A plane P contains the line x + 2y + 3z + 1 = 0 = x - y - z - 6, and is perpendicular to the plane - 2x + y + z + 8 = 0. Then which of the following points lies on P?

Solution

Maths sir · 3 steps

Any plane through that line is first plane plus λ times second; the perpendicular condition fixes λ.

1. Family of planes. Every plane containing the line of intersection has the form (x + 2y + 3z + 1) + λ(x - y - z - 6) = 0. Its normal is (1+λ, 2-λ, 3-λ).
2. Use perpendicularity. Perpendicular planes have perpendicular normals, so take the dot product with (-2, 1, 1) and set it to zero: -2(1+λ) + (2-λ) + (3-λ) = 3 - 4λ = 0, which gives λ = 3/4.

   7/4x + 5/4y + 9/4z - 14/4 = 0 ⇒ 7x + 5y + 9z - 14 = 0

   Point	7x + 5y + 9z - 14	On P?
   (a) (-1, 1, 2)	-7 + 5 + 18 - 14 = 2	no
   (b) (0, 1, 1)	0 + 5 + 9 - 14 = 0	yes
   (c) (1, 0, 1)	7 + 0 + 9 - 14 = 2	no
   (d) (2, -1, 1)	14 - 5 + 9 - 14 = 4	no

3. So. Only (0, 1, 1) satisfies 7x + 5y + 9z = 14, so the answer is (b).

The idea to keep. A plane through the intersection of two planes is P₁ + λ P₂ = 0. Use the one extra condition, here normals with dot product zero, to find λ.

Answer from the NTA final answer key. Solution by PrepWiser.