If the plane 2x + y - 5z = 0 is rotated about its line of intersection with the plane 3x - y + 4z - 7 = 0 by an angle of π/2, then the plane after the rotation passes through the point :

Solution

Maths sir · 4 steps

Turning a plane by 90° about a line gives a plane through that line, perpendicular to where it started.

1. Family of planes. Every plane through the line of intersection has the form (2x + y - 5z) + λ(3x - y + 4z - 7) = 0. Its normal is (2 + 3λ, 1 - λ, -5 + 4λ).
2. Perpendicular condition. A rotation by π/2 makes the new plane perpendicular to 2x + y - 5z = 0. So the new normal must have zero dot product with (2, 1, -5).

   2(2 + 3λ) + (1 - λ) - 5(-5 + 4λ) = 30 - 15λ = 0 ⇒ λ = 2

3. New plane. Put λ = 2: 2x + y - 5z + 6x - 2y + 8z - 14 = 0, that is, 8x - y + 3z - 14 = 0.

   Point	8x - y + 3z - 14	On the plane?
   (a) (2, -2, 0)	16 + 2 + 0 - 14 = 4	no
   (b) (-2, 2, 0)	-16 - 2 + 0 - 14 = -32	no
   (c) (1, 0, 2)	8 - 0 + 6 - 14 = 0	yes
   (d) (-1, 0, -2)	-8 - 0 - 6 - 14 = -28	no

4. So. The rotated plane 8x - y + 3z - 14 = 0 passes through (1, 0, 2), option (c).

The idea to keep. Rotating a plane about a line keeps it in the family P₁ + λ P₂ = 0. The rotation angle fixes λ: for 90°, set the dot product of the normals to zero.

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