{ }\mathrm{with}\mathrm{ }\stackrel{\to }{\mathrm{r}}=\stackrel{\to }{{\mathrm{a}}_{1}}+\mathrm{\lambda }\stackrel{\to }{{\mathrm{b}}_{1}}\mathrm{ }\mathrm{and}\mathrm{ }\mathrm{equation}\mathrm{ }\left(\mathrm{ii}\right)\\ \mathrm{with} \mathrm{ }\stackrel{\to }{\mathrm{r}}=\stackrel{\to }{{\mathrm{a}}_{2}}+\mathrm{\mu }\stackrel{\to }{{\mathrm{b}}_{2}},\mathrm{we}\mathrm{get}\\ \stackrel{\to }{{\mathrm{a}}_{1}}=\stackrel{^}{\mathrm{i}}+2\stackrel{^}{\mathrm{j}}+3\mathrm{ }\stackrel{^}{\mathrm{k}},\stackrel{\to }{{\mathrm{b}}_{1}}=\stackrel{^}{\mathrm{i}}-3\mathrm{ }\stackrel{^}{\mathrm{j}}+2\mathrm{ }\stackrel{^}{\mathrm{k}}\\ \mathrm{and} \mathrm{ }\stackrel{\to }{{\mathrm{a}}_{2}}=4\mathrm{ }\stackrel{^}{\mathrm{i}}+5\mathrm{ }\stackrel{^}{\mathrm{j}}+6\mathrm{ }\stackrel{^}{\mathrm{k}},\stackrel{\to }{{\mathrm{b}}_{2}}=2\mathrm{ }\stackrel{^}{\mathrm{i}}+3\mathrm{ }\stackrel{^}{\mathrm{j}}+\mathrm{ }\stackrel{^}{\mathrm{k}}\\ \mathrm{The}\mathrm{ }\mathrm{shortest}\mathrm{ }\mathrm{distance}\mathrm{ }\mathrm{between}\mathrm{ }\mathrm{the}\mathrm{ }\mathrm{lines}\\ \mathrm{d}=\left|\frac{\left(\stackrel{\to }{{\mathrm{b}}_{1}}×\stackrel{\to }{{\mathrm{b}}_{2}}\right).\left(\stackrel{\to }{{\mathrm{a}}_{2}}-\stackrel{\to }{{\mathrm{a}}_{1}}\right)}{\left|\stackrel{\to }{{\mathrm{b}}_{1}}×\stackrel{\to }{{\mathrm{b}}_{2}}\right|}\right|...\left(\mathrm{iii}\right)\\ \mathrm{So},\mathrm{ }\stackrel{\to }{{\mathrm{b}}_{1}}×\stackrel{\to }{{\mathrm{b}}_{2}}=\left|\begin{array}{ccc}\stackrel{^}{\mathrm{i}}& \stackrel{^}{\mathrm{j}}& \stackrel{^}{\mathrm{k}}\\ 1& -3& 2\\ 2& 3& 1\end{array}\right|\\ \mathrm{ }=\left(-3-6\right)\stackrel{^}{\mathrm{i}}-\left(1-4\right)\stackrel{^}{\mathrm{j}}+\left(3+6\right)\stackrel{^}{\mathrm{k}}\\ \mathrm{ }=-9\mathrm{ }\stackrel{^}{\mathrm{i}}+3\mathrm{ }\stackrel{^}{\mathrm{j}}+9\mathrm{ }\stackrel{^}{\mathrm{k}}\\ \left|\stackrel{\to }{{\mathrm{b}}_{1}}×\stackrel{\to }{{\mathrm{b}}_{2}}\right|=\left|-9\mathrm{ }\stackrel{^}{\mathrm{i}}+3\mathrm{ }\stackrel{^}{\mathrm{j}}+9\mathrm{ }\stackrel{^}{\mathrm{k}}\right|\\ \mathrm{ }=\sqrt{{\left(-9\right)}^{2}+{3}^{2}+{9}^{2}}\\ \mathrm{ }=\sqrt{81+9+81}\\ \mathrm{ }=\sqrt{171}=3\sqrt{19}\\ \mathrm{ }\stackrel{\to }{{\mathrm{a}}_{2}}-\stackrel{\to }{{\mathrm{a}}_{1}}=\left(4\mathrm{ }\stackrel{^}{\mathrm{i}}+5\mathrm{ }\stackrel{^}{\mathrm{j}}+6\mathrm{ }\stackrel{^}{\mathrm{k}}\right)-\left(\stackrel{^}{\mathrm{i}}+2\stackrel{^}{\mathrm{j}}+3\mathrm{ }\stackrel{^}{\mathrm{k}}\right)\\ =3\mathrm{ }\stackrel{^}{\mathrm{i}}+3\stackrel{^}{\mathrm{j}}+3\mathrm{ }\stackrel{^}{\mathrm{k}}\\ \mathrm{From}\mathrm{ }\mathrm{equation}\mathrm{ }\left(\mathrm{i}\right),\mathrm{ }\mathrm{we}\mathrm{ }\mathrm{get}\\ \mathrm{ }\mathrm{d}=\left|\frac{\left(-9\mathrm{ }\stackrel{^}{\mathrm{i}}+3\mathrm{ }\stackrel{^}{\mathrm{j}}+9\mathrm{ }\stackrel{^}{\mathrm{k}}\right).\left(3\mathrm{ }\stackrel{^}{\mathrm{i}}+3\stackrel{^}{\mathrm{j}}+3\mathrm{ }\stackrel{^}{\mathrm{k}}\right)}{3\sqrt{19}}\right|\\ \mathrm{ }=\left|\frac{3\left(-3\mathrm{ }\stackrel{^}{\mathrm{i}}+\mathrm{ }\stackrel{^}{\mathrm{j}}+3\mathrm{ }\stackrel{^}{\mathrm{k}}\right).\left(\mathrm{ }\stackrel{^}{\mathrm{i}}+\stackrel{^}{\mathrm{j}}+\mathrm{ }\stackrel{^}{\mathrm{k}}\right)}{\sqrt{19}}\right|\\ \mathrm{ }=\left|\frac{3\left(-3+1+3\right)}{\sqrt{19}}\right|\\ \mathrm{ }=\frac{3}{\sqrt{19}}×\frac{\sqrt{19}}{\sqrt{19}}\\ \mathrm{ }=\frac{3\sqrt{19}}{19}\\ \mathrm{So},\mathrm{ }\mathrm{the}\mathrm{ }\mathrm{distance}\mathrm{ }\mathrm{between}\mathrm{ }\mathrm{two}\mathrm{ }\mathrm{given}\mathrm{ }\mathrm{lines}\mathrm{ }\mathrm{is}\mathrm{ }\frac{3\sqrt{19}}{19}\mathrm{ }\mathrm{units}.\end{array}$

Q.22

$\begin{array}{l}\mathrm{Find}\mathrm{ }\mathrm{the}\mathrm{ }\mathrm{shortest}\mathrm{ }\mathrm{distance}\mathrm{ }\mathrm{between}\mathrm{ }\mathrm{the}\mathrm{ }\mathrm{lines}\mathrm{ }\mathrm{whose}\mathrm{ }\mathrm{vector}\\ \mathrm{equations}\mathrm{ }\mathrm{are}\\ \stackrel{\to }{\mathrm{r}}=\left(1-\mathrm{t}\right)\stackrel{^}{\mathrm{i}}+\left(\mathrm{t}-2\right)\stackrel{^}{\mathrm{j}}+\left(3-2\mathrm{t}\right)\stackrel{^}{\mathrm{k}}\mathrm{ }\mathrm{and}\\ \stackrel{\to }{\mathrm{r}}=\left(\mathrm{s}+1\right)\stackrel{^}{\mathrm{i}}+\left(2\mathrm{s}-1\right)\stackrel{^}{\mathrm{j}}-\left(2\mathrm{s}+1\right)\stackrel{^}{\mathrm{k}}\end{array}$

Ans.

$\begin{array}{l}\mathrm{The}\mathrm{ }\mathrm{equations}\mathrm{ }\mathrm{of}\mathrm{ }\mathrm{the}\mathrm{ }\mathrm{given}\mathrm{ }\mathrm{lines}\mathrm{ }\mathrm{are}:\\ \stackrel{\to }{\mathrm{r}}=\left(1-\mathrm{t}\right)\stackrel{^}{\mathrm{i}}+\left(\mathrm{t}-2\right)\stackrel{^}{\mathrm{j}}+\left(3-2\mathrm{t}\right)\stackrel{^}{\mathrm{k}}\\ \stackrel{\to }{\mathrm{r}}=\left(\stackrel{^}{\mathrm{i}}-2\mathrm{ }\stackrel{^}{\mathrm{j}}+3\mathrm{ }\stackrel{^}{\mathrm{k}}\right)+\mathrm{t}\left(-\stackrel{^}{\mathrm{i}}+\stackrel{^}{\mathrm{j}}-2\mathrm{ }\stackrel{^}{\mathrm{k}}\right)...\left(\mathrm{i}\right)\\ \stackrel{\to }{\mathrm{r}}=\left(\mathrm{s}+1\right)\stackrel{^}{\mathrm{i}}+\left(2\mathrm{s}-1\right)\stackrel{^}{\mathrm{j}}-\left(2\mathrm{s}+1\right)\stackrel{^}{\mathrm{k}}\\ \stackrel{\to }{\mathrm{r}}=\left(\stackrel{^}{\mathrm{i}}-\stackrel{^}{\mathrm{j}}-\stackrel{^}{\mathrm{k}}\right)+\mathrm{s}\left(\stackrel{^}{\mathrm{i}}+2\stackrel{^}{\mathrm{j}}-2\mathrm{ }\stackrel{^}{\mathrm{k}}\right) \mathrm{ }...\left(\mathrm{ii}\right)\\ \mathrm{Comparing}\mathrm{ }\mathrm{equatin}\mathrm{ }\left(\mathrm{i}\right)\mathrm{ }\mathrm{with}\mathrm{ }\stackrel{\to }{\mathrm{r}}=\stackrel{\to }{{\mathrm{a}}_{1}}+\mathrm{\lambda }\stackrel{\to }{{\mathrm{b}}_{1}}\mathrm{ }\mathrm{and}\mathrm{ }\mathrm{equation}\mathrm{ }\left(\mathrm{ii}\right)\\ \mathrm{with} \mathrm{ }\stackrel{\to }{\mathrm{r}}=\stackrel{\to }{{\mathrm{a}}_{2}}+\mathrm{\mu }\stackrel{\to }{{\mathrm{b}}_{2}},\mathrm{we}\mathrm{get}\\ \stackrel{\to }{{\mathrm{a}}_{1}}=\stackrel{^}{\mathrm{i}}-2\mathrm{ }\stackrel{^}{\mathrm{j}}+3\mathrm{ }\stackrel{^}{\mathrm{k}},\stackrel{\to }{{\mathrm{b}}_{1}}=-\stackrel{^}{\mathrm{i}}+\stackrel{^}{\mathrm{j}}-2\mathrm{ }\stackrel{^}{\mathrm{k}}\\ \mathrm{and} \mathrm{ }\stackrel{\to }{{\mathrm{a}}_{2}}=\stackrel{^}{\mathrm{i}}-\stackrel{^}{\mathrm{j}}-\stackrel{^}{\mathrm{k}},\stackrel{\to }{{\mathrm{b}}_{2}}=\stackrel{^}{\mathrm{i}}+2\stackrel{^}{\mathrm{j}}-2\mathrm{ }\stackrel{^}{\mathrm{k}}\\ \mathrm{The}\mathrm{ }\mathrm{shortest}\mathrm{ }\mathrm{distance}\mathrm{ }\mathrm{between}\mathrm{ }\mathrm{the}\mathrm{ }\mathrm{lines}\\ \mathrm{d}=\left|\frac{\left(\stackrel{\to }{{\mathrm{b}}_{1}}×\stackrel{\to }{{\mathrm{b}}_{2}}\right).\left(\stackrel{\to }{{\mathrm{a}}_{2}}-\stackrel{\to }{{\mathrm{a}}_{1}}\right)}{\left|\stackrel{\to }{{\mathrm{b}}_{1}}×\stackrel{\to }{{\mathrm{b}}_{2}}\right|}\right|...\left(\mathrm{iii}\right)\\ \mathrm{So},\stackrel{\to }{{\mathrm{b}}_{1}}×\stackrel{\to }{{\mathrm{b}}_{2}}=\left|\begin{array}{ccc}\stackrel{^}{\mathrm{i}}& \stackrel{^}{\mathrm{j}}& \stackrel{^}{\mathrm{k}}\\ -1& 1& -2\\ 1& 2& -2\end{array}\right|\\ \mathrm{ }=\left(-2+4\right)\stackrel{^}{\mathrm{i}}-\left(2+2\right)\stackrel{^}{\mathrm{j}}+\left(-2-1\right)\stackrel{^}{\mathrm{k}}\\ \mathrm{ }=2\mathrm{ }\stackrel{^}{\mathrm{i}}-4\stackrel{^}{\mathrm{j}}-3\mathrm{ }\stackrel{^}{\mathrm{k}}\\ \therefore \left|\stackrel{\to }{{\mathrm{b}}_{1}}×\stackrel{\to }{{\mathrm{b}}_{2}}\right|=\left|2\mathrm{ }\stackrel{^}{\mathrm{i}}-4\stackrel{^}{\mathrm{j}}-3\mathrm{ }\stackrel{^}{\mathrm{k}}\right|\\ \mathrm{ }=\sqrt{{2}^{2}+{\left(-4\right)}^{2}+{\left(-3\right)}^{2}}\\ \mathrm{ }=\sqrt{4+16+9}\\ \mathrm{ }=\sqrt{29}\\ \mathrm{ }\stackrel{\to }{{\mathrm{a}}_{2}}-\stackrel{\to }{{\mathrm{a}}_{1}}=\left(\mathrm{ }\stackrel{^}{\mathrm{i}}-\mathrm{ }\stackrel{^}{\mathrm{j}}-\mathrm{ }\stackrel{^}{\mathrm{k}}\right)-\left(\mathrm{ }\stackrel{^}{\mathrm{i}}-2\stackrel{^}{\mathrm{j}}+3\mathrm{ }\stackrel{^}{\mathrm{k}}\right)\\ =\stackrel{^}{\mathrm{j}}-4\stackrel{^}{\mathrm{k}}\\ \mathrm{From}\mathrm{ }\mathrm{equation}\mathrm{ }\left(\mathrm{i}\right),\mathrm{ }\mathrm{we}\mathrm{ }\mathrm{get}\\ \mathrm{ }\mathrm{d}=\left|\frac{\left(2\mathrm{ }\stackrel{^}{\mathrm{i}}-4\stackrel{^}{\mathrm{j}}-3\mathrm{ }\stackrel{^}{\mathrm{k}}\right).\left(\stackrel{^}{\mathrm{j}}-4\stackrel{^}{\mathrm{k}}\right)}{\sqrt{29}}\right|\\ \mathrm{ }=\left|\frac{-4+12}{\sqrt{29}}\right|\\ \mathrm{ }=\frac{8}{\sqrt{29}}×\frac{\sqrt{29}}{\sqrt{29}}\mathrm{ }=\frac{8\sqrt{29}}{29}\\ \mathrm{Therefore},\mathrm{ }\mathrm{the}\mathrm{ }\mathrm{shortest}\mathrm{ }\mathrm{distance}\mathrm{ }\mathrm{between}\mathrm{ }\mathrm{two}\mathrm{ }\mathrm{lines}\mathrm{ }\mathrm{is}\\ \frac{8\sqrt{29}}{29} \mathrm{units}\mathrm{.}\end{array}$

Q.23 In each of the following cases, determine the direction cosines of the normal to the plane and the distance from the origin.
(a) z = 2 (b) x + y + z = 1
(c) 2x + 3y – z = 5 (d) 5y + 8 = 0

Ans.

$\begin{array}{l}\left(\text{a}\right)\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{The equatin of the plane is z=2}⇒\text{0}\text{.x+0}\text{.y+z=2}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\dots \left(\text{i}\right)\\ \text{The direction ratios of normal are 0, 0 and 1}\text{.}\\ \therefore \sqrt{{\text{0}}^{\text{2}}{\text{+0}}^{\text{2}}{\text{+1}}^{\text{2}}}\text{=1}\\ \text{So, according to rule, dividing equation}\text{ }\left(\text{i}\right)\text{by 1, we get}\\ \text{0}\text{.x+0}\text{.y+z=2, which is in the form of lx+my+nz=d,}\\ \text{where direction cosines, l=0,m=0 and n=1 and the}\\ \text{distance of the plane from origin is 2 units}\text{.}\\ \left(\text{b}\right)\text{x+y+z=1}\text{}\dots \left(\text{i}\right)\\ \text{The direction ratios of normal are 1, 1 and 1}\text{.}\\ \text{}\sqrt{{\text{1}}^{\text{2}}{\text{+1}}^{\text{2}}{\text{+1}}^{\text{2}}}\text{=}\sqrt{\text{3}}\\ \text{Dividing equation}\left(\text{i}\right)\text{by}\sqrt{\text{3}}\text{, we get}\\ \text{}\frac{\text{x}}{\sqrt{\text{3}}}\text{+}\frac{\text{y}}{\sqrt{\text{3}}}\text{+}\frac{\text{z}}{\sqrt{\text{3}}}\text{=}\frac{\text{1}}{\sqrt{\text{3}}}\text{}\dots \left(\text{ii}\right)\\ \text{Comparing equation}\left(\text{ii}\right)\text{with the equation lx+my+nz=d,we get}\\ \text{Direction cosines:}\frac{\text{1}}{\sqrt{\text{3}}}\text{,}\frac{\text{1}}{\sqrt{\text{3}}}\text{,}\frac{\text{1}}{\sqrt{\text{3}}}\text{and the distance of plane from}\\ \text{the origin is}\frac{\text{1}}{\sqrt{\text{3}}}\text{units}\text{.}\\ \left(\text{c}\right)\text{2x+3y-z=5}\text{}\dots \left(\text{i}\right)\\ \text{The direction ratios of normal are 2, 3 and-1}\text{.}\\ \text{}\sqrt{{\text{2}}^{\text{2}}{\text{+3}}^{\text{2}}\text{+}{\left(\text{-1}\right)}^{\text{2}}}\text{=}\sqrt{\text{14}}\\ \text{Dividing equation}\left(\text{i}\right)\text{by}\sqrt{\text{14}}\text{, we get}\\ \text{}\frac{\text{2x}}{\sqrt{\text{14}}}\text{+}\frac{\text{3y}}{\sqrt{\text{14}}}\text{–}\frac{\text{z}}{\sqrt{\text{14}}}\text{=}\frac{\text{5}}{\sqrt{\text{14}}}\text{}\dots \left(\text{ii}\right)\\ \text{Comparing equation}\left(\text{ii}\right)\text{with the equation lx+my+nz=d,we get}\\ \text{Direction cosines:}\frac{\text{2}}{\sqrt{\text{14}}}\text{,}\frac{\text{3}}{\sqrt{\text{14}}}\text{,}\frac{\text{-1}}{\sqrt{\text{14}}}\text{and the distance of plane from}\\ \text{the origin is}\frac{\text{5}}{\sqrt{\text{14}}}\text{units}\text{.}\\ \left(\text{d}\right)\text{}\text{}\text{}\text{5y+8=0}\\ ⇒\text{-0x-5y-0}\text{\hspace{0.17em}}\text{z=8}\text{}\dots \left(\text{i}\right)\\ \text{The direction ratios of normal are 0, -5 and}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{0}\text{.}\\ \text{}\sqrt{{\text{0}}^{\text{2}}\text{+}{\left(\text{-5}\right)}^{\text{2}}{\text{+0}}^{\text{2}}}\text{=5}\\ \text{Dividing equation}\left(\text{i}\right)\text{by 5, we get}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{–}\frac{\text{0x}}{\text{5}}\text{–}\frac{\text{5y}}{\text{5}}\text{–}\frac{\text{0z}}{\text{5}}\text{=}\frac{\text{8}}{\text{5}}\text{}\dots \left(\text{ii}\right)\\ \text{Comparing equation}\left(\text{ii}\right)\text{with the equation lx+my+nz=d,we get}\\ \text{Direction cosines: 0,-1,0 and the distance of plane from}\\ \text{the origin is}\text{\hspace{0.17em}}\frac{\text{8}}{\text{5}}\text{}\text{\hspace{0.17em}}\text{units}\text{.}\end{array}$

Q.24 Find the vector equation of a plane which is at a distance of 7 units from the origin and normal to the vector

$3\text{\hspace{0.17em}}\stackrel{^}{i}+5\text{\hspace{0.17em}}\stackrel{^}{j}-6\text{\hspace{0.17em}}\stackrel{^}{\text{k}}.$

Ans.

$\begin{array}{l}\mathrm{Here},\mathrm{ }\stackrel{\to }{\mathrm{n}}=3\mathrm{ }\stackrel{^}{\mathrm{i}}+5\mathrm{ }\stackrel{^}{\mathrm{j}}-6\mathrm{ }\stackrel{^}{\mathrm{k}}\\ \mathrm{ }\stackrel{^}{\mathrm{n}}=\frac{\stackrel{\to }{\mathrm{n}}}{\left|\stackrel{\to }{\mathrm{n}}\right|}\\ =\frac{3\mathrm{ }\stackrel{^}{\mathrm{i}}+5\mathrm{ }\stackrel{^}{\mathrm{j}}-6\mathrm{ }\stackrel{^}{\mathrm{k}}}{\left|3\mathrm{ }\stackrel{^}{\mathrm{i}}+5\mathrm{ }\stackrel{^}{\mathrm{j}}-6\mathrm{ }\stackrel{^}{\mathrm{k}}\right|}\\ =\frac{3\mathrm{ }\stackrel{^}{\mathrm{i}}+5\mathrm{ }\stackrel{^}{\mathrm{j}}-6\mathrm{ }\stackrel{^}{\mathrm{k}}}{\sqrt{{3}^{2}+{5}^{2}+{\left(-6\right)}^{2}}}\\ =\frac{3\mathrm{ }\stackrel{^}{\mathrm{i}}+5\mathrm{ }\stackrel{^}{\mathrm{j}}-6\mathrm{ }\stackrel{^}{\mathrm{k}}}{\sqrt{9+25+36}}\\ =\frac{3\mathrm{ }\stackrel{^}{\mathrm{i}}+5\mathrm{ }\stackrel{^}{\mathrm{j}}-6\mathrm{ }\stackrel{^}{\mathrm{k}}}{\sqrt{70}}\\ \mathrm{The}\mathrm{ }\mathrm{equation}\mathrm{ }\mathrm{of}\mathrm{ }\mathrm{the}\mathrm{ }\mathrm{plane}\mathrm{ }\mathrm{with}\mathrm{ }\mathrm{position}\mathrm{ }\mathrm{vector}\mathrm{ }\stackrel{\to }{\mathrm{r}}\mathrm{ }\mathrm{is}\mathrm{ }\mathrm{given}\mathrm{ }\mathrm{by}\\ \mathrm{ }\stackrel{\to }{\mathrm{r}}.\stackrel{^}{\mathrm{n}}=\mathrm{d}\\ \mathrm{ }\stackrel{\to }{\mathrm{r}}\left(\frac{3\mathrm{ }\stackrel{^}{\mathrm{i}}+5\mathrm{ }\stackrel{^}{\mathrm{j}}-6\mathrm{ }\stackrel{^}{\mathrm{k}}}{\sqrt{70}}\right)=7\\ \mathrm{This}\mathrm{ }\mathrm{is}\mathrm{ }\mathrm{the}\mathrm{ }\mathrm{vector}\mathrm{ }\mathrm{eqution}\mathrm{ }\mathrm{of} \mathrm{the}\mathrm{ }\mathrm{plane}.\end{array}$

Q.25

$\begin{array}{l}\mathrm{Find}\mathrm{ }\mathrm{the}\mathrm{ }\mathrm{cartesian}\mathrm{ }\mathrm{equation}\mathrm{ }\mathrm{of}\mathrm{ }\mathrm{the}\mathrm{ }\mathrm{following} \mathrm{planes}:\\ \left(\mathrm{a}\right) \stackrel{\to }{\mathrm{r}}.\left(\stackrel{^}{\mathrm{i}}+\stackrel{^}{\mathrm{j}}-\stackrel{^}{\mathrm{k}}\right)=2\\ \left(\mathrm{b}\right)\mathrm{ }\stackrel{\to }{\mathrm{r}}.\left(2\mathrm{ }\stackrel{^}{\mathrm{i}}+3\stackrel{^}{\mathrm{j}}-4\stackrel{^}{\mathrm{k}}\right)=1\\ \left(\mathrm{c}\right)\stackrel{\to }{\mathrm{r}}.\left[\left(\mathrm{s}-2\mathrm{t}\right)\stackrel{^}{\mathrm{i}}+\left(3-\mathrm{t}\right)\stackrel{^}{\mathrm{j}}+\left(2\mathrm{s}+\mathrm{t}\right)\stackrel{^}{\mathrm{k}}\right]=15\end{array}$

Ans.

$\begin{array}{l}\left(\mathrm{a}\right)\mathrm{ }\mathrm{The}\mathrm{ }\mathrm{given}\mathrm{ }\mathrm{equation}\mathrm{ }\mathrm{of}\mathrm{ }\mathrm{the}\mathrm{ }\mathrm{plane}\mathrm{ }\mathrm{is}\\ \stackrel{\to }{\mathrm{r}}\left(\stackrel{^}{\mathrm{i}}+\stackrel{^}{\mathrm{j}}-\stackrel{^}{\mathrm{k}}\right)=2...\left(\mathrm{i}\right)\\ \mathrm{For}\mathrm{ }\mathrm{any}\mathrm{ }\mathrm{arbitrary}\mathrm{ }\mathrm{point}\mathrm{ }\mathrm{P}\left(\mathrm{x},\mathrm{y},\mathrm{z}\right)\mathrm{ }\mathrm{on}\mathrm{ }\mathrm{the}\mathrm{ }\mathrm{plane},\mathrm{ }\mathrm{position}\mathrm{ }\mathrm{vector}\\ \stackrel{\to }{\mathrm{r}}\mathrm{ }\mathrm{is}\mathrm{ }\mathrm{given}\mathrm{ }\mathrm{by},\mathrm{ }\stackrel{\to }{\mathrm{r}}=\mathrm{x}\stackrel{^}{\mathrm{i}}+\mathrm{y}\stackrel{^}{\mathrm{j}}+\mathrm{z}\mathrm{ }\stackrel{^}{\mathrm{k}}.\\ \mathrm{Substitutin}\mathrm{ }\mathrm{value}\mathrm{ }\mathrm{of}\mathrm{ }\stackrel{\to }{\mathrm{r}} \mathrm{in}\mathrm{ }\mathrm{equation}\mathrm{ }\left(\mathrm{i}\right),\mathrm{ }\mathrm{we}\mathrm{ }\mathrm{get}\\ \left(\mathrm{x}\stackrel{^}{\mathrm{i}}+\mathrm{y}\stackrel{^}{\mathrm{j}}+\mathrm{z}\stackrel{^}{\mathrm{k}}\right)\left(\stackrel{^}{\mathrm{i}}+\stackrel{^}{\mathrm{j}}-\stackrel{^}{\mathrm{k}}\right)=2\\ \mathrm{x}+\mathrm{y}-\mathrm{z}=2\\ \mathrm{This}\mathrm{ }\mathrm{is}\mathrm{ }\mathrm{the}\mathrm{ }\mathrm{cartesian}\mathrm{ }\mathrm{equation}\mathrm{ }\mathrm{of}\mathrm{ }\mathrm{the}\mathrm{ }\mathrm{plane}.\\ \left(\mathrm{b}\right)\mathrm{ }\mathrm{The}\mathrm{ }\mathrm{given}\mathrm{ }\mathrm{equation}\mathrm{ }\mathrm{of}\mathrm{ }\mathrm{the}\mathrm{ }\mathrm{plane}\mathrm{ }\mathrm{is}\\ \stackrel{\to }{\mathrm{r}}\left(2\mathrm{ }\stackrel{^}{\mathrm{i}}+3\mathrm{ }\stackrel{^}{\mathrm{j}}-4\mathrm{ }\stackrel{^}{\mathrm{k}}\right)=1...\left(\mathrm{i}\right)\\ \mathrm{For}\mathrm{ }\mathrm{any}\mathrm{ }\mathrm{arbitrary}\mathrm{ }\mathrm{point}\mathrm{ }\mathrm{P}\left(\mathrm{x},\mathrm{y},\mathrm{z}\right)\mathrm{ }\mathrm{on}\mathrm{ }\mathrm{the}\mathrm{ }\mathrm{plane},\mathrm{ }\mathrm{position}\mathrm{ }\mathrm{vector}\\ \stackrel{\to }{\mathrm{r}}\mathrm{ }\mathrm{is}\mathrm{ }\mathrm{given}\mathrm{ }\mathrm{by},\mathrm{ }\stackrel{\to }{\mathrm{r}}=\mathrm{x}\stackrel{^}{\mathrm{i}}+\mathrm{y}\stackrel{^}{\mathrm{j}}+\mathrm{z}\stackrel{^}{\mathrm{k}}\\ \mathrm{Substitutin}\mathrm{ }\mathrm{value}\mathrm{ }\mathrm{of}\mathrm{ }\stackrel{\to }{\mathrm{r}} \mathrm{in}\mathrm{ }\mathrm{equation}\mathrm{ }\left(\mathrm{i}\right),\mathrm{ }\mathrm{we}\mathrm{ }\mathrm{get}\\ \left(\mathrm{x}\stackrel{^}{\mathrm{i}}+\mathrm{y}\stackrel{^}{\mathrm{j}}+\mathrm{z}\stackrel{^}{\mathrm{k}}\right)\left(2\mathrm{ }\stackrel{^}{\mathrm{i}}+3\mathrm{ }\stackrel{^}{\mathrm{j}}-4\mathrm{ }\stackrel{^}{\mathrm{k}}\right)=1\\ 2\mathrm{x}+3\mathrm{y}-4\mathrm{z}=1\\ \mathrm{This}\mathrm{ }\mathrm{is}\mathrm{ }\mathrm{the}\mathrm{ }\mathrm{cartesian}\mathrm{ }\mathrm{equation}\mathrm{ }\mathrm{of}\mathrm{ }\mathrm{the}\mathrm{ }\mathrm{plane}.\\ \left(\mathrm{c}\right)\mathrm{ }\mathrm{The}\mathrm{ }\mathrm{given}\mathrm{ }\mathrm{equation}\mathrm{ }\mathrm{of}\mathrm{ }\mathrm{the}\mathrm{ }\mathrm{plane}\mathrm{ }\mathrm{is}\\ \stackrel{\to }{\mathrm{r}}.\left[\left(\mathrm{s}-2\mathrm{t}\right)\stackrel{^}{\mathrm{i}}+\left(3-\mathrm{t}\right)\stackrel{^}{\mathrm{j}}+\left(2\mathrm{s}+\mathrm{t}\right)\stackrel{^}{\mathrm{k}}\right]=15...\left(\mathrm{i}\right)\\ \mathrm{For}\mathrm{ }\mathrm{any}\mathrm{ }\mathrm{arbitrary}\mathrm{ }\mathrm{point}\mathrm{ }\mathrm{P}\left(\mathrm{x},\mathrm{y},\mathrm{z}\right)\mathrm{ }\mathrm{on}\mathrm{ }\mathrm{the}\mathrm{ }\mathrm{plane},\mathrm{ }\mathrm{position}\mathrm{ }\mathrm{vector}\\ \stackrel{\to }{\mathrm{r}}\mathrm{ }\mathrm{is}\mathrm{ }\mathrm{given}\mathrm{ }\mathrm{by},\mathrm{ }\stackrel{\to }{\mathrm{r}}=\mathrm{x}\stackrel{^}{\mathrm{i}}+\mathrm{y}\stackrel{^}{\mathrm{j}}+\mathrm{z}\stackrel{^}{\mathrm{k}}\\ \mathrm{Substitutin}\mathrm{ }\mathrm{value}\mathrm{ }\mathrm{of}\mathrm{ }\stackrel{\to }{\mathrm{r}} \mathrm{in}\mathrm{ }\mathrm{equation}\mathrm{ }\left(\mathrm{i}\right),\mathrm{ }\mathrm{we}\mathrm{ }\mathrm{get}\\ \left(\mathrm{x}\stackrel{^}{\mathrm{i}}+\mathrm{y}\stackrel{^}{\mathrm{j}}+\mathrm{z}\stackrel{^}{\mathrm{k}}\right).\left[\left(\mathrm{s}-2\mathrm{t}\right)\stackrel{^}{\mathrm{i}}+\left(3-\mathrm{t}\right)\stackrel{^}{\mathrm{j}}+\left(2\mathrm{s}+\mathrm{t}\right)\stackrel{^}{\mathrm{k}}\right]=1\\ \left(\mathrm{s}-2\mathrm{t}\right)\mathrm{x}+\left(3-\mathrm{t}\right)\mathrm{y}+\left(2\mathrm{s}+\mathrm{t}\right)\mathrm{z}=1\\ \mathrm{This}\mathrm{ }\mathrm{is}\mathrm{ }\mathrm{the}\mathrm{ }\mathrm{cartesian}\mathrm{ }\mathrm{equation}\mathrm{ }\mathrm{of}\mathrm{ }\mathrm{the}\mathrm{ }\mathrm{plane}.\end{array}$

Q.26 In the following cases, find the coordinates of the foot of the perpendicular drawn from the origin.
(a) 2x + 3y + 4z – 12= 0 (b) 3y + 4z – 6 = 0
(c) x + y + z = 1 (d) 5y + 8 = 0

Ans.

$\begin{array}{l}\left(a\right)Thegivenequationoftheplaneis\\ 2x+3y+4z-12=0\\ ⇒ 2x+3y+4z=12...\left(i\right)\\ Letthecoordinatesofthefootofperpendicularfromoriginto\\ theplaneis\left({x}_{1}, {y}_{1}, {z}_{1}\right).Thenpoint \left({x}_{1}, {y}_{1}, {z}_{1}\right)willsatisfy\\ equation\left(i\right),so\\ 2{x}_{1}+3{y}_{1}+4{z}_{1}=12\\ Then,thedirectionratiosoftheperpendicularare 2, 3, 4.\\ Dividingequation \left(i\right)by\sqrt{{2}^{2}+{3}^{2}+{4}^{2}}=\sqrt{\mathrm{29}},weget\\ ⇒ \frac{2}{\sqrt{\mathrm{29}}}x+\frac{3}{\sqrt{\mathrm{29}}}y+\frac{4}{\sqrt{\mathrm{29}}}z=\frac{\mathrm{12}}{\sqrt{\mathrm{29}}}\\ Thisequationisintheformoflx+my+nz=d,wherel,mandn\\ arethedirectioncosinesofnormaltotheplaneanddisthe\\ distanceofthenormalfromtheorigin.\\ Thecoordinatesofthefootoftheperpendiculararegivenby\\ \left(ld,md,nd\right).\\ Therefore,thecoordiantesofthefootoftheperpendicularare\\ \left(\frac{2}{\sqrt{\mathrm{29}}}.\frac{\mathrm{12}}{\sqrt{\mathrm{29}}},\frac{3}{\sqrt{\mathrm{29}}}.\frac{\mathrm{12}}{\sqrt{\mathrm{29}}},\frac{4}{\sqrt{\mathrm{29}}}.\frac{\mathrm{12}}{\sqrt{\mathrm{29}}}\right)=\left(\frac{\mathrm{24}}{\mathrm{29}},\frac{\mathrm{36}}{\mathrm{29}},\frac{\mathrm{48}}{\mathrm{29}}\right).\\ \left(b\right)Thegivenequationoftheplaneis\\ 3y+4z-6=0\\ ⇒ 0.x+3y+4z=12...\left(i\right)\\ Letthecoordinatesofthefootofperpendicularfromoriginto\\ theplaneis\left({x}_{1}, {y}_{1}, {z}_{1}\right).Thenpoint \left({x}_{1}, {y}_{1}, {z}_{1}\right)willsatisfy\\ equation\left(i\right),so\\ 0.{x}_{1}+3{y}_{1}+4{z}_{1}=6\\ Then,thedirectionratiosoftheperpendicularare 0, 3, 4.\\ Dividingequation \left(i\right)by\sqrt{{0}^{2}+{3}^{2}+{4}^{2}}=5,weget\\ ⇒ \frac{0}{5}x+\frac{3}{5}y+\frac{4}{5}z=\frac{6}{5}\\ Thisequationisintheformoflx+my+nz=d,wherel,mandn\\ arethedirectioncosinesofnormaltotheplaneanddisthe\\ distanceofthenormalfromtheorigin.\\ Thecoordinatesofthefootoftheperpendiculararegivenby\\ \left(ld,md,nd\right).\\ Therefore,thecoordiantesofthefootoftheperpendicularare\\ \left(\frac{0}{5}.\frac{6}{5},\frac{3}{5}.\frac{6}{5},\frac{4}{5}.\frac{6}{5}\right)=\left(0,\frac{\mathrm{18}}{\mathrm{25}},\frac{\mathrm{24}}{\mathrm{25}}\right).\\ \left(c\right)Thegivenequationoftheplaneis\\ x+y+z=1...\left(i\right)\\ Letthecoordinatesofthefootofperpendicularfromoriginto\\ theplaneis\left({x}_{1}, {y}_{1}, {z}_{1}\right).Thenpoint \left({x}_{1}, {y}_{1}, {z}_{1}\right)willsatisfy\\ equation\left(i\right),so\\ x+y+z=1\\ Then,thedirectionratiosoftheperpendicularare 1, 1, 1.\\ Dividingequation \left(i\right)by\sqrt{{1}^{2}+{1}^{2}+{1}^{2}}=\sqrt{3},weget\\ ⇒ \frac{1}{\sqrt{3}}x+\frac{1}{\sqrt{3}}y+\frac{1}{\sqrt{3}}z=\frac{1}{\sqrt{3}}\\ Thisequationisintheformoflx+my+nz=d,wherel,mandn\\ arethedirectioncosinesofnormaltotheplaneanddisthe\\ distanceofthenormalfromtheorigin.\\ Thecoordinatesofthefootoftheperpendiculararegivenby\\ \left(ld,md,nd\right).\\ Therefore,thecoordiantesofthefootoftheperpendicularare\\ \left(\frac{1}{\sqrt{3}}.\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}.\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}.\frac{1}{\sqrt{3}}\right)=\left(\frac{1}{3},\frac{1}{3},\frac{1}{3}\right).\\ \left(d\right)Thegivenequationoftheplaneis\\ 5y+8=0\\ 0.x-5y+0.z= 8...\left(i\right)\\ Letthecoordinatesofthefootofperpendicularfromoriginto\\ theplaneis\left({x}_{1}, {y}_{1}, {z}_{1}\right).Thenpoint \left({x}_{1}, {y}_{1}, {z}_{1}\right)willsatisfy\\ equation\left(i\right),so\\ 0.{x}_{1}–5{y}_{1}+0.{z}_{1}= 8\\ Then,thedirectionratiosoftheperpendicularare 0, 5, 0.\\ Dividingequation \left(i\right)by\sqrt{{0}^{2}+{\left(–5\right)}^{2}+{0}^{2}}=5,weget\\ ⇒ \frac{0}{5}x–\frac{5}{5}y+\frac{0}{5}z=\frac{8}{5}\\ Thisequationisintheformoflx+my+nz=d,wherel,mandn\\ arethedirectioncosinesofnormaltotheplaneanddisthe\\ distanceofthenormalfromtheorigin.\\ Thecoordinatesofthefootoftheperpendiculararegivenby\\ \left(ld,md,nd\right).\\ Therefore,thecoordiantesofthefootoftheperpendicularare\\ \left(0×\frac{8}{5},\frac{–5}{5}×\frac{8}{5},0×\frac{8}{5}\right)=\left(0,-\frac{8}{5},0\right).\end{array}$

Q.27

$\begin{array}{l}\mathrm{Find}\mathrm{ }\mathrm{the}\mathrm{ }\mathrm{vector}\mathrm{ }\mathrm{and}\mathrm{ }\mathrm{cartesian}\mathrm{ }\mathrm{equations}\mathrm{ }\mathrm{of}\mathrm{ }\mathrm{the}\mathrm{ }\mathrm{planes}\\ \left(\mathrm{a}\right)\mathrm{that}\mathrm{ }\mathrm{passes}\mathrm{ }\mathrm{through}\mathrm{ }\mathrm{the}\mathrm{ }\mathrm{point}\mathrm{ }\left(1,0,-2\right)\mathrm{ }\mathrm{and}\mathrm{ }\mathrm{the}\mathrm{ }\mathrm{normal}\\ \mathrm{to}\mathrm{ }\mathrm{the}\mathrm{ }\mathrm{plane}\mathrm{ }\mathrm{is}\mathrm{ }\stackrel{^}{\mathrm{i}}+\stackrel{^}{\mathrm{j}}-\stackrel{^}{\mathrm{k}}.\\ \left(\mathrm{b}\right)\mathrm{that}\mathrm{ }\mathrm{passes}\mathrm{ }\mathrm{through}\mathrm{ }\mathrm{the}\mathrm{ }\mathrm{point}\mathrm{ }\left(1,4,6\right)\mathrm{ }\mathrm{and}\mathrm{ }\mathrm{the}\mathrm{ }\mathrm{normal}\\ \mathrm{vector}\mathrm{ }\mathrm{to}\mathrm{ }\mathrm{the}\mathrm{ }\mathrm{plane}\mathrm{ }\mathrm{is}\mathrm{ }\stackrel{^}{\mathrm{i}}-2\stackrel{^}{\mathrm{j}}+\stackrel{^}{\mathrm{k}}.\end{array}$

Ans.

$\begin{array}{l}\left(\mathrm{a}\right)\mathrm{ }\mathrm{The}\mathrm{ }\mathrm{position}\mathrm{ }\mathrm{vector}\mathrm{ }\mathrm{of}\mathrm{ }\mathrm{point}\mathrm{ }\left(1,0,-2\right), \stackrel{\to }{\mathrm{a}}=\stackrel{^}{\mathrm{i}}+0.\stackrel{^}{\mathrm{j}}-2\mathrm{ }\stackrel{^}{\mathrm{k}}\mathrm{ }\mathrm{and}\mathrm{ }\mathrm{the}\\ \mathrm{normal}\mathrm{ }\mathrm{vector}\mathrm{ }\stackrel{\to }{\mathrm{N}}\mathrm{ }\mathrm{perpendicular}\mathrm{ }\mathrm{to}\mathrm{ }\mathrm{the}\mathrm{ }\mathrm{plane}\mathrm{ }\mathrm{as}\mathrm{ }\stackrel{\to }{\mathrm{N}}=\stackrel{^}{\mathrm{i}}+\stackrel{^}{\mathrm{j}}-\stackrel{^}{\mathrm{k}}\\ \mathrm{Therefore},\mathrm{ }\mathrm{the}\mathrm{ }\mathrm{vector}\mathrm{ }\mathrm{equation}\mathrm{ }\mathrm{ofthe}\mathrm{ }\mathrm{plane}\mathrm{ }\mathrm{is}\mathrm{ }\mathrm{given}\mathrm{ }\mathrm{by}\\ \mathrm{ }\left(\stackrel{\to }{\mathrm{r}}-\stackrel{\to }{\mathrm{a}}\right).\stackrel{\to }{\mathrm{N}}=0\\ \mathrm{or}\mathrm{ }\left[\stackrel{\to }{\mathrm{r}}-\left(\stackrel{^}{\mathrm{i}}+0.\stackrel{^}{\mathrm{j}}-2\mathrm{ }\stackrel{^}{\mathrm{k}}\right)\right].\left(\stackrel{^}{\mathrm{i}}+\stackrel{^}{\mathrm{j}}-\stackrel{^}{\mathrm{k}}\right)=0...\left(\mathrm{i}\right)\\ \stackrel{\to }{\mathrm{r}}\mathrm{ }\mathrm{is}\mathrm{ }\mathrm{the}\mathrm{ }\mathrm{position}\mathrm{ }\mathrm{vector}\mathrm{ }\mathrm{of}\mathrm{ }\mathrm{any}\mathrm{ }\mathrm{point}\mathrm{ }\mathrm{P}\mathrm{ }\left(\mathrm{x},\mathrm{y},\mathrm{z}\right)\mathrm{ }\mathrm{in}\mathrm{ }\mathrm{the}\mathrm{ }\mathrm{plane}.\\ \therefore \stackrel{\to }{\mathrm{r}}=\mathrm{x}\mathrm{ }\stackrel{^}{\mathrm{i}}+\mathrm{y}\mathrm{ }\stackrel{^}{\mathrm{j}}+\mathrm{z}\mathrm{ }\stackrel{^}{\mathrm{k}}\\ \mathrm{Therefore},\mathrm{ }\mathrm{equation}\mathrm{ }\left(\mathrm{i}\right)\mathrm{ }\mathrm{becomes}\\ \left[\mathrm{x}\mathrm{ }\stackrel{^}{\mathrm{i}}+\mathrm{y}\mathrm{ }\stackrel{^}{\mathrm{j}}+\mathrm{z}\mathrm{ }\stackrel{^}{\mathrm{k}}-\left(\stackrel{^}{\mathrm{i}}+0.\stackrel{^}{\mathrm{j}}-2\mathrm{ }\stackrel{^}{\mathrm{k}}\right)\right].\left(\stackrel{^}{\mathrm{i}}+\stackrel{^}{\mathrm{j}}-\stackrel{^}{\mathrm{k}}\right)=0\\ ⇒ \left[\left(\mathrm{x}-1\right)\stackrel{^}{\mathrm{i}}+\mathrm{y}\mathrm{ }\stackrel{^}{\mathrm{j}}+\left(\mathrm{z}+2\right)\stackrel{^}{\mathrm{k}}\right].\left(\stackrel{^}{\mathrm{i}}+\stackrel{^}{\mathrm{j}}-\stackrel{^}{\mathrm{k}}\right)=0\\ ⇒ \left(\mathrm{x}-1\right)+\mathrm{y}-\left(\mathrm{z}+2\right)=0\\ ⇒ \mathrm{x}+\mathrm{y}-\mathrm{z}=3\\ \mathrm{This}\mathrm{ }\mathrm{is}\mathrm{ }\mathrm{the}\mathrm{ }\mathrm{cartesian}\mathrm{ }\mathrm{equation}\mathrm{ }\mathrm{of}\mathrm{ }\mathrm{the}\mathrm{}\mathrm{required}\mathrm{ }\mathrm{plane}.\\ \left(\mathrm{b}\right)\mathrm{ }\mathrm{The}\mathrm{ }\mathrm{position}\mathrm{ }\mathrm{vector}\mathrm{ }\mathrm{of}\mathrm{ }\mathrm{point} \left(1,4,6\right), \stackrel{\to }{\mathrm{a}}=\stackrel{^}{\mathrm{i}}+4\mathrm{ }\stackrel{^}{\mathrm{j}}+6\mathrm{ }\stackrel{^}{\mathrm{k}}\mathrm{ }\mathrm{and}\mathrm{ }\mathrm{the}\\ \mathrm{normal}\mathrm{ }\mathrm{vector}\mathrm{ }\stackrel{\to }{\mathrm{N}}\mathrm{ }\mathrm{perpendicular}\mathrm{ }\mathrm{to}\mathrm{ }\mathrm{the}\mathrm{ }\mathrm{plane}\mathrm{ }\mathrm{as}\mathrm{ }\stackrel{\to }{\mathrm{N}}=\stackrel{^}{\mathrm{i}}-2\stackrel{^}{\mathrm{j}}+\stackrel{^}{\mathrm{k}}\\ \mathrm{Therefore},\mathrm{ }\mathrm{the}\mathrm{ }\mathrm{vector}\mathrm{ }\mathrm{equation}\mathrm{ }\mathrm{of}\mathrm{ }\mathrm{the}\mathrm{ }\mathrm{plane}\mathrm{ }\mathrm{is}\mathrm{ }\mathrm{given}\mathrm{ }\mathrm{by}\\ \mathrm{ }\left(\stackrel{\to }{\mathrm{r}}-\stackrel{\to }{\mathrm{a}}\right).\stackrel{\to }{\mathrm{N}}=0\\ \mathrm{or}\left[\stackrel{\to }{\mathrm{r}}-\left(\stackrel{^}{\mathrm{i}}+4\mathrm{ }\stackrel{^}{\mathrm{j}}+6\mathrm{ }\stackrel{^}{\mathrm{k}}\right)\right].\left(\stackrel{^}{\mathrm{i}}-2\stackrel{^}{\mathrm{j}}+\stackrel{^}{\mathrm{k}}\right)=0...\left(\mathrm{i}\right)\\ \stackrel{\to }{\mathrm{r}}\mathrm{ }\mathrm{is}\mathrm{ }\mathrm{the}\mathrm{ }\mathrm{position}\mathrm{ }\mathrm{vector}\mathrm{ }\mathrm{of}\mathrm{ }\mathrm{any}\mathrm{ }\mathrm{point}\mathrm{ }\mathrm{P}\left(\mathrm{x},\mathrm{y},\mathrm{z}\right)\mathrm{ }\mathrm{in}\mathrm{ }\mathrm{the}\mathrm{ }\mathrm{plane}.\\ \therefore \stackrel{\to }{\mathrm{r}}=\mathrm{x}\mathrm{ }\stackrel{^}{\mathrm{i}}+\mathrm{y}\mathrm{ }\stackrel{^}{\mathrm{j}}+\mathrm{z}\mathrm{ }\stackrel{^}{\mathrm{k}}\\ \mathrm{Therefore},\mathrm{ }\mathrm{equation}\mathrm{ }\left(\mathrm{i}\right)\mathrm{ }\mathrm{becomes}\\ \left[\mathrm{x}\mathrm{ }\stackrel{^}{\mathrm{i}}+\mathrm{y}\mathrm{ }\stackrel{^}{\mathrm{j}}+\mathrm{z}\mathrm{ }\stackrel{^}{\mathrm{k}}-\left(\stackrel{^}{\mathrm{i}}+4\mathrm{ }\stackrel{^}{\mathrm{j}}+6\mathrm{ }\stackrel{^}{\mathrm{k}}\right)\right].\left(\stackrel{^}{\mathrm{i}}-2\stackrel{^}{\mathrm{j}}+\stackrel{^}{\mathrm{k}}\right)=0\\ ⇒ \left[\left(\mathrm{x}-1\right)\stackrel{^}{\mathrm{i}}+\left(\mathrm{y}-4\right)\mathrm{ }\stackrel{^}{\mathrm{j}}+\left(\mathrm{z}-6\right)\stackrel{^}{\mathrm{k}}\right].\left(\stackrel{^}{\mathrm{i}}-2\stackrel{^}{\mathrm{j}}+\stackrel{^}{\mathrm{k}}\right)=0\\ ⇒ \mathrm{ }\left(\mathrm{x}-1\right)-2\left(\mathrm{y}-4\right)+\left(\mathrm{z}-6\right)=0\\ ⇒ \mathrm{x}-2\mathrm{y}+\mathrm{z}+1=0\\ \mathrm{This}\mathrm{ }\mathrm{is}\mathrm{ }\mathrm{the}\mathrm{ }\mathrm{cartesian}\mathrm{ }\mathrm{equation}\mathrm{ }\mathrm{of}\mathrm{ }\mathrm{the}\mathrm{ }\mathrm{required}\mathrm{ }\mathrm{plane}.\end{array}$

Q.28 Find the equations of the planes that passes through three points.
(a) (1, 1, – 1), (6, 4, –5), (–4, –2, 3)

(b) (1, 1, 0), (1, 2, 1), (–2, 2, –1)

Ans.

$\begin{array}{l}\left(a\right)Let\text{vector forms of the given three points are:}\\ \stackrel{\to }{a}=\stackrel{^}{i}+\stackrel{^}{j}-\stackrel{^}{k},\text{\hspace{0.17em}}\stackrel{\to }{b}=6\text{\hspace{0.17em}}\stackrel{^}{i}+4\text{\hspace{0.17em}}\stackrel{^}{j}-5\text{\hspace{0.17em}}\stackrel{^}{k}\text{and}\stackrel{\to }{c}=-\text{\hspace{0.17em}}4\text{\hspace{0.17em}}\stackrel{^}{i}-2\stackrel{^}{j}+3\text{\hspace{0.17em}}\stackrel{^}{k}\\ Then,\text{the vector equation of the plane passing through}\stackrel{\to }{a},\text{\hspace{0.17em}}\stackrel{\to }{b}\text{and}\\ \stackrel{\to }{c}\text{is given by}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left(\stackrel{\to }{r}-\stackrel{\to }{a}\right).\left[\left(\stackrel{\to }{b}-\stackrel{\to }{a}\right)×\left(\stackrel{\to }{c}-\stackrel{\to }{a}\right)\right]=0\\ i.e.,\text{}\left[\stackrel{\to }{r}-\left(\stackrel{^}{i}+\stackrel{^}{j}-\stackrel{^}{k}\right)\right].\left[\left(5\text{\hspace{0.17em}}\stackrel{^}{i}+3\text{\hspace{0.17em}}\stackrel{^}{j}-4\stackrel{^}{k}\right)×\left(-5\text{\hspace{0.17em}}\stackrel{^}{i}-3\stackrel{^}{j}+4\text{\hspace{0.17em}}\stackrel{^}{k}\right)\right]=0\\ ⇒\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left[\stackrel{\to }{r}-\left(\stackrel{^}{i}+\stackrel{^}{j}-\stackrel{^}{k}\right)\right].|\begin{array}{ccc}\text{\hspace{0.17em}}\stackrel{^}{i}& \stackrel{^}{j}& \stackrel{^}{k}\\ 5& 3& -4\\ -5& -3& 4\end{array}|=0\\ ⇒\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left[\stackrel{\to }{r}-\left(\stackrel{^}{i}+\stackrel{^}{j}-\stackrel{^}{k}\right)\right].\left(0.\stackrel{^}{i}-0\stackrel{^}{.j}+0.\stackrel{^}{k}\right)=0\\ ⇒\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}0=0\\ Since,\text{the given three points are collinear, so there will be}\\ \text{infininte numbers of planes passing through the given points}\text{.}\\ \left(b\right)Let\text{vector forms of the given three points are:}\\ \text{}\stackrel{\to }{a}=\stackrel{^}{i}+\stackrel{^}{j}+0\text{\hspace{0.17em}}\stackrel{^}{k},\text{\hspace{0.17em}}\stackrel{\to }{b}=\text{\hspace{0.17em}}\stackrel{^}{i}+2\text{\hspace{0.17em}}\stackrel{^}{j}+\text{\hspace{0.17em}}\stackrel{^}{k}\text{and}\stackrel{\to }{c}=-\text{\hspace{0.17em}}2\text{\hspace{0.17em}}\stackrel{^}{i}+2\stackrel{^}{j}-\text{\hspace{0.17em}}\stackrel{^}{k}\\ Then,\text{the vector equation of the plane passing through}\stackrel{\to }{a},\text{\hspace{0.17em}}\stackrel{\to }{b}\text{and}\\ \stackrel{\to }{c}\text{is given by}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left(\stackrel{\to }{r}-\stackrel{\to }{a}\right).\left[\left(\stackrel{\to }{b}-\stackrel{\to }{a}\right)×\left(\stackrel{\to }{c}-\stackrel{\to }{a}\right)\right]=0\\ i.e.,\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left[\stackrel{\to }{r}-\left(\stackrel{^}{i}+\stackrel{^}{j}\right)\right].\left[\left(0\text{\hspace{0.17em}}\text{\hspace{0.17em}}\stackrel{^}{i}+\text{\hspace{0.17em}}\stackrel{^}{j}+\stackrel{^}{k}\right)×\left(-3\text{\hspace{0.17em}}\stackrel{^}{i}+\stackrel{^}{j}-\text{\hspace{0.17em}}\stackrel{^}{k}\right)\right]=0\\ ⇒\text{}\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left[\stackrel{\to }{r}-\left(\stackrel{^}{i}+\stackrel{^}{j}\right)\right].|\begin{array}{ccc}\text{\hspace{0.17em}}\stackrel{^}{i}& \stackrel{^}{j}& \stackrel{^}{k}\\ 0& 1& 1\\ -3& 1& -1\end{array}|=0\\ ⇒\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left[x\text{\hspace{0.17em}}\stackrel{^}{i}+y\stackrel{^}{j}+z\text{\hspace{0.17em}}\stackrel{^}{k}-\left(\stackrel{^}{i}+\stackrel{^}{j}\right)\right].\left(-2\text{\hspace{0.17em}}\stackrel{^}{i}-3\stackrel{^}{\text{\hspace{0.17em}}j}+3\text{\hspace{0.17em}}\stackrel{^}{k}\right)=0\\ ⇒\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left[\left(x-1\right)\text{\hspace{0.17em}}\stackrel{^}{i}+\left(y-1\right)\stackrel{^}{j}+z\text{\hspace{0.17em}}\stackrel{^}{k}\right].\left(-2\text{\hspace{0.17em}}\stackrel{^}{i}-3\stackrel{^}{\text{\hspace{0.17em}}j}+3\text{\hspace{0.17em}}\stackrel{^}{k}\right)=0\\ ⇒\text{}\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}-2\left(x-1\right)-3\left(y-1\right)+3\text{\hspace{0.17em}}z=0\\ ⇒\text{}\text{}\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}-2x+2-3y+3+3\text{\hspace{0.17em}}z=0\\ ⇒\text{}\text{}\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}2x+3y-3\text{\hspace{0.17em}}z=5\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}This\text{is the cartesian equation of the required plane}\text{.}\end{array}$

Q.29 Find the intercepts cut off by the plane
2x + y – z = 5

Ans.

$\begin{array}{l}\mathrm{The}\mathrm{ }\mathrm{given}\mathrm{ }\mathrm{equation}\mathrm{ }\mathrm{of}\mathrm{ }\mathrm{the}\mathrm{ }\mathrm{plane}\mathrm{ }\mathrm{is}:\\ 2\mathrm{x}+\mathrm{y}-\mathrm{z}=5...\left(\mathrm{i}\right)\\ \mathrm{Dividing}\mathrm{ }\mathrm{both}\mathrm{ }\mathrm{sides}\mathrm{ }\mathrm{of}\mathrm{ }\mathrm{equation}\mathrm{ }\left(\mathrm{i}\right)\mathrm{ }\mathrm{by}\mathrm{ }5,\mathrm{ }\mathrm{we}\mathrm{ }\mathrm{get}\\ \frac{2\mathrm{x}}{5}+\frac{\mathrm{y}}{5}-\frac{\mathrm{z}}{5}=\frac{5}{5}\\ ⇒ \frac{\mathrm{x}}{\left(\frac{5}{2}\right)}+\frac{\mathrm{y}}{5}+\frac{\mathrm{z}}{\left(-5\right)}=1\\ \mathrm{Comparing}\mathrm{ }\mathrm{this}\mathrm{ }\mathrm{equation}\mathrm{ }\mathrm{of}\mathrm{ }\mathrm{the}\mathrm{ }\mathrm{plane}\mathrm{ }\mathrm{with}\mathrm{ }\frac{\mathrm{x}}{\mathrm{a}}+\frac{\mathrm{y}}{\mathrm{b}}+\frac{\mathrm{z}}{\mathrm{c}}=1,\mathrm{ }\mathrm{we}\mathrm{get}\\ \mathrm{intercepts}\mathrm{ }\mathrm{cut}\mathrm{ }\mathrm{by}\mathrm{ }\mathrm{plane}\mathrm{ }\mathrm{at}\mathrm{ }\mathrm{x},\mathrm{y}\mathrm{ }\mathrm{and}\mathrm{ }\mathrm{z}\mathrm{ }\mathrm{axes}:\\ \mathrm{a}=\frac{5}{2},\mathrm{ }\mathrm{b}=5\mathrm{ }\mathrm{and}\mathrm{ }\mathrm{c}=-\mathrm{5}\\ \mathrm{Thus},\mathrm{ }\mathrm{the}\mathrm{ }\mathrm{intercepts}\mathrm{ }\mathrm{cut}\mathrm{ }\mathrm{off}\mathrm{ }\mathrm{by}\mathrm{ }\mathrm{the}\mathrm{ }\mathrm{plane}\mathrm{ }\mathrm{are}\mathrm{ }\frac{5}{2},5\mathrm{ }\mathrm{and}\mathrm{ }-5.\end{array}$

Q.30 Find the equation of the plane with intercept 3 on the y-axis and parallel to ZOX plane.

Ans.

The equation of ZOX plane is y=0

Equation of any plane parallel to ZOX plane is

y = a

Since, y-intercept of the plane is 3, so

a = 3

Thus, the equation of the required plane is y = 3.

Q.31 Find the equation of the plane through the intersection of the planes 3x – y + 2z – 4 = 0 and x + y + z – 2 = 0 with (2, 2, 1).

Ans.

$\begin{array}{l}The\text{ }equations\text{ }of\text{ }given\text{ }planes\text{ }are:\\ 3x-y+2z-4=0andx+y+z-2=0\\ The\text{ }equation\text{ }of\text{ }plane\text{ }through\text{ }intersection\text{ }of\text{ }given\text{ }planes\text{ }is\\ \left(3x-y+2z-4\right)+\lambda \left(x+y+z-2\right)=0\text{\hspace{0.17em}}\text{ }where\text{ }\text{\hspace{0.17em}}\lambda \in R\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\dots \left(i\right)\\ The\text{ }plane\text{ }passes\text{ }through\text{ }the\text{ }point\text{ }\left(2,2,1\right),\text{ }so\text{ }from\text{ }equation\text{ }\left(i\right)\\ \left(3×2-2+2×1-4\right)+\lambda \left(2+2+1-2\right)=0\\ 2+\lambda \left(3\right)=0⇒\lambda =-\frac{2}{3}\\ Putting\text{ }value\text{ }of\text{ }\text{\hspace{0.17em}}\lambda \text{ }in\text{ }equation\text{ }\left(i\right),\text{ }we\text{ }get\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left(3x-y+2z-4\right)-\frac{2}{3}\left(x+y+z-2\right)=0\\ ⇒3\left(3x-y+2z-4\right)-2\left(x+y+z-2\right)=0\\ ⇒\text{\hspace{0.17em}}\text{\hspace{0.17em}}9x-3y+6z-12-2x-2y-2z+4=0\\ ⇒\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}7x-5y+4z-8=0\\ \text{This}\text{ }\text{is}\text{ }\text{the}\text{ }\text{required}\text{ }\text{equation}\text{ }\text{of}\text{ }\text{the}\text{ }\text{plane}\text{.}\end{array}$

Q.32n

$\begin{array}{l}\mathrm{Find}\mathrm{ }\mathrm{the}\mathrm{ }\mathrm{vector}\mathrm{ }\mathrm{equation}\mathrm{ }\mathrm{of}\mathrm{ }\mathrm{the}\mathrm{ }\mathrm{plane}\mathrm{ }\mathrm{passing} \mathrm{through}\mathrm{ }\mathrm{the}\\ \mathrm{intersection}\mathrm{ }\mathrm{of}\mathrm{ }\mathrm{the} \mathrm{planes} \stackrel{\to }{\mathrm{r}}.\left(2\mathrm{ }\stackrel{^}{\mathrm{i}}+2\stackrel{^}{\mathrm{j}}-3\mathrm{ }\stackrel{^}{\mathrm{k}}\right)=7,\\ \stackrel{\to }{\mathrm{r}}.\left(2\mathrm{ }\stackrel{^}{\mathrm{i}}+5\mathrm{ }\stackrel{^}{\mathrm{j}}+3\mathrm{ }\stackrel{^}{\mathrm{k}}\right)=9\mathrm{ }\mathrm{and}\mathrm{ }\mathrm{through}\mathrm{ }\mathrm{the}\mathrm{ }\mathrm{point}\mathrm{ }\left(2,1,3\right).\end{array}$

Ans.

$\begin{array}{l}\mathrm{The}\mathrm{givenen}\mathrm{equations}\mathrm{of}\mathrm{the}\mathrm{plane}\mathrm{are}:\\ \stackrel{\to }{\mathrm{r}}.\left(2\mathrm{ }\stackrel{^}{\mathrm{i}}+2\stackrel{^}{\mathrm{j}}-3\mathrm{ }\stackrel{^}{\mathrm{k}}\right)=7\mathrm{or}\stackrel{\to }{\mathrm{r}}.\left(2\mathrm{ }\stackrel{^}{\mathrm{i}}+2\stackrel{^}{\mathrm{j}}-3\mathrm{ }\stackrel{^}{\mathrm{k}}\right)-7=0...\left(\mathrm{i}\right)\\ \stackrel{\to }{\mathrm{r}}.\left(2\mathrm{ }\stackrel{^}{\mathrm{i}}+5\stackrel{^}{\mathrm{j}}+3\mathrm{ }\stackrel{^}{\mathrm{k}}\right)=9\mathrm{or}\stackrel{\to }{\mathrm{r}}.\left(2\mathrm{ }\stackrel{^}{\mathrm{i}}+5\stackrel{^}{\mathrm{j}}+3\mathrm{ }\stackrel{^}{\mathrm{k}}\right)-9=0...\left(\mathrm{ii}\right)\\ \mathrm{The}\mathrm{equation}\mathrm{of}\mathrm{any}\mathrm{plane}\mathrm{throught}\mathrm{the}\mathrm{intersection}\mathrm{of}\mathrm{the}\\ \mathrm{planes}\mathrm{given}\mathrm{in}\mathrm{equation}\left(\mathrm{i}\right)\mathrm{and}\left(\mathrm{ii}\right)\mathrm{is}\mathrm{given}\mathrm{by}\\ \left\{\stackrel{\to }{\mathrm{r}}.\left(2\mathrm{ }\stackrel{^}{\mathrm{i}}+2\stackrel{^}{\mathrm{j}}-3\mathrm{ }\stackrel{^}{\mathrm{k}}\right)-7\right\}+\mathrm{\lambda }\left\{\stackrel{\to }{\mathrm{r}}.\left(2\mathrm{ }\stackrel{^}{\mathrm{i}}+5\stackrel{^}{\mathrm{j}}+3\mathrm{ }\stackrel{^}{\mathrm{k}}\right)-9\right\}=0,\mathrm{\lambda }\in \mathrm{R}\\ \stackrel{\to }{\mathrm{r}}.\left[\left(2\mathrm{ }\stackrel{^}{\mathrm{i}}+2\stackrel{^}{\mathrm{j}}-3\mathrm{ }\stackrel{^}{\mathrm{k}}\right)+\mathrm{\lambda }\left(2\mathrm{ }\stackrel{^}{\mathrm{i}}+5\stackrel{^}{\mathrm{j}}+3\mathrm{ }\stackrel{^}{\mathrm{k}}\right)\right]=9\mathrm{\lambda }+7\\ \stackrel{\to }{\mathrm{r}}.\left[\left(2+2\mathrm{\lambda }\right)\mathrm{ }\stackrel{^}{\mathrm{i}}+\left(2+5\mathrm{\lambda }\right)\stackrel{^}{\mathrm{j}}-\left(3-3\mathrm{\lambda }\right)\mathrm{ }\stackrel{^}{\mathrm{k}}\right]=9\mathrm{\lambda }+7 \mathrm{ }...\left(\mathrm{iii}\right)\\ \mathrm{The}\mathrm{plane}\mathrm{passes}\mathrm{through}\mathrm{the}\mathrm{point}\left(2,1,3\right).\mathrm{Therefore},\mathrm{its}\\ \mathrm{position}\mathrm{vector}\mathrm{is}\mathrm{given}\mathrm{by},\\ \stackrel{\to }{\mathrm{r}}=2\mathrm{ }\stackrel{^}{\mathrm{i}}+\stackrel{^}{\mathrm{j}}+3\mathrm{ }\stackrel{^}{\mathrm{k}}\\ \mathrm{Substituting}\mathrm{value}\mathrm{of}\stackrel{\to }{\mathrm{r}}\mathrm{ }\mathrm{in}\mathrm{equation}\left(\mathrm{iii}\right),\mathrm{we}\mathrm{get}\\ \left(2\mathrm{ }\stackrel{^}{\mathrm{i}}+\stackrel{^}{\mathrm{j}}+3\mathrm{ }\stackrel{^}{\mathrm{k}}\right).\left[\left(2+2\mathrm{\lambda }\right)\mathrm{ }\stackrel{^}{\mathrm{i}}+\left(2+5\mathrm{\lambda }\right)\stackrel{^}{\mathrm{j}}-\left(3-3\mathrm{\lambda }\right)\mathrm{ }\stackrel{^}{\mathrm{k}}\right]=9\mathrm{\lambda }+7\\ ⇒2\left(2+2\mathrm{\lambda }\right)+\left(2+5\mathrm{\lambda }\right)-3\left(3-3\mathrm{\lambda }\right)=9\mathrm{\lambda }+7\\ ⇒ \mathrm{ }4+4\mathrm{\lambda }+2+5\mathrm{\lambda }-9+9\mathrm{\lambda }=9\mathrm{\lambda }+7\\ ⇒\mathrm{ }9\mathrm{\lambda }-3=7\\ ⇒\mathrm{\lambda }=\frac{10}{9}\\ \mathrm{Putting}\mathrm{}\mathrm{\lambda }=\frac{10}{9}\mathrm{in}\mathrm{equation}\left(\mathrm{iii}\right),\mathrm{we}\mathrm{get}\\ \mathrm{ }\stackrel{\to }{\mathrm{r}}.\left[\left(2+2×\frac{10}{9}\right)\mathrm{ }\stackrel{^}{\mathrm{i}}+\left(2+5×\frac{10}{9}\right)\stackrel{^}{\mathrm{j}}-\left(3-3×\frac{10}{9}\right)\mathrm{ }\stackrel{^}{\mathrm{k}}\right]=9×\frac{10}{9}+7\\ \mathrm{ }\stackrel{\to }{\mathrm{r}}.\left[38\mathrm{ }\stackrel{^}{\mathrm{i}}+68\mathrm{ }\stackrel{^}{\mathrm{j}}+3\mathrm{ }\stackrel{^}{\mathrm{k}}\right]=153\\ \mathrm{This}\mathrm{is}\mathrm{the}\mathrm{vector}\mathrm{equation}\mathrm{of}\mathrm{the}\mathrm{required}\mathrm{plane}\mathrm{.}\end{array}$

Q.33 Find the equation of the plane through the line of intersection of the planes x + y + z = 1 and 2x + 3y + 4z = 5 which is perpendicular to the plane x – y + z = 0.

Ans.

$\begin{array}{l}The\text{ }equations\text{ }of\text{ }given\text{ }planes\text{ }are:\\ x+y+z=1\text{ }and\text{ }2x+3y+4z=5\\ The\text{ }equation\text{ }of\text{ }plane\text{ }through\text{ }intersection\text{ }of\text{ }given\text{ }planes\text{ }is\\ \text{}\text{\hspace{0.17em}}\left(x+y+z-1\right)+\lambda \left(2x+3y+4z-5\right)=0\text{\hspace{0.17em}}\text{\hspace{0.17em}}where\text{\hspace{0.17em}}\text{\hspace{0.17em}}\lambda \in R\\ ⇒\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left(1+2\lambda \right)x+\left(1+3\lambda \right)y+\left(1+4\lambda \right)z=5\lambda +1\text{}\text{}\dots \left(i\right)\\ The\text{direction ratios of the plane are}\left(1+2\lambda \right),\text{\hspace{0.17em}}\left(1+3\lambda \right)\text{and}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left(1+4\lambda \right).\\ Plane\text{}\left(i\right)\text{is perpendicular to}x-y+z=0,\text{whose direction ratios}\\ \text{are 1,}-1,\text{\hspace{0.17em}}\text{\hspace{0.17em}}1.\\ Since,\text{plane}\left(\text{i}\right)\text{and}\left(\text{ii}\right)\text{are perpendicular, so}\\ \text{}\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{\text{a}}_{\text{1}}{\text{a}}_{\text{2}}{\text{+b}}_{\text{1}}{\text{b}}_{\text{2}}{\text{+c}}_{\text{1}}{\text{c}}_{\text{2}}=0\\ ⇒\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left(1+2\lambda \right)\left(1\right)\text{+}\left(1+3\lambda \right)\left(-1\right)\text{+}\left(1+4\lambda \right)\left(1\right)=0\\ ⇒\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}1+2\lambda -1-3\lambda +1+4\lambda =0\\ ⇒\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}3\lambda =-1\\ ⇒\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\lambda =-\frac{1}{3}\\ Substituting\text{​}\text{}\lambda =-\frac{1}{3}\text{in equation}\left(\text{i}\right)\text{, we get}\\ ⇒\left(1+2×-\frac{1}{3}\right)x+\left(1+3×-\frac{1}{3}\right)y+\left(1+4×-\frac{1}{3}\right)z=5×-\frac{1}{3}+1\\ ⇒\text{}\text{}\text{}\text{}\text{}\text{}\frac{1}{3}x+\left(0\right)y-\frac{1}{3}z=-\frac{2}{3}\\ \text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}x-z=-2\\ \text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}x-z+2=0\\ This\text{is the required equation of the plane}\text{.}\\ The\text{direction ratios of the plane are}\left(1+2\lambda \right),\text{\hspace{0.17em}}\left(1+3\lambda \right)\text{and}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left(1+4\lambda \right).\\ Plane\text{}\left(i\right)\text{is perpendicular to}x-y+z=0,\text{whose direction ratios}\\ \text{are 1,}-1,\text{\hspace{0.17em}}\text{\hspace{0.17em}}1.\\ Since,\text{plane}\left(\text{i}\right)\text{and}\left(\text{ii}\right)\text{are perpendicular, so}\\ \text{}\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{\text{a}}_{\text{1}}{\text{a}}_{\text{2}}{\text{+b}}_{\text{1}}{\text{b}}_{\text{2}}{\text{+c}}_{\text{1}}{\text{c}}_{\text{2}}=0\\ ⇒\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left(1+2\lambda \right)\left(1\right)\text{+}\left(1+3\lambda \right)\left(-1\right)\text{+}\left(1+4\lambda \right)\left(1\right)=0\\ ⇒\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}1+2\lambda -1-3\lambda +1+4\lambda =0\\ ⇒\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}3\lambda =-1\\ ⇒\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\lambda =-\frac{1}{3}\\ Substituting\text{​}\text{}\lambda =-\frac{1}{3}\text{in equation}\left(\text{i}\right)\text{, we get}\\ ⇒\left(1+2×-\frac{1}{3}\right)x+\left(1+3×-\frac{1}{3}\right)y+\left(1+4×-\frac{1}{3}\right)z=5×-\frac{1}{3}+1\\ ⇒\text{}\text{}\text{}\text{}\text{}\text{}\frac{1}{3}x+\left(0\right)y-\frac{1}{3}z=-\frac{2}{3}\\ \text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}x-z=-2\\ \text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}x-z+2=0\\ This\text{is the required equation of the plane}\text{.}\end{array}$

Q.34

$\begin{array}{l}\mathrm{Find}\mathrm{ }\mathrm{the}\mathrm{ }\mathrm{angle}\mathrm{ }\mathrm{between}\mathrm{ }\mathrm{the}\mathrm{ }\mathrm{planes}\mathrm{ }\mathrm{whose}\mathrm{ }\mathrm{vector}\mathrm{ }\mathrm{equations}\\ \mathrm{are}\mathrm{ }\stackrel{\to }{\mathrm{r}}\left(2\mathrm{ }\stackrel{^}{\mathrm{i}}+2\mathrm{ }\stackrel{^}{\mathrm{j}}-\mathrm{ }\stackrel{^}{\mathrm{k}}\right)=5\mathrm{ }\mathrm{and} \stackrel{\to }{\mathrm{r}}\left(3\mathrm{ }\stackrel{^}{\mathrm{i}}-3\mathrm{ }\stackrel{^}{\mathrm{j}}+5\mathrm{ }\stackrel{^}{\mathrm{k}}\right)=\end{array}$

Ans.

$\begin{array}{l}\mathrm{Theplaneswhosevectorequationsare}\\ \stackrel{\to }{\mathrm{r}}\left(2\mathrm{ }\stackrel{^}{\mathrm{i}}+2\mathrm{ }\stackrel{^}{\mathrm{j}}-3\mathrm{ }\stackrel{^}{\mathrm{k}}\right)=5\mathrm{and}\mathrm{ }\stackrel{\to }{\mathrm{r}}\left(3\mathrm{ }\stackrel{^}{\mathrm{i}}-3\mathrm{ }\stackrel{^}{\mathrm{j}}+5\mathrm{ }\stackrel{^}{\mathrm{k}}\right)=\mathrm{}3\\ \mathrm{comparing}\mathrm{with}\stackrel{\to }{\mathrm{r}}.\mathrm{ }\stackrel{\to }{{\mathrm{n}}_{1}}={\mathrm{d}}_{1}\mathrm{and}\stackrel{\to }{\mathrm{r}}.\mathrm{ }\stackrel{\to }{{\mathrm{n}}_{2}}={\mathrm{d}}_{2},\mathrm{then}\\ \stackrel{\to }{{\mathrm{n}}_{1}}=\left(2\mathrm{ }\stackrel{^}{\mathrm{i}}+2\mathrm{ }\stackrel{^}{\mathrm{j}}-3\mathrm{ }\stackrel{^}{\mathrm{k}}\right)\mathrm{and}\stackrel{\to }{{\mathrm{n}}_{2}}=\left(3\mathrm{ }\stackrel{^}{\mathrm{i}}-3\mathrm{ }\stackrel{^}{\mathrm{j}}+5\mathrm{ }\stackrel{^}{\mathrm{k}}\right)\\ \therefore \stackrel{\to }{{\mathrm{n}}_{1}}.\stackrel{\to }{{\mathrm{n}}_{2}}=\left(2\mathrm{ }\stackrel{^}{\mathrm{i}}+2\mathrm{ }\stackrel{^}{\mathrm{j}}-3\mathrm{ }\stackrel{^}{\mathrm{k}}\right).\left(3\mathrm{ }\stackrel{^}{\mathrm{i}}-3\mathrm{ }\stackrel{^}{\mathrm{j}}+5\mathrm{ }\stackrel{^}{\mathrm{k}}\right)\\ =6-6-15\\ =-15\\ \mathrm{ }|\stackrel{\to }{{\mathrm{n}}_{1}}|=|2\mathrm{ }\stackrel{^}{\mathrm{i}}+2\mathrm{ }\stackrel{^}{\mathrm{j}}-3\mathrm{ }\stackrel{^}{\mathrm{k}}|\\ =\sqrt{{2}^{2}+{2}^{2}+{\left(-3\right)}^{2}}\\ =\sqrt{4+4+9}=\sqrt{17}\\ \mathrm{ }|\stackrel{\to }{{\mathrm{n}}_{2}}|=|3\mathrm{ }\stackrel{^}{\mathrm{i}}-3\mathrm{ }\stackrel{^}{\mathrm{j}}+5\mathrm{ }\stackrel{^}{\mathrm{k}}|\\ =\sqrt{{3}^{2}+{\left(-3\right)}^{2}+{5}^{2}}\\ =\sqrt{9+9+25}=\sqrt{43}\\ \mathrm{Let}\mathrm{angle}\mathrm{between}\mathrm{two}\mathrm{planes}\mathrm{be}\mathrm{\theta },\mathrm{then}\\ \mathrm{ }\mathrm{cos\theta }=\frac{\stackrel{\to }{{\mathrm{n}}_{1}}.\stackrel{\to }{{\mathrm{n}}_{2}}}{|\stackrel{\to }{{\mathrm{n}}_{1}}||\stackrel{\to }{{\mathrm{n}}_{2}}|}\\ =\frac{-15}{\sqrt{17}\sqrt{43}}\\ =-\frac{15}{\sqrt{731}}\\ \mathrm{\theta }={\mathrm{cos}}^{-1}\left(-\frac{15}{\sqrt{731}}\right)\end{array}$

Q.35 In the following cases, determine whether
the given planes are parallel or perpendicular and in case they are neither, find the angle between them.
(a) 7x + 5y + 6z + 30 = 0 and 3x – y – 10z + 4 = 0
(b) 2x + y + 3z – 2 = 0 and x –2y + 5 = 0
(c) 2x – 2y + 4z + 5 = 0 and 3x – 3y +6 z – 1 = 0
(d) 2x – y + 3z – 1 = 0 and 2x – y +3z + 3 = 0
(e) 4x + 8y + z – 8 = 0 and y + z – 4 = 0

Ans.

$\begin{array}{l}\left(a\right)\text{}7x+5y+6z+30=0\text{}and\text{}3x-y-10z+4=0\\ Direction\text{ratios of the given planes are:}\\ {\text{a}}_{\text{1}}=\text{}7,\text{}{b}_{1}=\text{}5,\text{}{c}_{1}=\text{6}\\ {\text{a}}_{\text{2}}=\text{}3,\text{}{b}_{2}=\text{}-1,\text{}{c}_{2}=\text{\hspace{0.17em}}-10\\ {\text{a}}_{\text{1}}{\text{a}}_{\text{2}}+{b}_{1}{b}_{2}+{c}_{1}{c}_{2}=7×3+5×-1+6×-10\\ \text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=21-5-60=-44\ne 0\\ \text{Therefore, the given planes are not perpendicular to each other}\text{.}\\ \frac{{a}_{1}}{{a}_{2}}=\frac{7}{3},\text{\hspace{0.17em}}\text{\hspace{0.17em}}\frac{{b}_{1}}{{b}_{2}}=\frac{5}{-1}\text{and}\frac{{c}_{1}}{{c}_{2}}=\frac{6}{-10}=-\frac{3}{5}\\ ⇒\text{}\frac{{a}_{1}}{{a}_{2}}\ne \text{}\frac{{b}_{1}}{{b}_{2}}\ne \frac{{c}_{1}}{{c}_{2}}\\ \text{Therefore, the given planes are not parallel to each other}\text{.}\\ \text{The angle between two planes is}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\theta ={\text{cos}}^{-\text{1}}\left|\left(\frac{{\text{a}}_{\text{1}}{\text{a}}_{\text{2}}+{b}_{1}{b}_{2}+{c}_{1}{c}_{2}}{\sqrt{{a}_{1}{}^{2}+{b}_{1}{}^{2}+{c}_{1}{}^{2}}\sqrt{{a}_{2}{}^{2}+{b}_{2}{}^{2}+{c}_{2}{}^{2}}}\right)\right|\end{array}$ $\begin{array}{l}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}={\text{cos}}^{-\text{1}}\left|\left(\frac{7×3+5×-1+6×-10}{\sqrt{{7}^{2}+{5}^{2}+{6}^{2}}\sqrt{{3}^{2}+{\left(-1\right)}^{2}+{\left(-10\right)}^{2}}}\right)\right|\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}={\text{cos}}^{-\text{1}}\left|\left(\frac{-44}{\sqrt{110}\sqrt{110}}\right)\right|\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}={\text{cos}}^{-\text{1}}\left(\frac{44}{110}\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}={\text{cos}}^{-\text{1}}\left(\frac{2}{5}\right)\end{array}$ $\begin{array}{l}\left(b\right)\text{}2x+y+3z-2=0\text{}and\text{}x-2y+5=0\\ Direction\text{ratios of the given planes are:}\\ {\text{a}}_{\text{1}}=\text{}2,\text{}{b}_{1}=\text{}1,\text{}{c}_{1}=\text{3}\\ {\text{a}}_{\text{2}}=\text{}1,\text{}{b}_{2}=\text{}-2,\text{}{c}_{2}=\text{\hspace{0.17em}}0\\ {\text{a}}_{\text{1}}{\text{a}}_{\text{2}}+{b}_{1}{b}_{2}+{c}_{1}{c}_{2}=2×1+1×-2+3×0\\ \text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=2-2+0=0\\ \text{Therefore, the given planes are perpendicular to each other}\text{.}\\ \left(c\right)\text{}2x-2y+4z+5=0\text{}and\text{}3x-3y+6\text{}z-1=0\\ Direction\text{ratios of the given planes are:}\\ {\text{a}}_{\text{1}}=\text{}2,\text{}{b}_{1}=\text{}-2,\text{}{c}_{1}=\text{4}\\ {\text{a}}_{\text{2}}=\text{}3,\text{}{b}_{2}=\text{}-3,\text{}{c}_{2}=\text{\hspace{0.17em}}6\\ {\text{a}}_{\text{1}}{\text{a}}_{\text{2}}+{b}_{1}{b}_{2}+{c}_{1}{c}_{2}=2×3+-2×-3+4×6\\ \text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=6+6+24=36\ne 0\end{array}$ $\begin{array}{l}\text{Therefore, the given planes are not perpendicular to each other}\text{.}\\ \frac{{a}_{1}}{{a}_{2}}=\frac{2}{3},\text{\hspace{0.17em}}\text{\hspace{0.17em}}\frac{{b}_{1}}{{b}_{2}}=\frac{-2}{-3}=\frac{2}{3}\text{and}\frac{{c}_{1}}{{c}_{2}}=\frac{4}{6}=\frac{2}{3}\end{array}$ $\begin{array}{l}⇒\text{}\frac{{a}_{1}}{{a}_{2}}=\text{}\frac{{b}_{1}}{{b}_{2}}=\frac{{c}_{1}}{{c}_{2}}\\ \text{Therefore, the given planes are parallel to each other}\text{.}\\ \left(d\right)\text{}2x-y+3z-1=0\text{}and\text{}2x-y+3z+3=0\\ Direction\text{ratios of the given planes are:}\\ {\text{a}}_{\text{1}}=\text{}2,\text{}{b}_{1}=\text{}-1,\text{}{c}_{1}=\text{3}\\ {\text{a}}_{\text{2}}=\text{}2,\text{}{b}_{2}=\text{}-1,\text{}{c}_{2}=\text{\hspace{0.17em}}3\\ {\text{a}}_{\text{1}}{\text{a}}_{\text{2}}+{b}_{1}{b}_{2}+{c}_{1}{c}_{2}=2×2+-1×-1+3×3\\ \text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=4+1+9=14\ne 0\\ \text{Therefore, the given planes are not perpendicular to each other}\text{.}\\ \frac{{a}_{1}}{{a}_{2}}=\frac{2}{2}=1,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\frac{{b}_{1}}{{b}_{2}}=\frac{-1}{-1}=1\text{and}\frac{{c}_{1}}{{c}_{2}}=\frac{3}{3}=1\\ ⇒\text{}\frac{{a}_{1}}{{a}_{2}}=\text{}\frac{{b}_{1}}{{b}_{2}}=\frac{{c}_{1}}{{c}_{2}}\\ \text{Therefore, the given planes are parallel to each other}\text{.}\\ \left(e\right)\text{\hspace{0.17em}}\text{\hspace{0.17em}}4x+8y+z-8=0\text{}and\text{}y+z-4=0\\ Direction\text{ratios of the given planes are:}\\ {\text{a}}_{\text{1}}=\text{}4,\text{}{b}_{1}=\text{}8,\text{}{c}_{1}=\text{1}\\ {\text{a}}_{\text{2}}=\text{}0,\text{}{b}_{2}=\text{}1,\text{}{c}_{2}=\text{\hspace{0.17em}}1\\ {\text{a}}_{\text{1}}{\text{a}}_{\text{2}}+{b}_{1}{b}_{2}+{c}_{1}{c}_{2}=4×0+8×1+1×1\\ \text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=0+8+1=9\ne 0\end{array}$ $\begin{array}{l}\text{Therefore, the given planes are not perpendicular to each other}\text{.}\\ \frac{{a}_{1}}{{a}_{2}}=\frac{4}{0},\text{\hspace{0.17em}}\text{\hspace{0.17em}}\frac{{b}_{1}}{{b}_{2}}=\frac{8}{1}=8\text{and}\frac{{c}_{1}}{{c}_{2}}=\frac{1}{1}=1\\ ⇒\text{}\frac{{a}_{1}}{{a}_{2}}\ne \text{}\frac{{b}_{1}}{{b}_{2}}\ne \frac{{c}_{1}}{{c}_{2}}\end{array}$ $\begin{array}{l}\text{Therefore, the given planes are not}\text{\hspace{0.17em}}\text{parallel to each other}\text{.}\\ \text{The angle between two planes is}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\theta ={\text{cos}}^{-\text{1}}\left|\left(\frac{{\text{a}}_{\text{1}}{\text{a}}_{\text{2}}+{b}_{1}{b}_{2}+{c}_{1}{c}_{2}}{\sqrt{{a}_{1}{}^{2}+{b}_{1}{}^{2}+{c}_{1}{}^{2}}\sqrt{{a}_{2}{}^{2}+{b}_{2}{}^{2}+{c}_{2}{}^{2}}}\right)\right|\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}={\text{cos}}^{-\text{1}}\left|\left(\frac{4×0+8×1+1×1}{\sqrt{{4}^{2}+{8}^{2}+{1}^{2}}\sqrt{{0}^{2}+{1}^{2}+{1}^{2}}}\right)\right|\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}={\text{cos}}^{-\text{1}}\left|\left(\frac{9}{\sqrt{16+64+1}\sqrt{0+1+1}}\right)\right|\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}={\text{cos}}^{-\text{1}}\left|\left(\frac{9}{\sqrt{81}\sqrt{2}}\right)\right|\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}={\text{cos}}^{-\text{1}}\left|\frac{1}{\sqrt{2}}\right|=\frac{\pi }{4}\\ Thu\text{s, the angle between two planes is}\frac{\pi }{4}.\end{array}$

Q.36 In the following cases, find the distance of each of the given points from the corresponding given plane.

Point Plane

(a) (0, 0, 0) 3x – 4y + 12 z = 3

(b) (3, –2, 1) 2x – y + 2z + 3 = 0

(c) (2, 3, –5) x – 2y + 2z = 9

(d) (–6, 0, 0) 2x – 3y + 6z – 2 =0

Ans.

$\begin{array}{l}\left(a\right)\text{The given equation of plane is}3x-4y+12\text{}z=3\text{and point}\\ \text{is}\text{ }\left(0,0,0\right).\\ The\text{distance of plane from given point}\left(d\right)=\left|\frac{3\left(0\right)-4\left(0\right)+12\left(0\right)-3}{\sqrt{{3}^{2}+{\left(-4\right)}^{2}+{\left(12\right)}^{2}}}\right|\\ \text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{3}{\sqrt{9+16+144}}\\ \text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{3}{\sqrt{169}}\\ \text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{3}{13}\\ \left(b\right)\text{The given equation of plane is}2x-y+2z+3=0\text{and point}\\ \text{is}\text{ }\left(3,-2,1\right).\\ The\text{distance of plane from given point}\left(d\right)=\left|\frac{2\left(3\right)-\left(-2\right)+2\left(1\right)+3}{\sqrt{{2}^{2}+{\left(-1\right)}^{2}+{\left(2\right)}^{2}}}\right|\\ \text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{13}{\sqrt{4+1+4}}\\ \text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{13}{\sqrt{9}}\\ \text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{13}{3}\\ \left(c\right)\text{ }\text{The given equation of plane is}x-2y+2z-9=0\text{and point}\\ \text{is}\text{ }\left(2,3,-5\right).\\ The\text{distance of plane from given point}\left(d\right)=\left|\frac{\left(2\right)-2\left(3\right)+2\left(-5\right)-9}{\sqrt{{1}^{2}+{\left(-2\right)}^{2}+{\left(2\right)}^{2}}}\right|\\ \text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{23}{\sqrt{1+4+4}}\\ \text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{23}{\sqrt{9}}\\ \text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{23}{3}\\ \left(d\right)\text{ }The\text{ }given\text{ }equation\text{ }of\text{ }plane\text{ }is\text{ }2x-3y+6z-2=0\text{ }and\text{ }point\\ is\text{ }\left(-6,0,0\right).\\ The\text{ }distance\text{ }ofplane\text{ }from\text{ }given\text{ }point\text{ }\left(d\right)\text{ }=\left|\frac{2\left(-6\right)-3\left(0\right)+6\left(0\right)-2}{\sqrt{{2}^{2}+{\left(-3\right)}^{2}+{6}^{2}}}\right|\\ \text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{14}{\sqrt{4+9+36}}\\ \text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{14}{\sqrt{49}}=\frac{14}{7}=2\\ \end{array}$

Q.37 Show that the line joining the origin to the point (2, 1, 1) is perpendicular to the line determined by the points (3, 5, –1), (4, 3, –1).

Ans.

Let OA be the line joining the origin, O (0, 0, 0) and the point, A (2, 1, 1). Then, the direction ratios of OA are: (2 – 0), (1 – 0), (1 – 0) i.e., 2, 1, 1.
Let BC be the line joining the origin, B (3, 5, –1) and the point, C (4, 3,–1).Then, the direction ratios of BC are: (4 – 3), (3 – 5), (–1+1) i.e., 1, –2, 0.

OA is perpendicular to BC, if a1a2+b1b2+c1c2= 0

a1a2+b1b2+c1c2 = (2)(1)+(1)(–2)+(1)(0)
= 2 – 2+0 = 0
Thus, OA is perpendicular to BC.

Q.38 If l1, m1, n1 and l2, m2, n2 are the direction cosines of two mutually perpendicular lines, show that the direction cosines of the line perpendicular to both of these are m1n2 – m2n1, n1l2 – n2l1, l1m2l2 m1.

Ans.

$\begin{array}{l}Since\text{\hspace{0.17em}}{l}_{1},{m}_{1},{n}_{1}\text{and}{l}_{2},{m}_{2},{n}_{2}\text{are the direction cosines of two}\\ \text{mutually perpendicular lines}\text{. Therefore,}\\ {l}_{1}{l}_{2}+{m}_{1}{m}_{2}+{n}_{1}{n}_{2}=0\text{}\dots \left(i\right)\\ {l}_{1}{}^{2}+{m}_{1}{}^{2}+{n}_{1}{}^{2}=1\text{}\text{}\dots \left(ii\right)\\ {l}_{2}{}^{2}+{m}_{2}{}^{2}+{n}_{2}{}^{2}=1\text{}\text{}\dots \left(iii\right)\\ Let\text{l,m,n be the direction cosines of a line which is perpendicular}\\ \text{to each given line with direction cosines}{l}_{1},{m}_{1},{n}_{1}\text{and}{l}_{2},{m}_{2},{n}_{2}\text{.}\\ \therefore \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{l}_{1}l+{m}_{1}m+{n}_{1}n=0\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}l{l}_{2}+m{m}_{2}+n{n}_{2}=0\\ \therefore \text{\hspace{0.17em}}\text{\hspace{0.17em}}\frac{l}{{m}_{1}{n}_{2}-{m}_{2}{n}_{1}}=\frac{m}{{n}_{1}{l}_{2}-{n}_{2}{l}_{1}}=\frac{n}{{l}_{1}{m}_{2}-{m}_{1}{l}_{2}}\\ ⇒\frac{{l}^{2}}{{\left({m}_{1}{n}_{2}-{m}_{2}{n}_{1}\right)}^{2}}=\frac{{m}^{2}}{{\left({n}_{1}{l}_{2}-{n}_{2}{l}_{1}\right)}^{2}}=\frac{{n}^{2}}{{\left({l}_{1}{m}_{2}-{m}_{1}{l}_{2}\right)}^{2}}\\ \text{}=\frac{{l}^{2}+{m}^{2}+{n}^{2}}{{\left({m}_{1}{n}_{2}-{m}_{2}{n}_{1}\right)}^{2}+{\left({n}_{1}{l}_{2}-{n}_{2}{l}_{1}\right)}^{2}+{\left({l}_{1}{m}_{2}-{m}_{1}{l}_{2}\right)}^{2}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\dots \left(iv\right)\\ Since,\text{}l\text{,m,n are the direction cosines of line}\text{.}\\ \therefore \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{l}^{\text{2}}\text{\hspace{0.17em}}{\text{+ m}}^{\text{2}}\text{\hspace{0.17em}}\text{+}\text{\hspace{0.17em}}{\text{n}}^{\text{2}}=1\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\dots \left(v\right)\\ It\text{is known that}\\ \left({l}_{1}{}^{2}+{m}_{1}{}^{2}+{n}_{1}{}^{2}\right)\left({l}_{2}{}^{2}+{m}_{2}{}^{2}+{n}_{2}{}^{2}\right)-{\left({l}_{1}{l}_{2}+{m}_{1}{m}_{2}+{n}_{1}{n}_{2}\right)}^{2}\\ \text{}\text{}\text{}={\left({m}_{1}{n}_{2}-{m}_{2}{n}_{1}\right)}^{2}+{\left({n}_{1}{l}_{2}-{n}_{2}{l}_{1}\right)}^{2}+{\left({l}_{1}{m}_{2}-{m}_{1}{l}_{2}\right)}^{2}\\ From\text{equations}\left(i\right),\left(ii\right)\text{and}\left(iii\right),\text{we get}\\ ⇒1.1-0={\left({m}_{1}{n}_{2}-{m}_{2}{n}_{1}\right)}^{2}+{\left({n}_{1}{l}_{2}-{n}_{2}{l}_{1}\right)}^{2}+{\left({l}_{1}{m}_{2}-{m}_{1}{l}_{2}\right)}^{2}\\ \therefore {\left({m}_{1}{n}_{2}-{m}_{2}{n}_{1}\right)}^{2}+{\left({n}_{1}{l}_{2}-{n}_{2}{l}_{1}\right)}^{2}+{\left({l}_{1}{m}_{2}-{m}_{1}{l}_{2}\right)}^{2}=1\text{}\dots \left(vi\right)\\ Substituting\text{the values from equations}\left(v\right)\text{and}\left(vi\right)\text{in equation}\\ \left(iv\right)\text{, we get}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\frac{{l}^{2}}{{\left({m}_{1}{n}_{2}-{m}_{2}{n}_{1}\right)}^{2}}=\frac{{m}^{2}}{{\left({n}_{1}{l}_{2}-{n}_{2}{l}_{1}\right)}^{2}}=\frac{{n}^{2}}{{\left({l}_{1}{m}_{2}-{m}_{1}{l}_{2}\right)}^{2}}=\frac{1}{1}\\ ⇒\frac{{l}^{2}}{{\left({m}_{1}{n}_{2}-{m}_{2}{n}_{1}\right)}^{2}}=\frac{{m}^{2}}{{\left({n}_{1}{l}_{2}-{n}_{2}{l}_{1}\right)}^{2}}=\frac{{n}^{2}}{{\left({l}_{1}{m}_{2}-{m}_{1}{l}_{2}\right)}^{2}}=1\\ ⇒l={m}_{1}{n}_{2}-{m}_{2}{n}_{1},\text{m}={n}_{1}{l}_{2}-{n}_{2}{l}_{1}\text{and n}={l}_{1}{m}_{2}-{m}_{1}{l}_{2}\\ Thus,\text{the direction cosines of the required line are}\\ {m}_{1}{n}_{2}-{m}_{2}{n}_{1},{n}_{1}{l}_{2}-{n}_{2}{l}_{1}\text{and}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{l}_{1}{m}_{2}-{m}_{1}{l}_{2}.\end{array}$

Q.39 Find the angle between the lines whose direction ratios are a, b, c and b – c, c – a, a – b.

Ans.

$\begin{array}{l}\text{L}\text{e}\text{t}\text{t}he\text{}angle\text{}between\text{}the\text{}lines\text{}whose\text{}direction\text{}ratios\text{}are\text{}\\ a,\text{}b,\text{}c\text{}and\text{}b-c,\text{}c-a,\text{}a-b\text{be}\theta \text{, then}\\ \text{cos}\theta =|\frac{a\left(b-c\right)+b\left(c-a\right)+c\left(a-b\right)}{\sqrt{{a}^{2}+{b}^{2}+{c}^{2}}\sqrt{{\left(b-c\right)}^{2}+{\left(c-a\right)}^{2}+{\left(a-b\right)}^{2}}}|\\ \text{}=|\frac{ab-ac+bc-ba+ca-bc}{\sqrt{{a}^{2}+{b}^{2}+{c}^{2}}\sqrt{{\left(b-c\right)}^{2}+{\left(c-a\right)}^{2}+{\left(a-b\right)}^{2}}}|\\ \text{}=0\\ \text{cos}\theta =\mathrm{cos}90°⇒\theta =90°\\ \text{Thus, the angle between the lines is}90°.\end{array}$

Q.40 Find the equation of a line parallel to x-axis and passing through the origin.

Ans.

$\begin{array}{l}\text{S}\text{i}\text{n}\text{c}\text{e},\text{the line parallel to x-axis and passing through the origin}\\ \text{is x-axis itself}\text{. Let A}\left(a,0,0\right)\text{be a point on x-axis, where a}\in \text{R}\text{.}\\ \text{So, direction cosines of OA are}\left(a-0\right),\left(0-0\right),\left(0-0\right)i.e.,\text{a,0,0}\text{.}\\ \text{The equation of OA is given as}\\ \frac{x-0}{a}=\frac{y-0}{0}=\frac{z-0}{0}⇒\frac{x}{1}=\frac{y}{0}=\frac{z}{0}=a\\ \text{T}\text{h}\text{u}\text{s},\text{the equation of line parallel to x-axis and passing through}\\ \text{origin is}\frac{x}{1}=\frac{y}{0}=\frac{z}{0}.\end{array}$

Q.41 If the coordinates of the points A, B, C, D be (1, 2, 3), (4, 5, 7), (– 4, 3, – 6) and (2, 9, 2) respectively, then find the angle between the lines AB and CD.

Ans.

$\begin{array}{l}\text{The coordinates of the points A, B, C, D be are}\\ \left(\text{1, 2, 3}\right)\text{,}\left(\text{4, 5, 7}\right)\text{,}\left(-\text{4,3,}-\text{6}\right)\text{and}\left(\text{2, 9, 2}\right)\text{respectivley}\text{.}\\ The\text{direction ratios of AB are:}\left(4-1\right),\left(5-2\right),\left(7-3\right)i.e,\text{3, 3, 4}\text{.}\\ The\text{direction ratios of CD are:}\left(2+4\right),\left(9-3\right),\left(2+6\right)\text{\hspace{0.17em}}i.e,\text{6, 6, 8}\text{.}\\ Then,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\frac{{a}_{1}}{{a}_{2}}=\frac{3}{6}=\frac{1}{2}\\ \text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\frac{{b}_{1}}{{b}_{2}}=\frac{3}{6}=\frac{1}{2}\\ \text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\frac{{c}_{1}}{{c}_{2}}=\frac{4}{8}=\frac{1}{2}\\ So,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\frac{{a}_{1}}{{a}_{2}}=\frac{{b}_{1}}{{b}_{2}}=\frac{{c}_{1}}{{c}_{2}}\\ Thus,\text{AB is parallel to CD}\text{.}\\ \text{Therefore, the angle between AB and CD is either 0° or 180°}\text{.}\end{array}$

Q.42

$\begin{array}{l}\mathrm{If}\mathrm{ }\mathrm{the}\mathrm{ }\mathrm{lines}\mathrm{ }\frac{\mathrm{x}-1}{-3}=\frac{\mathrm{y}-2}{2\mathrm{k}}=\frac{\mathrm{z}-3}{2}\mathrm{ }\mathrm{and}\mathrm{ }\frac{\mathrm{x}-1}{3\mathrm{k}}=\frac{\mathrm{y}-1}{1}=\frac{\mathrm{z}-6}{-5}\mathrm{ }\mathrm{are}\\ \mathrm{perpendicular},\mathrm{ }\mathrm{find}\mathrm{ }\mathrm{the}\mathrm{ }\mathrm{value}\mathrm{ }\mathrm{of}\mathrm{ }\mathrm{k}.\end{array}$

Ans.

$\begin{array}{l}\mathrm{Thegivenlinesare}\frac{\mathrm{x}-1}{-3}=\frac{\mathrm{y}-2}{2\mathrm{k}}=\frac{\mathrm{z}-3}{2}\mathrm{and}\frac{\mathrm{x}-1}{3\mathrm{k}}=\frac{\mathrm{y}-1}{1}=\frac{\mathrm{z}-6}{-5}.\\ \mathrm{The}\mathrm{direction}\mathrm{ratios}\mathrm{of}\mathrm{given}\mathrm{lines}\mathrm{are}:\\ {\mathrm{a}}_{\mathrm{1}}=-3,{\mathrm{b}}_{1}=2\mathrm{k}, {\mathrm{c}}_{1}=2\mathrm{and}{\mathrm{a}}_{\mathrm{2}}=3\mathrm{k},{\mathrm{b}}_{2}=1, {\mathrm{c}}_{2}=-5\\ \mathrm{Then}, {\mathrm{a}}_{1}{\mathrm{a}}_{2}+{\mathrm{b}}_{1}{\mathrm{b}}_{2}+{\mathrm{c}}_{1}{\mathrm{c}}_{2}=0\\ -3×3\mathrm{k}+2\mathrm{k}×1+2×-5=0\\ -9\mathrm{k}+2\mathrm{k}-10=0\\ -7\mathrm{k}-13=0\\ \mathrm{ }\mathrm{k}=-\frac{13}{7}\\ \mathrm{Therefore},\mathrm{given}\mathrm{lines}\mathrm{are}\mathrm{perpendicular}\mathrm{for}\mathrm{k}=-\frac{13}{7}.\end{array}$

Q.43

$\begin{array}{l}\mathrm{Find}\mathrm{ }\mathrm{the}\mathrm{ }\mathrm{vector}\mathrm{ }\mathrm{equation}\mathrm{ }\mathrm{of}\mathrm{ }\mathrm{the}\mathrm{ }\mathrm{line}\mathrm{ }\mathrm{passing}\mathrm{ }\mathrm{through}\mathrm{ }\left(1,2,3\right)\\ \mathrm{and}\mathrm{ }\mathrm{perpendicular}\mathrm{ }\mathrm{to}\mathrm{ }\mathrm{the}\mathrm{ }\mathrm{plane}\mathrm{ }\stackrel{\to }{\mathrm{r}}\left(\stackrel{^}{\mathrm{i}}+2\stackrel{^}{\mathrm{j}}-5\mathrm{ }\stackrel{^}{\mathrm{k}}\right)+9=0.\end{array}$

Ans.

$\begin{array}{l}\mathrm{The}\mathrm{position}\mathrm{vector}\mathrm{of}\mathrm{point}\left(1,2,3\right)=\stackrel{^}{\mathrm{i}}+2\mathrm{ }\stackrel{^}{\mathrm{j}}+3\mathrm{ }\stackrel{^}{\mathrm{k}}\\ \mathrm{The}\mathrm{perpendicular}\mathrm{plane}\mathrm{is}\stackrel{\to }{\mathrm{r}}.\left(\stackrel{^}{\mathrm{i}}+2\stackrel{^}{\mathrm{j}}-5\mathrm{ }\stackrel{^}{\mathrm{k}}\right)+9=0\mathrm{then}\\ \mathrm{direction}\mathrm{ratios}\mathrm{to}\mathrm{the}\mathrm{plane}\mathrm{are}1,2,-5\mathrm{and}\mathrm{normal}\mathrm{vector}\mathrm{is}\\ \stackrel{\to }{\mathrm{N}}=\stackrel{^}{\mathrm{i}}+2\mathrm{ }\stackrel{^}{\mathrm{j}}-5\mathrm{ }\stackrel{^}{\mathrm{k}}\\ \mathrm{The}\mathrm{equation}\mathrm{of}\mathrm{line}\mathrm{passing}\mathrm{through}\mathrm{a}\mathrm{point}\mathrm{and}\mathrm{perpendicular}\\ \mathrm{to}\mathrm{the}\mathrm{given}\mathrm{plane}\mathrm{is}\\ \stackrel{\to }{\mathrm{l}}=\stackrel{\to }{\mathrm{r}}+\mathrm{\lambda }\stackrel{\to }{\mathrm{N}}, \mathrm{\lambda }\in \mathrm{R}\\ \stackrel{\to }{\mathrm{l}}=\left(\stackrel{^}{\mathrm{i}}+2\mathrm{ }\stackrel{^}{\mathrm{j}}+3\mathrm{ }\stackrel{^}{\mathrm{k}}\right)+\mathrm{\lambda }\left(\stackrel{^}{\mathrm{i}}+2\mathrm{ }\stackrel{^}{\mathrm{j}}-5\mathrm{ }\stackrel{^}{\mathrm{k}}\right)\end{array}$

Q.44

$\begin{array}{l}\mathrm{Find}\mathrm{ }\mathrm{the}\mathrm{ }\mathrm{equation}\mathrm{ }\mathrm{of}\mathrm{ }\mathrm{the}\mathrm{ }\mathrm{plane}\mathrm{ }\mathrm{passing}\mathrm{ }\mathrm{through}\mathrm{ }\left(\mathrm{a},\mathrm{b},\mathrm{c}\right)\mathrm{ }\mathrm{and}\\ \mathrm{parallel}\mathrm{ }\mathrm{to}\mathrm{ }\mathrm{the}\mathrm{ }\mathrm{plane}\mathrm{ }\stackrel{\to }{\mathrm{r}}\left(\stackrel{^}{\mathrm{i}}+\stackrel{^}{\mathrm{j}}+\stackrel{^}{\mathrm{k}}\right)=2.\end{array}$

Ans.

$\begin{array}{l}\mathrm{Theequationoftheplanepassingthrough}\left(\mathrm{a},\mathrm{b},\mathrm{c}\right)\mathrm{and}\\ \mathrm{paralleltotheplane}\stackrel{\to }{\mathrm{r}}\left(\stackrel{^}{\mathrm{i}}+\stackrel{^}{\mathrm{j}}+\stackrel{^}{\mathrm{k}}\right)=2.\\ \mathrm{Equation}\mathrm{of}\mathrm{the}\mathrm{plane}\mathrm{parallel}\mathrm{to}\mathrm{the}\mathrm{plane}\stackrel{\to }{\mathrm{r}}\left(\stackrel{^}{\mathrm{i}}+\stackrel{^}{\mathrm{j}}+\stackrel{^}{\mathrm{k}}\right)=2\mathrm{ }\mathrm{is}:\\ \stackrel{\to }{\mathrm{r}}\left(\stackrel{^}{\mathrm{i}}+\stackrel{^}{\mathrm{j}}+\stackrel{^}{\mathrm{k}}\right)=\mathrm{\lambda } \mathrm{ }...\left(\mathrm{i}\right)\\ \mathrm{The}\mathrm{plane}\mathrm{passes}\mathrm{through}\mathrm{the}\mathrm{point}\left(\mathrm{a},\mathrm{b},\mathrm{c}\right).\mathrm{Therefore},\mathrm{the}\\ \mathrm{position}\mathrm{vector}\stackrel{\to }{\mathrm{r}}\mathrm{of}\mathrm{this}\mathrm{point}\mathrm{is}\stackrel{\to }{\mathrm{r}}=\mathrm{a}\mathrm{ }\stackrel{^}{\mathrm{i}}+\mathrm{b}\mathrm{ }\stackrel{^}{\mathrm{j}}+\mathrm{c}\mathrm{ }\stackrel{^}{\mathrm{k}}\\ \mathrm{Substituting}\mathrm{the}\mathrm{value}\mathrm{of}\stackrel{\to }{\mathrm{r}}\mathrm{in}\mathrm{equation}\left(\mathrm{i}\right),\mathrm{we}\mathrm{get}\\ \left(\mathrm{a}\mathrm{ }\stackrel{^}{\mathrm{i}}+\mathrm{b}\mathrm{ }\stackrel{^}{\mathrm{j}}+\mathrm{c}\mathrm{ }\stackrel{^}{\mathrm{k}}\right)\left(\stackrel{^}{\mathrm{i}}+\stackrel{^}{\mathrm{j}}+\stackrel{^}{\mathrm{k}}\right)=\mathrm{\lambda }\\ ⇒\mathrm{a}+\mathrm{b}+\mathrm{c}=\mathrm{\lambda }\\ \mathrm{So},\mathrm{ }\mathrm{from}\mathrm{equation}\left(\mathrm{i}\right),\mathrm{the}\mathrm{equation}\mathrm{of}\mathrm{required}\mathrm{plane}\mathrm{is}\\ \stackrel{\to }{\mathrm{r}}\left(\stackrel{^}{\mathrm{i}}+\stackrel{^}{\mathrm{j}}+\stackrel{^}{\mathrm{k}}\right)=\mathrm{a}+\mathrm{b}+\mathrm{c}...\left(\mathrm{ii}\right)\\ \mathrm{This}\mathrm{is}\mathrm{the}\mathrm{vector}\mathrm{equation}\mathrm{of}\mathrm{required}\mathrm{plane}\mathrm{.}\\ \mathrm{Substituting}\stackrel{\to }{\mathrm{r}}=\mathrm{x}\mathrm{ }\stackrel{^}{\mathrm{i}}+\mathrm{y}\mathrm{ }\stackrel{^}{\mathrm{j}}+\mathrm{z}\mathrm{ }\stackrel{^}{\mathrm{k}}\mathrm{in}\mathrm{equation}\left(\mathrm{ii}\right),\mathrm{we}\mathrm{get}\\ \left(\mathrm{x}\mathrm{ }\stackrel{^}{\mathrm{i}}+\mathrm{y}\mathrm{ }\stackrel{^}{\mathrm{j}}+\mathrm{z}\mathrm{ }\stackrel{^}{\mathrm{k}}\right)\left(\stackrel{^}{\mathrm{i}}+\stackrel{^}{\mathrm{j}}+\stackrel{^}{\mathrm{k}}\right)=\mathrm{a}+\mathrm{b}+\mathrm{c}\\ ⇒ \mathrm{ }\mathrm{x}+\mathrm{y}+\mathrm{z}=\mathrm{a}+\mathrm{b}+\mathrm{c}\end{array}$

Q.45

$\begin{array}{l}\mathrm{Find}\mathrm{ }\mathrm{the}\mathrm{ }\mathrm{shortest}\mathrm{ }\mathrm{distance}\mathrm{ }\mathrm{between}\mathrm{ }\mathrm{lines}\\ \stackrel{\to }{\mathrm{r}}=6\stackrel{^}{\mathrm{i}}+2\stackrel{^}{\mathrm{j}}+2\stackrel{^}{\mathrm{k}}+\mathrm{\lambda }\left(\stackrel{^}{\mathrm{i}}-2\stackrel{^}{\mathrm{j}}+2\stackrel{^}{\mathrm{k}}\right)\mathrm{ }\mathrm{and}\\ \stackrel{\to }{\mathrm{r}}=-4\mathrm{ }\stackrel{^}{\mathrm{i}}-\stackrel{^}{\mathrm{k}}+\mathrm{\mu }\left(3\mathrm{ }\stackrel{^}{\mathrm{i}}-2\stackrel{^}{\mathrm{j}}-2\stackrel{^}{\mathrm{k}}\right).\end{array}$

Ans.

$\begin{array}{l}\mathrm{The}\mathrm{equations}\mathrm{of}\mathrm{given}\mathrm{lines}\mathrm{are}:\\ \stackrel{\to }{\mathrm{r}}=6\stackrel{^}{\mathrm{i}}+2\stackrel{^}{\mathrm{j}}+2\stackrel{^}{\mathrm{k}}+\mathrm{\lambda }\left(\stackrel{^}{\mathrm{i}}-2\stackrel{^}{\mathrm{j}}+2\stackrel{^}{\mathrm{k}}\right)...\left(\mathrm{i}\right)\\ \stackrel{\to }{\mathrm{r}}=-4\mathrm{ }\stackrel{^}{\mathrm{i}}-\stackrel{^}{\mathrm{k}}+\mathrm{\mu }\left(3\mathrm{ }\stackrel{^}{\mathrm{i}}-2\stackrel{^}{\mathrm{j}}-2\stackrel{^}{\mathrm{k}}\right)...\left(\mathrm{ii}\right)\\ \mathrm{Comparing}\mathrm{equation}\left(\mathrm{i}\right)\mathrm{and}\mathrm{equation}\left(\mathrm{ii}\right)\mathrm{with}\stackrel{\to }{\mathrm{r}}={\mathrm{a}}_{1}+{\mathrm{\lambda b}}_{1}\mathrm{and}\\ \stackrel{\to }{\mathrm{r}}={\mathrm{a}}_{2}+{\mathrm{\lambda b}}_{2},\mathrm{we}\mathrm{get}\\ {\mathrm{a}}_{1}=6\stackrel{^}{\mathrm{i}}+2\stackrel{^}{\mathrm{j}}+2\stackrel{^}{\mathrm{k}},{\mathrm{b}}_{1}=\stackrel{^}{\mathrm{i}}-2\stackrel{^}{\mathrm{j}}+2\stackrel{^}{\mathrm{k}}\\ {\mathrm{a}}_{2}=-4\mathrm{ }\stackrel{^}{\mathrm{i}}-\stackrel{^}{\mathrm{k}}{\mathrm{b}}_{2}=3\mathrm{ }\stackrel{^}{\mathrm{i}}-2\stackrel{^}{\mathrm{j}}-2\stackrel{^}{\mathrm{k}}\\ \mathrm{The}\mathrm{shortest}\mathrm{distance}\mathrm{between}\mathrm{two}\mathrm{lines}\left(\mathrm{d}\right)=|\frac{\left(\stackrel{\to }{{\mathrm{b}}_{1}}×\stackrel{\to }{{\mathrm{b}}_{2}}\right).\left(\stackrel{\to }{{\mathrm{a}}_{2}}-\stackrel{\to }{{\mathrm{a}}_{1}}\right)}{|\stackrel{\to }{{\mathrm{b}}_{1}}×\stackrel{\to }{{\mathrm{b}}_{2}}|}|\\ \stackrel{\to }{{\mathrm{b}}_{1}}×\stackrel{\to }{{\mathrm{b}}_{2}}=|\begin{array}{ccc}\mathrm{ }\stackrel{^}{\mathrm{i}}& \stackrel{^}{\mathrm{j}}& \stackrel{^}{\mathrm{k}}\\ 1& -2& 2\\ 3& -2& -2\end{array}|\\ =\left(4+4\right)\stackrel{^}{\mathrm{i}}-\left(-2-6\right)\stackrel{^}{\mathrm{j}}+\left(-2+6\right)\stackrel{^}{\mathrm{k}}\\ =8\mathrm{ }\stackrel{^}{\mathrm{i}}+8\mathrm{ }\stackrel{^}{\mathrm{j}}+4\mathrm{ }\stackrel{^}{\mathrm{k}}\\ |\mathrm{ }\stackrel{\to }{{\mathrm{b}}_{1}}×\stackrel{\to }{{\mathrm{b}}_{2}}|=\sqrt{{8}^{2}+{8}^{2}+{4}^{2}}\\ =12\\ \mathrm{ }{\mathrm{a}}_{2}-{\mathrm{a}}_{1}=\left(-4\mathrm{ }\stackrel{^}{\mathrm{i}}-\stackrel{^}{\mathrm{k}}\right)-\left(6\stackrel{^}{\mathrm{i}}+2\stackrel{^}{\mathrm{j}}+2\stackrel{^}{\mathrm{k}}\right)\\ =-10\mathrm{ }\stackrel{^}{\mathrm{i}}-2\mathrm{ }\stackrel{^}{\mathrm{j}}-3\mathrm{ }\stackrel{^}{\mathrm{k}}\\ \therefore \mathrm{d}\mathrm{ }=|\frac{\left(8\mathrm{ }\stackrel{^}{\mathrm{i}}+8\mathrm{ }\stackrel{^}{\mathrm{j}}+4\mathrm{ }\stackrel{^}{\mathrm{k}}\right).\left(-10\mathrm{ }\stackrel{^}{\mathrm{i}}-2\mathrm{ }\stackrel{^}{\mathrm{j}}-3\mathrm{ }\stackrel{^}{\mathrm{k}}\right)}{12}|\\ =|\frac{-80-16-12}{12}|\\ =\frac{108}{12}=9\\ \mathrm{Therefore},\mathrm{the}\mathrm{shortest}\mathrm{distance}\mathrm{between}\mathrm{two}\mathrm{lines}\mathrm{is}9\mathrm{units}\mathrm{.}\end{array}$

Q.46 Find the coordinates of the point where the line through (5, 1, 6) and (3, 4, 1) crosses the YZ-plane.

Ans.

$\begin{array}{l}The\text{equation of the line passing through the points}\left({x}_{1},{y}_{1},{z}_{1}\right)\\ and\text{}\left({x}_{2},{y}_{2},{z}_{2}\right)\text{is}\\ \text{}\text{}\frac{x-{x}_{1}}{{x}_{2}-{x}_{1}}=\frac{y-{y}_{1}}{{y}_{2}-{y}_{1}}=\frac{z-{z}_{1}}{{z}_{2}-{z}_{1}}\\ The\text{equation of the line passing through the points}\left(5,1,6\right)\\ and\text{}\left(3,4,1\right)\text{is given as}\\ \text{}\text{}\frac{x-5}{3-5}=\frac{y-1}{4-1}=\frac{z-6}{1-6}\\ ⇒\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\frac{x-5}{-2}=\frac{y-1}{3}=\frac{z-6}{-5}=k\left(let\right)\\ ⇒x=5-2k,\text{\hspace{0.17em}}\text{\hspace{0.17em}}y=1+3k\text{and z}=6-5k\\ Any\text{point on the line is of the form}\left(5-2k,\text{\hspace{0.17em}}1+3k,\text{\hspace{0.17em}}6-5k\right).\\ The\text{equation of YZ-plane is x}=\text{0}\\ \text{Since, the line passes through YZ-plane,}\\ \therefore \text{}\text{}5-2k=0⇒k=\frac{5}{2}\\ So,\text{\hspace{0.17em}}y=1+3×\frac{5}{2}=\frac{17}{2}\text{and z}=6-5×\frac{5}{2}=\frac{-13}{2}\\ Therefore,\text{the required point is}\left(0,\text{\hspace{0.17em}}\frac{17}{2},\frac{-13}{2}\right).\end{array}$

Q.47 Find the coordinates of the point where the line through (5, 1, 6) and (3, 4, 1) crosses the ZX-plane.

Ans.

$\begin{array}{l}The\text{equation of the line passing through the points}\left({x}_{1},{y}_{1},{z}_{1}\right)\\ and\text{}\left({x}_{2},{y}_{2},{z}_{2}\right)\text{is}\\ \text{}\text{}\frac{x-{x}_{1}}{{x}_{2}-{x}_{1}}=\frac{y-{y}_{1}}{{y}_{2}-{y}_{1}}=\frac{z-{z}_{1}}{{z}_{2}-{z}_{1}}\\ The\text{equation of the line passing through the points}\left(5,1,6\right)\\ and\text{}\left(3,4,1\right)\text{is given as}\\ \text{}\text{}\frac{x-5}{3-5}=\frac{y-1}{4-1}=\frac{z-6}{1-6}\\ ⇒\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\frac{x-5}{-2}=\frac{y-1}{3}=\frac{z-6}{-5}=k\left(let\right)\\ ⇒x=5-2k,\text{\hspace{0.17em}}\text{\hspace{0.17em}}y=1+3k\text{and z}=6-5k\\ Any\text{point on the line is of the form}\left(5-2k,\text{\hspace{0.17em}}1+3k,\text{\hspace{0.17em}}6-5k\right).\\ The\text{equation of ZX-plane is y}=\text{0}\\ \text{Since, the line passes through ZX-plane,}\\ \therefore \text{}\text{}1+3k=0⇒k=-\frac{1}{3}\\ So,\text{\hspace{0.17em}}x=5-2×-\frac{1}{3}=\frac{17}{3}\text{and z}=6-5×-\frac{1}{3}=\frac{23}{3}\\ Therefore,\text{the required point is}\left(\frac{17}{3},\text{\hspace{0.17em}}0,\frac{23}{3}\right).\end{array}$

Q.48 Find the coordinates of the point where the line through (3, – 4, – 5) and (2, – 3, 1) crosses the plane 2x + y + z = 7.

Ans.

$\begin{array}{l}The\text{equation of the line passing through the points}\left({x}_{1},{y}_{1},{z}_{1}\right)\\ and\text{}\left({x}_{2},{y}_{2},{z}_{2}\right)\text{is}\\ \text{}\text{}\frac{x-{x}_{1}}{{x}_{2}-{x}_{1}}=\frac{y-{y}_{1}}{{y}_{2}-{y}_{1}}=\frac{z-{z}_{1}}{{z}_{2}-{z}_{1}}\\ The\text{equation of the line passing through the points}\left(3,-4,-5\right)\\ and\text{}\left(2,-3,1\right)\text{is given as}\\ \text{}\text{}\frac{x-3}{2-3}=\frac{y+4}{-3+4}=\frac{z+5}{1+5}\\ ⇒\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\frac{x-3}{-1}=\frac{y+4}{1}=\frac{z+5}{6}=k\left(let\right)\\ ⇒x=3-k,\text{\hspace{0.17em}}\text{\hspace{0.17em}}y=k-4\text{and z}=6k-5\\ Any\text{point on the line is of the form}\left(3-k,\text{\hspace{0.17em}}k-4,\text{\hspace{0.17em}}6\text{\hspace{0.17em}}k-5\right).\\ Since,\text{point lies on the plane,}2x+y+z=7\\ \therefore \text{}2\left(3-k\right)+k-4+6\text{\hspace{0.17em}}k-5=7\\ ⇒\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}6-2k+k-4+6k-5=7\\ ⇒\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}5k-3=7\\ ⇒\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}k=\frac{10}{5}=2\\ So,\text{\hspace{0.17em}}x=3-2=1,\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}y=2-4=-2\text{and}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{z}=6\text{\hspace{0.17em}}×2-5=7\\ Therefore,\text{the required point is}\left(1,\text{\hspace{0.17em}}-2,7\right).\end{array}$

Q.49 Find the equation of the plane passing through the point (–1, 3, 2) and perpendicular to each of the planes x + 2y + 3z = 5 and 3x + 3y + z = 0

Ans.

$\begin{array}{l}\text{T}\text{h}\text{e}\text{equation of the plane passing through the point}\left(-1,3,2\right)\text{is}\\ \text{a}\left(x+1\right)+b\left(y-3\right)+c\left(z-2\right)=0\text{}\text{}\dots \left(i\right)\\ \text{w}\text{h}\text{e}\text{r}\text{e}\text{a, b, c are the direction ratios of the plane}\text{.}\\ \text{Since, two planes are perpendicular to each other, if}\\ {a}_{1}{a}_{2}+{b}_{1}{b}_{2}+{c}_{1}{c}_{2}=\text{0}\\ \text{Plane}\left(i\right)\text{is perpendicular to the plane, x}+2y+3z=5\\ \therefore \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}a.1+b.2+c.3=\text{0}\\ ⇒\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}a+2b+3c=0\text{}\text{}\text{}\text{}\dots \left(ii\right)\\ \text{Plane}\left(i\right)\text{is perpendicular to the plane, 3x}+3y+z=0\\ \therefore \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}a.3+b.3+c.1=\text{0}\\ ⇒\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}3a+3\text{\hspace{0.17em}}b+c=0\text{}\text{}\text{}\text{}\dots \left(iii\right)\\ From\text{equation}\left(ii\right)\text{and equation}\left(iii\right),\text{we have}\\ \text{}\frac{a}{2-9}=\frac{-b}{1-9}=\frac{c}{3-6}\\ ⇒\text{}\frac{a}{-7}=\frac{b}{8}=\frac{c}{-3}=k\left(let\right)\\ ⇒a=-7k,\text{b}=\text{\hspace{0.17em}}8k\text{and c}=-3k\\ \text{S}\text{u}\text{b}\text{s}\text{t}\text{i}\text{t}\text{u}\text{t}\text{i}\text{n}\text{g}\text{the values of a, b and c in equation}\left(i\right),\text{we get}\\ -7k\left(x+1\right)+\text{\hspace{0.17em}}8k\left(y-3\right)-3k\left(z-2\right)=0\\ ⇒\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}7x+7-8y+24+3z-6=0\\ ⇒\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}7x-8y+3z+25=0\\ \text{This is the required equation of the plane}\text{.}\end{array}$

Q.50

$\begin{array}{l}\mathrm{If}\mathrm{ }\mathrm{the}\mathrm{ }\mathrm{points}\mathrm{ }\left(1,1,\mathrm{p}\right) \mathrm{and}\mathrm{ }\left(-3,0,1\right)\mathrm{ }\mathrm{be}\mathrm{ }\mathrm{equidistant}\mathrm{ }\mathrm{from}\mathrm{ }\mathrm{the}\\ \mathrm{plane}\mathrm{ }\stackrel{\to }{\mathrm{r}}.\left(3\mathrm{ }\stackrel{^}{\mathrm{i}}+4\mathrm{ }\stackrel{^}{\mathrm{j}}-12\mathrm{ }\stackrel{^}{\mathrm{k}}\right)+13=0,\mathrm{ }\mathrm{then}\mathrm{ }\mathrm{find}\mathrm{ }\mathrm{the}\mathrm{ }\mathrm{value}\mathrm{ }\mathrm{of}\mathrm{ }\mathrm{p}.\end{array}$

Ans.

$\begin{array}{l}\mathrm{The}\mathrm{position}\mathrm{vector}\mathrm{through}\mathrm{the}\mathrm{point}\left(1,\mathrm{ }1,\mathrm{p}\right)\mathrm{is}\stackrel{\to }{{\mathrm{a}}_{1}}=\stackrel{^}{\mathrm{i}}+\stackrel{^}{\mathrm{j}}+\mathrm{p}\stackrel{^}{\mathrm{k}}\\ \mathrm{Similarly},\mathrm{the}\mathrm{position}\mathrm{vector}\mathrm{through}\mathrm{the}\mathrm{point}\left(-3,0,1\right)\mathrm{is}\\ \stackrel{\to }{{\mathrm{a}}_{2}}=-3\mathrm{ }\stackrel{^}{\mathrm{i}}+\stackrel{^}{\mathrm{k}}\\ \mathrm{The}\mathrm{equation}\mathrm{of}\mathrm{the}\mathrm{given}\mathrm{plane}\mathrm{is}\stackrel{\to }{\mathrm{r}}.\left(3\mathrm{ }\stackrel{^}{\mathrm{i}}+4\mathrm{ }\stackrel{^}{\mathrm{j}}-12\mathrm{ }\stackrel{^}{\mathrm{k}}\right)+13=0\\ \mathrm{Since},\mathrm{the}\mathrm{perpendicular}\mathrm{distance}\mathrm{of}\mathrm{a}\mathrm{point}\mathrm{and}\mathrm{the}\mathrm{plane}\mathrm{is}\\ \mathrm{d}=|\frac{\stackrel{\to }{\mathrm{a}}.\stackrel{\to }{\mathrm{N}}-\mathrm{d}}{|\stackrel{\to }{\mathrm{N}}|}|\\ \mathrm{Here},\mathrm{ }\stackrel{\to }{\mathrm{N}}=3\mathrm{ }\stackrel{^}{\mathrm{i}}+4\mathrm{ }\stackrel{^}{\mathrm{j}}-12\mathrm{ }\stackrel{^}{\mathrm{k}}\mathrm{and}\mathrm{d}=-\mathrm{13}\\ \mathrm{Therefore},\mathrm{the}\mathrm{distance}\mathrm{between}\mathrm{point}\left(1,\mathrm{ }1,\mathrm{p}\right)\mathrm{and}\mathrm{plane}\\ \stackrel{\to }{\mathrm{r}}.\left(3\mathrm{ }\stackrel{^}{\mathrm{i}}+4\mathrm{ }\stackrel{^}{\mathrm{j}}-12\mathrm{ }\stackrel{^}{\mathrm{k}}\right)+13=0\mathrm{is}\\ {\mathrm{d}}_{\mathrm{1}}=|\frac{\left(\stackrel{^}{\mathrm{i}}+\stackrel{^}{\mathrm{j}}+\mathrm{p}\stackrel{^}{\mathrm{k}}\right).\left(3\mathrm{ }\stackrel{^}{\mathrm{i}}+4\mathrm{ }\stackrel{^}{\mathrm{j}}-12\mathrm{ }\stackrel{^}{\mathrm{k}}\right)+13}{|3\mathrm{ }\stackrel{^}{\mathrm{i}}+4\mathrm{ }\stackrel{^}{\mathrm{j}}-12\mathrm{ }\stackrel{^}{\mathrm{k}}|}|\\ =|\frac{3+4-12\mathrm{p}+13}{\sqrt{{3}^{2}+{4}^{2}+{\left(-12\right)}^{2}}}|\\ \mathrm{ }{\mathrm{d}}_{\mathrm{1}}=\frac{|20-12\mathrm{p}|}{13}\\ \mathrm{Similarly},\mathrm{the}\mathrm{distance}\mathrm{between}\mathrm{point}\left(-3,0,1\right)\mathrm{and}\mathrm{plane}\\ \stackrel{\to }{\mathrm{r}}.\left(3\mathrm{ }\stackrel{^}{\mathrm{i}}+4\mathrm{ }\stackrel{^}{\mathrm{j}}-12\mathrm{ }\stackrel{^}{\mathrm{k}}\right)+13=0\mathrm{is}\\ {\mathrm{d}}_{\mathrm{2}}=|\frac{\left(-3\mathrm{ }\stackrel{^}{\mathrm{i}}+\stackrel{^}{\mathrm{k}}\right).\left(3\mathrm{ }\stackrel{^}{\mathrm{i}}+4\mathrm{ }\stackrel{^}{\mathrm{j}}-12\mathrm{ }\stackrel{^}{\mathrm{k}}\right)+13}{|3\mathrm{ }\stackrel{^}{\mathrm{i}}+4\mathrm{ }\stackrel{^}{\mathrm{j}}-12\mathrm{ }\stackrel{^}{\mathrm{k}}|}|\\ =|\frac{-\mathrm{ }9-12+13}{\sqrt{{3}^{2}+{4}^{2}+{\left(-12\right)}^{2}}}|\\ \mathrm{ }{\mathrm{d}}_{\mathrm{2}}=\frac{|-8|}{13}=\frac{8}{13}\\ \mathrm{Since},\mathrm{distance}\mathrm{of}\mathrm{points}\left(1,\mathrm{ }1,\mathrm{p}\right)\mathrm{ }\mathrm{and}\left(-3,0,1\right)\mathrm{from}\mathrm{plane}\\ \stackrel{\to }{\mathrm{r}}.\left(3\mathrm{ }\stackrel{^}{\mathrm{i}}+4\mathrm{ }\stackrel{^}{\mathrm{j}}-12\mathrm{ }\stackrel{^}{\mathrm{k}}\right)+13=0\mathrm{is}\mathrm{equal}\mathrm{.}\\ \therefore {\mathrm{d}}_{\mathrm{1}}={\mathrm{d}}_{\mathrm{2}}\\ ⇒\frac{|20-12\mathrm{p}|}{13}=\frac{8}{13}\\ ⇒\mathrm{Either}\mathrm{}\left(20-12\mathrm{p}\right)=8\mathrm{or}-\left(20-12\mathrm{p}\right)=8\\ ⇒\mathrm{p}=1\mathrm{or}\frac{7}{3}\end{array}$

Q.51

$\begin{array}{l}\mathrm{Find}\mathrm{ }\mathrm{the}\mathrm{ }\mathrm{equation}\mathrm{ }\mathrm{of}\mathrm{ }\mathrm{the}\mathrm{ }\mathrm{plane}\mathrm{ }\mathrm{passing}\mathrm{ }\mathrm{through}\mathrm{ }\mathrm{the}\mathrm{ }\mathrm{line}\mathrm{ }\mathrm{of}\\ \mathrm{intersection}\mathrm{ }\mathrm{of}\mathrm{ }\mathrm{the}\mathrm{ }\mathrm{planes}\mathrm{ }\stackrel{\to }{\mathrm{r}}.\left(\stackrel{^}{\mathrm{i}}+\stackrel{^}{\mathrm{j}}+\stackrel{^}{\mathrm{k}}\right)=1\mathrm{ }\mathrm{and}\\ \stackrel{\to }{\mathrm{r}}.\left(2\mathrm{ }\stackrel{^}{\mathrm{i}}+3\stackrel{^}{\mathrm{j}}-\stackrel{^}{\mathrm{k}}\right)+4=0\mathrm{ }\mathrm{and}\mathrm{ }\mathrm{parallel}\mathrm{ }\mathrm{to}\mathrm{ }\mathrm{x}\mathrm{ }–\mathrm{ }\mathrm{axis}.\end{array}$

Ans.

$\begin{array}{l}\mathrm{Theequationoftheplanepassingthroughthelineof}\mathrm{intersection}\\ \mathrm{oftheplanes}\stackrel{\to }{\mathrm{r}}.\left(\stackrel{^}{\mathrm{i}}+\stackrel{^}{\mathrm{j}}+\stackrel{^}{\mathrm{k}}\right)=1\mathrm{and}\stackrel{\to }{\mathrm{r}}.\left(2\mathrm{ }\stackrel{^}{\mathrm{i}}+3\stackrel{^}{\mathrm{j}}-\stackrel{^}{\mathrm{k}}\right)+4=0\mathrm{is}\therefore \\ \mathrm{ }\left[\stackrel{\to }{\mathrm{r}}.\left(\stackrel{^}{\mathrm{i}}+\stackrel{^}{\mathrm{j}}+\stackrel{^}{\mathrm{k}}\right)-1\right]+\mathrm{\lambda }\left[\stackrel{\to }{\mathrm{r}}.\left(2\mathrm{ }\stackrel{^}{\mathrm{i}}+3\stackrel{^}{\mathrm{j}}-\stackrel{^}{\mathrm{k}}\right)+4\right]=0\\ \mathrm{ }\stackrel{\to }{\mathrm{r}}\left[\left(\stackrel{^}{\mathrm{i}}+\stackrel{^}{\mathrm{j}}+\stackrel{^}{\mathrm{k}}\right)+\mathrm{\lambda }\left(2\mathrm{ }\stackrel{^}{\mathrm{i}}+3\stackrel{^}{\mathrm{j}}-\stackrel{^}{\mathrm{k}}\right)\right]+\left(4\mathrm{\lambda }-1\right)=0\\ \stackrel{\to }{\mathrm{r}}\left[\left(1+2\mathrm{\lambda }\right)\stackrel{^}{\mathrm{i}}+\left(1+3\mathrm{\lambda }\right)\stackrel{^}{\mathrm{j}}+\left(1-\mathrm{\lambda }\right)\stackrel{^}{\mathrm{k}}\right]+\left(4\mathrm{\lambda }-1\right)=0...\left(\mathrm{i}\right)\\ \mathrm{The}\mathrm{direction}\mathrm{cosines}\mathrm{of}\mathrm{plane}\mathrm{are}\left(1+2\mathrm{\lambda }\right),\left(1+3\mathrm{\lambda }\right),\left(1-\mathrm{\lambda }\right).\\ \mathrm{The}\mathrm{required}\mathrm{plane}\mathrm{is}\mathrm{parallel}\mathrm{to}\mathrm{x}–\mathrm{axis}.\mathrm{Therefore},\mathrm{its}\mathrm{normal}\\ \mathrm{is}\mathrm{perpendicular}\mathrm{to}\mathrm{x}–\mathrm{axis}\mathrm{.}\\ \mathrm{The}\mathrm{direction}\mathrm{ratios}\mathrm{of}\mathrm{x}–\mathrm{axis}\mathrm{are}1, 0, 0\mathrm{.}\\ \because {\mathrm{a}}_{1}{\mathrm{a}}_{2}+{\mathrm{b}}_{1}{\mathrm{b}}_{2}+{\mathrm{c}}_{1}{\mathrm{c}}_{2}=0\left[\mathrm{For}\mathrm{perpendicular}\mathrm{line}\mathrm{on}\mathrm{plane}\mathrm{.}\right]\\ \therefore \left(1+2\mathrm{\lambda }\right).1+\left(1+3\mathrm{\lambda }\right).0+\left(1-\mathrm{\lambda }\right).0=0\\ ⇒ 1+2\mathrm{\lambda }=0\\ ⇒ \mathrm{ }\mathrm{\lambda }=-\frac{1}{2}\\ \mathrm{Substituting}\mathrm{}\mathrm{\lambda }=-\frac{1}{2}\mathrm{in}\mathrm{equation}\left(\mathrm{i}\right),\mathrm{we}\mathrm{get}\\ \stackrel{\to }{\mathrm{r}}\left[\left(1+2×-\frac{1}{2}\right)\stackrel{^}{\mathrm{i}}+\left(1+3×-\frac{1}{2}\right)\stackrel{^}{\mathrm{j}}+\left(1+\frac{1}{2}\right)\stackrel{^}{\mathrm{k}}\right]+\left(4×-\frac{1}{2}-1\right)=0\\ ⇒ \stackrel{\to }{\mathrm{r}}\left[\left(1-1\right)\stackrel{^}{\mathrm{i}}+\left(1-\frac{3}{2}\right)\stackrel{^}{\mathrm{j}}+\frac{3}{2}\stackrel{^}{\mathrm{k}}\right]+\left(-2-1\right)=0\\ ⇒ \stackrel{\to }{\mathrm{r}}\left[-\frac{1}{2}\stackrel{^}{\mathrm{j}}+\frac{3}{2}\stackrel{^}{\mathrm{k}}\right]-3=0\\ ⇒ \stackrel{\to }{\mathrm{r}}\left[-\stackrel{^}{\mathrm{j}}+3\stackrel{^}{\mathrm{k}}\right]-6=0\\ \mathrm{Putting}\mathrm{}\stackrel{\to }{\mathrm{r}}=\mathrm{x}\mathrm{ }\stackrel{^}{\mathrm{i}}+\mathrm{y}\mathrm{ }\stackrel{^}{\mathrm{j}}+\mathrm{z}\mathrm{ }\stackrel{^}{\mathrm{k}},\mathrm{we}\mathrm{get}\\ \left(\mathrm{x}\mathrm{ }\stackrel{^}{\mathrm{i}}+\mathrm{y}\mathrm{ }\stackrel{^}{\mathrm{j}}+\mathrm{z}\mathrm{ }\stackrel{^}{\mathrm{k}}\right)\left[-\stackrel{^}{\mathrm{j}}+3\stackrel{^}{\mathrm{k}}\right]-6=0\\ -\mathrm{y}+3\mathrm{z}-6=0\mathrm{or}\mathrm{y}-3\mathrm{z}+6=0\\ \mathrm{This}\mathrm{is}\mathrm{the}\mathrm{equation}\mathrm{of}\mathrm{the}\mathrm{required}\mathrm{plane}\mathrm{.}\end{array}$

Q.52 If O be the origin and the coordinates of P be (1, 2, – 3), then find the equation of the plane passing through P and perpendicular to OP.

Ans.

The coordinates of the origin, O and point P are (0, 0, 0) and (1, 2, –3) respectively.

So, the direction ratios of OP are: 1–0, 2 – 0, –3 – 0, i.e., 1, 2, –3.

The equation of plane passing through (1, 2, –3) and perpendicular to OP having direction ratios 1, 2, –3 is

1(x – 1) + 2(y – 2) – 3 (z + 3) = 0
x – 1 + 2y – 4 –3z –9 = 0
x + 2y – 3z – 14 = 0

This is the required equation of the plane.

Q.53

$\begin{array}{l}\mathrm{Find}\mathrm{ }\mathrm{the}\mathrm{ }\mathrm{equation}\mathrm{ }\mathrm{of}\mathrm{ }\mathrm{the}\mathrm{ }\mathrm{plane}\mathrm{ }\mathrm{which}\mathrm{ }\mathrm{contains}\mathrm{ }\mathrm{the}\mathrm{ }\mathrm{line}\mathrm{ }\mathrm{of}\\ \mathrm{intersection}\mathrm{ }\mathrm{of}\mathrm{ }\mathrm{the}\mathrm{ }\mathrm{planes}\mathrm{ }\stackrel{\to }{\mathrm{r}}.\left(\stackrel{^}{\mathrm{i}}+2\mathrm{ }\stackrel{^}{\mathrm{j}}+3\mathrm{ }\stackrel{^}{\mathrm{k}}\right)-4=0,\\ \stackrel{\to }{\mathrm{r}}.\left(2\mathrm{ }\stackrel{^}{\mathrm{i}}+\stackrel{^}{\mathrm{j}}-\stackrel{^}{\mathrm{k}}\right)+5=0\mathrm{ }\mathrm{and}\mathrm{ }\mathrm{which}\mathrm{ }\mathrm{is}\mathrm{ }\mathrm{perpendicular}\mathrm{ }\mathrm{to}\mathrm{ }\mathrm{the}\\ \mathrm{plane}\mathrm{ }\stackrel{\to }{\mathrm{r}}.\left(5\mathrm{ }\stackrel{^}{\mathrm{i}}+3\mathrm{ }\stackrel{^}{\mathrm{j}}-6\mathrm{ }\stackrel{^}{\mathrm{k}}\right)+8=0.\end{array}$

Ans.

$\begin{array}{l}\mathrm{The}\mathrm{equation}\mathrm{of}\mathrm{the}\mathrm{planes}\mathrm{are}:\\ \stackrel{\to }{\mathrm{r}}.\left(\stackrel{^}{\mathrm{i}}+2\mathrm{ }\stackrel{^}{\mathrm{j}}+3\mathrm{ }\stackrel{^}{\mathrm{k}}\right)-4=0...\left(\mathrm{i}\right)\\ \stackrel{\to }{\mathrm{r}}.\left(2\mathrm{ }\stackrel{^}{\mathrm{i}}+\stackrel{^}{\mathrm{j}}-\stackrel{^}{\mathrm{k}}\right)+5=0...\left(\mathrm{ii}\right)\\ \mathrm{The}\mathrm{equation}\mathrm{of}\mathrm{the}\mathrm{plane}\mathrm{passing}\mathrm{through}\mathrm{the}\mathrm{intersection}\mathrm{of}\\ \mathrm{the}\mathrm{planes} \left(\mathrm{i}\right)\mathrm{and}\left(\mathrm{ii}\right)\mathrm{is}\\ \left[\stackrel{\to }{\mathrm{r}}.\left(\stackrel{^}{\mathrm{i}}+2\mathrm{ }\stackrel{^}{\mathrm{j}}+3\mathrm{ }\stackrel{^}{\mathrm{k}}\right)-4\right]+\mathrm{\lambda }\left[\stackrel{\to }{\mathrm{r}}.\left(2\mathrm{ }\stackrel{^}{\mathrm{i}}+\stackrel{^}{\mathrm{j}}-\stackrel{^}{\mathrm{k}}\right)+5\right]=0\\ \stackrel{\to }{\mathrm{r}}\left[\left(1+2\mathrm{\lambda }\right)\stackrel{^}{\mathrm{i}}+\left(2+\mathrm{\lambda }\right)\stackrel{^}{\mathrm{j}}+\left(3-\mathrm{\lambda }\right)\stackrel{^}{\mathrm{k}}\right]+5\mathrm{\lambda }-4=0 \mathrm{ }...\left(\mathrm{iii}\right)\\ \mathrm{Plane}\mathrm{}\left(\mathrm{iii}\right)\mathrm{is}\mathrm{perpendicular}\mathrm{to}\mathrm{the}\mathrm{plane}\stackrel{\to }{\mathrm{r}}.\left(5\mathrm{ }\stackrel{^}{\mathrm{i}}+3\mathrm{ }\stackrel{^}{\mathrm{j}}-6\mathrm{ }\stackrel{^}{\mathrm{k}}\right)+8=0,\\ \therefore \mathrm{ }\left(1+2\mathrm{\lambda }\right).5+\left(2+\mathrm{\lambda }\right).3+\left(3-\mathrm{\lambda }\right).\left(-6\right)=0\\ ⇒ 5+10\mathrm{\lambda }+6+3\mathrm{\lambda }-18+6\mathrm{\lambda }=0\\ ⇒ 19\mathrm{\lambda }\mathrm{ }=7\\ ⇒ \mathrm{\lambda }\mathrm{ }=\frac{7}{19}\\ \mathrm{Substituting}\mathrm{}\mathrm{\lambda }\mathrm{ }=\frac{7}{19}\mathrm{in}\mathrm{equation}\left(\mathrm{iii}\right),\mathrm{we}\mathrm{get}\\ \stackrel{\to }{\mathrm{r}}\left[\left(1+2×\frac{7}{19}\right)\stackrel{^}{\mathrm{i}}+\left(2+\frac{7}{19}\right)\stackrel{^}{\mathrm{j}}+\left(3-\frac{7}{19}\right)\stackrel{^}{\mathrm{k}}\right]+5×\frac{7}{19}-4=0\\ \stackrel{\to }{\mathrm{r}}\left[\frac{33}{19}\stackrel{^}{\mathrm{i}}+\frac{45}{19}\stackrel{^}{\mathrm{j}}+\frac{50}{19}\stackrel{^}{\mathrm{k}}\right]-\frac{41}{19}=0\\ ⇒ \mathrm{ }\stackrel{\to }{\mathrm{r}}\left(33\mathrm{ }\stackrel{^}{\mathrm{i}}+45\stackrel{^}{\mathrm{j}}+50\mathrm{ }\stackrel{^}{\mathrm{k}}\right)-41=0\dots \left(\mathrm{iv}\right)\\ \mathrm{This}\mathrm{is}\mathrm{the}\mathrm{vector}\mathrm{equation}\mathrm{of}\mathrm{the} \mathrm{required}\mathrm{plane}\mathrm{.}\\ \mathrm{Substituting}\stackrel{\to }{\mathrm{r}}=\mathrm{x}\mathrm{ }\stackrel{^}{\mathrm{i}}+\mathrm{y}\mathrm{ }\stackrel{^}{\mathrm{j}}+\mathrm{z}\mathrm{ }\stackrel{^}{\mathrm{k}}\mathrm{in}\mathrm{equation} \left(\mathrm{iv}\right),\mathrm{we}\mathrm{get}\\ \left(\mathrm{x}\mathrm{ }\stackrel{^}{\mathrm{i}}+\mathrm{y}\mathrm{ }\stackrel{^}{\mathrm{j}}+\mathrm{z}\mathrm{ }\stackrel{^}{\mathrm{k}}\right)\left(33\mathrm{ }\stackrel{^}{\mathrm{i}}+45\stackrel{^}{\mathrm{j}}+50\mathrm{ }\stackrel{^}{\mathrm{k}}\right)-41=0\\ ⇒ 33\mathrm{ }\mathrm{x}+45\mathrm{ }\mathrm{y}+50\mathrm{ }\mathrm{z}-41=0\\ \mathrm{This}\mathrm{is}\mathrm{the}\mathrm{cartesian}\mathrm{equation}\mathrm{of}\mathrm{the} \mathrm{required}\mathrm{plane}\mathrm{.}\end{array}$

Q.54

$\begin{array}{l}\mathrm{Find}\mathrm{ }\mathrm{the}\mathrm{ }\mathrm{distance}\mathrm{ }\mathrm{of}\mathrm{ }\mathrm{the}\mathrm{ }\mathrm{point}\mathrm{ }\left(-1,-5,-10\right)\mathrm{ }\mathrm{from}\mathrm{ }\mathrm{the}\mathrm{ }\mathrm{point}\\ \mathrm{of}\mathrm{ }\mathrm{intersection}\mathrm{ }\mathrm{of}\mathrm{ }\mathrm{the}\mathrm{ }\mathrm{line}\mathrm{ }\stackrel{\to }{\mathrm{r}}=2\mathrm{ }\stackrel{^}{\mathrm{i}}-\stackrel{^}{\mathrm{j}}+2\mathrm{ }\stackrel{^}{\mathrm{k}}+\mathrm{\lambda }\left(3\mathrm{ }\stackrel{^}{\mathrm{i}}+4\mathrm{ }\stackrel{^}{\mathrm{j}}+2\mathrm{ }\stackrel{^}{\mathrm{k}}\right)\\ \mathrm{and}\mathrm{ }\mathrm{the}\mathrm{ }\mathrm{plane}\mathrm{ }\stackrel{\to }{\mathrm{r}}.\left(\stackrel{^}{\mathrm{i}}-\stackrel{^}{\mathrm{j}}+\stackrel{^}{\mathrm{k}}\right)=5.\end{array}$

Ans.

$\begin{array}{l}\mathrm{The}\mathrm{equation}\mathrm{of}\mathrm{line}\mathrm{is}:\\ \stackrel{\to }{\mathrm{r}}=2\mathrm{ }\stackrel{^}{\mathrm{i}}-\stackrel{^}{\mathrm{j}}+2\mathrm{ }\stackrel{^}{\mathrm{k}}+\mathrm{\lambda }\left(3\mathrm{ }\stackrel{^}{\mathrm{i}}+4\mathrm{ }\stackrel{^}{\mathrm{j}}+2\mathrm{ }\stackrel{^}{\mathrm{k}}\right)...\left(\mathrm{i}\right)\\ \mathrm{Equation}\mathrm{of}\mathrm{plane}\mathrm{is}:\\ \stackrel{\to }{\mathrm{r}}.\left(\stackrel{^}{\mathrm{i}}-\stackrel{^}{\mathrm{j}}+\stackrel{^}{\mathrm{k}}\right)=5\mathrm{}...\left(\mathrm{ii}\right)\\ \mathrm{Putting}\mathrm{}\stackrel{\to }{\mathrm{r}}\mathrm{in}\mathrm{equation}\left(\mathrm{ii}\right),\mathrm{we}\mathrm{get}\\ \left[2\mathrm{ }\stackrel{^}{\mathrm{i}}-\stackrel{^}{\mathrm{j}}+2\mathrm{ }\stackrel{^}{\mathrm{k}}+\mathrm{\lambda }\left(3\mathrm{ }\stackrel{^}{\mathrm{i}}+4\mathrm{ }\stackrel{^}{\mathrm{j}}+2\mathrm{ }\stackrel{^}{\mathrm{k}}\right)\right].\left(\stackrel{^}{\mathrm{i}}-\stackrel{^}{\mathrm{j}}+\stackrel{^}{\mathrm{k}}\right)=5\\ ⇒ \mathrm{ }2+1+2+\mathrm{\lambda }\left(3-4+2\right)=5\\ ⇒ \mathrm{\lambda }=0\\ \mathrm{Putting}\mathrm{}\mathrm{\lambda }=0\mathrm{in}\mathrm{equation}\left(\mathrm{i}\right)\mathrm{},\mathrm{we}\mathrm{get}\\ \mathrm{ }\stackrel{\to }{\mathrm{r}}=2\mathrm{ }\stackrel{^}{\mathrm{i}}-\stackrel{^}{\mathrm{j}}+2\mathrm{ }\stackrel{^}{\mathrm{k}}+0.\left(3\mathrm{ }\stackrel{^}{\mathrm{i}}+4\mathrm{ }\stackrel{^}{\mathrm{j}}+2\mathrm{ }\stackrel{^}{\mathrm{k}}\right)\\ =2\mathrm{ }\stackrel{^}{\mathrm{i}}-\stackrel{^}{\mathrm{j}}+2\mathrm{ }\stackrel{^}{\mathrm{k}}\\ \mathrm{Then},\mathrm{the}\mathrm{position}\mathrm{vector}\mathrm{of}\mathrm{the}\mathrm{point}\mathrm{of}\mathrm{intersection}\mathrm{of}\mathrm{the}\mathrm{line}\\ \mathrm{and}\mathrm{the}\mathrm{plane}\mathrm{is}\mathrm{ }\stackrel{\to }{\mathrm{r}}=2\mathrm{ }\stackrel{^}{\mathrm{i}}-\stackrel{^}{\mathrm{j}}+2\mathrm{ }\stackrel{^}{\mathrm{k}}.\\ \mathrm{The}\mathrm{coordinates}\mathrm{of}\mathrm{the}\mathrm{point}\mathrm{of}\mathrm{intersection}\mathrm{of}\mathrm{line}\mathrm{and}\mathrm{the}\mathrm{plane}\\ \mathrm{is}\left(2,-1,2\right).\mathrm{ }\mathrm{So},\mathrm{the}\mathrm{distance}\mathrm{of}\mathrm{the}\mathrm{point}\left(-1,-5,-10\right)\mathrm{from}\mathrm{the}\\ \mathrm{point}\mathrm{of}\mathrm{intersection}\mathrm{i}.\mathrm{e}.,\left(2,-1,2\right)=\sqrt{{\left(2+1\right)}^{2}+{\left(-1+5\right)}^{2}+{\left(2+10\right)}^{2}}\\ \mathrm{ }=\sqrt{9+16+144}\\ \mathrm{ }=\sqrt{169} =13\mathrm{units}\mathrm{.}\\ \mathrm{Thus},\mathrm{the}\mathrm{required}\mathrm{distance}\mathrm{is}13\mathrm{units}\mathrm{.}\end{array}$

Q.55

$\begin{array}{l}\mathrm{Find}\mathrm{ }\mathrm{the}\mathrm{ }\mathrm{vector}\mathrm{ }\mathrm{equation}\mathrm{ }\mathrm{of}\mathrm{ }\mathrm{the}\mathrm{ }\mathrm{line}\mathrm{ }\mathrm{passing}\mathrm{ }\mathrm{through}\mathrm{ }\left(1,2,3\right)\\ \mathrm{and}\mathrm{ }\mathrm{parallel}\mathrm{ }\mathrm{to}\mathrm{ }\mathrm{the}\mathrm{ }\mathrm{planes}\mathrm{ }\stackrel{\to }{\mathrm{r}}.\mathrm{ }\left(\stackrel{^}{\mathrm{i}}-\stackrel{^}{\mathrm{j}}+2\mathrm{ }\stackrel{^}{\mathrm{k}}\right)=\mathrm{ }5\mathrm{ }\mathrm{and}\mathrm{ }\stackrel{\to }{\mathrm{r}}.\mathrm{ }\left(3\mathrm{ }\stackrel{^}{\mathrm{i}}+\stackrel{^}{\mathrm{j}}+\stackrel{^}{\mathrm{k}}\right)\mathrm{ }=\mathrm{ }6.\end{array}$

Ans.

$\begin{array}{l}\mathrm{Here},\mathrm{a}\mathrm{line}\mathrm{through}\left(1,2,3\right)\mathrm{is}\mathrm{parallel}\mathrm{to}\mathrm{the}\mathrm{two}\mathrm{planes}\\ \stackrel{\to }{\mathrm{r}}.\left(\stackrel{^}{\mathrm{i}}-\stackrel{^}{\mathrm{j}}+2\mathrm{ }\stackrel{^}{\mathrm{k}}\right)=5\mathrm{and}\stackrel{\to }{\mathrm{r}}.\left(3\mathrm{ }\stackrel{^}{\mathrm{i}}+\stackrel{^}{\mathrm{j}}+\stackrel{^}{\mathrm{k}}\right)=6,\mathrm{so}\mathrm{given}\mathrm{line}\mathrm{will}\mathrm{be}\\ \mathrm{perpendicular}\mathrm{to}\mathrm{the}\mathrm{normals}\mathrm{on}\mathrm{the}\mathrm{given}\mathrm{planes}.\mathrm{Then},\\ \mathrm{Position}\mathrm{vector}\mathrm{of}\mathrm{the}\mathrm{point}\left(1,2,3\right), \stackrel{\to }{\mathrm{a}}=\stackrel{^}{\mathrm{i}}+2\mathrm{ }\stackrel{^}{\mathrm{j}}+3\mathrm{ }\stackrel{^}{\mathrm{k}}\\ \mathrm{Let}\mathrm{vector}\mathrm{parallel}\mathrm{to}\mathrm{required}\mathrm{line}\mathrm{be}\\ \stackrel{\to }{\mathrm{b}}={\mathrm{b}}_{1}\mathrm{ }\stackrel{^}{\mathrm{i}}+{\mathrm{b}}_{2}\mathrm{ }\stackrel{^}{\mathrm{j}}+{\mathrm{b}}_{3}\mathrm{ }\stackrel{^}{\mathrm{k}}\\ \mathrm{Comparing}\mathrm{given}\mathrm{planes}\mathrm{with}\stackrel{\to }{\mathrm{r}}\mathrm{ }\stackrel{\to }{{\mathrm{N}}_{1}}-\mathrm{d}=0\mathrm{and}\stackrel{\to }{\mathrm{r}}\mathrm{ }\stackrel{\to }{{\mathrm{N}}_{2}}-\mathrm{d}=0,\mathrm{we}\mathrm{get}\\ \stackrel{\to }{{\mathrm{N}}_{1}}=\stackrel{^}{\mathrm{i}}-\stackrel{^}{\mathrm{j}}+2\mathrm{ }\stackrel{^}{\mathrm{k}}\mathrm{and}\stackrel{\to }{{\mathrm{N}}_{2}}=3\mathrm{ }\stackrel{^}{\mathrm{i}}+\stackrel{^}{\mathrm{j}}+\stackrel{^}{\mathrm{k}}\\ \stackrel{\to }{\mathrm{b}}.\stackrel{\to }{{\mathrm{N}}_{1}}=0\mathrm{and}\stackrel{\to }{\mathrm{b}}.\stackrel{\to }{{\mathrm{N}}_{2}}=0\\ ⇒\left({\mathrm{b}}_{1}\mathrm{ }\stackrel{^}{\mathrm{i}}+{\mathrm{b}}_{2}\mathrm{ }\stackrel{^}{\mathrm{j}}+{\mathrm{b}}_{3}\mathrm{ }\stackrel{^}{\mathrm{k}}\right).\left(\stackrel{^}{\mathrm{i}}-\stackrel{^}{\mathrm{j}}+2\mathrm{ }\stackrel{^}{\mathrm{k}}\right)=0\mathrm{}⇒{\mathrm{b}}_{1}\mathrm{ }-{\mathrm{b}}_{2}+2\mathrm{ }{\mathrm{b}}_{3}=0\mathrm{ }...\left(\mathrm{i}\right)\\ \mathrm{ }\left({\mathrm{b}}_{1}\mathrm{ }\stackrel{^}{\mathrm{i}}+{\mathrm{b}}_{2}\mathrm{ }\stackrel{^}{\mathrm{j}}+{\mathrm{b}}_{3}\mathrm{ }\stackrel{^}{\mathrm{k}}\right).\left(3\mathrm{ }\stackrel{^}{\mathrm{i}}+\stackrel{^}{\mathrm{j}}+\stackrel{^}{\mathrm{k}}\right)=0 ⇒3\mathrm{ }{\mathrm{b}}_{1}\mathrm{ }+{\mathrm{b}}_{2}+\mathrm{ }{\mathrm{b}}_{3}=0 ...\left(\mathrm{ii}\right)\\ \mathrm{On}\mathrm{}\mathrm{solving}\mathrm{equation}\left(\mathrm{i}\right)\mathrm{and}\left(\mathrm{ii}\right),\mathrm{we}\mathrm{get}\\ \frac{{\mathrm{b}}_{1}}{-1×1-2×1}=\frac{-{\mathrm{b}}_{2}}{1×1-2×3}=\frac{{\mathrm{b}}_{3}}{1×1+1×3}\\ ⇒ \frac{{\mathrm{b}}_{1}}{-3}=\frac{-{\mathrm{b}}_{2}}{-5}=\frac{{\mathrm{b}}_{3}}{4}\\ \mathrm{Therefore},\mathrm{the}\mathrm{direction}\mathrm{ratios}\mathrm{of}\stackrel{\to }{\mathrm{b}}\mathrm{are}-3,-5\mathrm{and}4.\\ \therefore \stackrel{\to }{\mathrm{b}}=-3\mathrm{ }\stackrel{^}{\mathrm{i}}-5\mathrm{ }\stackrel{^}{\mathrm{j}}+4\mathrm{ }\stackrel{^}{\mathrm{k}}\\ \mathrm{So},\mathrm{the}\mathrm{eqution}\mathrm{of}\mathrm{the}\mathrm{line}\mathrm{passing}\mathrm{through}\left(1,2,3\right)\mathrm{and}\mathrm{parallel}\\ \mathrm{to}\mathrm{vector}\stackrel{\to }{\mathrm{b}}\mathrm{is}\mathrm{given}\mathrm{as}:\\ \stackrel{\to }{\mathrm{r}}=\left(\stackrel{^}{\mathrm{i}}+2\mathrm{ }\stackrel{^}{\mathrm{j}}+3\mathrm{ }\stackrel{^}{\mathrm{k}}\right)+\mathrm{\lambda }\left(-3\mathrm{ }\stackrel{^}{\mathrm{i}}-5\mathrm{ }\stackrel{^}{\mathrm{j}}+4\mathrm{ }\stackrel{^}{\mathrm{k}}\right)\end{array}$

Q.56

$\begin{array}{l}\mathrm{Find}\mathrm{ }\mathrm{the}\mathrm{ }\mathrm{vector}\mathrm{ }\mathrm{equation}\mathrm{ }\mathrm{of}\mathrm{ }\mathrm{the}\mathrm{ }\mathrm{line}\mathrm{ }\mathrm{passing}\mathrm{ }\mathrm{through}\mathrm{ }\mathrm{the}\\ \mathrm{point}\mathrm{ }\left(1,2,-4\right)\mathrm{ }\mathrm{and}\mathrm{ }\mathrm{perpendicular}\mathrm{ }\mathrm{to}\mathrm{ }\mathrm{the}\mathrm{ }\mathrm{two}\mathrm{ }\mathrm{lines}:\\ \frac{\mathrm{x}-8}{3}=\frac{\mathrm{y}+19}{-16}=\frac{\mathrm{z}-10}{7}\mathrm{ }\mathrm{and}\mathrm{ }\frac{\mathrm{x}-15}{3}=\frac{\mathrm{y}-29}{8}=\frac{\mathrm{z}-5}{-5}.\end{array}$

Ans.

$\begin{array}{l}\mathrm{The}\mathrm{equations}\mathrm{of}\mathrm{lines}\mathrm{are}\\ \frac{\mathrm{x}-8}{3}=\frac{\mathrm{y}+19}{-16}=\frac{\mathrm{z}-10}{7}...\left(\mathrm{i}\right)\\ \frac{\mathrm{x}-15}{3}=\frac{\mathrm{y}-29}{8}=\frac{\mathrm{z}-5}{-5}...\left(\mathrm{ii}\right)\\ \mathrm{Let}\mathrm{vector}\stackrel{\to }{\mathrm{b}}={\mathrm{b}}_{1}\stackrel{^}{\mathrm{i}}+{\mathrm{b}}_{2}\stackrel{^}{\mathrm{j}}+{\mathrm{b}}_{3}\mathrm{ }\stackrel{^}{\mathrm{k}},\mathrm{be}\mathrm{parallel}\mathrm{to}\mathrm{required}\mathrm{line}\mathrm{.}\\ \mathrm{The}\mathrm{equation}\mathrm{of}\mathrm{line}\mathrm{passing}\mathrm{through}\mathrm{point}\left(1,2,-4\right)\mathrm{and}\mathrm{parallel}\\ \mathrm{to}\mathrm{vector}\stackrel{\to }{\mathrm{b}}\mathrm{is}:\\ \stackrel{\to }{\mathrm{r}}=\stackrel{^}{\mathrm{i}}+2\mathrm{ }\stackrel{^}{\mathrm{j}}-4\mathrm{ }\stackrel{^}{\mathrm{k}}+\mathrm{\lambda }\left({\mathrm{b}}_{1}\stackrel{^}{\mathrm{i}}+{\mathrm{b}}_{2}\stackrel{^}{\mathrm{j}}+{\mathrm{b}}_{3}\mathrm{ }\stackrel{^}{\mathrm{k}}\right)...\left(\mathrm{iii}\right)\\ \mathrm{Since},\mathrm{}\stackrel{\to }{\mathrm{b}}\mathrm{is}\mathrm{perpendicular}\mathrm{to}\mathrm{line}\left(\mathrm{i}\right),\mathrm{then}\\ 3\mathrm{ }{\mathrm{b}}_{1}-16{\mathrm{b}}_{2}+7\mathrm{ }{\mathrm{b}}_{3}=0...\left(\mathrm{iv}\right)\\ \mathrm{And},\mathrm{}\stackrel{\to }{\mathrm{b}}\mathrm{is}\mathrm{perpendicular}\mathrm{to}\mathrm{line}\left(\mathrm{ii}\right),\mathrm{then}\\ 3\mathrm{ }{\mathrm{b}}_{1}+8{\mathrm{b}}_{2}-5\mathrm{ }{\mathrm{b}}_{3}=0...\left(\mathrm{v}\right)\\ \mathrm{From}\mathrm{equation}\left(\mathrm{iv}\right)\mathrm{and}\left(\mathrm{v}\right),\mathrm{we}\mathrm{have}\\ \frac{{\mathrm{b}}_{1}}{80-56}=\frac{{\mathrm{b}}_{2}}{-15-21}=\frac{{\mathrm{b}}_{3}}{24+48}\\ ⇒\frac{{\mathrm{b}}_{1}}{24}=\frac{{\mathrm{b}}_{2}}{-36}=\frac{{\mathrm{b}}_{3}}{72}\\ ⇒\frac{{\mathrm{b}}_{1}}{2}=\frac{{\mathrm{b}}_{2}}{-3}=\frac{{\mathrm{b}}_{3}}{6}\\ \therefore \mathrm{Direction}\mathrm{ratios}\mathrm{of}\stackrel{\to }{\mathrm{b}}\mathrm{are}2,-3\mathrm{and}6\mathrm{.}\\ \mathrm{So}, \stackrel{\to }{\mathrm{b}}=2\stackrel{^}{\mathrm{i}}-3\stackrel{^}{\mathrm{j}}+6\mathrm{ }\stackrel{^}{\mathrm{k}}\\ \mathrm{From}\mathrm{equation}\left(\mathrm{iii}\right),\mathrm{we}\mathrm{have}\\ \stackrel{\to }{\mathrm{r}}=\stackrel{^}{\mathrm{i}}+2\mathrm{ }\stackrel{^}{\mathrm{j}}-4\mathrm{ }\stackrel{^}{\mathrm{k}}+\mathrm{\lambda }\left(2\stackrel{^}{\mathrm{i}}-3\stackrel{^}{\mathrm{j}}+6\mathrm{ }\stackrel{^}{\mathrm{k}}\right)\\ \mathrm{This}\mathrm{is}\mathrm{the}\mathrm{equation}\mathrm{of}\mathrm{the}\mathrm{required}\mathrm{line}\mathrm{.}\end{array}$

Q.57

$\begin{array}{l}\mathrm{Prove}\mathrm{ }\mathrm{that}\mathrm{ }\mathrm{if}\mathrm{ }\mathrm{a}\mathrm{ }\mathrm{plane}\mathrm{ }\mathrm{has}\mathrm{ }\mathrm{the}\mathrm{ }\mathrm{intercepts}\mathrm{ }\mathrm{a},\mathrm{b},\mathrm{c}\mathrm{ }\mathrm{and}\mathrm{ }\mathrm{isat}\mathrm{ }\mathrm{a}\\ \mathrm{distance}\mathrm{ }\mathrm{of}\mathrm{ }\mathrm{p}\mathrm{ }\mathrm{units}\mathrm{ }\mathrm{from}\mathrm{ }\mathrm{the}\mathrm{ }\mathrm{origin},\mathrm{ }\mathrm{then}\mathrm{ }\frac{1}{{\mathrm{a}}^{2}}+\frac{1}{{\mathrm{b}}^{2}}+\frac{1}{{\mathrm{c}}^{2}}=\frac{1}{{\mathrm{p}}^{2}}.\end{array}$

Ans.

$\begin{array}{l}\mathrm{The}\mathrm{equation}\mathrm{of}\mathrm{a}\mathrm{plane}\mathrm{having}\mathrm{intercepts}\mathrm{a},\mathrm{b},\mathrm{c}\mathrm{on}\mathrm{x},\mathrm{y}\mathrm{and}\mathrm{z}\\ \mathrm{axis}\mathrm{respectively}\mathrm{is}\mathrm{given}\mathrm{below}:\\ \frac{\mathrm{x}}{\mathrm{a}}+\frac{\mathrm{y}}{\mathrm{b}}+\frac{\mathrm{z}}{\mathrm{c}}=1...\left(\mathrm{i}\right)\\ \mathrm{The}\mathrm{distance}\left(\mathrm{p}\right)\mathrm{of}\mathrm{the}\mathrm{plane}\mathrm{from}\mathrm{origin}\mathrm{is}:\\ \mathrm{p}=|\frac{\frac{0}{\mathrm{a}}+\frac{0}{\mathrm{b}}+\frac{0}{\mathrm{c}}-1}{\sqrt{{\left(\frac{1}{\mathrm{a}}\right)}^{2}+{\left(\frac{1}{\mathrm{b}}\right)}^{2}+{\left(\frac{1}{\mathrm{c}}\right)}^{2}}}|\\ \mathrm{p}=\frac{1}{\sqrt{{\left(\frac{1}{\mathrm{a}}\right)}^{2}+{\left(\frac{1}{\mathrm{b}}\right)}^{2}+{\left(\frac{1}{\mathrm{c}}\right)}^{2}}}\\ \mathrm{Squaring}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get}\\ {\mathrm{p}}^{\mathrm{2}}=\frac{1}{\frac{1}{{\mathrm{a}}^{2}}+\frac{1}{{\mathrm{b}}^{2}}+\frac{1}{{\mathrm{c}}^{2}}}\\ ⇒ \frac{1}{{\mathrm{p}}^{2}}=\frac{1}{{\mathrm{a}}^{2}}+\frac{1}{{\mathrm{b}}^{2}}+\frac{1}{{\mathrm{c}}^{2}}\end{array}$

Q.58 Distance between the two planes:

2x + 3y + 4z = 4 and 4x + 6y + 8z = 12 is

(A) 2 units (B) 4 units
(C) 8 units (D)

$\frac{2}{\sqrt{29}}units$ 

Ans.

$\begin{array}{l}\text{The equations of two planes}:\\ \text{2x}+\text{3y}+\text{4z}=\text{4}\text{}\text{}\dots \left(i\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{}\text{\hspace{0.17em}}\text{4x}+\text{6y}+\text{8z}=\text{12}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{}\text{\hspace{0.17em}}\text{2x}+\text{3y}+\text{4z}=\text{6}\text{}\text{}\dots \left(ii\right)\\ \text{W}\text{e}\text{can see that both the planes are parallel, so}\\ \text{distance between parallel planes}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{D}=|\frac{{d}_{2}-{d}_{1}}{\sqrt{{a}^{2}+{b}^{2}+{c}^{2}}}|\\ \text{}=|\frac{6-4}{\sqrt{{2}^{2}+{3}^{2}+{4}^{2}}}|\\ \text{}=\frac{2}{\sqrt{4+9+16}}\\ \text{}=\frac{2}{\sqrt{29}}\\ \text{Thus, the distance between two parallel lines is}\frac{2}{\sqrt{29}}.\\ \text{S}\text{o},\text{the correct option is D}\text{.}\end{array}$

Q.59 The planes: 2x – y + 4z = 5 and 5x – 2.5y + 10z = 6 are

(A) Perpendicular

(B) Parallel
(C) intersect y-axis
(D) passes through

$\left(0,0,\frac{5}{4}\right)$

Ans.

$\begin{array}{l}\text{T}\text{h}\text{e}\text{equations of the planes are:}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{2x}-\text{y}+\text{4z}=\text{5}\text{}\dots \left(i\right)\\ \text{5x}-\text{2}.\text{5y}+\text{1}0\text{z}=\text{6}\text{}\dots \left(ii\right)\\ So,\text{}\frac{{a}_{1}}{{a}_{2}}=\frac{2}{5},\text{\hspace{0.17em}}\text{}\frac{{b}_{1}}{{b}_{2}}=\frac{-1}{-2.5}=\frac{10}{25}=\frac{2}{5}\\ \text{a}\text{n}\text{d}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\frac{{c}_{1}}{{c}_{2}}=\frac{4}{10}=\frac{2}{5}\\ \therefore \text{}\frac{{a}_{1}}{{a}_{2}}=\frac{{b}_{1}}{{b}_{2}}=\frac{{c}_{1}}{{c}_{2}}\\ \text{T}\text{h}\text{e}\text{r}\text{e}\text{f}\text{o}\text{r}\text{e},\text{the planes are parallel}\text{.}\\ \text{Thus, the correct option is B}\text{.}\end{array}$