# Class 11 Maths NCERT Solutions for Chapter 12 Introduction to Three Dimensional Geometry Exercise 12.2

### Three Dimensional Geometry Exercise 12.2 Solutions

**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) **

**Solution**

The distance between points P(x_{1}, y_{1}, z_{1}) and P(x_{2}, y_{2}, z_{2}) is given by

(i) Distance between points (2, 3, 5) and (4, 3, 1)

(ii) Distance between points (-3, 7, 2) and (2, 4, -1)

(iii) Distance between points (-1, 3, -4) and (1, -3, 4)

(iv) Distance between points (2, -1, 3) and (-2, 1, 3)

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

**Solution**

Let points (–2, 3, 5), (1, 2, 3), and (7, 0, –1) be denoted by P, Q, and R respectively.

Points P, Q, and R are collinear if they lie on a line.

Here, PQ + QR = √14 + 2√14 = 3√14 = PR

Hence, points P(-2, 3, 5), Q(1, 2, 3), and R(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.**

**Solution**

(i) Let points (0, 7, -10), (1, 6, -6), and (4, 9, -6) be denoted by A, B, and C respectively.

Here, AB = BC ≠ CA

Thus, the given points are the vertices of an isosceles triangle.

(ii) Let (0, 7, 10), (–1, 6, 6), and (–4, 9, 6) be denoted by A, B, and C respectively.

Now, AB^{2} + BC^{2} = (3√2)^{2} + (3√2)^{2} = 18 + 18 = 36 = AC^{2}.

Therefore, by Pythagoras theorem, ABC is a right triangle.

Hence, the given points are the vertices of a right - angled triangle.

(iii) Let (-1, 2, 1), (1, -2, 5), (4, -7, 8), and (2, -3, 4) be denoted by A, B, C, and D respectively.

Here, AB = CD = 6, BC = AD = √43

Hence, the opposite sides of quadrilateral ABCD, whose vertices are taken in order, are equal.

Therefore, ABCD is a parallelogram.

Hence, the given points are the vertices of a parallelogram.

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

**Solution**

Let P (x, y, z) be the point that is equidistant from points A(1, 2, 3) and B(3, 2, –1).

Accordingly, PA = PB

⇒ PA^{2} = PB^{2}

⇒ (x - 1)^{2} + (y - 2)^{2} + (z - 3)^{2} = (x -3)^{2} + (y - 2)^{2} + (z + 1)^{2}

⇒ x^{2} - 2x + 1 + y^{2} - 4y + 4 +z^{2} - 6z + 9 = x^{2} - 6x + 9 + y^{2} - 4y + 4 + z^{2} + 2z + 1

⇒ -2x - 4y - 6z + 14 = -6x - 4y + 2z + 14

⇒ -2x - 6z + 6x - 2z = 0

⇒ 4x - 8z = 0

⇒ x - 2z = 0

Thus, the required equation is x - 2z = 0.

**5. 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.**

**Solution**

Let the coordinates of P be (x, y, z).

The coordinates of points A and B are (4, 0, 0) and (–4, 0, 0) respectively.

It is given that PA + PB = 10.

On squaring both sides, we obtain

On squaring both sides again, we obtain

25 (x^{2} + 8x + 16 + y^{2} + z^{2}) = 625 + 16x^{2} + 200x

⇒ 25x^{2} + 200x + 400 + 25y^{2} + 25z^{2} = 625 + 16x^{2} + 200x

⇒ 9x^{2} + 25y^{2} + 25z^{2} – 225 = 0

Thus, the required equation is 9x^{2} + 25y^{2} + 25z^{2} – 225 = 0.