Question

Find the equation of the plane passing through the point (1, 1, –1) and perpendicular to the planes x + 2y + 3z = 7 and 2x – 3y + 4z = 0.

Answer 1

Equation of the plane passing through A(1, 1, –1) is

a(x – 1) + b(y – 1) + c(z + 1) = 0   ...(1)

Where a, b, c are d.r.’s of the normal to plane.

∵ This plane is perpendicular to the planes

x + 2y + 3z = 7 and 2x – 3y + 4z = 0

∵ a + 2b + 3c = 0   ...(2)

and 2a – 3b + 4c = 0   ...(3)

Now solving (2) and (3),

`a/(8 - (-9)) = (-b)/(4 - 6) = c/(-3 - 4)`

`a/17 = b/2 = c/-7`

Put these values in (1), we get

17(x – 1) + 2(y – 1) – 7(z + 1) = 0

`\implies` 17x + 2y – 7z = 26

Which is the required plane.