Question

Find the equation of the plane which contains the line \(\frac{x-1}{2}=\frac{y+1}{-1}=\frac{z-3}{4}\) perpendicular to the plane x + 2y + z = 12.

Answer 1

Equation of plane passing through (1,-1,3) is

A(x-1) + B (y+1) + C (z-3) = 0

This is to the line and given plane,

∴  2A - B + 4C = 0 

and A + 2B + C = 0 

Eliminating A, B and C from equations, we get

∴ (x-1) (-9) - (y+1) (-2) + (z-3)5 = 0 

-9x + 9 + 2y + 2 + 5z - 15 = 0 

-9x + 2y + 5z - 4 = 0 

9x - 2y - 5z + 4 = 0