First slide

Planes in 3D

Question

The distance of the point (1,-5,9) from the plane $x - y + z = 5$ measured along a straight line $x = y = z$ is

Moderate

A

$3 \sqrt{5}$
B

$10 \sqrt{3}$
C

$5 \sqrt{3}$
D

$3 \sqrt{10}$

Solution

Let $P = (1, - 5, 9)$
Let Q be a point on the given plane such that
PQ is parallel to given line
The equation of the line PQ is $\frac{x - 1}{1} = \frac{y + 5}{1} = \frac{z - 9}{1}$
Let $Q = (1 + t, t - 5, t + 9)$
Sub Q in the given plane, $t = - 10$
$∴ Q (- 9, - 15, - 1) P Q = \sqrt{300} = 10 \sqrt{3}$

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