Question

1. Find the equation of the plane through the point(1, 2, 3) and perpendicular to the plane x – y + z = 2 and 2x + y – 3z = 5 
2. Find the distance between the planes x – 2y + 2z – 8 = 0 and 6y – 3x – 6z = 57

Answer 1

1. Required equation is

\(\begin{vmatrix} x-1 & y-2 & z-3 \\ 1 & -1 & 1 \\ 2 & 1 & -3 \end{vmatrix}\)

(x – 1)(3 – 1) – (y – 2)(-3 – 2) + (z – 3)(1 + 2) = 0
2(x – 1) + 5(y – 2) + 3(z – 3) = 0
2x + 5y + 3z – 2 – 10 – 9 = 0
2x + 5y + 3z – 21 = 0

2. The planes are
x – 2y + 2z – 8 = 0 and 3x – 6y + 6z + 57 = 0
ie, 3x – 6y + 6z – 24 = 0 and 3x – 6y + 6z + 57 = 0
Distance