Each exercise deals with different topics. The 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 which divides the line joining two points in a ratio, internally or externally. To learn more about section formula, practice is mandatory. Hence, solving the problems given in this exercise will provide the students with some questions that can help them in understanding the concept better. Similarly, practicing all the problems of NCERT Solutions for Class 11 maths will help the students in scoring high marks in the examinations.

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

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### Solutions for Class 11 Maths Chapter 12 â€“ Exercise 12.3

**1. Find the coordinates of the point which divides the line segment joining the points (â€“ 2, 3, 5) and (1, â€“ 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 section formula,

We know that the coordinates of the point R which divides the line segment joining two points P (x_{1}, y_{1}, z_{1}) and Q (x_{2}, y_{2}, z_{2}) internally in the ratio m: n is given by:

Upon comparing we have

x_{1}Â = -2, y_{1}Â = 3, z_{1}Â = 5;

x_{2}Â = 1, y_{2}Â = -4, z_{2}Â = 6 and

m = 2, n = 3

So, theÂ coordinates of the point which divides the line segment joining the points P (â€“ 2, 3, 5) and Q (1, â€“ 4, 6) in the ratio 2 : 3 internally is 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 section formula,

We know that the coordinates of the point R which divides the line segment joining two points P (x_{1}, y_{1}, z_{1}) and Q (x_{2}, y_{2}, z_{2}) externally in the ratio m: n is given by:

Upon comparing we have

x_{1}Â = -2, y_{1}Â = 3, z_{1}Â = 5;

x_{2}Â = 1, y_{2}Â = -4, z_{2}Â = 6 and

m = 2, n = 3

So, theÂ coordinates of the point which divides the line segment joining the points P (â€“ 2, 3, 5) and Q (1, â€“ 4, 6) in the ratio 2: 3 externally is given by:

âˆ´ The co-ordinates of the point which divides the line segment joining the points (-2, 3, 5) and (1, -4, 6) is (-8, 17, 3).

**2. Given that P (3, 2, â€“ 4), Q (5, 4, â€“ 6) and R (9, 8, â€“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 section formula,

We know that the coordinates of the point R which divides the line segment joining two points P (x_{1}, y_{1}, z_{1}) and Q (x_{2}, y_{2}, z_{2}) internally in the ratio m : n is given by:

Upon comparing we have,

x_{1}Â = 3, y_{1}Â = 2, z_{1}Â = -4;

x_{2}Â = 9, y_{2}Â = 8, z_{2}Â = -10 and

m = k, n = 1

So, we have

9k + 3 = 5 (k+1)Â

9k + 3 = 5k + 5Â

9k â€“ 5k = 5 â€“ 3

4k = 2

k = 2/4

= Â½

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 (â€“2, 4, 7) and (3, â€“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) divides the line segment PQ in the ratio k: 1.

Then,

Upon comparing we have,

x_{1}Â = -2, y_{1}Â = 4, z_{1}Â = 7;

x_{2}Â = 3, y_{2}Â = -5, z_{2}Â = 8 and

m = k, n = 1

By using the section formula,

We know that the coordinates of the point R which divides the line segment joining two points P (x_{1}, y_{1}, z_{1}) and Q (x_{2}, y_{2}, z_{2}) internally in the ratio m: n is given by:

So we have,

3k â€“ 2 = 0

3k = 2

k = 2/3

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

**4. Using section formula, show that the points A (2, â€“3, 4), B (â€“1, 2, 1) and C (0, 1/3, 2)Â are collinear.**

**Solution:**

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

Upon comparing we have,

x_{1}Â = 2, y_{1}Â = -3, z_{1}Â = 4;

x_{2}Â = -1, y_{2}Â = 2, z_{2}Â = 1 and

m = k, n = 1

By using section formula,

We know that the coordinates of the point R which divides the line segment joining two points P (x_{1}, y_{1}, z_{1}) and Q (x_{2}, y_{2}, z_{2}) internally in the ratio m : n is given by:

So we have,

Now, we check if for some value of k, the point coincides with the 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

âˆ´ C (0, 1/3, 2)Â is a point which divides AB in the ratio 2: 1 and is 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, â€“ 6) and Q (10, â€“16, 6).**

**Solution:**

Let A (x_{1}, y_{1}, z_{1}) and B (x_{2}, y_{2}, z_{2}) 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,

x_{1}Â = 4, y_{1}Â = 2, z_{1}Â = -6;

x_{2}Â = 10, y_{2}Â = -16, z_{2}Â = 6 and

m = 1, n = 2

By using the section formula,

We know that the coordinates of the point R which divides the line segment joining two points P (x_{1}, y_{1}, z_{1}) and Q (x_{2}, y_{2}, z_{2}) internally in the ratio m : n is given by:

So we have,

= (18/3, -12/3, -6/3)

= (6, -4, -2)

Similarly, we know that B divides the line segment PQ in the ratio 2: 1.

Upon comparing we have,

x_{1}Â = 4, y_{1}Â = 2, z_{1}Â = -6;

x_{2}Â = 10, y_{2}Â = -16, z_{2}Â = 6 and

m = 2, n = 1

By using the section formula,

We know that the coordinates of the point R which divides the line segment joining two points P (x_{1}, y_{1}, z_{1}) and Q (x_{2}, y_{2}, z_{2}) internally in the ratio m: n is given by:

So we have,

= (24/3, -30/3, 6/3)

= (8, -10, 2)

âˆ´ The coordinates of the points which trisect the line segment joining the points P (4, 2, â€“ 6) and Q (10, â€“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