# NCERT Solutions Class 11 Maths Chapter 12

## NCERT Solutions for Class 11 Mathematics Chapter 12 – Introduction to Three Dimensional Geometry

Mathematics is used almost everywhere, right from buying regular groceries to planning your family finances. Thus, it requires proper understanding and conceptual clarity. This can only be attained if the basics of Mathematics are clear. Hence, students are advised to practice as many problems as possible to excel in Mathematics. The more you practice, the easier it will get!

Geometry is an essential section of Mathematics. Students find it difficult only if the basics are unclear. As a result, the chapter introduction to 3D Geometry has been included in the NCERT Class 11 Mathematics textbook. It covers all the vital concepts like coordinate axis and coordinate plane in three-dimensional space, what are direction cosines and direction ratios, how to calculate the distance between two points in three-dimensional Geometry and the coordinates of a point in space.

Generally, students are confused between the terminologies, direction cosines and direction ratios.  Taking note of it, NCERT Solutions for Class 11 Mathematics Chapter 12 provides a clear-cut difference between the two terms for students to grasp the two concepts quickly. Moreover, it has also provided ways to calculate the distance between the two points in 3D Geometry.

After studying from the NCERT Solutions, students will find this chapter relatively easier and scoring. They are capable of solving different types of questions covered in the exercises of the NCERT textbook. They can get the NCERT Solutions for Class 11 Mathematics Chapter 12 from the Extramarks’ website and be sure of their success.

### Key Topics Covered In Class 11 Mathematics Chapter 12

The study of three-dimensional Geometry has a great scope in architecture, planning and engineering fields. The essential work processes carried out in these fields are incomplete without it. As a result, it has become one of the crucial chapters of Class 11 Mathematics. All the methodologies covered in this chapter will go a long way in higher education.

The chapter introduces you to three-dimensional Geometry, followed by describing the coordinate axis and coordinate plane. Further, it helps you learn the differentiation between the direction cosines and direction ratios, methods to calculate the distance between the two points and how to locate the coordinates in space. The complete chapter is covered in detail in the NCERT Solutions for Class 11 Mathematics Chapter 12, available on the Extramarks’ website.

After completing this chapter, students will be able to correlate all the terminologies and methodologies covered in the chapter without getting confused, anxious or stressed.

Introduction

In this chapter, we will study three-dimensional space. We will also read about coordinate planes, coordinate axes, the coordination of a point in space and the distance between two points in space.

Besides additional questions to practice, the entire chapter notes and their exercise solutions have been covered in the NCERT Solutions for Class 11 Mathematics Chapter 12 available on the Extramarks’ website.

Coordinate Axes and Coordinate Plane in the Three-Dimensional Space

In the section of this chapter, we will study the coordinate axes and coordinate planes in three-dimensional space.

Coordinate axes are the intersecting lines passing through the coordinates in a three-dimensional space, whereas a coordinate plane is the collection of all the points on a coordinate in the three-dimensional space.

The three-dimensional space has three planes, namely the plane of the X-axis, the plane of the Y-axis and the plane of the Z-axis. In the three-dimensional space, all the three planes are mutually perpendicular to each other.

Students can know more about the coordinate axes and the coordinate plane in the three-dimensional space covered in the NCERT Solutions for Class 11 Mathematics Chapter 12 available on the Extramarks’ website.

Coordinates of a Point in Space

In the section of this chapter, we will study the coordination of a point in space.

The coordinates of a point in space have a point randomly arranged in space and help to locate coordinates in space.

Three-dimensional space has three axes. We will locate the coordinates of a point on these axes in space.

All the highlight points on the coordinates of a point in space and points to ponder are included in the NCERT Solutions for Class 11 Mathematics Chapter 12 available on the Extramarks’ website.

Distance between Two Points

In this section of the chapter, we will get to know how to calculate the distance between two points.

When two points are placed randomly in the three-dimensional space, calculating the distance between two points becomes a complex topic. But if students know a few methodologies and formulas to find the distance, they can quickly find the t answers.

The formula derived for finding the distance between two points is given by:

PQ = √(x1+x2)2+(y1+y2)2+(z1+z2)2.

Students can find a lot of easy as well as challenging questions to practice on this topic in the NCERT Solutions for Class 11 Mathematics Chapter 12 available on the Extramarks’ website.

Section Formula

In this section, you will learn about calculating the midpoint of a point using the section formula.

It is divided into two cases. They  are as follows:

• Case 1

Coordinates of the midpoint:

In case R is the midpoint of PQ, then m: n = 1: 1 such that

x = (x1+x2)/2

y = (y1+y2)/2

z = (z1+z2)/2

These are the coordinates of the mid point of the segment joining P (x1, y1, z1 ) and Q (x2, y2, z2 )

• Case 2

The coordinates of the point R, which divides PQ in the ratio k:1 are obtained by taking k = m,/n, which are as given below:

[(k.x1+x2)/(1+k), (k.y1+y2)/(1+k), (k.z1+z2)/(1+k)]

Generally, the above result is used in solving problems involving a general point on the line passing through two given points.

Students can find more information on it in the NCERT Solutions for Class 11 Mathematics Chapter 12 available on the Extramarks’ website.

### NCERT Solutions for Class 11 Mathematics Chapter 12 Exercise &  Solutions

NCERT Solutions for Class 11 Mathematics Chapter 12 has theoretical explanations and step-by-step solutions to all questions from the NCERT textbook. The experienced subject matter experts  of Extramarks have created it to help the students to excel in their examinations.

It contains all the important concepts and techniques covered in a well-structured format. You can look for multiple ways to solve a question and choose the best solution that suits you. It also includes a detailed analysis of the chapter along with revision notes. This will help students to score well in school as well as in competitive examinations.

You can avail of NCERT Solutions for Class 11 Mathematics Chapter 12 from the Extramarks’ website.

Click on the  links below to view exercise-specific questions and solutions for NCERT Solutions for Class 11 Mathematics Chapter 12:

• Class 11 Mathematics Chapter 12: Exercise 12.1
• Class 11 Mathematics Chapter 12: Exercise 12.2
• Class 11 Mathematics Chapter 12: Exercise 12.3.
• Class 11 Mathematics Chapter 12: Miscellaneous Exercise.

Along with Class 11 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 Class 11 Mathematics

NCERT Exemplar Class 11 Mathematics book has created a significant impact on students preparing for their school examinations and other competitive examinations as the questions are designed while adhering to the  CBSE  guidelines.

It helps students develop confidence during their preparation as they have basic as well as advanced level questions. As a result, students learn to solve advanced-level questions after using Exemplar.   As a result, they can face their examinations confidently.

The useful tips provided in this book will make a difference in their preparation. Hence, the students will develop a more practical and logical way of thinking and will be able to approach questions in a better way. By studying from the Exemplar, students can constantly assure themselves to be among the top rankers.

### Key Features for NCERT Solutions for Class 11 Mathematics Chapter 12

To complete the entire syllabus within the time frame, students must know how to manage their time effectively. Hence, NCERT Solutions for Class 11 Mathematics Chapter 12 teaches students time management. The key features are as follows:

• Students will be able to grasp all the concepts in less time once they study from our NCERT Solutions for Class 11 Mathematics Chapter 12 and will have enough time for other subjects as well.
• It also provides easy tips and tricks to solve lengthy problems in less time.  This will help the students to speed up their mathematical calculations and complete their question paper on time and to check their answers as well.
• After completing the NCERT Solutions for Class 11 Mathematics Chapter 12, students will get good command over all the basics related to 3D Geometry.

Q.1 A point is on the x -axis. What are its y-coordinate and z-coordinates?

Ans

The coordinate of the point on x-axis is (x, 0, 0). The y-coordinate of this point is 0 and that of z-coordinate is also 0.

Q.2 A point is in the XZ-plane. What can you say about its y-coordinate?

Ans

The y-coordinate of a point in XZ-plane is 0.

Q.3 Name the octants in which the following points lie: (1, 2, 3), (4, –2, 3), (4, –2, –5), (4, 2, –5), (–4, 2, –5), (– 4, 2, 5), (–3, –1, 6) (2, – 4, –7).

Ans

 Coordinate Name of octants (1, 2, 3) I (4, –2, 3) IV (4, –2, –5) VIII (4, 2, –5) V (–4, 2, –5) VI (–4, 2, 5) II (–3, –1, 6) III (2, –4, –7) VIII

Q.4 Fill in the blanks:

(i) The x-axis and y-axis taken together determine a plane known as_______.
(ii) The coordinates of points in the XY-plane are of the form _______.
(iii) Coordinate planes divide the space into ______ octants.

Ans

(i) The x-axis and y-axis taken together determine a plane known as XY-plane.
(ii) The coordinates of points in the XY-plane are of the form (x, y, 0).
(iii) Coordinate planes divide the space into eight octants.

Q.5 Find the distance between the following pairs of points:

(i) (2, 3, 5) and (4, 3, 1)
(ii) (–3, 7, 2) and (2, 4, –1)
(iii) (–1, 3, – 4) and (1, –3, 4)
(iv) (2, –1, 3) and (–2, 1, 3).

Ans

$\begin{array}{l}\text{Distance between two points}\left({\mathrm{x}}_{1},{\mathrm{y}}_{1},{\mathrm{z}}_{1}\right)\text{and}\left({\mathrm{x}}_{2},{\mathrm{y}}_{2},{\mathrm{z}}_{2}\right),\\ \mathrm{d}=\sqrt{{\left({\mathrm{x}}_{2}-{\mathrm{x}}_{1}\right)}^{2}+{\left({\mathrm{y}}_{2}-{\mathrm{y}}_{1}\right)}^{2}+{\left({\mathrm{z}}_{2}-{\mathrm{z}}_{1}\right)}^{2}}\\ \left(\mathrm{i}\right)\left(2,\text{}3,\text{}5\right)\text{}\mathrm{and}\text{}\left(4,\text{}3,\text{}1\right)\\ \mathrm{d}=\sqrt{{\left(4-2\right)}^{2}+{\left(3-3\right)}^{2}+{\left(1-5\right)}^{2}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\sqrt{4+0+16}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\sqrt{20}\text{\hspace{0.17em}}\mathrm{unit}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=2\sqrt{5}\text{\hspace{0.17em}}\mathrm{unit}\end{array}$ $\begin{array}{l}\left(\mathrm{ii}\right)\left(–3,\text{}7,\text{}2\right)\text{}\mathrm{and}\text{}\left(2,\text{}4,\text{}–1\right)\\ \mathrm{d}=\sqrt{{\left(2+3\right)}^{2}+{\left(4-7\right)}^{2}+{\left(2+1\right)}^{2}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\sqrt{25+9+9}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\sqrt{43}\text{\hspace{0.17em}}\mathrm{unit}\\ \left(\mathrm{iii}\right)\text{\hspace{0.17em}}\left(-1,\text{}3,-\text{\hspace{0.17em}}4\right)\text{}\mathrm{and}\text{}\left(1,-3,\text{}4\right)\\ \mathrm{d}=\sqrt{{\left(1+1\right)}^{2}+{\left(-3-3\right)}^{2}+{\left(4+4\right)}^{2}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\sqrt{4+36+64}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\sqrt{104}\text{\hspace{0.17em}}\mathrm{unit}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=2\sqrt{26}\text{\hspace{0.17em}}\mathrm{unit}\\ \left(\mathrm{iv}\right)\left(2,-1,3\right)\text{}\mathrm{and}\text{}\left(-2,1,3\right)\\ \mathrm{d}=\sqrt{{\left(-2-2\right)}^{2}+{\left(1+1\right)}^{2}+{\left(3-3\right)}^{2}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\sqrt{16+4+0}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\sqrt{20}\text{\hspace{0.17em}}\mathrm{unit}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=2\sqrt{5}\text{\hspace{0.17em}}\mathrm{unit}\end{array}$

Q.6 Show that the points (–2, 3, 5), (1, 2, 3) and (7, 0, –1) are collinear.

Ans

$\begin{array}{l}\mathrm{Let}\text{three points be A}\left(-2,3,5\right),\mathrm{B}\left(\text{1},\text{2},\text{3}\right)\text{and}\mathrm{}\text{\hspace{0.17em}}\mathrm{C}\left(7,0,–1\right).\mathrm{Then},\\ \mathrm{AB}=\sqrt{{\left(1+2\right)}^{2}+{\left(2-3\right)}^{2}+{\left(3-5\right)}^{2}}\\ =\sqrt{9+1+4}\\ =\sqrt{14}\\ =3.741\end{array}$ $\begin{array}{l}\mathrm{BC}=\sqrt{{\left(7-1\right)}^{2}+{\left(0-2\right)}^{2}+{\left(-1-3\right)}^{2}}\\ =\sqrt{36+4+16}\\ =\sqrt{56}\\ =14.966\\ \mathrm{CA}=\sqrt{{\left(-2-7\right)}^{2}+{\left(0-3\right)}^{2}+{\left(-1-5\right)}^{2}}\\ =\sqrt{81+9+36}\\ =\sqrt{126}\\ =11.225\\ \therefore \mathrm{AB}+\mathrm{CA}=3.741+11.225\\ =14.966=\mathrm{BC}\\ \mathrm{Thus},\text{the points A, B and C are collinear.}\end{array}$

Q.7 Verify the following:
(i) (0, 7, –10), (1, 6, – 6) and (4, 9, – 6) are the vertices of an isosceles triangle.
(ii) (0, 7, 10), (–1, 6, 6) and (– 4, 9, 6) are the vertices of a right angled triangle.
(iii) (–1, 2, 1), (1, –2, 5), (4, –7, 8) and (2, –3, 4) are the vertices of a parallelogram.

Ans

$\begin{array}{l}\left(\text{i}\right)\text{ Let three points be A}\left(\text{0, 7,–10}\right)\text{,B}\left(\text{1, 6,– 6}\right)\text{ and C}\left(\text{4, 9,– 6}\right)\text{.}\\ \text{Then,}\\ \text{ AB = }\sqrt{{\left(\text{1-0}\right)}^{\text{2}}\text{+}{\left(\text{6-7}\right)}^{\text{2}}\text{+}{\left(\text{-\hspace{0.17em}6+10}\right)}^{\text{2}}}\\ \text{ = }\sqrt{\text{1+1+16}}\\ \text{ = }\sqrt{\text{18}}\\ \text{ BC = }\sqrt{{\left(\text{4-1}\right)}^{\text{2}}\text{+}{\left(\text{9-6}\right)}^{\text{2}}\text{+}{\left(\text{-\hspace{0.17em}6+6}\right)}^{\text{2}}}\end{array}$ $\begin{array}{l}\text{ = }\sqrt{\text{9+9+0}}\\ \text{ = }\sqrt{\text{18}}\\ \text{ CA = }\sqrt{{\left(\text{4-0}\right)}^{\text{2}}\text{+}{\left(\text{9-7}\right)}^{\text{2}}\text{+}{\left(\text{-\hspace{0.17em}6+10}\right)}^{\text{2}}}\\ \text{ = }\sqrt{\text{16+4+16}}\\ \text{ = }\sqrt{\text{36}}\\ \text{Since, AB = BC = }\sqrt{\text{18}}\text{.}\\ \text{So, }\left(\text{0,7,–10}\right)\text{,}\left(\text{1,6,– 6}\right)\text{ and }\left(\text{4,9,– 6}\right)\text{ are the vertices of}\\ \text{an isosceles triangle.}\\ \left(\text{ii}\right)\text{ Let three points be A}\left(\text{0, 7, 10}\right)\text{,B}\left(\text{–1, 6, 6}\right)\text{ and C}\left(\text{– 4, 9, 6}\right)\text{. }\\ \text{Then,}\\ \text{ AB = }\sqrt{{\left(\text{-1-0}\right)}^{\text{2}}\text{+}{\left(\text{6-7}\right)}^{\text{2}}\text{+}{\left(\text{6-10}\right)}^{\text{2}}}\\ \text{ = }\sqrt{\text{1+1+16}}\\ \text{ = }\sqrt{\text{18}}\\ \text{ BC = }\sqrt{{\left(\text{-\hspace{0.17em}4+1}\right)}^{\text{2}}\text{+}{\left(\text{9-6}\right)}^{\text{2}}\text{+}{\left(\text{\hspace{0.17em}6-6}\right)}^{\text{2}}}\\ \text{ = }\sqrt{\text{9+9+0}}\\ \text{ = }\sqrt{\text{18}}\\ \text{ CA = }\sqrt{{\left(\text{-\hspace{0.17em}4-0}\right)}^{\text{2}}\text{+}{\left(\text{9-7}\right)}^{\text{2}}\text{+}{\left(\text{\hspace{0.17em}6-10}\right)}^{\text{2}}}\\ \text{ = }\sqrt{\text{16+4+16}}\\ \text{ = }\sqrt{\text{36}}\\ \text{Since,}\\ {\text{AB}}^{\text{2}}{\text{+BC}}^{\text{2}}\text{ = 18+18}\\ \text{ = 36}\\ {\text{ = CA}}^{\text{2}}\end{array}$ $\begin{array}{l}\text{Therefore, \hspace{0.17em}​}\left(\text{0, 7, 10}\right)\text{,}\left(\text{–1, 6, 6}\right)\text{ and }\left(\text{– 4, 9, 6}\right)\text{ are the \hspace{0.17em}}\\ \text{vertices of a right angled triangle.}\\ \left(\text{iii}\right)\text{ Let four points be A}\left(\text{–1, 2, 1}\right)\text{, B}\left(\text{1,–2, 5}\right)\text{, C}\left(\text{4,–7, 8}\right)\text{}\\ \text{and D}\left(\text{2,–3,4}\right)\text{. Then,}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}AB = }\sqrt{{\left(\text{1+1}\right)}^{\text{2}}\text{+}{\left(\text{-2-2}\right)}^{\text{2}}\text{+}{\left(\text{5-1}\right)}^{\text{2}}}\\ \text{= }\sqrt{\text{4+16+16}}\\ \text{= }\sqrt{\text{36}}\text{=6}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}BC = }\sqrt{{\left(\text{4-1}\right)}^{\text{2}}\text{+}{\left(\text{-7+2}\right)}^{\text{2}}\text{+}{\left(\text{8-5}\right)}^{\text{2}}}\\ \text{= }\sqrt{\text{9+25+9}}\\ \text{= }\sqrt{\text{43}}\\ \text{\hspace{0.17em}\hspace{0.17em}CD = }\sqrt{{\left(\text{2-4}\right)}^{\text{2}}\text{+}{\left(\text{-3+7}\right)}^{\text{2}}\text{+}{\left(\text{4-8}\right)}^{\text{2}}}\\ \text{= }\sqrt{\text{4+16+16}}\\ \text{= }\sqrt{\text{36}}\\ \text{\hspace{0.17em}\hspace{0.17em}DA = }\sqrt{{\left(\text{-1-2}\right)}^{\text{2}}\text{+}{\left(\text{2+3}\right)}^{\text{2}}\text{+}{\left(\text{1-4}\right)}^{\text{2}}}\\ \text{= }\sqrt{\text{9+25+9}}\\ \text{= }\sqrt{\text{43}}\\ \text{Since, opposite sides AB and CD, BC and DA are equal.}\\ \text{So, ABCD is a parallelogram.}\\ \text{Therefore, }\left(\text{–1,2,1}\right)\text{,}\left(\text{1,–2,5}\right)\text{,}\left(\text{4,–7,8}\right)\text{ and }\left(\text{2,–3,4}\right)\text{}\\ \text{are the vertices of a parallelogram.}\end{array}$

Q.8 Find the equation of the set of points which are equidistant from the points (1, 2, 3) and (3, 2, –1).

Ans

$\begin{array}{l}\mathrm{Let}\text{the point P}\left(\mathrm{x},\mathrm{y},\mathrm{z}\right)\text{are equidistant from the points Q}\left(\text{1},\text{2},\text{3}\right)\text{}\\ \text{and R}\left(\text{3},\text{2},\text{}–\text{1}\right).\text{Then,}\\ \mathrm{PQ}=\sqrt{{\left(\mathrm{x}-1\right)}^{2}+{\left(\mathrm{y}-2\right)}^{2}+{\left(\mathrm{z}-3\right)}^{2}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\sqrt{{\mathrm{x}}^{2}-2\mathrm{x}+1+{\mathrm{y}}^{2}-4\mathrm{y}+4+{\mathrm{z}}^{2}-6\mathrm{z}+9}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\sqrt{{\mathrm{x}}^{2}-2\mathrm{x}+{\mathrm{y}}^{2}-4\mathrm{y}+{\mathrm{z}}^{2}-6\mathrm{z}+14}\\ \mathrm{PR}=\sqrt{{\left(\mathrm{x}-3\right)}^{2}+{\left(\mathrm{y}-2\right)}^{2}+{\left(\mathrm{z}+1\right)}^{2}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\sqrt{{\mathrm{x}}^{2}-6\mathrm{x}+9+{\mathrm{y}}^{2}-4\mathrm{y}+4+{\mathrm{z}}^{2}+2\mathrm{z}+1}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\sqrt{{\mathrm{x}}^{2}-6\mathrm{x}+{\mathrm{y}}^{2}-4\mathrm{y}+{\mathrm{z}}^{2}+2\mathrm{z}+14}\\ \mathrm{According}\text{to given condition,}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} PQ}=\text{PR}\\ ⇒{\text{\hspace{0.17em}PQ}}^{\text{2}}={\text{PR}}^{2}\\ {\mathrm{x}}^{2}-2\mathrm{x}+{\mathrm{y}}^{2}-4\mathrm{y}+{\mathrm{z}}^{2}-6\mathrm{z}+14\\ ={\mathrm{x}}^{2}-6\mathrm{x}+{\mathrm{y}}^{2}-4\mathrm{y}+{\mathrm{z}}^{2}+2\mathrm{z}+14\\ 6\mathrm{x}-2\mathrm{x}-2\mathrm{z}-6\mathrm{z}=0\\ 4\mathrm{x}-8\mathrm{z}=0\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{x}-2\mathrm{z}=0,\text{which is the required equation of the set of points}\\ \text{which are equidistant from the points}\left(1,2,3\right)\text{and}\left(3,2,-1\right).\end{array}$

Q.9 Find the equation of the set of points P, the sum of whose distances from A (4, 0, 0) and B (– 4, 0, 0) is equal to 10.

Ans

$\mathrm{Let}\text{the sum of distances of point P}\left(\mathrm{x},\mathrm{y},\mathrm{z}\right)\text{from the points Q}\left(\text{4},\text{}0,\text{}0\right)\text{and R}\left(\text{}–\text{4},\text{}0,\text{}0,\right)\text{is 1}0.\text{Then}$ $\begin{array}{l}\mathrm{PQ}=\sqrt{{\left(\mathrm{x}-4\right)}^{2}+{\left(\mathrm{y}-0\right)}^{2}+{\left(\mathrm{z}-0\right)}^{2}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\sqrt{{\mathrm{x}}^{2}-8\mathrm{x}+16+{\mathrm{y}}^{2}+{\mathrm{z}}^{2}}\\ \mathrm{PR}=\sqrt{{\left(\mathrm{x}+4\right)}^{2}+{\left(\mathrm{y}-0\right)}^{2}+{\left(\mathrm{z}-0\right)}^{2}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\sqrt{{\mathrm{x}}^{2}+8\mathrm{x}+16+{\mathrm{y}}^{2}+{\mathrm{z}}^{2}}\\ \mathrm{According}\text{to given condition,}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} PQ}+\text{PR}=\text{10}\\ ⇒{\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}PQ}}^{\text{2}}={\left(10-\text{PR}\right)}^{2}\\ ⇒{\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}PQ}}^{\text{2}}=100-20\text{\hspace{0.17em}}\mathrm{PR}+{\mathrm{PR}}^{2}\\ 20\text{\hspace{0.17em}}\mathrm{PR}-100={\mathrm{x}}^{2}+8\mathrm{x}+16+{\mathrm{y}}^{2}+{\mathrm{z}}^{2}-{\mathrm{x}}^{2}+8\mathrm{x}-16-{\mathrm{y}}^{2}-{\mathrm{z}}^{2}\\ 20\text{\hspace{0.17em}}\mathrm{PR}-100=16\mathrm{x}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}20\text{\hspace{0.17em}}\mathrm{PR}=100+16\mathrm{x}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}5\text{\hspace{0.17em}}\mathrm{PR}=25+4\mathrm{x}\\ \mathrm{Squarring}\text{both sides, we get}\\ \text{\hspace{0.17em}}{\left(5\text{\hspace{0.17em}}\mathrm{PR}\right)}^{2}={\left(25+4\mathrm{x}\right)}^{2}\\ \text{\hspace{0.17em}\hspace{0.17em}}25\text{\hspace{0.17em}}{\mathrm{PR}}^{2}=625+200\mathrm{x}+16{\mathrm{x}}^{2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}25\left({\mathrm{x}}^{2}+8\mathrm{x}+16+{\mathrm{y}}^{2}+{\mathrm{z}}^{2}\right)=625+200\mathrm{x}+16{\mathrm{x}}^{2}\\ 25{\mathrm{x}}^{2}+200\mathrm{x}+400+25{\mathrm{y}}^{2}+25{\mathrm{z}}^{2}=625+200\mathrm{x}+16{\mathrm{x}}^{2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}9{\mathrm{x}}^{2}+25{\mathrm{y}}^{2}+25{\mathrm{z}}^{2}-225=0\end{array}$

Q.10 Find the coordinates of the point which divides the line segment joining the points (– 2, 3, 5) and (1, – 4, 6) in the ratio

(i) 2 : 3 internally, (ii) 2 : 3 externally.

Ans

$\begin{array}{l}\left(\mathrm{i}\right)\mathrm{Let}\text{point P}\left(\mathrm{x},\mathrm{y},\mathrm{z}\right)\text{divides the line segment joining the points}\\ \left(-\text{\hspace{0.17em}2},\text{3},\text{5}\right)\text{and}\left(\text{1},-\text{\hspace{0.17em}4},\text{6}\right)\text{in the ratio 2:3 internally.Then,}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{x}=\frac{{\mathrm{mx}}_{2}+{\mathrm{nx}}_{1}}{\mathrm{m}+\mathrm{n}}\\ =\frac{2\left(1\right)+3\left(-2\right)}{2+3}=\frac{-4}{5}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{y}=\frac{{\mathrm{my}}_{2}+{\mathrm{ny}}_{1}}{\mathrm{m}+\mathrm{n}}\\ =\frac{2\left(-4\right)+3\left(3\right)}{2+3}\\ =\frac{1}{5}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{z}=\frac{{\mathrm{mz}}_{2}+{\mathrm{nz}}_{1}}{\mathrm{m}+\mathrm{n}}\\ =\frac{2\left(6\right)+3\left(5\right)}{2+3}\\ =\frac{27}{5}\\ \mathrm{Therefore},\text{the coordinates of the point P are}\left(-\frac{4}{5},\frac{1}{5},\frac{27}{5}\right).\\ \left(\mathrm{ii}\right)\mathrm{Let}\text{point P}\left(\mathrm{x},\mathrm{y},\mathrm{z}\right)\text{divides the line segment joining the points}\\ \left(-\text{\hspace{0.17em}2},\text{3},\text{5}\right)\text{and}\left(\text{1},-\text{\hspace{0.17em}4},\text{6}\right)\text{in the ratio 2:3 externally.Then,}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{x}=\frac{{\mathrm{mx}}_{2}-{\mathrm{nx}}_{1}}{\mathrm{m}-\mathrm{n}}\\ =\frac{2\left(1\right)-3\left(-2\right)}{2-3}\end{array}$ $\begin{array}{l}=\frac{8}{-1}=-8\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{y}=\frac{{\mathrm{my}}_{2}-{\mathrm{ny}}_{1}}{\mathrm{m}-\mathrm{n}}\\ =\frac{2\left(-4\right)-3\left(3\right)}{2-3}\\ =\frac{-17}{-1}=17\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{z}=\frac{{\mathrm{mz}}_{2}-{\mathrm{nz}}_{1}}{\mathrm{m}-\mathrm{n}}\\ =\frac{2\left(6\right)-3\left(5\right)}{2-3}\\ =\frac{-3}{-1}=3\\ \mathrm{Therefore},\text{the coordinates of the point P are}\left(-8,17,3\right).\end{array}$

Q.11 Given that P (3, 2, – 4), Q (5, 4, – 6) and R (9, 8, –10) are collinear. Find the ratio in which Q divides PR.

Ans

$\begin{array}{l}\mathrm{P}\text{oint}\mathrm{Q}\left(5,\text{}4,-6\right)\text{divides the line segment joining the points}\\ \mathrm{P}\left(3,\text{}2,-\text{\hspace{0.17em}}4\right)\mathrm{}\text{and}\mathrm{R}\left(9,\text{}8,-10\right)\text{in the ratio k:1 internally.Then,}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{x}=\frac{{\mathrm{mx}}_{2}+{\mathrm{nx}}_{1}}{\mathrm{m}+\mathrm{n}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}5=\frac{\mathrm{k}\left(9\right)+1\left(3\right)}{\mathrm{k}+1}\\ 5\left(\mathrm{k}+1\right)=9\text{\hspace{0.17em}}\mathrm{k}+3\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}5\text{\hspace{0.17em}}\mathrm{k}+5=9\text{\hspace{0.17em}}\mathrm{k}+3\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}5-3=9\mathrm{k}-5\mathrm{k}\end{array}$ $\begin{array}{l}⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}2=4\text{\hspace{0.17em}}\mathrm{k}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{k}=\frac{2}{4}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{k}=\frac{1}{2}\\ \mathrm{The}\text{required ratio is}\frac{1}{2}:1\text{i.e., 1:2.}\end{array}$

Q.12 Find the ratio in which the YZ-plane divides the line segment formed by joining the points (–2, 4, 7) and (3, –5, 8).

Ans

$\begin{array}{l}\mathrm{Plane}\text{\hspace{0.17em}}\mathrm{YZ}\text{divides the line segment joining the points}\\ \mathrm{Q}\left(-2,\text{}4,\text{\hspace{0.17em}\hspace{0.17em}}7\right)\mathrm{}\text{\hspace{0.17em}and}\mathrm{R}\left(3,-5,\text{\hspace{0.17em}\hspace{0.17em}}8\right)\text{in the ratio k:1 internally.Then,}\\ \text{let coordinates of point P in YZ plane:}\left(0,\mathrm{y},\mathrm{z}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{x}=\frac{{\mathrm{mx}}_{2}+{\mathrm{nx}}_{1}}{\mathrm{m}+\mathrm{n}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}0=\frac{\mathrm{k}\left(3\right)+1\left(-2\right)}{\mathrm{k}+1}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}0=3\text{\hspace{0.17em}}\mathrm{k}-2\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}2=3\text{\hspace{0.17em}}\mathrm{k}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{k}=\frac{2}{3}\\ \mathrm{The}\text{required ratio is}\frac{2}{3}:1\text{i.e., 2:3.}\end{array}$

Q.13

$\begin{array}{l}\mathbf{\text{Using section formula, show that the points A(2,}}\mathbf{-}\mathbf{\text{3, 4), B(}}\mathbf{-}\mathbf{\text{1, 2, 1)}}\\ \mathbf{\text{and C}}\left(\text{0,}\frac{\text{1}}{\text{3}}\text{,2}\right)\mathbf{\text{are collinear.}}\end{array}$

Ans

$\begin{array}{l}\text{The given points are A(2,}-\text{3, 4), B(}-\text{1, 2, 1) and C}\left(\text{0,}\frac{\text{1}}{\text{3}}\text{, 2}\right)\text{.}\\ \text{Let a point P divides AB​in the ratio k:1.}\\ \text{Hence, by the section formula, the coordinates of P ​are:}\\ \left(\frac{{\text{mx}}_{\text{2}}{\text{+nx}}_{\text{1}}}{\text{m+n}}\text{,}\frac{{\text{my}}_{\text{2}}{\text{+ny}}_{\text{1}}}{\text{m+n}}\text{,}\frac{{\text{mz}}_{\text{2}}{\text{+nz}}_{\text{1}}}{\text{m+n}}\right)\\ \left(\frac{\text{k}\left(-\text{1}\right)\text{+2}}{\text{k+1}}\text{,}\frac{\text{k}\left(\text{2}\right)-\text{3}}{\text{k+1}}\text{,}\frac{\text{k}\left(\text{1}\right)\text{+4}}{\text{k+1}}\right)\\ \text{Now, we find the value of k at which P coincides with point C.}\\ \text{So, by taking}\frac{-\text{k+2}}{\text{k+1}}=0\\ ⇒\text{k =2}\\ \text{For k=2, the coordinates of P ​are:}\left(\text{0,}\frac{\text{1}}{\text{3}}\text{, 2}\right)\\ \text{Therefore, C}\left(\text{0,}\frac{\text{1}}{\text{3}}\text{,2}\right)\text{​​ \hspace{0.17em}\hspace{0.17em}is a point that divides AB ​externally in ratio 2:1.}\\ \text{Hence, points A, B​ and C are collinear.}\end{array}$

Q.14 Find the coordinates of the points which trisect the line segment joining the points P(4, 2, – 6) and Q(10, –16, 6).

Ans

Let points A and B trisect the line segment joining the points P(4, 2, – 6) and Q(10, –16, 6). Point A divides the line segment in the ratio of 1:2 and point B divides the line segment in the ratio of 2:1.

$\begin{array}{l}\mathrm{coordinates}\text{of point A}=\left\{\frac{1\left(10\right)+2\left(4\right)}{1+2},\frac{1\left(-16\right)+2\left(2\right)}{1+2},\frac{1\left(6\right)+2\left(-\text{\hspace{0.17em}}6\right)}{1+2}\right\}\\ =\left(\frac{10+8}{3},\frac{-16+4}{3},\frac{6-12}{3}\right)\\ =\left(6,-4,-2\right)\\ \mathrm{coordinates}\text{of point B}=\left\{\frac{2\left(10\right)+1\left(4\right)}{1+2},\frac{2\left(-16\right)+1\left(2\right)}{1+2},\frac{2\left(6\right)+1\left(-\text{\hspace{0.17em}}6\right)}{1+2}\right\}\\ =\left(\frac{20+4}{3},\frac{-32+2}{3},\frac{12-6}{3}\right)\\ =\left(8,-10,2\right)\\ \mathrm{Thus},\text{points}\left(6,-4,-2\right)\text{and}\left(8,-10,2\right)\text{trisect the line segment}\\ \text{formed by the given points.}\end{array}$

Q.15 Three vertices of a parallelogram ABCD are A(3, – 1, 2), B (1, 2, – 4) and C (– 1, 1, 2). Find the coordinates of the fourth vertex.

Ans

$\begin{array}{l}\text{Three vertices of a parallelogram ABCD are}\mathrm{}\text{\hspace{0.17em}A}\left(\text{3},-\text{1},\text{2}\right),\text{}\\ \text{B}\left(\text{1},\text{2},-\text{\hspace{0.17em}4}\right)\text{and C}\left(-\text{1},\text{1},\text{2}\right).\text{\hspace{0.17em}}\mathrm{L}\text{et fourth vertex be D}\left(\mathrm{a},\mathrm{b},\mathrm{c}\right).\\ \mathrm{Since},\text{diagonals of a parallelogram bisect each other.}\\ \text{So, \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} Mid-point of AC}=\mathrm{Mid}\text{–}\mathrm{point}\text{of BD}\\ ⇒\left(\frac{3-1}{2},\frac{-1+1}{2},\frac{2+2}{2}\right)=\left(\frac{1+\mathrm{a}}{2},\frac{2+\mathrm{b}}{2},\frac{-4+\mathrm{c}}{2}\right)\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left(\frac{2}{2},\frac{0}{2},\frac{4}{2}\right)=\left(\frac{1+\mathrm{a}}{2},\frac{2+\mathrm{b}}{2},\frac{-4+\mathrm{c}}{2}\right)\\ ⇒\frac{2}{2}=\frac{1+\mathrm{a}}{2},\text{\hspace{0.17em}}\frac{0}{2}=\frac{2+\mathrm{b}}{2}\text{and}\frac{4}{2}=\frac{-4+\mathrm{c}}{2}\\ ⇒2=1+\mathrm{a},\text{\hspace{0.17em}}0=2+\mathrm{b}\text{and 4}=-\text{\hspace{0.17em}}4+\mathrm{c}\\ ⇒1=\mathrm{a},\text{\hspace{0.17em}}-2=\mathrm{b}\text{and c}=\text{8}\\ \mathrm{Therefore},\text{the coordinates of vertes D is}\left(1,-2,8\right).\end{array}$

Q.16 Find the lengths of the medians of the triangle with vertices A (0, 0, 6), B (0, 4, 0) and C (6, 0, 0).

Ans

$\begin{array}{l}\text{The triangle with vertices A}\left(\text{0, 0, 6}\right)\text{, B}\left(\text{0, 4, 0}\right)\text{and C}\left(\text{6, 0, 0}\right)\text{}\\ \text{is given.}\\ \text{Mid-point of AB}=\left(\frac{0+0}{2},\frac{0+4}{2},\frac{6+0}{2}\right)\end{array}$ $\begin{array}{l}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\left(0,2,3\right)\\ \text{Mid-point of BC}=\left(\frac{0+6}{2},\frac{4+0}{2},\frac{0+0}{2}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\left(3,2,0\right)\\ \text{Mid-point of CA}=\left(\frac{6+0}{2},\frac{0+0}{2},\frac{0+6}{2}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\left(3,0,3\right)\end{array}$ $\begin{array}{l}\end{array}$ $\begin{array}{l}\mathrm{Length}\text{of median AD}=\sqrt{{\left(0-3\right)}^{2}+{\left(0-2\right)}^{2}+{\left(6-0\right)}^{2}}\\ \text{\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\sqrt{9+4+36}\\ \text{\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=7\\ \mathrm{Length}\text{of median BE}=\sqrt{{\left(0-3\right)}^{2}+{\left(4-0\right)}^{2}+{\left(0-3\right)}^{2}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\sqrt{9+16+9}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\sqrt{34}\\ \mathrm{Length}\text{of median CF}=\sqrt{{\left(6-0\right)}^{2}+{\left(0-2\right)}^{2}+{\left(0-3\right)}^{2}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\sqrt{36+4+9}\end{array}$ $\begin{array}{l}\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\end{array}$ $\begin{array}{l}=\sqrt{49}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=7\\ \mathrm{Thus},\text{the length of medians are 7,}\sqrt{34}\text{and 7 unit.}\end{array}$

Q.17 If the origin is the centroid of the triangle PQR with vertices P(2a, 2, 6), Q(– 4, 3b, –10) and R(8, 14, 2c), then find the values of a, b and c.

Ans

$\begin{array}{l}\mathrm{Since},\text{the origin}\left(0,0,0\right)\text{is the centroid of the triangle PQR}\mathrm{}\text{with}\\ \text{vertices P}\left(\text{2a},\text{2},\text{6}\right),\text{Q}\left(-\text{\hspace{0.17em}4},\text{3b},-\text{1}0\right)\text{and R}\left(\text{8},\text{14},\text{2c}\right).\\ \mathrm{Then},\\ \left\{\frac{2\mathrm{a}+\left(-4\right)+8}{3},\frac{2+3\mathrm{b}+14}{3},\frac{6+\left(-10\right)+2\mathrm{c}}{3}\right\}=\left(0,0,0\right)\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left\{\frac{2\mathrm{a}-4+8}{3},\frac{2+3\mathrm{b}+14}{3},\frac{6-10+2\mathrm{c}}{3}\right\}=\left(0,0,0\right)\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left\{\frac{2\mathrm{a}+4}{3},\frac{3\mathrm{b}+16}{3},\frac{2\mathrm{c}-4}{3}\right\}=\left(0,0,0\right)\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{2\mathrm{a}+4}{3}=0,\frac{3\mathrm{b}+16}{3}=0\text{and}\frac{2\mathrm{c}-4}{3}=0\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{a}=-\text{\hspace{0.17em}}2,\text{\hspace{0.17em}}\mathrm{b}=-\frac{16}{3}\text{and c}=2\end{array}$

Q.18

$\begin{array}{l}\text{Find the coordinates of a point on\hspace{0.17em}\hspace{0.17em}y-axis which\hspace{0.17em}are at a distance of}\\ \text{5}\sqrt{\text{2}}\text{\hspace{0.17em}\hspace{0.17em}from\hspace{0.17em}the point P}\left(\text{3,}-\text{2,5}\right)\text{.}\end{array}$

Ans

$\mathrm{Let}\text{the point on y-axis be Q}\left(0,\mathrm{b},0\right)\text{and distance of Q from the point P}\left(3,-2,5\right)\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{is}\text{5}\sqrt{2}.\text{Then,}$ $\begin{array}{l}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{PQ}=\sqrt{{\left(3-0\right)}^{2}+{\left(-2-\mathrm{b}\right)}^{2}+{\left(5-0\right)}^{2}}\\ ⇒\text{\hspace{0.17em}5}\sqrt{2}=\sqrt{9+{\left(-2-\mathrm{b}\right)}^{2}+25}\\ \mathrm{Squaring}\text{both sides, we get}\\ ⇒\text{50}=34+{\left(-2-\mathrm{b}\right)}^{2}\\ ⇒16={\left(2+\mathrm{b}\right)}^{2}\\ ⇒{4}^{2}={\left(2+\mathrm{b}\right)}^{2}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}4=2+\mathrm{b}\text{or}-\text{4}=2+\mathrm{b}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{b}=2,\text{\hspace{0.17em}}-6\\ \mathrm{Therefore},\text{coordinates of the point Q\hspace{0.17em}on y-axis are}\\ \text{}\left(0,2,0\right)\text{\hspace{0.17em}\hspace{0.17em}or\hspace{0.17em}\hspace{0.17em}}\left(0,-6,0\right).\end{array}$

Q.19 A point R with x-coordinate 4 lies on the line segment joining the points P(2, –3, 4) and Q (8, 0, 10). Find the coordinates of the point R.

Ans

$\begin{array}{l}\mathrm{The}\text{coordinates of points P}\left(\text{2},-\text{3},\text{4}\right)\text{and}\mathrm{}\text{\hspace{0.17em}Q}\left(\text{8},0,\text{1}0\right)\text{of line}\\ \text{segment PQ. Let x-cordinate 4 divides PQ in the ratio of k:1.}\\ \text{Then, x}=\frac{{\mathrm{mx}}_{2}+{\mathrm{nx}}_{1}}{\mathrm{m}+\mathrm{n}},\text{\hspace{0.17em}\hspace{0.17em}y}=\frac{{\mathrm{my}}_{2}+{\mathrm{ny}}_{1}}{\mathrm{m}+\mathrm{n}},\text{z}=\frac{{\mathrm{mz}}_{2}+{\mathrm{nz}}_{1}}{\mathrm{m}+\mathrm{n}}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}4=\frac{\mathrm{m}\left(8\right)+\mathrm{n}\left(2\right)}{\mathrm{m}+\mathrm{n}}\\ ⇒4\left(\mathrm{m}+\mathrm{n}\right)=8\text{\hspace{0.17em}}\mathrm{m}+2\text{\hspace{0.17em}}\mathrm{n}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}4\mathrm{m}+4\mathrm{n}=8\text{\hspace{0.17em}}\mathrm{m}+2\text{\hspace{0.17em}}\mathrm{n}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}2\mathrm{n}=4\mathrm{m}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{\mathrm{m}}{\mathrm{n}}=\frac{2}{4}\end{array}$ $\begin{array}{l}⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{m}:\mathrm{n}=1:2\\ \therefore \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} y}=\frac{1\left(0\right)+2\left(-3\right)}{1+2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{-6}{3}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=-2\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} z}=\frac{1\left(10\right)+2\left(4\right)}{1+2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{18}{3}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=6\\ \mathrm{Therefore},\text{the coordinates of point R are}\left(4,-2,6\right).\end{array}$

Q.20 If A and B be the points (3, 4, 5) and (–1, 3, –7), respectively, find the equation of the set of points P such that PA2 + PB2 = k2, where k is a constant.

Ans

$\begin{array}{l}\text{Let the coordinates of point P be}\left(\text{x,y,z}\right)\text{.}\\ \text{Coordinates of point A and B are}\left(3,4,5\right)\text{and}\\ \left(-1,3,-7\right)\text{respectively.}\\ {\text{\hspace{0.17em}\hspace{0.17em}PA}}^{\text{2}}={\left(\mathrm{x}-3\right)}^{2}+{\left(\mathrm{y}-4\right)}^{2}+{\left(\mathrm{z}-5\right)}^{2}\\ ={\mathrm{x}}^{2}-6\mathrm{x}+9+{\mathrm{y}}^{2}-8\mathrm{y}+16+{\mathrm{z}}^{2}-10\mathrm{z}+25\\ ={\mathrm{x}}^{2}+{\mathrm{y}}^{2}+{\mathrm{z}}^{2}-6\mathrm{x}-8\mathrm{y}-10\mathrm{z}+50\\ {\text{\hspace{0.17em}\hspace{0.17em}PB}}^{\text{2}}={\left(\mathrm{x}+1\right)}^{2}+{\left(\mathrm{y}-3\right)}^{2}+{\left(\mathrm{z}+7\right)}^{2}\\ ={\mathrm{x}}^{2}+2\mathrm{x}+1+{\mathrm{y}}^{2}-6\mathrm{y}+9+{\mathrm{z}}^{2}+14\text{\hspace{0.17em}}\mathrm{z}+49\\ ={\mathrm{x}}^{2}+{\mathrm{y}}^{2}+{\mathrm{z}}^{2}+2\mathrm{x}-6\mathrm{y}+14\text{\hspace{0.17em}}\mathrm{z}+59\\ \text{According to given condition:}\\ {\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}PA}}^{\text{2}}+{\mathrm{PB}}^{2}={\mathrm{k}}^{\text{2}}\\ {\mathrm{x}}^{2}+{\mathrm{y}}^{2}+{\mathrm{z}}^{2}-6\mathrm{x}-8\mathrm{y}-10\mathrm{z}+50+{\mathrm{x}}^{2}+{\mathrm{y}}^{2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}+\text{\hspace{0.17em}}{\mathrm{z}}^{2}+2\mathrm{x}-6\mathrm{y}+14\text{\hspace{0.17em}}\mathrm{z}+59={\mathrm{k}}^{2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}2{\mathrm{x}}^{2}+2{\mathrm{y}}^{2}+2{\mathrm{z}}^{2}-4\mathrm{x}-14\mathrm{y}+4\mathrm{z}+109={\mathrm{k}}^{2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{x}}^{2}+{\mathrm{y}}^{2}+{\mathrm{z}}^{2}-2\mathrm{x}-7\mathrm{y}+2\mathrm{z}=\frac{{\mathrm{k}}^{2}-109}{2},\text{\hspace{0.17em}}\\ \mathrm{which}\text{is the required equation.}\end{array}$

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