Question

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

Answer 1

Let the equation of plane passing through point (1, 1, -1) be

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

Since (i) is perpendicular to the plane x + 2y + 3z -7 = 0

Again plane (i) is perpendicular to the plane 2x - 3y + 4z = 0

From (ii) and (iii), we get

Putting the value of a, b, c in (i), we get

[Note: The equation of plane passing through \((x_1,y_1,z_1)\) is given by \(a(x - x_1) + b(y - y_1) + c(z - z_1) = 0\), where a, b, c are direction ratios of normal of plane.]