First slide

Planes in 3D

Question

Let A (1, 1, 1), B (2,3,5) and C (-1 ,0,2) be three points, then equation of a plane parallel to the plane ABC which is at distance 2 is

Moderate

A

$2 x - 3 y + z + 2 \sqrt{14} = 0$
B

$2 x - 3 y + z - \sqrt{14} = 0$
C

$2 x - 3 y + z + 2 = 0$
D

$2 x - 3 y + z - 2 = 0$

Solution

$A (1, 1, 1), B (2, 3, 5), C (- 1, 0, 2) direction ratios of A B are < 1, 2, 4 ⟩$
$Direction ratios of A C are < - 2, - 1, 1 > .$
$Therefore, direction ratios of normal to plane A B C are < 2, - 3, 1>$
$As a result, equation of the plane A B C is 2 x - 3 y + z = 0 .$
$Let the equation of the required plane be 2 x - 3 y + z = k . Then$
$|\frac{k}{\sqrt{4 + 9 + 1}}| = 2 or k = \pm 2 \sqrt{14}$
$Hence, equation of the required plane is 2 x - 3 y + z + 2 \sqrt{14} = 0$

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