Chapter 12 Introduction to Three Dimensional Geometry Ex-12.1 |
Chapter 12 Introduction to Three Dimensional Geometry Ex-12.3 |

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)

**Answer
1** :

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)

**Answer
2** :

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 = PR

Hence, points P(–2, 3, 5), Q(1, 2, 3), and R(7, 0, –1) are collinear.

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.

**Answer
3** :

(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.

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 =

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.

**Answer
4** :

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

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

⇒ –2*x* –*4y* –6*z* + 14 = –6*x* – 4*y* + 2*z *+ 14

⇒ – 2*x* – 6*z* +6*x* – 2*z* = 0

⇒ 4*x* – 8*z* =0

⇒ *x* – 2*z* =0

Thus, the requiredequation is *x* – 2*z* = 0.

**Answer
5** :

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.

Onsquaring both sides, we obtain

Onsquaring both sides again, we obtain

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

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

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

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

Name:

Email:

Copyright 2017, All Rights Reserved. A Product Design BY CoreNet Web Technology