Solveeit Logo
Login

Question

Question: The equation of the plane through the three points (1, 1, 1), (1, –1, 1) and (–7,–3,–5), is...

The equation of the plane through the three points (1, 1, 1), (1, –1, 1) and (–7,–3,–5), is

A

3x−4z+1=03x - 4z + 1 = 0

B

3x−4y+1=03x - 4y + 1 = 0

C

3x+4y+1=03x + 4y + 1 = 0

D

None of these

Answer

3x−4z+1=03x - 4z + 1 = 0

Explanation

Solution

Plane passing through (1, 1, 1) is

a(x−1)+b(y−1)+c(z−1)=0a ( x - 1 ) + b ( y - 1 ) + c ( z - 1 ) = 0

It also passes through (1, – 1, 1) and (–7, – 3, – 5), then b=0,a=−34cb = 0 , a = - \frac { 3 } { 4 } c

Hence the required equation is 3x−4z+1=03 x - 4 z + 1 = 0.

Trick : Since the plane 3x−4z+1=03 x - 4 z + 1 = 0 passes through given three points.