NCERT Solutions for Class 12 Maths Chapter 11 Three-Dimensional Geometry Updated For 2022-23

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

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.

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.

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.

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.

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.

NCERT Solutions Class 12 Mathematics Chapter 11 - Exercises and Answer Solutions.

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 11.1 – 5 Questions (Short Questions)

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

Exercise 11.2 – 17 Questions (9 Short Questions and 8 Long Questions)

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

Exercise 11.3 – 14 Questions (5 Short Questions and 9 Long Questions)

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.

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

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

Ans.

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

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

Ans.

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

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.

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

Q.6

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

Ans.

$\begin{array}{l} T w o lines with direction cosines l_{1} {,m}_{1} {,n}_{1} and l_{2} {,m}_{2} {,n}_{2} are \\ perpendicular to each other if l_{1} l_{2} + m_{1} m_{2} + n_{1} n_{2} = 0 \\ (i) For the lines with direction cosines, \frac{12}{13}, \frac{- 3}{13}, \frac{- 4}{13} and \frac{4}{13}, \frac{12}{13}, \frac{3}{13}, \\ w e have \\ l_{1} l_{2} + m_{1} m_{2} + n_{1} n_{2} = \frac{12}{13} . \frac{4}{13} + \frac{- 3}{13} . \frac{12}{13} + \frac{- 4}{13} . \frac{3}{13} \\ = \frac{48}{169} - \frac{36}{169} - \frac{12}{169} = 0 \\ T h u s, lines are perpendicular . \\ (i i) For the lines with direction cosines \frac{4}{13}, \frac{12}{13}, \frac{3}{13} and \frac{3}{13}, \frac{- 4}{13}, \frac{12}{13}, \\ w e have \\ l_{1} l_{2} + m_{1} m_{2} + n_{1} n_{2} = \frac{4}{13} . \frac{3}{13} + \frac{12}{13} . (\frac{- 4}{13}) + \frac{3}{13} . \frac{12}{13} \\ = \frac{12}{169} - \frac{48}{169} + \frac{36}{169} = 0 \\ T h u s, lines are perpendicular . \\ (i i i) For the lines with direction cosines \frac{3}{13}, \frac{- 4}{13}, \frac{12}{13} and \frac{12}{13}, \frac{- 3}{13}, \frac{- 4}{13}, \\ w e have \\ l_{1} l_{2} + m_{1} m_{2} + n_{1} n_{2} = \frac{3}{13} . \frac{12}{13} + \frac{- 4}{13} . (\frac{- 3}{13}) + \frac{12}{13} . (\frac{- 4}{13}) \\ = \frac{36}{169} + \frac{12}{169} - \frac{48}{169} = 0 \\ T h u s, lines are perpendicular . \\ Therefore, all lines are mutually perpendicular . \end{array}$

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.

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

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.

$\begin{array}{l} Let \vec{a} =2 \hat{i} - \hat{j} +4 \hat{k} \\ \vec{b} = \hat{i} +2 \hat{j} – \hat{k} \\ Since, the line passes through the point with position vector \vec{a} \\ and parallel to vector \vec{b} is given by \\ \vec{r} = \vec{a} +λ \vec{b}, where λ is a constant . \\ \Rightarrow \vec{r} = (2 \hat{i} - \hat{j} +4 \hat{k}) +λ (\hat{i} +2 \hat{j} - \hat{k}) … (i) \\ This is the required equation of the line in vector form . \\ Putting \vec{r} =x \hat{i} +y \hat{j} +z \hat{k} in equation (i), we get \\ x \hat{i} +y \hat{j} +z \hat{k} = (2 \hat{i} - \hat{j} +4 \hat{k}) +λ (\hat{i} +2 \hat{j} - \hat{k}) \\ = (2+λ) \hat{i} - (1 - 2λ) \hat{j} + (4 - λ) \hat{k} \\ \Rightarrow x=2+λ, y=2λ-1 and z=4-λ \\ \Rightarrow λ=x-2, λ= \frac{y+1}{2} and λ=4-z \\ \Rightarrow \frac{x - 2}{1} = \frac{y+1}{2} = \frac{z - 4}{- 1} \\ This is the required equation of the given line in cartesian form . \end{array}$

Q.11 Find the cartesian equation of the line which passes through the point −2, 4, −5 and parallel to the line given by x+3 3 = y−4 5 = z+8 6 .

Ans.

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

Q.12

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

Ans.

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

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

Ans.

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

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

Ans.

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

Q.15

$\begin{array}{l} Find the angle between the following pairs of lines : \\ (i) \vec{r} = 2 \hat{i} - 5 \hat{j} + \hat{k} + λ (3 \hat{i} + 2 \hat{j} + 6 \hat{k}) and \\ \vec{r} = 7 \hat{i} - 6 \hat{k} + μ (\hat{i} + 2 \hat{j} + 2 \hat{k}) \\ (ii) \vec{r} = 3 \hat{i} + \hat{j} - 2 \hat{k} + λ (\hat{i} - \hat{j} - 2 \hat{k}) and \\ \vec{r} = 2 \hat{i} - \hat{j} - 56 \hat{k} + μ (3 \hat{i} - 5 \hat{j} - 4 \hat{k}) \end{array}$

Ans.

$\begin{array}{l} (i) Here, \vec{b_{1}} = 3 \hat{i} + 2 \hat{j} + 6 \hat{k} and \vec{b_{2}} = \hat{i} + 2 \hat{j} + 2 \hat{k} \\ Then angle θ between the two lines is given by \\ cosθ = |\frac{\vec{b_{1}} . \vec{b_{2}}}{|\vec{b_{1}}| |\vec{b_{1}}|}| \\ = |\frac{(3 \hat{i} + 2 \hat{j} + 6 \hat{k}) . (\hat{i} + 2 \hat{j} + 2 \hat{k})}{|3 \hat{i} + 2 \hat{j} + 6 \hat{k}| |\hat{i} + 2 \hat{j} + 2 \hat{k}|}| \\ = |\frac{3 + 4 + 12}{\sqrt{3^{2} + 2^{2} + 6^{2}} \sqrt{1^{2} + 2^{2} + 2^{2}}}| \\ = |\frac{19}{\sqrt{9 + 4 + 36} \sqrt{1 + 4 + 4}}| \\ = \frac{19}{7 \times 3} \\ = \frac{19}{21} \\ ∴ θ = \cos^{- 1} (\frac{19}{21}) \\ (ii) Here, \vec{b_{1}} = \hat{i} - \hat{j} - 2 \hat{k} and \vec{b_{2}} = 3 \hat{i} - 5 \hat{j} - 4 \hat{k} \\ Then angle θ between the two lines is given by \\ cosθ = |\frac{\vec{b_{1}} . \vec{b_{2}}}{|\vec{b_{1}}| |\vec{b_{1}}|}| \\ = |\frac{(\hat{i} - \hat{j} - 2 \hat{k}) . (3 \hat{i} - 5 \hat{j} - 4 \hat{k})}{|\hat{i} - \hat{j} - 2 \hat{k}| |3 \hat{i} - 5 \hat{j} - 4 \hat{k}|}| \\ = |\frac{3 + 5 + 8}{\sqrt{1^{2} + {(- 1)}^{2} + {(- 2)}^{2}} \sqrt{3^{2} + {(- 5)}^{2} + {(- 4)}^{2}}}| \\ = |\frac{16}{\sqrt{1 + 1 + 4} \sqrt{9 + 25 + 16}}| \\ = \frac{16}{\sqrt{6} \times \sqrt{50}} \\ = \frac{16}{10 \sqrt{3}} \\ ∴ θ = \cos^{- 1} (\frac{8}{5 \sqrt{3}}) \end{array}$

Q.16

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

Ans.

$\begin{array}{l} (i) \frac{x - 2}{2} = \frac{y - 1}{5} = \frac{z + 3}{- 3} and \frac{x + 2}{- 1} = \frac{y - 4}{8} = \frac{z - 5}{4} \\ The direction ratios of the first line are 2,5, - 3 and the direction \\ ratios of the second line are - 1,8,4 . If θ is the angle between \\ them, then \\ cosθ = | \frac{a_{1} a_{2} + b_{1} b_{2} + c_{1} c_{2}}{\sqrt{{a_{1}}^{2} + {b_{1}}^{2} + {c_{1}}^{2}} . \sqrt{{a_{2}}^{2} + {b_{2}}^{2} + {c_{2}}^{2}}} | \\ = | \frac{2 \times - 1 + 5 \times 8 + (- 3) \times 4}{\sqrt{{(2)}^{2} + {(5)}^{2} + {(- 3)}^{2}} . \sqrt{{(- 1)}^{2} + 8^{2} + 4^{2}}} | \\ = | \frac{- 2 + 40 - 12}{\sqrt{4 + 25 + 9} . \sqrt{1 + 64 + 16}} | \\ = | \frac{- 2 + 40 - 12}{\sqrt{38} . \sqrt{81}} | \\ θ = \cos^{- 1} (\frac{26}{9 \sqrt{38}}) \\ (ii) \frac{x}{2} = \frac{y}{2} = \frac{z}{1} and \frac{x - 5}{4} = \frac{y - 2}{1} = \frac{z - 3}{8} \\ The direction ratios of the first line are 2, 2, 1 and the direction \\ ratios of the second line are 4, 1, 8 . If θ is the angle between \\ them, then \\ cosθ = | \frac{a_{1} a_{2} + b_{1} b_{2} + c_{1} c_{2}}{\sqrt{{a_{1}}^{2} + {b_{1}}^{2} + {c_{1}}^{2}} . \sqrt{{a_{2}}^{2} + {b_{2}}^{2} + {c_{2}}^{2}}} | \\ = | \frac{2 \times 4 + 2 \times 1 + 1 \times 8}{\sqrt{{(2)}^{2} + {(2)}^{2} + {(1)}^{2}} . \sqrt{{(4)}^{2} + 1^{2} + 8^{2}}} | \\ = | \frac{8 + 2 + 8}{\sqrt{4 + 4 + 1} . \sqrt{16 + 1 + 64}} | \\ = | \frac{18}{\sqrt{9} . \sqrt{81}} | = | \frac{18}{3 \times 9} | \\ θ = \cos^{- 1} (\frac{2}{3}) \end{array}$

Q.17

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

Ans.

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

Q.18

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

Ans.

$\begin{array}{l} The eqution of given lines are \\ \frac{x - 5}{7} = \frac{x + 2}{- 5} = \frac{z}{1} and \frac{x}{1} = \frac{y}{2} = \frac{z}{3} \\ The direction ratios of the first line are 7, - 5, 1 and the direction \\ ratios of the second line are 1, 2, 3 . So, \\ a_{1} a_{2} + b_{1} b_{2} + c_{1} c_{2} = 7 .1 + (- 5) . 2 + 1 .3 \\ = 7 - 10 + 3 \\ = 0 \\ Thus, both lines are perpendicular to each other . \end{array}$

Q.19

$\begin{array}{l} Find the shortest distance between the lines \\ \vec{r} = (\hat{i} + 2 \hat{j} + \hat{k}) + λ (\hat{i} - \hat{j} + \hat{k}) and \\ \vec{r} = (2 \hat{i} - \hat{j} + \hat{k}) + μ (2 \hat{i} + \hat{j} + 2 \hat{k}) \end{array}$

Ans.

$\begin{array}{l} The equations of the given lines are : \\ \vec{r} = (\hat{i} + 2 \hat{j} + \hat{k}) + λ (\hat{i} - \hat{j} + \hat{k}) . .. (i) \\ \vec{r} = (2 \hat{i} - \hat{j} - \hat{k}) + μ (2 \hat{i} + \hat{j} + 2 \hat{k}) . .. (ii) \\ Comparing equatin (i) with \vec{r} = \vec{a_{1}} + λ \vec{b_{1}} and equation (ii) \\ with \vec{r} = \vec{a_{2}} + μ \vec{b_{2}}, we get \\ \vec{a_{1}} = \hat{i} + 2 \hat{j} + \hat{k}, \vec{b_{1}} = \hat{i} - \hat{j} + \hat{k} \\ and \vec{a_{2}} = 2 \hat{i} - \hat{j} - \hat{k}, \vec{b_{2}} = 2 \hat{i} + \hat{j} + 2 \hat{k} \\ The shortest distance between the lines \\ d = |\frac{(\vec{b_{1}} \times \vec{b_{2}}) . (\vec{a_{2}} - \vec{a_{1}})}{|\vec{b_{1}} \times \vec{b_{2}}|}| . .. (iii) \\ So, \vec{b_{1}} \times \vec{b_{2}} = |\begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & - 1 & 1 \\ 2 & 1 & 2 \end{matrix}| \\ = (- 2 - 1) \hat{i} - (2 - 2) \hat{j} + (1 + 2) \hat{k} \\ = - 3 \hat{i} - 0 . \hat{j} + 3 \hat{k} \\ |\vec{b_{1}} \times \vec{b_{2}}| = |- 3 \hat{i} - 0 . \hat{j} + 3 \hat{k}| \\ = \sqrt{{(- 3)}^{2} + 3^{2}} \\ = \sqrt{9 + 9} \\ = 3 \sqrt{2} \\ \vec{a_{2}} - \vec{a_{1}} = (2 \hat{i} - \hat{j} - \hat{k}) - (\hat{i} + 2 \hat{j} + \hat{k}) \\ = \hat{i} - 3 \hat{j} - 2 \hat{k} \\ From equation (i), we get \\ d = |\frac{(- 3 \hat{i} + 3 \hat{k}) . (\hat{i} - 3 \hat{j} - 2 \hat{k})}{3 \sqrt{2}}| \\ = |\frac{- 3 - 6}{3 \sqrt{2}}| \\ = \frac{3}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} \\ = \frac{3 \sqrt{2}}{2} \\ Thus, the distance between the given two lines is \frac{3 \sqrt{2}}{2} units . \end{array}$

Q.20

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

Ans.

$\begin{array}{l} The equations of the given lines are : \\ \frac{X + 1}{7} = \frac{Y + 1}{- 6} = \frac{Z - 1}{1} and \frac{X - 3}{1} = \frac{Y - 5}{- 2} = \frac{Z - 7}{1} \\ Comparing both the equations with \\ \frac{X - X_{1}}{a_{1}} = \frac{Y - Y_{1}}{b_{1}} = \frac{Z - Z_{1}}{C_{1}} and \frac{X - X_{2}}{a_{2}} = \frac{Y - 5}{b_{2}} = \frac{Z - 7}{C_{2}} respectively . \\ X_{1} = - 1, X_{2} = 3 \\ Y_{1} = - 1, Y_{2} = 5 \\ Z_{1} = 1, Z_{2} = 7 \\ a_{1} = 7, a_{2} = 1 \\ b_{1} = - 6, b_{2} = - 2 \\ c_{1} = 1, c_{2} = 1 \\ Distance between two lines is given by \\ d = |\frac{|\begin{matrix} x_{2} - x_{1} & y_{2} - y_{1} & z_{2} - z_{1} \\ a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2} \end{matrix}|}{\sqrt{{(b_{1} c_{2} - b_{2} c_{1})}^{2} + {(c_{1} a_{2} - c_{2} a_{1})}^{2} + {(a_{1} b_{2} - a_{2} b_{1})}^{2}}}| \\ = |\frac{|\begin{matrix} 3 - (- 1) & 5 - (- 1) & 7 - (- 1) \\ 7 & - 6 & 1 \\ 1 & - 2 & 1 \end{matrix}|}{\sqrt{{\{- 6 \times 1 - (- 2) . 1\}}^{2} + {(1 \times 1 - 1 \times 7)}^{2} + {(7 \times - 2 - 1 \times - 6)}^{2}}}| \\ d = |\frac{|\begin{matrix} 4 & 6 & 8 \\ 7 & - 6 & 1 \\ 1 & - 2 & 1 \end{matrix}|}{\sqrt{{(- 4)}^{2} + {(- 6)}^{2} + {(- 8)}^{2}}}| \\ = |\frac{4 (- 6 + 2) - 6 (7 - 1) + 8 (- 14 + 6)}{\sqrt{16 + 36 + 64}}| \\ = |\frac{- 16 - 36 - 64}{\sqrt{116}}| \\ = |\frac{- 116}{2 \sqrt{29}}| \\ = |- 2 \sqrt{29}| \\ So, the distance between two given lines is 2 \sqrt{29} units . \end{array}$

Q.21

$\begin{array}{l} Find the shortest distance between the lines whose vector \\ equations are \\ \vec{r} = (\hat{i} + 2 \hat{j} + 3 \hat{k}) + λ (\hat{i} - \hat{j} + 2 \hat{k}) and \\ \vec{r} = (4 \hat{i} + \hat{j} + 6 \hat{k}) + μ (2 \hat{i} + 3 \hat{j} + \hat{k}) \end{array}$

Ans.

$\begin{array}{l} The equations of the given lines are : \\ \vec{r} = (\hat{i} + 2 \hat{j} + 3 \hat{k}) + λ (\hat{i} - 3 \hat{j} + 2 \hat{k}) . .. (i) \\ \vec{r} = (4 \hat{i} + 5 \hat{j} + 6 \hat{k}) + μ (2 \hat{i} + 3 \hat{j} + \hat{k}) . .. (ii) \\ Comparing equatin (i) with \vec{r} = \vec{a_{1}} + λ \vec{b_{1}} and equation (ii) \\ with \vec{r} = \vec{a_{2}} + μ \vec{b_{2}}, we get \\ \vec{a_{1}} = \hat{i} + 2 \hat{j} + 3 \hat{k}, \vec{b_{1}} = \hat{i} - 3 \hat{j} + 2 \hat{k} \\ and \vec{a_{2}} = 4 \hat{i} + 5 \hat{j} + 6 \hat{k}, \vec{b_{2}} = 2 \hat{i} + 3 \hat{j} + \hat{k} \\ The shortest distance between the lines \\ d = |\frac{(\vec{b_{1}} \times \vec{b_{2}}) . (\vec{a_{2}} - \vec{a_{1}})}{|\vec{b_{1}} \times \vec{b_{2}}|}| . .. (iii) \\ So, \vec{b_{1}} \times \vec{b_{2}} = |\begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & - 3 & 2 \\ 2 & 3 & 1 \end{matrix}| \\ = (- 3 - 6) \hat{i} - (1 - 4) \hat{j} + (3 + 6) \hat{k} \\ = - 9 \hat{i} + 3 \hat{j} + 9 \hat{k} \\ |\vec{b_{1}} \times \vec{b_{2}}| = |- 9 \hat{i} + 3 \hat{j} + 9 \hat{k}| \\ = \sqrt{{(- 9)}^{2} + 3^{2} + 9^{2}} \\ = \sqrt{81 + 9 + 81} \\ = \sqrt{171} = 3 \sqrt{19} \\ \vec{a_{2}} - \vec{a_{1}} = (4 \hat{i} + 5 \hat{j} + 6 \hat{k}) - (\hat{i} + 2 \hat{j} + 3 \hat{k}) \\ = 3 \hat{i} + 3 \hat{j} + 3 \hat{k} \\ From equation (i), we get \\ d = |\frac{(- 9 \hat{i} + 3 \hat{j} + 9 \hat{k}) . (3 \hat{i} + 3 \hat{j} + 3 \hat{k})}{3 \sqrt{19}}| \\ = |\frac{3 (- 3 \hat{i} + \hat{j} + 3 \hat{k}) . (\hat{i} + \hat{j} + \hat{k})}{\sqrt{19}}| \\ = |\frac{3 (- 3 + 1 + 3)}{\sqrt{19}}| \\ = \frac{3}{\sqrt{19}} \times \frac{\sqrt{19}}{\sqrt{19}} \\ = \frac{3 \sqrt{19}}{19} \\ So, the distance between two given lines is \frac{3 \sqrt{19}}{19} units . \end{array}$

Q.22

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

Ans.

$\begin{array}{l} The equations of the given lines are : \\ \vec{r} = (1 - t) \hat{i} + (t - 2) \hat{j} + (3 - 2 t) \hat{k} \\ \vec{r} = (\hat{i} - 2 \hat{j} + 3 \hat{k}) + t (- \hat{i} + \hat{j} - 2 \hat{k}) . .. (i) \\ \vec{r} = (s + 1) \hat{i} + (2 s - 1) \hat{j} - (2 s + 1) \hat{k} \\ \vec{r} = (\hat{i} - \hat{j} - \hat{k}) + s (\hat{i} + 2 \hat{j} - 2 \hat{k}) . .. (ii) \\ Comparing equatin (i) with \vec{r} = \vec{a_{1}} + λ \vec{b_{1}} and equation (ii) \\ with \vec{r} = \vec{a_{2}} + μ \vec{b_{2}}, we get \\ \vec{a_{1}} = \hat{i} - 2 \hat{j} + 3 \hat{k}, \vec{b_{1}} = - \hat{i} + \hat{j} - 2 \hat{k} \\ and \vec{a_{2}} = \hat{i} - \hat{j} - \hat{k}, \vec{b_{2}} = \hat{i} + 2 \hat{j} - 2 \hat{k} \\ The shortest distance between the lines \\ d = |\frac{(\vec{b_{1}} \times \vec{b_{2}}) . (\vec{a_{2}} - \vec{a_{1}})}{|\vec{b_{1}} \times \vec{b_{2}}|}| . .. (iii) \\ So, \vec{b_{1}} \times \vec{b_{2}} = |\begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ - 1 & 1 & - 2 \\ 1 & 2 & - 2 \end{matrix}| \\ = (- 2 + 4) \hat{i} - (2 + 2) \hat{j} + (- 2 - 1) \hat{k} \\ = 2 \hat{i} - 4 \hat{j} - 3 \hat{k} \\ ∴ |\vec{b_{1}} \times \vec{b_{2}}| = |2 \hat{i} - 4 \hat{j} - 3 \hat{k}| \\ = \sqrt{2^{2} + {(- 4)}^{2} + {(- 3)}^{2}} \\ = \sqrt{4 + 16 + 9} \\ = \sqrt{29} \\ \vec{a_{2}} - \vec{a_{1}} = (\hat{i} - \hat{j} - \hat{k}) - (\hat{i} - 2 \hat{j} + 3 \hat{k}) \\ = \hat{j} - 4 \hat{k} \\ From equation (i), we get \\ d = |\frac{(2 \hat{i} - 4 \hat{j} - 3 \hat{k}) . (\hat{j} - 4 \hat{k})}{\sqrt{29}}| \\ = |\frac{- 4 + 12}{\sqrt{29}}| \\ = \frac{8}{\sqrt{29}} \times \frac{\sqrt{29}}{\sqrt{29}} = \frac{8 \sqrt{29}}{29} \\ Therefore, the shortest distance between two lines is \\ \frac{8 \sqrt{29}}{29} units . \end{array}$

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.

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

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

$3 \hat{i} + 5 \hat{j} - 6 \hat{k} .$

Ans.

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

Q.25

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

Ans.

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

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} (a) T h e g i v e n e q u a t i o n o f t h e p l a n e i s \\ 2 x +3 y +4 z -12=0 \\ \Rightarrow 2 x +3 y +4 z =12 . .. (i) \\ L e t t h e c o o r d i n a t e s o f t h e f o o t o f p e r p e n d i c u l a r f r o m o r i g i n t o \\ t h e p l a n e i s (x_{1}, y_{1}, z_{1}) . T h e n p o i n t (x_{1}, y_{1}, z_{1}) w i l l s a t i s f y \\ e q u a t i o n (i), s o \\ 2 x_{1} + 3 y_{1} + 4 z_{1} = 12 \\ T h e n, t h e d i r e c t i o n r a t i o s o f t h e p e r p e n d i c u l a r a r e 2, 3, 4 . \\ D i v i d i n g e q u a t i o n (i) b y \sqrt{2^{2} + 3^{2} + 4^{2}} = \sqrt{29}, w e g e t \\ \Rightarrow \frac{2}{\sqrt{29}} x + \frac{3}{\sqrt{29}} y + \frac{4}{\sqrt{29}} z = \frac{12}{\sqrt{29}} \\ T h i s e q u a t i o n i s i n t h e f o r m o f l x + m y + n z = d, w h e r e l, m a n d n \\ a r e t h e d i r e c t i o n c o s i n e s o f n o r m a l t o t h e p l a n e a n d d i s t h e \\ d i s t a n c e o f t h e n o r m a l f r o m t h e o r i g i n . \\ T h e c o o r d i n a t e s o f t h e f o o t o f t h e p e r p e n d i c u l a r a r e g i v e n b y \\ (l d, m d, n d) . \\ T h e r e f o r e, t h e c o o r d i a n t e s o f t h e f o o t o f t h e p e r p e n d i c u l a r a r e \\ (\frac{2}{\sqrt{29}} . \frac{12}{\sqrt{29}}, \frac{3}{\sqrt{29}} . \frac{12}{\sqrt{29}}, \frac{4}{\sqrt{29}} . \frac{12}{\sqrt{29}}) = (\frac{24}{29}, \frac{36}{29}, \frac{48}{29}) . \\ (b) T h e g i v e n e q u a t i o n o f t h e p l a n e i s \\ 3 y +4 z -6=0 \\ \Rightarrow 0 . x +3 y +4 z =12 . .. (i) \\ L e t t h e c o o r d i n a t e s o f t h e f o o t o f p e r p e n d i c u l a r f r o m o r i g i n t o \\ t h e p l a n e i s (x_{1}, y_{1}, z_{1}) . T h e n p o i n t (x_{1}, y_{1}, z_{1}) w i l l s a t i s f y \\ e q u a t i o n (i), s o \\ 0 . x_{1} + 3 y_{1} + 4 z_{1} = 6 \\ T h e n, t h e d i r e c t i o n r a t i o s o f t h e p e r p e n d i c u l a r a r e 0, 3, 4 . \\ D i v i d i n g e q u a t i o n (i) b y \sqrt{0^{2} + 3^{2} + 4^{2}} = 5, w e g e t \\ \Rightarrow \frac{0}{5} x + \frac{3}{5} y + \frac{4}{5} z = \frac{6}{5} \\ T h i s e q u a t i o n i s i n t h e f o r m o f l x + m y + n z = d, w h e r e l, m a n d n \\ a r e t h e d i r e c t i o n c o s i n e s o f n o r m a l t o t h e p l a n e a n d d i s t h e \\ d i s t a n c e o f t h e n o r m a l f r o m t h e o r i g i n . \\ T h e c o o r d i n a t e s o f t h e f o o t o f t h e p e r p e n d i c u l a r a r e g i v e n b y \\ (l d, m d, n d) . \\ T h e r e f o r e, t h e c o o r d i a n t e s o f t h e f o o t o f t h e p e r p e n d i c u l a r a r e \\ (\frac{0}{5} . \frac{6}{5}, \frac{3}{5} . \frac{6}{5}, \frac{4}{5} . \frac{6}{5}) = (0, \frac{18}{25}, \frac{24}{25}) . \\ (c) T h e g i v e n e q u a t i o n o f t h e p l a n e i s \\ x + y + z =1 . .. (i) \\ L e t t h e c o o r d i n a t e s o f t h e f o o t o f p e r p e n d i c u l a r f r o m o r i g i n t o \\ t h e p l a n e i s (x_{1}, y_{1}, z_{1}) . T h e n p o i n t (x_{1}, y_{1}, z_{1}) w i l l s a t i s f y \\ e q u a t i o n (i), s o \\ x + y + z =1 \\ T h e n, t h e d i r e c t i o n r a t i o s o f t h e p e r p e n d i c u l a r a r e 1, 1, 1 . \\ D i v i d i n g e q u a t i o n (i) b y \sqrt{1^{2} + 1^{2} + 1^{2}} = \sqrt{3}, w e g e t \\ \Rightarrow \frac{1}{\sqrt{3}} x + \frac{1}{\sqrt{3}} y + \frac{1}{\sqrt{3}} z = \frac{1}{\sqrt{3}} \\ T h i s e q u a t i o n i s i n t h e f o r m o f l x + m y + n z = d, w h e r e l, m a n d n \\ a r e t h e d i r e c t i o n c o s i n e s o f n o r m a l t o t h e p l a n e a n d d i s t h e \\ d i s t a n c e o f t h e n o r m a l f r o m t h e o r i g i n . \\ T h e c o o r d i n a t e s o f t h e f o o t o f t h e p e r p e n d i c u l a r a r e g i v e n b y \\ (l d, m d, n d) . \\ T h e r e f o r e, t h e c o o r d i a n t e s o f t h e f o o t o f t h e p e r p e n d i c u l a r a r e \\ (\frac{1}{\sqrt{3}} . \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}} . \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}} . \frac{1}{\sqrt{3}}) = (\frac{1}{3}, \frac{1}{3}, \frac{1}{3}) . \\ (d) T h e g i v e n e q u a t i o n o f t h e p l a n e i s \\ 5 y +8=0 \\ 0 . x -5 y +0 . z = 8 . .. (i) \\ L e t t h e c o o r d i n a t e s o f t h e f o o t o f p e r p e n d i c u l a r f r o m o r i g i n t o \\ t h e p l a n e i s (x_{1}, y_{1}, z_{1}) . T h e n p o i n t (x_{1}, y_{1}, z_{1}) w i l l s a t i s f y \\ e q u a t i o n (i), s o \\ 0 . x_{1} - 5 y_{1} + 0 . z_{1} = 8 \\ T h e n, t h e d i r e c t i o n r a t i o s o f t h e p e r p e n d i c u l a r a r e 0, 5, 0 . \\ D i v i d i n g e q u a t i o n (i) b y \sqrt{0^{2} + {(- 5)}^{2} + 0^{2}} = 5, w e g e t \\ \Rightarrow \frac{0}{5} x - \frac{5}{5} y + \frac{0}{5} z = \frac{8}{5} \\ T h i s e q u a t i o n i s i n t h e f o r m o f l x + m y + n z = d, w h e r e l, m a n d n \\ a r e t h e d i r e c t i o n c o s i n e s o f n o r m a l t o t h e p l a n e a n d d i s t h e \\ d i s t a n c e o f t h e n o r m a l f r o m t h e o r i g i n . \\ T h e c o o r d i n a t e s o f t h e f o o t o f t h e p e r p e n d i c u l a r a r e g i v e n b y \\ (l d, m d, n d) . \\ T h e r e f o r e, t h e c o o r d i a n t e s o f t h e f o o t o f t h e p e r p e n d i c u l a r a r e \\ (0 \times \frac{8}{5}, \frac{- 5}{5} \times \frac{8}{5}, 0 \times \frac{8}{5}) = (0,- \frac{8}{5}, 0) . \end{array}$

Q.27

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

Ans.

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

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} (a) L e t vector forms of the given three points are: \\ \vec{a} = \hat{i} + \hat{j} - \hat{k}, \vec{b} = 6 \hat{i} + 4 \hat{j} - 5 \hat{k} and \vec{c} = - 4 \hat{i} - 2 \hat{j} + 3 \hat{k} \\ T h e n, the vector equation of the plane passing through \vec{a}, \vec{b} and \\ \vec{c} is given by (\vec{r} - \vec{a}) . [(\vec{b} - \vec{a}) \times (\vec{c} - \vec{a})] = 0 \\ i . e ., [\vec{r} - (\hat{i} + \hat{j} - \hat{k})] . [(5 \hat{i} + 3 \hat{j} - 4 \hat{k}) \times (- 5 \hat{i} - 3 \hat{j} + 4 \hat{k})] = 0 \\ \Rightarrow [\vec{r} - (\hat{i} + \hat{j} - \hat{k})] . | \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 5 & 3 & - 4 \\ - 5 & - 3 & 4 \end{matrix} | = 0 \\ \Rightarrow [\vec{r} - (\hat{i} + \hat{j} - \hat{k})] . (0. \hat{i} - 0 \hat{. j} + 0. \hat{k}) = 0 \\ \Rightarrow 0 = 0 \\ S i n c e, the given three points are collinear, so there will be \\ infininte numbers of planes passing through the given points . \\ (b) L e t vector forms of the given three points are: \\ \vec{a} = \hat{i} + \hat{j} + 0 \hat{k}, \vec{b} = \hat{i} + 2 \hat{j} + \hat{k} and \vec{c} = - 2 \hat{i} + 2 \hat{j} - \hat{k} \\ T h e n, the vector equation of the plane passing through \vec{a}, \vec{b} and \\ \vec{c} is given by (\vec{r} - \vec{a}) . [(\vec{b} - \vec{a}) \times (\vec{c} - \vec{a})] = 0 \\ i . e ., [\vec{r} - (\hat{i} + \hat{j})] . [(0 \hat{i} + \hat{j} + \hat{k}) \times (- 3 \hat{i} + \hat{j} - \hat{k})] = 0 \\ \Rightarrow [\vec{r} - (\hat{i} + \hat{j})] . | \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 0 & 1 & 1 \\ - 3 & 1 & - 1 \end{matrix} | = 0 \\ \Rightarrow [x \hat{i} + y \hat{j} + z \hat{k} - (\hat{i} + \hat{j})] . (- 2 \hat{i} - 3 \hat{j} + 3 \hat{k}) = 0 \\ \Rightarrow [(x - 1) \hat{i} + (y - 1) \hat{j} + z \hat{k}] . (- 2 \hat{i} - 3 \hat{j} + 3 \hat{k}) = 0 \\ \Rightarrow - 2 (x - 1) - 3 (y - 1) + 3 z = 0 \\ \Rightarrow - 2 x + 2 - 3 y + 3 + 3 z = 0 \\ \Rightarrow 2 x + 3 y - 3 z = 5 \\ T h i s is the cartesian equation of the required plane . \end{array}$

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

Ans.

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

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.

$\begin{array}{l} T h e e q u a t i o n s o f g i v e n p l a n e s a r e : \\ 3 x - y + 2 z - 4 = 0 a n d x + y + z - 2 = 0 \\ T h e e q u a t i o n o f p l a n e t h r o u g h i n t e r s e c t i o n o f g i v e n p l a n e s i s \\ (3 x - y + 2 z - 4) + λ (x + y + z - 2) = 0 w h e r e λ \in R … (i) \\ T h e p l a n e p a s s e s t h r o u g h t h e p o i n t (2, 2, 1), s o f r o m e q u a t i o n (i) \\ (3 \times 2 - 2 + 2 \times 1 - 4) + λ (2 + 2 + 1 - 2) = 0 \\ 2 + λ (3) = 0 \Rightarrow λ = - \frac{2}{3} \\ P u t t i n g v a l u e o f λ i n e q u a t i o n (i), w e g e t \\ (3 x - y + 2 z - 4) - \frac{2}{3} (x + y + z - 2) = 0 \\ \Rightarrow 3 (3 x - y + 2 z - 4) - 2 (x + y + z - 2) = 0 \\ \Rightarrow 9 x - 3 y + 6 z - 12 - 2 x - 2 y - 2 z + 4 = 0 \\ \Rightarrow 7 x - 5 y + 4 z - 8 = 0 \\ This is the required equation of the plane . \end{array}$

Q.32n

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

Ans.

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

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.

$\begin{array}{l} T h e e q u a t i o n s o f g i v e n p l a n e s a r e : \\ x + y + z = 1 a n d 2 x + 3 y + 4 z = 5 \\ T h e e q u a t i o n o f p l a n e t h r o u g h i n t e r s e c t i o n o f g i v e n p l a n e s i s \\ (x + y + z - 1) + λ (2 x + 3 y + 4 z - 5) = 0 w h e r e λ \in R \\ \Rightarrow (1 + 2 λ) x + (1 + 3 λ) y + (1 + 4 λ) z = 5 λ + 1 … (i) \\ T h e direction ratios of the plane are (1 + 2 λ), (1 + 3 λ) and (1 + 4 λ) . \\ P l a n e (i) is perpendicular to x - y + z = 0, whose direction ratios \\ are 1, - 1, 1. \\ S i n c e, plane (i) and (ii) are perpendicular, so \\ a_{1} a_{2} {+b}_{1} b_{2} {+c}_{1} c_{2} = 0 \\ \Rightarrow (1 + 2 λ) (1) + (1 + 3 λ) (- 1) + (1 + 4 λ) (1) = 0 \\ \Rightarrow 1 + 2 λ - 1 - 3 λ + 1 + 4 λ = 0 \\ \Rightarrow 3 λ = - 1 \\ \Rightarrow λ = - \frac{1}{3} \\ S u b s t i t u t i n g λ = - \frac{1}{3} in equation (i), we get \\ \Rightarrow (1 + 2 \times - \frac{1}{3}) x + (1 + 3 \times - \frac{1}{3}) y + (1 + 4 \times - \frac{1}{3}) z = 5 \times - \frac{1}{3} + 1 \\ \Rightarrow \frac{1}{3} x + (0) y - \frac{1}{3} z = - \frac{2}{3} \\ x - z = - 2 \\ x - z + 2 = 0 \\ T h i s is the required equation of the plane . \\ T h e direction ratios of the plane are (1 + 2 λ), (1 + 3 λ) and (1 + 4 λ) . \\ P l a n e (i) is perpendicular to x - y + z = 0, whose direction ratios \\ are 1, - 1, 1. \\ S i n c e, plane (i) and (ii) are perpendicular, so \\ a_{1} a_{2} {+b}_{1} b_{2} {+c}_{1} c_{2} = 0 \\ \Rightarrow (1 + 2 λ) (1) + (1 + 3 λ) (- 1) + (1 + 4 λ) (1) = 0 \\ \Rightarrow 1 + 2 λ - 1 - 3 λ + 1 + 4 λ = 0 \\ \Rightarrow 3 λ = - 1 \\ \Rightarrow λ = - \frac{1}{3} \\ S u b s t i t u t i n g λ = - \frac{1}{3} in equation (i), we get \\ \Rightarrow (1 + 2 \times - \frac{1}{3}) x + (1 + 3 \times - \frac{1}{3}) y + (1 + 4 \times - \frac{1}{3}) z = 5 \times - \frac{1}{3} + 1 \\ \Rightarrow \frac{1}{3} x + (0) y - \frac{1}{3} z = - \frac{2}{3} \\ x - z = - 2 \\ x - z + 2 = 0 \\ T h i s is the required equation of the plane . \end{array}$

Q.34

$\begin{array}{l} Find the angle between the planes whose vector equations \\ are \vec{r} (2 \hat{i} + 2 \hat{j} - \hat{k}) = 5 and \vec{r} (3 \hat{i} - 3 \hat{j} + 5 \hat{k}) = \end{array}$

Ans.

$\begin{array}{l} Theplaneswhosevectorequationsare \\ \vec{r} (2 \hat{i} + 2 \hat{j} - 3 \hat{k}) = 5 and \vec{r} (3 \hat{i} - 3 \hat{j} + 5 \hat{k}) = 3 \\ comparing with \vec{r} . \vec{n_{1}} = d_{1} and \vec{r} . \vec{n_{2}} = d_{2}, then \\ \vec{n_{1}} = (2 \hat{i} + 2 \hat{j} - 3 \hat{k}) and \vec{n_{2}} = (3 \hat{i} - 3 \hat{j} + 5 \hat{k}) \\ ∴ \vec{n_{1}} . \vec{n_{2}} = (2 \hat{i} + 2 \hat{j} - 3 \hat{k}) . (3 \hat{i} - 3 \hat{j} + 5 \hat{k}) \\ = 6 - 6 - 15 \\ = - 15 \\ | \vec{n_{1}} | = | 2 \hat{i} + 2 \hat{j} - 3 \hat{k} | \\ = \sqrt{2^{2} + 2^{2} + {(- 3)}^{2}} \\ = \sqrt{4 + 4 + 9} = \sqrt{17} \\ | \vec{n_{2}} | = | 3 \hat{i} - 3 \hat{j} + 5 \hat{k} | \\ = \sqrt{3^{2} + {(- 3)}^{2} + 5^{2}} \\ = \sqrt{9 + 9 + 25} = \sqrt{43} \\ Let angle between two planes be θ, then \\ cosθ = \frac{\vec{n_{1}} . \vec{n_{2}}}{| \vec{n_{1}} | | \vec{n_{2}} |} \\ = \frac{- 15}{\sqrt{17} \sqrt{43}} \\ = - \frac{15}{\sqrt{731}} \\ θ = \cos^{- 1} (- \frac{15}{\sqrt{731}}) \end{array}$

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.

$\begin{array}{l} (a) 7 x + 5 y + 6 z + 30 = 0 a n d 3 x - y - 10 z + 4 = 0 \\ D i r e c t i o n ratios of the given planes are: \\ a_{1} = 7, b_{1} = 5, c_{1} = 6 \\ a_{2} = 3, b_{2} = - 1, c_{2} = - 10 \\ a_{1} a_{2} + b_{1} b_{2} + c_{1} c_{2} = 7 \times 3 + 5 \times - 1 + 6 \times - 10 \\ = 21 - 5 - 60 = - 44 \neq 0 \\ Therefore, the given planes are not perpendicular to each other . \\ \frac{a_{1}}{a_{2}} = \frac{7}{3}, \frac{b_{1}}{b_{2}} = \frac{5}{- 1} and \frac{c_{1}}{c_{2}} = \frac{6}{- 10} = - \frac{3}{5} \\ \Rightarrow \frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}} \\ Therefore, the given planes are not parallel to each other . \\ The angle between two planes is \\ θ = {cos}^{- 1} |(\frac{a_{1} a_{2} + b_{1} b_{2} + c_{1} c_{2}}{\sqrt{a_{1}^{2} + b_{1}^{2} + c_{1}^{2}} \sqrt{a_{2}^{2} + b_{2}^{2} + c_{2}^{2}}})| \end{array}$ $\begin{array}{l} = {cos}^{- 1} |(\frac{7 \times 3 + 5 \times - 1 + 6 \times - 10}{\sqrt{7^{2} + 5^{2} + 6^{2}} \sqrt{3^{2} + {(- 1)}^{2} + {(- 10)}^{2}}})| \\ = {cos}^{- 1} |(\frac{- 44}{\sqrt{110} \sqrt{110}})| \\ = {cos}^{- 1} (\frac{44}{110}) \\ = {cos}^{- 1} (\frac{2}{5}) \end{array}$ $\begin{array}{l} (b) 2 x + y + 3 z - 2 = 0 a n d x - 2 y + 5 = 0 \\ D i r e c t i o n ratios of the given planes are: \\ a_{1} = 2, b_{1} = 1, c_{1} = 3 \\ a_{2} = 1, b_{2} = - 2, c_{2} = 0 \\ a_{1} a_{2} + b_{1} b_{2} + c_{1} c_{2} = 2 \times 1 + 1 \times - 2 + 3 \times 0 \\ = 2 - 2 + 0 = 0 \\ Therefore, the given planes are perpendicular to each other . \\ (c) 2 x - 2 y + 4 z + 5 = 0 a n d 3 x - 3 y + 6 z - 1 = 0 \\ D i r e c t i o n ratios of the given planes are: \\ a_{1} = 2, b_{1} = - 2, c_{1} = 4 \\ a_{2} = 3, b_{2} = - 3, c_{2} = 6 \\ a_{1} a_{2} + b_{1} b_{2} + c_{1} c_{2} = 2 \times 3 + - 2 \times - 3 + 4 \times 6 \\ = 6 + 6 + 24 = 36 \neq 0 \end{array}$ $\begin{array}{l} Therefore, the given planes are not perpendicular to each other . \\ \frac{a_{1}}{a_{2}} = \frac{2}{3}, \frac{b_{1}}{b_{2}} = \frac{- 2}{- 3} = \frac{2}{3} and \frac{c_{1}}{c_{2}} = \frac{4}{6} = \frac{2}{3} \end{array}$ $\begin{array}{l} \Rightarrow \frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}} \\ Therefore, the given planes are parallel to each other . \\ (d) 2 x - y + 3 z - 1 = 0 a n d 2 x - y + 3 z + 3 = 0 \\ D i r e c t i o n ratios of the given planes are: \\ a_{1} = 2, b_{1} = - 1, c_{1} = 3 \\ a_{2} = 2, b_{2} = - 1, c_{2} = 3 \\ a_{1} a_{2} + b_{1} b_{2} + c_{1} c_{2} = 2 \times 2 + - 1 \times - 1 + 3 \times 3 \\ = 4 + 1 + 9 = 14 \neq 0 \\ Therefore, the given planes are not perpendicular to each other . \\ \frac{a_{1}}{a_{2}} = \frac{2}{2} = 1, \frac{b_{1}}{b_{2}} = \frac{- 1}{- 1} = 1 and \frac{c_{1}}{c_{2}} = \frac{3}{3} = 1 \\ \Rightarrow \frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}} \\ Therefore, the given planes are parallel to each other . \\ (e) 4 x + 8 y + z - 8 = 0 a n d y + z - 4 = 0 \\ D i r e c t i o n ratios of the given planes are: \\ a_{1} = 4, b_{1} = 8, c_{1} = 1 \\ a_{2} = 0, b_{2} = 1, c_{2} = 1 \\ a_{1} a_{2} + b_{1} b_{2} + c_{1} c_{2} = 4 \times 0 + 8 \times 1 + 1 \times 1 \\ = 0 + 8 + 1 = 9 \neq 0 \end{array}$ $\begin{array}{l} Therefore, the given planes are not perpendicular to each other . \\ \frac{a_{1}}{a_{2}} = \frac{4}{0}, \frac{b_{1}}{b_{2}} = \frac{8}{1} = 8 and \frac{c_{1}}{c_{2}} = \frac{1}{1} = 1 \\ \Rightarrow \frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}} \end{array}$ $\begin{array}{l} Therefore, the given planes are not parallel to each other . \\ The angle between two planes is \\ θ = {cos}^{- 1} |(\frac{a_{1} a_{2} + b_{1} b_{2} + c_{1} c_{2}}{\sqrt{a_{1}^{2} + b_{1}^{2} + c_{1}^{2}} \sqrt{a_{2}^{2} + b_{2}^{2} + c_{2}^{2}}})| \\ = {cos}^{- 1} |(\frac{4 \times 0 + 8 \times 1 + 1 \times 1}{\sqrt{4^{2} + 8^{2} + 1^{2}} \sqrt{0^{2} + 1^{2} + 1^{2}}})| \\ = {cos}^{- 1} |(\frac{9}{\sqrt{16 + 64 + 1} \sqrt{0 + 1 + 1}})| \\ = {cos}^{- 1} |(\frac{9}{\sqrt{81} \sqrt{2}})| \\ = {cos}^{- 1} |\frac{1}{\sqrt{2}}| = \frac{π}{4} \\ T h u s, the angle between two planes is \frac{π}{4} . \end{array}$

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.

$\begin{array}{l} (a) The given equation of plane is 3 x - 4 y + 12 z = 3 and point \\ is (0, 0, 0) . \\ T h e distance of plane from given point (d) = |\frac{3 (0) - 4 (0) + 12 (0) - 3}{\sqrt{3^{2} + {(- 4)}^{2} + {(12)}^{2}}}| \\ = \frac{3}{\sqrt{9 + 16 + 144}} \\ = \frac{3}{\sqrt{169}} \\ = \frac{3}{13} \\ (b) The given equation of plane is 2 x - y + 2 z + 3 = 0 and point \\ is (3, - 2, 1) . \\ T h e distance of plane from given point (d) = |\frac{2 (3) - (- 2) + 2 (1) + 3}{\sqrt{2^{2} + {(- 1)}^{2} + {(2)}^{2}}}| \\ = \frac{13}{\sqrt{4 + 1 + 4}} \\ = \frac{13}{\sqrt{9}} \\ = \frac{13}{3} \\ (c) The given equation of plane is x - 2 y + 2 z - 9 = 0 and point \\ is (2, 3, - 5) . \\ T h e distance of plane from given point (d) = |\frac{(2) - 2 (3) + 2 (- 5) - 9}{\sqrt{1^{2} + {(- 2)}^{2} + {(2)}^{2}}}| \\ = \frac{23}{\sqrt{1 + 4 + 4}} \\ = \frac{23}{\sqrt{9}} \\ = \frac{23}{3} \\ (d) T h e g i v e n e q u a t i o n o f p l a n e i s 2 x - 3 y + 6 z - 2 = 0 a n d p o i n t \\ i s (- 6, 0, 0) . \\ T h e d i s t a n c e o f p l a n e f r o m g i v e n p o i n t (d) = |\frac{2 (- 6) - 3 (0) + 6 (0) - 2}{\sqrt{2^{2} + {(- 3)}^{2} + 6^{2}}}| \\ = \frac{14}{\sqrt{4 + 9 + 36}} \\ = \frac{14}{\sqrt{49}} = \frac{14}{7} = 2 \end{array}$

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, l₁m₂ – l₂ m₁.

Ans.

$\begin{array}{l} S i n c e l_{1}, m_{1}, n_{1} and l_{2}, m_{2}, n_{2} are the direction cosines of two \\ mutually perpendicular lines . Therefore, \\ l_{1} l_{2} + m_{1} m_{2} + n_{1} n_{2} = 0 … (i) \\ l_{1}^{2} + m_{1}^{2} + n_{1}^{2} = 1 … (i i) \\ l_{2}^{2} + m_{2}^{2} + n_{2}^{2} = 1 … (i i i) \\ L e t l,m,n be the direction cosines of a line which is perpendicular \\ to each given line with direction cosines l_{1}, m_{1}, n_{1} and l_{2}, m_{2}, n_{2} . \\ ∴ l_{1} l + m_{1} m + n_{1} n = 0 \\ l l_{2} + m m_{2} + n n_{2} = 0 \\ ∴ \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}} \\ \Rightarrow \frac{l^{2}}{{(m_{1} n_{2} - m_{2} n_{1})}^{2}} = \frac{m^{2}}{{(n_{1} l_{2} - n_{2} l_{1})}^{2}} = \frac{n^{2}}{{(l_{1} m_{2} - m_{1} l_{2})}^{2}} \\ = \frac{l^{2} + m^{2} + n^{2}}{{(m_{1} n_{2} - m_{2} n_{1})}^{2} + {(n_{1} l_{2} - n_{2} l_{1})}^{2} + {(l_{1} m_{2} - m_{1} l_{2})}^{2}} … (i v) \\ S i n c e, l,m,n are the direction cosines of line . \\ ∴ l^{2} {+ m}^{2} + n^{2} = 1 … (v) \\ I t is known that \\ (l_{1}^{2} + m_{1}^{2} + n_{1}^{2}) (l_{2}^{2} + m_{2}^{2} + n_{2}^{2}) - {(l_{1} l_{2} + m_{1} m_{2} + n_{1} n_{2})}^{2} \\ = {(m_{1} n_{2} - m_{2} n_{1})}^{2} + {(n_{1} l_{2} - n_{2} l_{1})}^{2} + {(l_{1} m_{2} - m_{1} l_{2})}^{2} \\ F r o m equations (i), (i i) and (i i i), we get \\ \Rightarrow 1.1 - 0 = {(m_{1} n_{2} - m_{2} n_{1})}^{2} + {(n_{1} l_{2} - n_{2} l_{1})}^{2} + {(l_{1} m_{2} - m_{1} l_{2})}^{2} \\ ∴ {(m_{1} n_{2} - m_{2} n_{1})}^{2} + {(n_{1} l_{2} - n_{2} l_{1})}^{2} + {(l_{1} m_{2} - m_{1} l_{2})}^{2} = 1 … (v i) \\ S u b s t i t u t i n g the values from equations (v) and (v i) in equation \\ (i v), we get \\ \frac{l^{2}}{{(m_{1} n_{2} - m_{2} n_{1})}^{2}} = \frac{m^{2}}{{(n_{1} l_{2} - n_{2} l_{1})}^{2}} = \frac{n^{2}}{{(l_{1} m_{2} - m_{1} l_{2})}^{2}} = \frac{1}{1} \\ \Rightarrow \frac{l^{2}}{{(m_{1} n_{2} - m_{2} n_{1})}^{2}} = \frac{m^{2}}{{(n_{1} l_{2} - n_{2} l_{1})}^{2}} = \frac{n^{2}}{{(l_{1} m_{2} - m_{1} l_{2})}^{2}} = 1 \\ \Rightarrow l = m_{1} n_{2} - m_{2} n_{1}, m = n_{1} l_{2} - n_{2} l_{1} and n = l_{1} m_{2} - m_{1} l_{2} \\ T h u s, the direction cosines of the required line are \\ m_{1} n_{2} - m_{2} n_{1}, n_{1} l_{2} - n_{2} l_{1} and 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.

$\begin{array}{l} L e t t h e a n g l e b e t w e e n t h e l i n e s w h o s e d i r e c t i o n r a t i o s a r e \\ a, b, c a n d b - c, c - a, a - b be θ, then \\ cos θ = | \frac{a (b - c) + b (c - a) + c (a - b)}{\sqrt{a^{2} + b^{2} + c^{2}} \sqrt{{(b - c)}^{2} + {(c - a)}^{2} + {(a - b)}^{2}}} | \\ = | \frac{a b - a c + b c - b a + c a - b c}{\sqrt{a^{2} + b^{2} + c^{2}} \sqrt{{(b - c)}^{2} + {(c - a)}^{2} + {(a - b)}^{2}}} | \\ = 0 \\ cos θ = \cos 90 ° \Rightarrow θ = 90 ° \\ Thus, the angle between the lines is 90 ° . \end{array}$

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

Ans.

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

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.

$\begin{array}{l} The coordinates of the points A, B, C, D be are \\ (1, 2, 3), (4, 5, 7), (- 4,3, - 6) and (2, 9, 2) respectivley . \\ T h e direction ratios of AB are: (4 - 1), (5 - 2), (7 - 3) i . e, 3, 3, 4 . \\ T h e direction ratios of CD are: (2 + 4), (9 - 3), (2 + 6) i . e, 6, 6, 8 . \\ T h e n, \frac{a_{1}}{a_{2}} = \frac{3}{6} = \frac{1}{2} \\ \frac{b_{1}}{b_{2}} = \frac{3}{6} = \frac{1}{2} \\ \frac{c_{1}}{c_{2}} = \frac{4}{8} = \frac{1}{2} \\ S o, \frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}} \\ T h u s, AB is parallel to CD . \\ Therefore, the angle between AB and CD is either 0° or 180° . \end{array}$

Q.42

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

Ans.

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

Q.43

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

Ans.

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

Q.44

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

Ans.

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

Q.45

$\begin{array}{l} Find the shortest distance between lines \\ \vec{r} = 6 \hat{i} + 2 \hat{j} + 2 \hat{k} + λ (\hat{i} - 2 \hat{j} + 2 \hat{k}) and \\ \vec{r} = - 4 \hat{i} - \hat{k} + μ (3 \hat{i} - 2 \hat{j} - 2 \hat{k}) . \end{array}$

Ans.

$\begin{array}{l} The equations of given lines are : \\ \vec{r} = 6 \hat{i} + 2 \hat{j} + 2 \hat{k} + λ (\hat{i} - 2 \hat{j} + 2 \hat{k}) . .. (i) \\ \vec{r} = - 4 \hat{i} - \hat{k} + μ (3 \hat{i} - 2 \hat{j} - 2 \hat{k}) . .. (ii) \\ Comparing equation (i) and equation (ii) with \vec{r} = a_{1} + {λb}_{1} and \\ \vec{r} = a_{2} + {λb}_{2}, we get \\ a_{1} = 6 \hat{i} + 2 \hat{j} + 2 \hat{k}, b_{1} = \hat{i} - 2 \hat{j} + 2 \hat{k} \\ a_{2} = - 4 \hat{i} - \hat{k} b_{2} = 3 \hat{i} - 2 \hat{j} - 2 \hat{k} \\ The shortest distance between two lines (d) = | \frac{(\vec{b_{1}} \times \vec{b_{2}}) . (\vec{a_{2}} - \vec{a_{1}})}{| \vec{b_{1}} \times \vec{b_{2}} |} | \\ \vec{b_{1}} \times \vec{b_{2}} = | \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & - 2 & 2 \\ 3 & - 2 & - 2 \end{matrix} | \\ = (4 + 4) \hat{i} - (- 2 - 6) \hat{j} + (- 2 + 6) \hat{k} \\ = 8 \hat{i} + 8 \hat{j} + 4 \hat{k} \\ | \vec{b_{1}} \times \vec{b_{2}} | = \sqrt{8^{2} + 8^{2} + 4^{2}} \\ = 12 \\ a_{2} - a_{1} = (- 4 \hat{i} - \hat{k}) - (6 \hat{i} + 2 \hat{j} + 2 \hat{k}) \\ = - 10 \hat{i} - 2 \hat{j} - 3 \hat{k} \\ ∴ d = | \frac{(8 \hat{i} + 8 \hat{j} + 4 \hat{k}) . (- 10 \hat{i} - 2 \hat{j} - 3 \hat{k})}{12} | \\ = | \frac{- 80 - 16 - 12}{12} | \\ = \frac{108}{12} = 9 \\ Therefore, the shortest distance between two lines is 9 units . \end{array}$

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

Ans.

$\begin{array}{l} T h e equation of the line passing through the points (x_{1}, y_{1}, z_{1}) \\ a n d (x_{2}, y_{2}, z_{2}) is \\ \frac{x - x_{1}}{x_{2} - x_{1}} = \frac{y - y_{1}}{y_{2} - y_{1}} = \frac{z - z_{1}}{z_{2} - z_{1}} \\ T h e equation of the line passing through the points (5, 1, 6) \\ a n d (3, 4, 1) is given as \\ \frac{x - 5}{3 - 5} = \frac{y - 1}{4 - 1} = \frac{z - 6}{1 - 6} \\ \Rightarrow \frac{x - 5}{- 2} = \frac{y - 1}{3} = \frac{z - 6}{- 5} = k (l e t) \\ \Rightarrow x = 5 - 2 k, y = 1 + 3 k and z = 6 - 5 k \\ A n y point on the line is of the form (5 - 2 k, 1 + 3 k, 6 - 5 k) . \\ T h e equation of YZ-plane is x = 0 \\ Since, the line passes through YZ-plane, \\ ∴ 5 - 2 k = 0 \Rightarrow k = \frac{5}{2} \\ S o, y = 1 + 3 \times \frac{5}{2} = \frac{17}{2} and z = 6 - 5 \times \frac{5}{2} = \frac{- 13}{2} \\ T h e r e f o r e, the required point is (0, \frac{17}{2}, \frac{- 13}{2}) . \end{array}$

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

Ans.

$\begin{array}{l} T h e equation of the line passing through the points (x_{1}, y_{1}, z_{1}) \\ a n d (x_{2}, y_{2}, z_{2}) is \\ \frac{x - x_{1}}{x_{2} - x_{1}} = \frac{y - y_{1}}{y_{2} - y_{1}} = \frac{z - z_{1}}{z_{2} - z_{1}} \\ T h e equation of the line passing through the points (5, 1, 6) \\ a n d (3, 4, 1) is given as \\ \frac{x - 5}{3 - 5} = \frac{y - 1}{4 - 1} = \frac{z - 6}{1 - 6} \\ \Rightarrow \frac{x - 5}{- 2} = \frac{y - 1}{3} = \frac{z - 6}{- 5} = k (l e t) \\ \Rightarrow x = 5 - 2 k, y = 1 + 3 k and z = 6 - 5 k \\ A n y point on the line is of the form (5 - 2 k, 1 + 3 k, 6 - 5 k) . \\ T h e equation of ZX-plane is y = 0 \\ Since, the line passes through ZX-plane, \\ ∴ 1 + 3 k = 0 \Rightarrow k = - \frac{1}{3} \\ S o, x = 5 - 2 \times - \frac{1}{3} = \frac{17}{3} and z = 6 - 5 \times - \frac{1}{3} = \frac{23}{3} \\ T h e r e f o r e, the required point is (\frac{17}{3}, 0, \frac{23}{3}) . \end{array}$

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.

$\begin{array}{l} T h e equation of the line passing through the points (x_{1}, y_{1}, z_{1}) \\ a n d (x_{2}, y_{2}, z_{2}) is \\ \frac{x - x_{1}}{x_{2} - x_{1}} = \frac{y - y_{1}}{y_{2} - y_{1}} = \frac{z - z_{1}}{z_{2} - z_{1}} \\ T h e equation of the line passing through the points (3, - 4, - 5) \\ a n d (2, - 3, 1) is given as \\ \frac{x - 3}{2 - 3} = \frac{y + 4}{- 3 + 4} = \frac{z + 5}{1 + 5} \\ \Rightarrow \frac{x - 3}{- 1} = \frac{y + 4}{1} = \frac{z + 5}{6} = k (l e t) \\ \Rightarrow x = 3 - k, y = k - 4 and z = 6 k - 5 \\ A n y point on the line is of the form (3 - k, k - 4, 6 k - 5) . \\ S i n c e, point lies on the plane, 2 x + y + z = 7 \\ ∴ 2 (3 - k) + k - 4 + 6 k - 5 = 7 \\ \Rightarrow 6 - 2 k + k - 4 + 6 k - 5 = 7 \\ \Rightarrow 5 k - 3 = 7 \\ \Rightarrow k = \frac{10}{5} = 2 \\ S o, x = 3 - 2 = 1, \\ y = 2 - 4 = - 2 and \\ z = 6 \times 2 - 5 = 7 \\ T h e r e f o r e, the required point is (1, - 2, 7) . \end{array}$

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.

$\begin{array}{l} T h e equation of the plane passing through the point (- 1, 3, 2) is \\ a (x + 1) + b (y - 3) + c (z - 2) = 0 … (i) \\ w h e r e a, b, c are the direction ratios of the plane . \\ Since, two planes are perpendicular to each other, if \\ a_{1} a_{2} + b_{1} b_{2} + c_{1} c_{2} = 0 \\ Plane (i) is perpendicular to the plane, x + 2 y + 3 z = 5 \\ ∴ a .1 + b .2 + c .3 = 0 \\ \Rightarrow a + 2 b + 3 c = 0 … (i i) \\ Plane (i) is perpendicular to the plane, 3x + 3 y + z = 0 \\ ∴ a .3 + b .3 + c .1 = 0 \\ \Rightarrow 3 a + 3 b + c = 0 … (i i i) \\ F r o m equation (i i) and equation (i i i), we have \\ \frac{a}{2 - 9} = \frac{- b}{1 - 9} = \frac{c}{3 - 6} \\ \Rightarrow \frac{a}{- 7} = \frac{b}{8} = \frac{c}{- 3} = k (l e t) \\ \Rightarrow a = - 7 k, b = 8 k and c = - 3 k \\ S u b s t i t u t i n g the values of a, b and c in equation (i), we get \\ - 7 k (x + 1) + 8 k (y - 3) - 3 k (z - 2) = 0 \\ \Rightarrow 7 x + 7 - 8 y + 24 + 3 z - 6 = 0 \\ \Rightarrow 7 x - 8 y + 3 z + 25 = 0 \\ This is the required equation of the plane . \end{array}$

Q.50

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

Ans.

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

Q.51

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

Ans.

$\begin{array}{l} Theequationoftheplanepassingthroughthelineof intersection \\ oftheplanes \vec{r} . (\hat{i} + \hat{j} + \hat{k}) = 1 and \vec{r} . (2 \hat{i} + 3 \hat{j} - \hat{k}) + 4 = 0 is ∴ \\ [\vec{r} . (\hat{i} + \hat{j} + \hat{k}) - 1] + λ [\vec{r} . (2 \hat{i} + 3 \hat{j} - \hat{k}) + 4] = 0 \\ \vec{r} [(\hat{i} + \hat{j} + \hat{k}) + λ (2 \hat{i} + 3 \hat{j} - \hat{k})] + (4 λ - 1) = 0 \\ \vec{r} [(1 + 2 λ) \hat{i} + (1 + 3 λ) \hat{j} + (1 - λ) \hat{k}] + (4 λ - 1) = 0 . .. (i) \\ The direction cosines of plane are (1 + 2 λ), (1 + 3 λ), (1 - λ) . \\ The required plane is parallel to x - axis . Therefore, its normal \\ is perpendicular to x - axis . \\ The direction ratios of x - axis are 1, 0, 0 . \\ ∵ a_{1} a_{2} + b_{1} b_{2} + c_{1} c_{2} = 0 [For perpendicular line on plane .] \\ ∴ (1 + 2 λ) . 1 + (1 + 3 λ) . 0 + (1 - λ) . 0 = 0 \\ \Rightarrow 1 + 2 λ = 0 \\ \Rightarrow λ = - \frac{1}{2} \\ Substituting λ = - \frac{1}{2} in equation (i), we get \\ \vec{r} [(1 + 2 \times - \frac{1}{2}) \hat{i} + (1 + 3 \times - \frac{1}{2}) \hat{j} + (1 + \frac{1}{2}) \hat{k}] + (4 \times - \frac{1}{2} - 1) = 0 \\ \Rightarrow \vec{r} [(1 - 1) \hat{i} + (1 - \frac{3}{2}) \hat{j} + \frac{3}{2} \hat{k}] + (- 2 - 1) = 0 \\ \Rightarrow \vec{r} [- \frac{1}{2} \hat{j} + \frac{3}{2} \hat{k}] - 3 = 0 \\ \Rightarrow \vec{r} [- \hat{j} + 3 \hat{k}] - 6 = 0 \\ Putting \vec{r} = x \hat{i} + y \hat{j} + z \hat{k}, we get \\ (x \hat{i} + y \hat{j} + z \hat{k}) [- \hat{j} + 3 \hat{k}] - 6 = 0 \\ - y + 3 z - 6 = 0 or y - 3 z + 6 = 0 \\ This is the equation of the required plane . \end{array}$

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

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

Ans.

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

Q.54

$\begin{array}{l} Find the distance of the point (- 1, - 5, - 10) from the point \\ of intersection of the line \vec{r} = 2 \hat{i} - \hat{j} + 2 \hat{k} + λ (3 \hat{i} + 4 \hat{j} + 2 \hat{k}) \\ and the plane \vec{r} . (\hat{i} - \hat{j} + \hat{k}) = 5 . \end{array}$

Ans.

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

Q.55

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

Ans.

$\begin{array}{l} Here, a line through (1, 2, 3) is parallel to the two planes \\ \vec{r} . (\hat{i} - \hat{j} + 2 \hat{k}) = 5 and \vec{r} . (3 \hat{i} + \hat{j} + \hat{k}) = 6, so given line will be \\ perpendicular to the normals on the given planes . Then, \\ Position vector of the point (1, 2, 3), \vec{a} = \hat{i} + 2 \hat{j} + 3 \hat{k} \\ Let vector parallel to required line be \\ \vec{b} = b_{1} \hat{i} + b_{2} \hat{j} + b_{3} \hat{k} \\ Comparing given planes with \vec{r} \vec{N_{1}} - d = 0 and \vec{r} \vec{N_{2}} - d = 0, we get \\ \vec{N_{1}} = \hat{i} - \hat{j} + 2 \hat{k} and \vec{N_{2}} = 3 \hat{i} + \hat{j} + \hat{k} \\ \vec{b} . \vec{N_{1}} = 0 and \vec{b} . \vec{N_{2}} = 0 \\ \Rightarrow (b_{1} \hat{i} + b_{2} \hat{j} + b_{3} \hat{k}) . (\hat{i} - \hat{j} + 2 \hat{k}) = 0 \Rightarrow b_{1} - b_{2} + 2 b_{3} = 0 . .. (i) \\ (b_{1} \hat{i} + b_{2} \hat{j} + b_{3} \hat{k}) . (3 \hat{i} + \hat{j} + \hat{k}) = 0 \Rightarrow 3 b_{1} + b_{2} + b_{3} = 0 . .. (ii) \\ On solving equation (i) and (ii), we get \\ \frac{b_{1}}{- 1 \times 1 - 2 \times 1} = \frac{- b_{2}}{1 \times 1 - 2 \times 3} = \frac{b_{3}}{1 \times 1 + 1 \times 3} \\ \Rightarrow \frac{b_{1}}{- 3} = \frac{- b_{2}}{- 5} = \frac{b_{3}}{4} \\ Therefore, the direction ratios of \vec{b} are - 3, - 5 and 4 . \\ ∴ \vec{b} = - 3 \hat{i} - 5 \hat{j} + 4 \hat{k} \\ So, the eqution of the line passing through (1, 2, 3) and parallel \\ to vector \vec{b} is given as : \\ \vec{r} = (\hat{i} + 2 \hat{j} + 3 \hat{k}) + λ (- 3 \hat{i} - 5 \hat{j} + 4 \hat{k}) \end{array}$

Q.56

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

Ans.

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

Q.57

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

Ans.

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

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)

$\frac{2}{\sqrt{29}} u n i t s$

Ans.

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

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

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

Ans.

$\begin{array}{l} T h e equations of the planes are: \\ 2x - y + 4z = 5 … (i) \\ 5x - 2 . 5y + 10 z = 6 … (i i) \\ S o, \frac{a_{1}}{a_{2}} = \frac{2}{5}, \frac{b_{1}}{b_{2}} = \frac{- 1}{- 2.5} = \frac{10}{25} = \frac{2}{5} \\ a n d \frac{c_{1}}{c_{2}} = \frac{4}{10} = \frac{2}{5} \\ ∴ \frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}} \\ T h e r e f o r e, the planes are parallel . \\ Thus, the correct option is B . \end{array}$

NCERT Solutions Class 12 Mathematics Chapter 11

NCERT Solutions Class 12 Mathematics Chapter 11 Three-Dimensional Geometry

Key Topics Covered In NCERT Class 12 Mathematics Chapter 11

Introduction

Coordinate Axes and Coordinate Planes in Three-Dimensional Space

Coordinates of a Point in Space

Distance Between Two Points

Section Formula

NCERT Solutions Class 12 Mathematics Chapter 11 - Exercises and Answer Solutions.

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Key Features of NCERT Solutions for Class 12 Mathematics Chapter 11

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