Class 11 Mathematics

Chapter 11 Solutions — Introduction to Three Dimensional Geometry

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Solutions

Overview

Step-by-step NCERT solutions for Introduction to Three Dimensional Geometry (Chapter 11, NCERT Class 11 Mathematics) — every question and answer worked out in full, not just the final result. You can also read the Introduction to Three Dimensional Geometry textbook chapter.

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What these solutions cover

All 13 questions in Introduction to Three Dimensional Geometry are solved in the PDF. Here's what's inside, exercise by exercise:

Exercise 11.1

A point is on the x-axis. What are its y-coordinate and z-coordinates?
A point is in the XZ-plane. What can you say about its y-coordinate?
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).
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.

Exercise 11.2

1.
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).
Show that the points (–2, 3, 5), (1, 2, 3) and (7, 0, –1) are collinear.
3.
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.
Find the equation of the set of points which are equidistant from the points (1, 2, 3) and (3, 2, –1).
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.

Miscellaneous Exercise

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

Sample

Worked examples

3 questions from Introduction to Three Dimensional Geometry, solved step by step.

Exercise 11.1

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

Any point on the x-axis lies on both the XY-plane and the XZ-plane simultaneously.

A point on the x-axis has the form (x, 0, 0): its perpendicular distance from the ZX-plane is 0 (so y = 0) and its perpendicular distance from the XY-plane is 0 (so z = 0).

Answer: The y-coordinate and z-coordinate are both 0.

Exercise 11.2

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

Let A = (–2, 3, 5), B = (1, 2, 3), C = (7, 0, –1).

AB = √[(1–(–2))² + (2–3)² + (3–5)²]

= √[9 + 1 + 4] = √14

BC = √[(7–1)² + (0–2)² + (–1–3)²]

= √[36 + 4 + 16] = √56 = 2√14

AC = √[(7–(–2))² + (0–3)² + (–1–5)²]

= √[81 + 9 + 36] = √126 = 3√14

Since AB + BC = √14 + 2√14 = 3√14 = AC,

point B lies between A and C on the segment AC.

Therefore, A, B and C are collinear.

Answer: AB = √14, BC = 2√14, AC = 3√14. Since AB + BC = AC, the three points are collinear.

Miscellaneous Exercise

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.

In a parallelogram, the diagonals bisect each other.

So the midpoint of diagonal AC equals the midpoint of diagonal BD.

Midpoint of AC = ((3 + (–1))/2, (–1 + 1)/2, (2 + 2)/2)

= (2/2, 0/2, 4/2) = (1, 0, 2)

Let D = (x, y, z). Midpoint of BD = ((1 + x)/2, (2 + y)/2, (–4 + z)/2).

Setting midpoint of BD = (1, 0, 2):

(1 + x)/2 = 1 ⇒ x = 1

(2 + y)/2 = 0 ⇒ y = –2

(–4 + z)/2 = 2 ⇒ z = 8

Therefore, D = (1, –2, 8).

Verification: Midpoint of BD = ((1+1)/2, (2–2)/2, (–4+8)/2) = (1, 0, 2) = Midpoint of AC. ✓

Answer: The fourth vertex D is (1, –2, 8).

All 13 questions in Introduction to Three Dimensional Geometry are solved step by step in the Solutions PDF above.

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Chapter 11 Solutions — Introduction to Three Dimensional Geometry

Overview

What these solutions cover

Exercise 11.1

Exercise 11.2

Miscellaneous Exercise

Worked examples

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

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

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.

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