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

Edited By Komal Miglani | Updated on May 06, 2025 04:31 PM IST

Have you ever looked at a building or a Rubik’s cube and wondered how we can describe its exact position in space? This is where we need three-dimensional geometry! 3D geometry uses three axes- x, y and z to describe the position of points in space. This exercise will put together all the important concepts you have studied so far like finding the distance between two points, identifying coordinates and working with 3D figures. These questions are simple yet crucial for understanding real-world applications.

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NCERT Solutions Class 11 Maths Chapter 11 Miscellaneous Exercise
Topics covered in Chapter 11 Introduction to Three-dimensional Geometry: Miscellaneous Exercise
NCERT Solutions of Class 11 Subject Wise
Subject-Wise NCERT Exemplar Solutions

The NCERT Solutions for Chapter 11 Miscellaneous Exercise is a powerful guide for students who are preparing for their exams. These NCERT solutions will offer detailed explanations that will help you improve your speed and accuracy. The solutions are a great resource to help you master the topics and for quick revisions. Students can also access NCERT notes for feasible learning.

NCERT Solutions Class 11 Maths Chapter 11 Miscellaneous Exercise

Question 1: 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.

Answer:

Given

The three vertices of the parallelogram ABCD are

A(3, – 1, 2), B (1, 2, – 4) and C (– 1, 1, 2).

Let the fourth vertex be (a, b, c )

Now, as we know, the concept that the diagonals of the parallelogram bisect each other,

Hence, here in parallelogram ABCD, the midpoint of the line segment AC will be equal to the midpoint of the line segment BD.

So,

$(\frac{3 - 1}{2}, \frac{- 1 + 1}{2}, \frac{2 + 2}{2}) = (\frac{1 + a}{2}, \frac{2 + b}{2}, \frac{- 4 + c}{2})$

On comparing both points, we get

$a = 1, b = - 2 a n d c = 8$

Hence the fourth vertex of the is (1,-2,8)

Question 2: Find the lengths of the medians of the triangle with vertices A (0, 0, 6), B (0,4, 0) and (6, 0, 0).

Answer:

Given,

Three vertices of the triangle are A (0, 0, 6), B (0,4, 0), and C(6, 0, 0).

Now,

Let D be the midpoint of AB, E be the midpoint of BC, and F be the midpoint of AC.

Vertex of the D =

$(\frac{0 + 0}{2}, \frac{0 + 4}{2}, \frac{6 + 0}{2}) = (0, 2, 3)$

Vertices of E =

$(\frac{0 + 6}{2}, \frac{4 + 0}{2}, \frac{0 + 0}{2}) = (3, 2, 0)$

Vertices of the F =

$(\frac{0 + 6}{2}, \frac{0 + 0}{2}, \frac{6 + 0}{2}) = (3, 0, 3)$

Now, the Medians of the triangle are CD, AE, and BF

So the lengths of the medians are

$C D = \sqrt{}$ $C D = \sqrt{(6 - 0)^{2} + (0 - 2)^{2} + (0 - 3)^{2}} = \sqrt{36 + 4 + 9} = \sqrt{49} = 7$

$A E = \sqrt{(0 - 3)^{2} + (0 - 2)^{2} + (6 - 0)^{2}} = \sqrt{9 + 4 + 36} = \sqrt{49} = 7$

$B F = \sqrt{(3 - 0)^{2} + (0 - 4)^{2} + (3 - 0)^{2}} = \sqrt{9 + 16 + 9} = \sqrt{34}$

Hence, the median lengths are $7, 7 a n d \sqrt{34}$ .

Question 3: 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.

Answer:

Given,

Triangle PQR with vertices P (2a, 2, 6), Q (– 4, 3b, –10) and R(8, 14, 2c),

Now, as we know,

The centroid of a triangle is given by

$(\frac{x_{1} + x_{2} + x_{3}}{3}, \frac{y_{1} + y_{2} + y_{3}}{3}, \frac{z_{1} + z_{2} + z_{3}}{3})$

Where coordinates of vertices of the triangle are $(x_{1}, x_{2}, x_{3}), (y_{1}, y_{2}, y_{3}) a n d (z_{1}, z_{2}, z_{3})$

Since the Centroid of the triangle, PQR, is the origin (0,0,0),

$(\frac{2 a - 4 + 8}{3}, \frac{2 + 3 b + 14}{3}, \frac{6 - 10 + 2 c}{3}) = (0, 0, 0)$

On equating both coordinates, we get

$a = - 2, b = \frac{- 16}{3} a n d c = 2$

Question 4: If A and B are the points (3, 4, 5) and (–1, 3, –7), respectively, find the equation of the set of points P such that $P A^{2} + P B^{2} = k^{2}$ , where k is a constant.

Answer:

Given Points,

A (3, 4, 5) and B (–1, 3, –7),

Let the coordinates of point P be (x,y,z)

Now,

Given condition :

$P B^{2} = (x - (- 1))^{2} + (y - (3))^{2} + (z - (- 7))^{2}$

$P B^{2} = x^{2} + 2 x + 1 + y^{2} - 6 y + 9 + z^{2} + 14 z + 49$

$P B^{2} = x^{2} + y^{2} + z^{2} + 2 x - 6 y + 14 z + 59$

And

$P A^{2} = (x - 3)^{2} + (y - 4)^{2} + (z - 5)^{2}$

$P A^{2} = x^{2} + y^{2} + z^{2} - 6 x - 8 y - 10 z + 50$

Now, Given the Condition

$P A^{2} + P B^{2} = k^{2}$

$x^{2} + y^{2} + z^{2} + 2 x - 6 y + 14 z + 59 + x^{2} + y^{2} + z^{2} - 6 x - 8 y - 10 z + 50 = k^{2}$

$2 x^{2} + 2 y^{2} + 2 z^{2} - 4 x - 14 y + 4 z + 109 = k^{2}$

Hence Equation of the set of points P is

$2 x^{2} + 2 y^{2} + 2 z^{2} - 4 x - 14 y + 4 z + 109 = k^{2}$ .

Also read

Topics covered in Chapter 11 Introduction to Three-dimensional Geometry: Miscellaneous Exercise

1. Coordinate System in 3D Space
In 3D geometry, a point is represented by an ordered triplet (x, y, z) where x, y and z are the distances from the three mutually perpendicular axes- X-axis, Y-axis, and Z-axis.

2. Distance Formula in 3D
The distance between two points $A (x_{1}, y_{1}, z_{1})$ and $B (x_{2}, y_{2}, z_{2})$ is given by
$\sqrt{{(x_{2} - x_{1})}^{2} + {(y_{2} - y_{1})}^{2} + {(z_{2} - z_{1})}^{2}}$
Using the above formula you can find how far two points are from each other in space.
3. Section Formula in 3D
It is used to find the coordinates of a point dividing a line segment joining two points in a given ratio. If a point divides the segment AB internally in ratio $m : n$ then its coordinates are-
$(\frac{m x_{2} + n x_{1}}{m + n}, \frac{m y_{2} + n y_{1}}{m + n}, \frac{m z_{2} + n z_{1}}{m + n})$

4. Midpoint Formula in 3D
The midpoint of a line joining $A (x_{1}, y_{1}, z_{1})$ and $B (x_{2}, y_{2}, z_{2})$ is given by
$(\frac{x_{1} + x_{2}}{2}, \frac{y_{1} + y_{2}}{2}, \frac{z_{1} + z_{2}}{2})$

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NCERT Solutions of Class 11 Subject Wise

Go through the links below and get solved NCERT exercise questions for all the subjects.

NCERT Solutions for Class 11 Maths

NCERT Solutions for Class 11 Physics

NCERT Solutions for Class 11 Chemistry

NCERT Solutions for Class 11 Biology

Subject-Wise NCERT Exemplar Solutions

Upgrade your exam preparations by following our NCERT exemplar solutions that are designed to give you a deeper understanding of the concepts.

NCERT Exemplar Solutions for Class 11 Maths

NCERT Exemplar Solutions for Class 11 Physics

NCERT Exemplar Solutions for Class 11 Chemistry

NCERT Exemplar Solutions for Class 11 Biology

Frequently Asked Questions (FAQs)

1. What is the weighatge of the three-dimensional geometry in the JEE Main exam ?

Generally, two questions from three-dimensional geometry are asked in the JEE Main exam which means it has 6.6% weightage in the JEE Main maths exam.

2. Which subject has more weightage in the JEE Main exam ?

All three subjects (Physics, Chemistry, Maths) have an equal weightage in the JEE Main exam.

3. What is weightage of Class 11 and Class 12 in JEE Main exam ?

Class 11 and Class 12 has almost equal weightage in the JEE Main exam.

4. What is weightage of modern Physics in the JEE Main exam ?

Modern Physics has 20-25% weightage in the JEE Main Physics exam.

5. What is weightage of mole concept in JEE Main Chemistry exam ?

Generally, one question is asked from the mole concept in JEE Main Chemistry exam.

6. Do we need to buy NCERT solutions book ?

No, you don't need to buy an NCERT solutions book for any class. You can find NCERT solutions online for free.

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NCERT Exemplar Class 11 Chemistry Solutions Chapter 1 Some Basic Concepts of Chemistry

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