NCERT Solutions for Class 11 Maths Chapter 11 Exercise 11.2 - Introduction to Three Dimensional Geometry

NCERT Solutions for Class 11 Maths Chapter 11 Exercise 11.2 - Introduction to Three Dimensional Geometry

Komal MiglaniUpdated on 06 May 2025, 04:30 PM IST

Have you ever noticed how a bird flies, not just sideways or forward, but also upward? To describe its position accurately, we need all three directions- length, width, and height. That is the core of three-dimensional geometry, which uses the three coordinate axes (x, y, and z) to teach us about the points, lines, and shapes in space. This concept is essential for understanding how far objects are from each other in 3D space and has wide applications in physics, engineering, architecture, animation, and 3D modeling.

In NCERT, you will learn how to calculate the distance between two points in space using their coordinates. The NCERT Solutions for Chapter 11 Exercise 11.2 will simplify this process by offering detailed explanations and solved examples that will be helpful in building your 3D concepts. If you are looking for NCERT Solutions, scroll down.

This Story also Contains

  1. Class 11 Maths Chapter 11 Exercise 11.2 Solutions - Download PDF
  2. NCERT Solutions Class 11 Maths Chapter 11 Exercise 11.2
  3. Topics covered in Chapter 11 Introduction to Three-Dimensional Geometry Exercise 11.2
  4. Class 11 Subject-Wise Solutions
NCERT Solutions for Class 11 Maths Chapter 11 Exercise 11.2 - Introduction to Three Dimensional Geometry
12.2

Class 11 Maths Chapter 11 Exercise 11.2 Solutions - Download PDF

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NCERT Solutions Class 11 Maths Chapter 11 Exercise 11.2

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

Answer:

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

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

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

$d=\sqrt{20}$

$d=2\sqrt{5}$

Question 1: (ii) Find the distance between the following pairs of points: (–3, 7, 2) and (2, 4, –1)

Answer:

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

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

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

$d=\sqrt{43}$

Question 1: (iii) Find the distance between the following pairs of points:(–1, 3, – 4) and (1, –3, 4)

Answer:

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

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

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

$d=\sqrt{104}$

$d=2\sqrt{26}$

Question 1: (iv) Find the distance between the following pairs of points: (2, –1, 3) and (–2, 1, 3).

Answer:

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

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

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

$d=\sqrt{20}$

$d=2\sqrt{5}$

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

Answer:

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.

Question 3: (i) Verify the following: (0, 7, –10), (1, 6, – 6) and (4, 9, – 6) are the vertices of an isosceles triangle.

Answer:

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.

Answer:

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.

Question 3:(iii) Verify the following: (–1, 2, 1), (1, –2, 5), (4, –7, 8) and (2, –3, 4) are the vertices of a parallelogram.

Answer:

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.

Question 4: Find the equation of the set of points which are equidistant from the points (1, 2, 3) and (3, 2, –1).

Answer:

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 $x-2z=0$.

Question 5: 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

Answer:

Given,

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

Let the point P(x,y,z) be a point 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 sides:

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

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

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

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

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

Also Read

Topics covered in Chapter 11 Introduction to Three-Dimensional Geometry Exercise 11.2

1. Distance formula in three-dimensional space
The distance formula is used to find the straight-line distance between two points in 3D space. You can measure how far two objects are from each other in space. It is given by

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

2. Derivation of the 3D distance formula
The formula is derived using the Pythagorean Theorem in three steps- first between x and y, then y and finally z coordinates. It extends the 2D formula by including the difference in z-coordinates. This will give the actual length of the diagonal connecting the two points in space.

3. Finding the distance between two points in space
We will use the 3D distance formula by substituting the coordinates of the two points. It will give the shortest distance between them.

4. Application of the distance formula in solving numerical problems
The formula will help to solve geometry problems like finding lengths of edges in 3D shapes or locating objects.

Also Read

Class 11 Subject-Wise Solutions

The table below contains the links for the NCERT exemplar and exercise solutions, which will help you in understanding the concepts.

NCERT Solutions of Class 11 Subject Wise

Subject Wise NCERT Exampler Solutions


Frequently Asked Questions (FAQs)

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

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

Q: What is the weightage of introduction to three dimensional geometry in CBSE Class 11 Maths ?
A:

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

Q: What is the weightage of introduction to three dimensional geometry in the JEE Main exam ?
A:

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

Q: Is our world is 3-D ?
A:

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

Q: What is the use of three dimensional geometry ?
A:

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.

Q: Who is the father of geometry ?
A:

Euclid is know as the father of geometry.

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