First slide

Introduction to 3-D Geometry

Question

It A and B be the points (3, 4,5) and (- 1, 3, - 7) respectively, find the equation of the set of points P such that (PA)² + (PB)² = K², where K is a constant.

Easy

A

$2 (x^{2} + y^{2} + z^{2}) + 4 x + 14 y + 4 z + 109 - K^{2} = 0$
B

$2 (x^{2} + y^{2} + z^{2}) - 4 x - 14 y + 4 z + 109 - K^{2} = 0$
C

$x^{2} + y^{2} + z^{2} + 4 x + 14 y + 4 z + 109 - K^{2} = 0$
D

None of the above

Solution

Let the point P is (x, y, z).
Given (PA)² + (PB)² = K²
$\Rightarrow (x - 3)^{2} + (y - 4)^{2} + (z - 5)^{2} + (x + 1)^{2} + (y - 3)^{2} + (z + 7)^{2} = K^{2}$
$\begin{array}{r} \Rightarrow x^{2} + 9 - 6 x + y^{2} + 16 - 8 y + z^{2} + 25 - 10 z + x^{2} + 1 + 2 x + y^{2} + 9 - 6 y + z^{2} + 49 + 14 z = K^{2} \end{array}$
$\begin{aligned} \Rightarrow 2 x^{2} + 2 y^{2} + 2 z^{2} - 4 x - 14 y + 4 z + 109 - K^{2} = 0 \\ \Rightarrow 2 (x^{2} + y^{2} + z^{2}) - 4 x - 14 y + 4 z + 109 - K^{2} = 0 \end{aligned}$
which is the required equation.

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App