Find the equation of the plane through the line of intersection of the planes 2x + 3y - z + 1 = 0 and x + y - 2z + 3 = 0 which is perpendicular to the plane 3x – y - 2z - 4 = 0?
Here, we have to find the equation of plane passing through the line of intersection of planes 2x + 3y - z + 1 = 0 and x + y - 2z + 3 = 0 which is perpendicular to plane 3x - y - 2z - 4 = 0 As we know that, equation of a plane passing through the intersection of these planes is given by: (a x + b y + c z + d) + λ (a x + b y + c z + d) = 0, where λ is a scalar ⇒ (2x + 3y - z + 1) + λ (x + y - 2z + 3) = 0 ⇒ (2 + λ)x + (3 + λ)y + (-1 - 2λ)z + 1 + 3λ = 0 .......(1) The direction ratios of the plane represented by (1) are: (2 + λ), (3 + λ), (-1 - 2λ) The direction ratios of the plane 3x – y - 2z - 4 = 0 are: 3, - 1, - 2 ∵ plane represented by (1) is perpendicular to plane 3x – y - 2z - 4 = 0 ⇒ 6 + 3λ - 3 - λ + 1 + 2 + 4λ = 0 ⇒ 6λ + 5 = 0 ⇒ λ = - 5/6 By substituting λ = - 5/6 in equation (1), we get ⇒ 7x + 13y + 4z - 9 = 0 Hence, equation of the required plane is 7x + 13y + 4z - 9 = 0