NCERT Solutions for Exercise 12.2 Class 11 Maths Chapter 12 - Introduction to Three Dimensional Geometry

# NCERT Solutions for Exercise 12.2 Class 11 Maths Chapter 12 - Introduction to Three Dimensional Geometry

Edited By Vishal kumar | Updated on Nov 14, 2023 09:23 AM IST

## NCERT Solutions for Class 11 Maths Chapter 12: Introduction to Three Dimensional Geometry Exercise 12.2- Download Free PDF

NCERT Solutions for Class 11 Maths Chapter 12: Exercise 12.2- In the previous exercise, you have already learned about the coordinate axes, coordinate planes, and coordinates of points in three-dimensional geometry. In the NCERT solutions for Class 11 Maths chapter 12 exercise 12.2, you will learn to find the distance between two points in three-dimensional geometry. You already know the distance formula for points in two-dimensional geometry. The distance formula for points in three-dimensional geometry is very similar to two-dimensional geometry, so you will easily understand this concept.

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You need to solve problems from NCERT book exercise 12.2 Class 11 Maths to get conceptual clarity. This ex 12.2 class 11 is very basic and small which won't take much effort to solve but it is very important to understand the upcoming exercises of this chapter. You can go through the Class 11 Maths chapter 12 exercise 12.2 solutions if you are facing any difficulty while solving the problem from Class 11 Maths chapter 12 exercise 12.2. You can click on the NCERT Solutions link if you are looking for NCERT solutions for Classes 6 to 12 at one place.

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**This chapter has been renumbered as Chapter 11 in the CBSE Syllabus for the academic year 2023-24.

## Question:1 (i) Find the distance between the following pairs of points: (2, 3, 5) and (4, 3, 1)

The distance between two points $(x_1,y_1,z_1)$ and $(x_2,y_2,z_2)$ is given by

$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}$

So, distance between (2, 3, 5) and (4, 3, 1) is given by

$d=\sqrt{(4-2)^2+(3-3)^2+(1-5)^2}$

$d=\sqrt{4+0+16}$

$d=\sqrt{20}$

$d=2\sqrt{5}$

The distance between two points $(x_1,y_1,z_1)$ and $(x_2,y_2,z_2)$ is given by

$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}$

So, distance between (–3, 7, 2) and (2, 4, –1) is given by

$d=\sqrt{(2-(-3))^2+(4-7)^2+(-1-2)^2}$

$d=\sqrt{25+9+9}$

$d=\sqrt{43}$

The distance between two points $(x_1,y_1,z_1)$ and $(x_2,y_2,z_2)$ is given by

$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}$

So, distance between (–1, 3, – 4) and (1, –3, 4) is given by

$d=\sqrt{(1-(-1))^2+(-3-3)^2+(4-(-4))^2}$

$d=\sqrt{4+36+64}$

$d=\sqrt{104}$

$d=2\sqrt{26}$

The distance between two points $(x_1,y_1,z_1)$ and $(x_2,y_2,z_2)$ is given by

$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}$

So, distance between (2, –1, 3) and (–2, 1, 3).) is given by

$d=\sqrt{(-2-(-2))^2+(1-(-1))^2+(3-3)^2}$

$d=\sqrt{16+4+0}$

$d=\sqrt{20}$

$d=2\sqrt{5}$

Given,theree points A=(–2, 3, 5), B=(1, 2, 3) and C=(7, 0, –1)

The distance between two points $(x_1,y_1,z_1)$ and $(x_2,y_2,z_2)$ is given by

$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}$

The distance AB :

$AB=\sqrt{(1-(-2))^2+(2-3)^2+(3-5)^2}$

$AB=\sqrt{(3)^2+(-1)^2+(-2)^2}$

$AB=\sqrt{9+1+4}$

$AB=\sqrt{14}$

The distance BC:

$BC=\sqrt{(7-1)^2+(0-2)^2+(-1-3)^2}$

$BC=\sqrt{36+4+16}$

$BC=\sqrt{56}$

$BC=2\sqrt{14}$

The distance CA

$CA=\sqrt{(7-(-2))^2+(0-3)^2+(-1-5)^2}$

$CA=\sqrt{81+9+36}$

$CA=\sqrt{126}$

$CA=3\sqrt{14}$

As we can see here,

$AB+BC=\sqrt{14}+2\sqrt{14}=3\sqrt{14}=AC$

Hence we can say that point A,B and C are colinear.

Given Three points A=(0, 7, –10), B=(1, 6, – 6) and C=(4, 9, – 6)

The distance between two points $(x_1,y_1,z_1)$ and $(x_2,y_2,z_2)$ is given by

$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}$

The distance AB

$AB=\sqrt{(1-0)^2+(6-7)^2+(-6-(-10))^2}$

$AB=\sqrt{1+1+16}$

$AB=\sqrt{18}$

The distance BC

$BC=\sqrt{(4-1)^2+(9-6)^2+(-6-(-6))^2}$

$BC=\sqrt{9+9+0}$

$BC=\sqrt{18}$

The distance CA

$CA=\sqrt{(4-0)^2+(9-7)^2+(-6-(-10))^2}$

$CA=\sqrt{16+4+16}$

$CA=\sqrt{36}$

$CA=6$

As we can see $AB=BC\neq CA$

Hence we can say that ABC is an isosceles triangle.

Question:3(ii) Verify the following

(0, 7, 10), (- 1, 6, 6) and (- 4, 9, 6) are the vertices of right angled triangle.

Given Three points A=(0, 7, 10), B=(-1, 6, 6) and C=(-4, 9, 6)

The distance between two points $(x_1,y_1,z_1)$ and $(x_2,y_2,z_2)$ is given by

$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}$

The distance AB

$AB=\sqrt{(-1-0)^2+(6-7)^2+(6-10)^2}$

$AB=\sqrt{1+1+16}$

$AB=\sqrt{18}$

The distance BC

$BC=\sqrt{(-4-(-1))^2+(9-6)^2+(6-6)^2}$

$BC=\sqrt{9+9+0}$

$BC=\sqrt{18}$

The distance CA

$CA=\sqrt{(-4-0)^2+(9-7)^2+(6-10)^2}$

$CA=\sqrt{16+4+16}$

$CA=\sqrt{36}$

$CA=6$

As we can see

$(AB)^2+(BC)^2=18+18=36=(CA)^2$

Since this follows pythagorus theorem, we can say that ABC is a right angle triangle.

Given A=(–1, 2, 1), B=(1, –2, 5), C=(4, –7, 8) and D=(2, –3, 4)

Given Three points A=(0, 7, –10), B=(1, 6, – 6) and C=(4, 9, – 6)

The distance between two points $(x_1,y_1,z_1)$ and $(x_2,y_2,z_2)$ is given by

$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}$

The distance AB

$AB=\sqrt{(1-(-1))^2+(-2-2)^2+(5-1)^2}$

$AB=\sqrt{4+16+16}$

$AB=\sqrt{36}$

$AB=6$

The distance BC

$BC=\sqrt{(4-1)^2+(-7-(-2))^2+(8-5)^2}$

$BC=\sqrt{9+25+9}$

$BC=\sqrt{43}$

The distance CD

$CD=\sqrt{(2-4)^2+(-3-(-7))^2+(4-8)^2}$

$CA=\sqrt{4+16+16}$

$CA=\sqrt{36}$

$CA=6$

The distance DA

$DA=\sqrt{(-1-2)^2+(2-(-3))^2+(1-4)^2}$

$DA=\sqrt{9+25+9}$

$DA=\sqrt{43}$

Here As we can see

$AB=6=CA$ And $BC=\sqrt{43}=DA$

As the opposite sides of quadrilateral are equal, we can say that ABCD is a parallelogram.

Given, two points A=(1, 2, 3) and B=(3, 2, –1).

Let the point P= (x,y,z) be a point which is equidistance from the points A and B.

so,

The distance PA= The distance PB

$\sqrt{(x-1)^2+(y-2)^2+(z-3)^2}=\sqrt{(x-3)^2+(y-2)^2+(z-(-1))^2}$

${(x-1)^2+(y-2)^2+(z-3)^2}={(x-3)^2+(y-2)^2+(z-(-1))^2}$

$\left [ (x-1)^2-(x-3)^2 \right ]+\left [ (y-2)^2-(y-2)^2 \right ]+\left [ (z-3)^2-(z+1)^2 \right ]=0$

Now lets apply the simplification property,

$a^2-b^2=(a+b)(a-b)$

$\left [ (2)(2x-4) \right ]+0+\left [ (-4)(2z-2) \right ]=0$

$4x-8-8z+8=0$

$4x-8z=0$

$x-2z=0$

Hence locus of the point which is equidistant from A and B is $x-2z=0$.

Given,

Two points A (4, 0, 0) and B (– 4, 0, 0)

let the point P(x,y,z) be a point sum of whose distance from A and B is 10.

So,

The distance PA+The distance PB=10

$\sqrt{(x-4)^2+(y-0)^2+(z-0)^2}+\sqrt{(x-(-4))^2+(y)^2+(z)^2}=10$

$\sqrt{(x-4)^2+(y)^2+(z)^2}+\sqrt{(x+4)^2+(y)^2+(z)^2}=10$

$\sqrt{(x-4)^2+(y)^2+(z)^2}=10-\sqrt{(x+4)^2+(y)^2+(z)^2}$

Squaring on both side :

${(x-4)^2+(y)^2+(z)^2}=100-20\sqrt{(x+4)^2+(y)^2+(z)^2}+{(x+4)^2+(y)^2+(z)^2}$

${(x-4)^2-(x+4)^2=100-20\sqrt{(x+4)^2+(y)^2+(z)^2}$

$-16x=100-20\sqrt{(x+4)^2+(y)^2+(z)^2}$

$20\sqrt{(x+4)^2+(y)^2+(z)^2}=100+16x$

$5\sqrt{(x+4)^2+(y)^2+(z)^2}=25+4x$

Now again squaring both sides,

$25\left ( {(x+4)^2+(y)^2+(z)^2} \right )=625+200x+16x^2$

$25x^2+200x+400+25y^2+25z^2=625+200x+16x^2$

$9x^2+25y^2+25z^2-225=0$

Hence the equation of the set of points P, the sum of whose distances from A and B is equal to 10 is $9x^2+25y^2+25z^2-225=0$.

## More About NCERT Solutions for Class 11 Maths Chapter 12 Exercise 12.2:-

Class 11 Maths chapter 12 exercise 12.2 consists of questions related to finding the distance between a pair of points in three-dimensional space, checking collinearity of three-point in the three-dimensional space. There are a few examples given before the exercise 12.2 Class 11 Maths that you can solve before solving this exercise.

## Benefits of NCERT Solutions for Class 11 Maths Chapter 12 Exercise 12.2:-

• Class 11 Maths chapter 12 exercise 12.2 solutions are explained in a step-by-step manner which is very easy to understand.
• You can use Class 11 Maths chapter 12 exercise 12.2 solutions for reference while the NCERT problems.
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## Key Features of NCERT Class 11 Maths Ex 12.2 Solution

1. Conceptual Clarity: The ex 12.2 class 11 solutions provide a detailed explanation of the concepts involved in finding the distance between two points in three-dimensional geometry.

2. Step-by-Step Solutions: Each problem in exercise 12.2 class 11 maths is solved step by step, making it easier for students to follow and understand the solution methodology.

3. Application of Formulas: The class 11 ex 12.2 solutions illustrate the application of the distance formula in three-dimensional geometry. The distance formula for points in three-dimensional space is explained in a manner that connects with the familiar formula used in two-dimensional geometry.

4. Emphasis on Basics: While the class 11 maths chapter 12 exercise 12.2 is considered basic, it lays a foundation for understanding more complex concepts in later exercises and chapters. The solutions emphasize the importance of grasping fundamental ideas.

5. Importance of Practice: The class 11 maths ex 12.2 solutions highlight the significance of solving problems from Exercise 12.2 for students to reinforce their understanding and prepare for more advanced exercises.

6. Reference for Difficulties: If students face challenges in solving problems independently, the solutions serve as a reference guide, offering insights into problem-solving approaches.

7. Link to NCERT Syllabus: The 11th class maths exercise 12.2 answers are aligned with the NCERT syllabus for Class 11 Maths, ensuring that students cover the prescribed topics.

8. Comprehensive Resource: For a broader range of NCERT solutions spanning Classes 6 to 12, a link is provided, offering a comprehensive resource for students seeking solutions across various subjects.

## Subject Wise NCERT Exampler Solutions

Happy learning!!!

1. Find the distance between point A( 1,2,0) and point B(1,5,0) ?

The distance AB = ( (1-1)^2 + (5-2)^2 + (0-0)^2)^(1/2) = 3 unit

2. What is the weightage of introduction to three dimensional geometry in CBSE Class 11 Maths ?

The weightage of the unit coordinate geometry is 10 marks in CBSE Class 11 Maths.

3. What is the weightage of introduction to three dimensional geometry in the JEE Main exam ?

Generally two questions are asked in the JEE Main exam which means it has 6.6% weightage in the JEE Main Maths paper.

4. Is our world is 3-D ?

Yes, Our world is 3D and every object around us is in 3D from.

5. What is the use of three dimensional geometry ?

The use of three-dimensional geometry is wide. It is useful in the design of electronics gadgets to the construction of the building, dams, rivers, roads, temples, etc.

6. Who is the father of geometry ?

Euclid is know as the father of geometry.

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