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

### Three Dimensional Geometry Exercise 12.3 Solutions

**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**

(i) The coordinates of point R that divides the line segment joining points P (x_{1}, y_{1}, z_{1}) and Q (x_{2}, y_{2}, z_{2}) internally in the ratio m: n are

Let R (*x*,* y*, *z*) be the point that divides the line segment joining points(–2, 3, 5) and (1, –4, 6) internally in the ratio 2:3

Thus, the coordinates of the required point are (-4/5, 1/5, 27/5).

(ii) The coordinates of point R that divides the line segment joining points P (x_{1}, y_{1}, z_{1}) and Q (x_{2}, y_{2}, z_{2}) externally in the ratio m: n are

Let R(x, y, z) be the point that divides the line segment joining points (-2, 3, 5) and (1, -4, 6) externally in the ratio 2 : 3

i.e., x = -8, y = 17, and z = 3

Thus, the coordinates of the required point 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 point Q (5, 4, –6) divide the line segment joining points P (3, 2, –4) and R (9, 8, –10) in the ratio k:1.

Therefore, by section formula,

⇒ 9k + 3 = 5k + 5

⇒ 4k = 2

⇒ k = 2/4 = 1/2

Thus, point Q divides PR in the ratio 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 YZ planedivide the line segment joining points (–2, 4, 7) and (3, –5, 8) in the ratio k:1.

Hence, by section formula, the coordinates of point of intersection are given by

On the YZ plane, the x - coordinate of any point is zero.

(3k - 2)/(k + 1) = 0

⇒ 3k -2 = 0

⇒ k = 2/3

Thus, the YZ plane divides the line segment formed by joining the given points in the ratio 2 : 3.

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

**Solution**

The given points are A (2, –3, 4), B (–1, 2, 1), and C(0, 1/3, 2). .

Let P be a point that divides AB in the ratio* k*:1.

Hence, by section formula, the coordinates of P are given by

Now, we find the value of k at which point P coincides with point C.

By taking (-k + 2)/(k + 1) = 0 , we obtain k = 2.

For k = 2, the coordinates of point P are (0, 1/3, 2).

i.e, C(0, 1/3, 2) is a point that divides AB externally in the ratio 2 : 1 and is the same as point P .

Hence, points A, B, and 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 and B be the points that trisect the line segment joining points P (4, 2, –6) and Q (10, –16, 6)

Point A divides PQ in the ratio 1 : 2. Therefore, by section formula, the coordinates of point A are given by

Point B divides PQ in the ratio 2 : 1. Therefore, by section formula, the coordinates of point B are given by

Thus, (6, -4, -2) and (8, -10, 2) are the points that trisect the line segment joining points P(4, 2, -6) and Q(10, -16, 6).