First slide

Application of vectors in geometry

Question

$Vertices of a parallelogram taken in order are A (2, - 1, 4),$ $B (1, 0, - 1), C (1, 2, 3) and D . Distance of the point$ $P (8, 2, - 12) from the plane of the parallelogram is -$

Moderate

A

$\frac{4 \sqrt{6}}{9}$
B

$\frac{32 \sqrt{6}}{9}$
C

$\frac{16 \sqrt{6}}{9}$
D

None of these

Solution

$\vec{n} = 7 \hat{i} + 2 \hat{j} - \hat{k} is normal to plane , this is AB X AC$
$(Assuming n = a \hat{i} + b \hat{j} + c \hat{k} and using$
$\bar{AP} = 6 \hat{i} + 3 \hat{j} - 16 \hat{k}$
$∴ Distance$
$d = |\frac{\bar{AP} \cdot \vec{n}}{| \vec{n} |}| = |\frac{42 + 6 + 16}{\sqrt{49 + 4 + 1}}| = \frac{64}{\sqrt{54}} = \frac{64}{3 \sqrt{6}} = \frac{64 \sqrt{6}}{18} = \frac{32 \sqrt{6}}{9}$

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