Let the plane 2x + 3y + z + 20 = 0 be rotated through a right angle about its line of intersection with the plane x - 3y + 5z = 8. If the mirror image of the point ( 2, - 1/2,2 ) in the rotated plane is B(a, b, c), then :

Solution

Maths sir · 5 steps

The rotated plane keeps the common line, so take P₁ + λ P₂ = 0 perpendicular to P₁.

1. Planes through the line. Any plane through the line of intersection is (2x + 3y + z + 20) + λ(x - 3y + 5z - 8) = 0. Its normal is (2+λ, 3-3λ, 1+5λ).
2. Right-angle condition. A rotation through a right angle makes the new plane perpendicular to 2x + 3y + z + 20 = 0. So the dot product of the two normals is zero: 2(2+λ) + 3(3-3λ) + (1+5λ) = 14 - 2λ = 0, which gives λ = 7.
3. Rotated plane. With λ = 7 the plane is 9x - 18y + 36z - 36 = 0, that is x - 2y + 4z - 4 = 0, with normal (1, -2, 4) and |n|² = 21. Now apply the mirror-image formula to (2, -1/2, 2).

   (a-2)/1 = (b+1/2)/-2 = (c-2)/4 = -2(2 + 1 + 8 - 4)/21 = -2/3

4. Find the image. So a = 2 - 2/3 = 4/3, b = -1/2 + 4/3 = 5/6 and c = 2 - 8/3 = -2/3.
5. So. a : b : c = 4/3 : 5/6 : -2/3 = 8 : 5 : -4. Each of a/8, b/5, c/-4 equals 1/6, so the answer is (a).

The idea to keep. Rotating a plane about a line keeps the line, so use the family P₁ + λ P₂ = 0. The rotation angle then fixes λ. For a right angle, the dot product of the normals is zero.

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