 # NCERT Solutions for Class 11 Maths Chapter 12 - Introduction to Three Dimensional Geometry Exercise 12.3

Chapter 12 Introduction to Three Dimensional Geometry of Class 11 Maths is categorised under the CBSE Syllabus for the session 2022-23. Each exercise of this chapter deals with different topics. Exercise 12.3 of NCERT Solutions for Class 11 Maths Chapter 12 – Introduction to Three Dimensional Geometry is based on the topic Section Formula. The section formula gives the coordinates of a point that divides the line joining two points in a ratio, internally or externally.

To learn more about the section formula, practice is mandatory. Hence, solving the problems given in this exercise will provide the students with some questions that can help them understand the concept better. Similarly, practising all the problems of NCERT Solutions for Class 11 Maths will help the students in scoring high marks in the examinations.

## Download the PDF of NCERT Solutions for Class 11 Maths Chapter 12 – Introduction to Three Dimensional Geometry Exercise 12.3     ### Solutions for Class 11 Maths Chapter 12 – Exercise 12.3

1. 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.

Solution:

Let the line segment joining the points P (-2, 3, 5) and Q (1, -4, 6) be PQ.

(i) 2: 3 internally

By using the section formula,

We know that the coordinates of the point R, which divide the line segment joining two points, P (x1, y1, z1) and Q (x2, y2, z2), internally in the ratio m: n, are given by Upon comparing, we have

x1 = -2, y1 = 3, z1 = 5;

x2 = 1, y2 = -4, z2 = 6 and

m = 2, n = 3

So, the coordinates of the point which divide the line segment joining the points P (– 2, 3, 5) and Q (1, – 4, 6) in the ratio 2 : 3 internally are given by Hence, the coordinates of the point which divides the line segment joining the points (-2, 3, 5) and (1, -4, 6) is (-4/5, 1/5, 27/5).

(ii) 2: 3 externally

By using the section formula,

We know that the coordinates of the point R, which divide the line segment joining two points, P (x1, y1, z1) and Q (x2, y2, z2), externally in the ratio m: n, are given by Upon comparing, we have

x1 = -2, y1 = 3, z1 = 5;

x2 = 1, y2 = -4, z2 = 6 and

m = 2, n = 3

So, the coordinates of the point which divide the line segment joining the points P (– 2, 3, 5) and Q (1, – 4, 6) in the ratio 2: 3 externally are given by ∴ The coordinates of the point which divides the line segment joining the points (-2, 3, 5) and (1, -4, 6) are (-8, 17, 3).

2. 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.

Solution:

Let us consider Q divides PR in the ratio k: 1.

By using the section formula,

We know that the coordinates of the point R, which divides the line segment joining two points, P (x1, y1, z1) and Q (x2, y2, z2), internally in the ratio m : n, are given by Upon comparing, we have

x1 = 3, y1 = 2, z1 = -4;

x2 = 9, y2 = 8, z2 = -10 and

m = k, n = 1

So, we have 9k + 3 = 5 (k+1)

9k + 3 = 5k + 5

9k – 5k = 5 – 3

4k = 2

k = 2/4

= ½

Hence, the ratio in which Q divides PR is 1: 2.

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

Solution:

Let the line segment formed by joining the points P (-2, 4, 7) and Q (3, -5, 8) be PQ.

We know that any point on the YZ-plane is of the form (0, y, z).

So now, let R (0, y, z) divide the line segment PQ in the ratio k: 1.

Then,

Upon comparing, we have

x1 = -2, y1 = 4, z1 = 7

x2 = 3, y2 = -5, z2 = 8 and

m = k, n = 1

By using the section formula,

We know that the coordinates of the point R, which divide the line segment joining two points P (x1, y1, z1) and Q (x2, y2, z2) internally in the ratio m: n are given by So we have, 3k – 2 = 0

3k = 2

k = 2/3

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

4. Using the section formula, show that the points A (2, –3, 4), B (–1, 2, 1) and C (0, 1/3, 2) are collinear.

Solution:

Let the point P divides AB in the ratio k: 1.

Upon comparing, we have

x1 = 2, y1 = -3, z1 = 4;

x2 = -1, y2 = 2, z2 = 1 and

m = k, n = 1

By using the section formula,

We know that the coordinates of the point R, which divide the line segment joining two points P (x1, y1, z1) and Q (x2, y2, z2) internally in the ratio m : n are given by So we have, Now, we check if, for some value of k, the point coincides with point C.

Put (-k+2)/(k+1) = 0

-k + 2 = 0

k = 2

When k = 2, then (2k-3)/(k+1) = (2(2)-3)/(2+1)

= (4-3)/3

= 1/3

And, (k+4)/(k+1) = (2+4)/(2+1)

= 6/3

= 2

∴ C (0, 1/3, 2) is a point which divides AB in the ratio 2: 1 and is the same as P.

Hence, A, B, C are collinear.

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

Solution:

Let A (x1, y1, z1) and B (x2, y2, z2) trisect the line segment joining the points P (4, 2, -6) and Q (10, -16, 6).

A divides the line segment PQ in the ratio 1: 2.

Upon comparing, we have

x1 = 4, y1 = 2, z1 = -6

x2 = 10, y2 = -16, z2 = 6 and

m = 1, n = 2

By using the section formula,

We know that the coordinates of the point R, which divide the line segment joining two points, P (x1, y1, z1) and Q (x2, y2, z2), internally in the ratio m : n, are given by So we have, Similarly, we know that B divides the line segment PQ in the ratio 2: 1.

Upon comparing, we have

x1 = 4, y1 = 2, z1 = -6;

x2 = 10, y2 = -16, z2 = 6 and

m = 2, n = 1

By using the section formula,

We know that the coordinates of the point R, which divide the line segment joining two points P (x1, y1, z1) and Q (x2, y2, z2) internally in the ratio m: n are given by So we have, ∴ The coordinates of the points which trisect the line segment joining the points P (4, 2, – 6) and Q (10, –16, 6) are (6, -4, -2) and (8, -10, 2).

### Access Other Exercise Solutions of Class 11 Maths Chapter 12 – Introduction to Three Dimensional Geometry

Exercise 12.1 Solutions: 4 Questions

Exercise 12.2 Solutions: 5 Questions

Miscellaneous Exercise on Chapter 12 Solutions: 6 Questions