Question

Find an equation of the plane. The plane that passes through the point (?1, 1, 1) and contains the line of intersection of the planes x + y ? z = 2 and 4x ? y + 5z = 3

Answer 1

We have two planes, x + y - z - 2 = 0............(i)

4x - y + 5z - 3 = 0.........(ii)

Equation of the plane passing through intersection of the two planes,

(x + y - z - 2) + K (4x - y + 5z - 3) = 0

=> x + y - z - 2 + 4Kx - Ky + 5Kz - 3K = 0

=> x( 1 + 4K ) + y( 1 - K ) + z ( 5K - 1 ) - 3K - 2 = 0 ...............(iii)

The plane passes through the point ( -1, 1, 1) . Put this point in equation (iii) we get the value of K,

(-1)( 1 + 4K ) + (1) (1 -K) + (1) (5K -1) - 3K - 2 = 0

=> -1 - 4K + 1 - K + 5K - 1 - 3K - 2 = 0

=> - 3K -3 = 0 => K = -1

Put the value of K in equation (iii) we get the equation of plane,

x ( 1 + 4(-1) ) + y ( 1 - (-1) ) + z ( 5 (-1) - 1 ) - 3 (-1) - 2 = 0

=> -3x + 2y - 6z + 1 = 0

=> 3x - 2y + 6z = 1 [Answer]