Home > NCERT Solutions > NCERT Solutions For Class 11 Maths Miscellaneous Exercise Chapter 11 Introduction To Three Dimensional Geometry

NCERT Solutions For Class 11 Maths Miscellaneous Exercise Chapter 11 Introduction To Three Dimensional Geometry

NCERT Solutions for Class 11 Maths Chapter 11 Exercise 11.2 – Introduction to Three Dimensional Geometry covers the distance formula in 3D space and its applications. This exercise includes problems on finding the distance between two points in 3D, verifying collinearity of points, identifying types of triangles and quadrilaterals, and finding equations of sets of points satisfying given distance conditions.

Prepared according to the latest CBSE Class 11 Maths syllabus 2025-26, Exercise 11.2 helps students apply the 3D distance formula to solve a variety of geometric problems. These concepts are foundational for understanding vectors and 3D geometry in Class 12.

NCERT Solutions For Class 11 Maths Miscellaneous Exercise Chapter 11 Introduction To Three Dimensional Geometry

Introduction to Three Dimensional Geometry

Easy

Q.

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

Introduction to Three Dimensional Geometry

Medium

Q.

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

Introduction to Three Dimensional Geometry

Difficult

Q.

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.

Introduction to Three Dimensional Geometry

Difficult

Q.

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

Introduction to Three Dimensional Geometry

Medium

Q.

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.

Introduction to Three Dimensional Geometry

Difficult

Q.

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

Introduction to Three Dimensional Geometry

Medium

Q.

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.

Introduction to Three Dimensional Geometry

Difficult

Q.

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

Introduction to Three Dimensional Geometry

Medium

Q.

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

Introduction to Three Dimensional Geometry

Medium

Q.

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.

Introduction to Three Dimensional Geometry

Medium

Q.

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

Introduction to Three Dimensional Geometry

Medium

Q.

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.

Introduction to Three Dimensional Geometry

Medium

Q.

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.

Introduction to Three Dimensional Geometry

Medium

Q.

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

Introduction to Three Dimensional Geometry

Difficult

Q.

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.

Introduction to Three Dimensional Geometry

Difficult

Q.

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

Introduction to Three Dimensional Geometry

Medium

Q.

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).

Introduction to Three Dimensional Geometry

Easy

Q.

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.

Introduction to Three Dimensional Geometry

Medium

Q.

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).

Introduction to Three Dimensional Geometry

Medium

Q.

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.

NCERT Solutions For Class 11 Maths Miscellaneous Exercise Chapter 11 Introduction To Three Dimensional Geometry

Q1. 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:
Let the fourth vertex be D(x, y, z).
Property: Diagonals of a parallelogram bisect each other.
Therefore, the midpoint of AC = Midpoint of BD.

Midpoint of AC:
(1, 0, 2)

Midpoint of BD:
(1, −2, 8)

Answer: D = (1, −2, 8)

Q2. Find the lengths of the medians of the triangle with vertices A(0, 0, 6), B(0, 4, 0), and C(6, 0, 0).

Answer:
Median AD = 7
Median BE = √34
Median CF = 7

Answer: The lengths of the medians are 7, √34, and 7.

Q3. If the origin is the centroid of triangle PQR with vertices P(2a, 2, 6), Q(−4, 3b, −10), and R(8, 14, 2c), find the values of a, b, and c.

Answer:
Centroid Formula:
G =

$\left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3}, \frac{z_1 + z_2 + z_3}{3} \right)$

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

For x-coordinate:

$3a + (-4) + 8 = 0 \Rightarrow a = -2$

$3 a + (- 4) + 8 = 0 \Rightarrow a = - 2$

For y-coordinate:

$3b + 16 = 0 \Rightarrow b = -\frac{16}{3}$

$3 b + 16 = 0 \Rightarrow b = - 3 16$

For z-coordinate:

$2c - 4 = 0 \Rightarrow c = 2$

$2 c - 4 = 0 \Rightarrow c = 2$

Answer: a = −2, b = −16/3, c = 2

Q4. If A and B are 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.

Answer:
Let P(x, y, z).

Finding PA²:
PA² =

$(x - 3)² + (y - 4)² + (z - 5)²$

$(x - 3)^{2} + (y - 4)^{2} + (z - 5)^{2}$
= x² + y² + z² - 6x - 8y - 10z + 50

Finding PB²:
PB² =

$(x + 1)² + (y - 3)² + (z + 7)²$

$(x + 1)^{2} + (y - 3)^{2} + (z + 7)^{2}$
= x² + y² + z² + 2x - 6y + 14z + 59

Given:
PA² + PB² = k²

$(x² + y² + z² - 6x - 8y - 10z + 50) + (x² + y² + z² + 2x - 6y + 14z + 59) = k²$

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

$2x² + 2y² + 2z² - 4x - 14y + 4z + 109 = k²$

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

Answer:
x² + y² + z² - 2x - 7y + 2z = 2k² - 109

FAQs – Miscellaneous Exercise

How to find the 4th vertex of a parallelogram?
Use the diagonal bisection property: Midpoint of AC = Midpoint of BD.
What is a median?
A median is the line segment from a vertex to the midpoint of the opposite side.
What is the centroid formula?
The centroid G of a triangle with vertices $(x_1, y_1, z_1)$

$(x_{1}, y_{1}, z_{1})$ ,

$(x_2, y_2, z_2)$

$(x_{2}, y_{2}, z_{2})$ , and

$(x_3, y_3, z_3)$

$(x_{3}, y_{3}, z_{3})$ is given by:

$G = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3}, \frac{z_1 + z_2 + z_3}{3} \right)$

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

What is a locus?
A locus is the set of all points satisfying a given condition.