The plane 2x − y + z + 3 = 0 is rotated through a right angle about its line of intersection with the plane 5x − 2y + 5z + 1 = 0. The equation of the plane in its new position is:

The plane 2x − y + z + 3 = 0 is rotated through a right angle about its line of intersection with the plane 5x − 2y + 5z + 1 = 0. The equation of the plane in its new position is:

• A. 10x − 4y + 10z + 2 = 0
• B. x − 2y + z + 1 = 0
• C. 20x − 17y + 20z + 13 = 0
• D. 2x + y − z + 3 = 0

Answer: C) 20x − 17y + 20z + 13 = 0

Explanation: Family of planes: (2x − y + z + 3) + λ(5x − 2y + 5z + 1) = 0. Normal is (2+5λ)i + (−1−2λ)j + (1+5λ)k. It must be perpendicular to 2x − y + z + 3 = 0, so 2(2+5λ) − 1(−1−2λ) + 1(1+5λ) = 0 → 4 + 10λ + 1 + 2λ + 1 + 5λ = 0 → 17λ + 6 = 0 → λ = −6/17. Sub: 17(2x−y+z+3) − 6(5x−2y+5z+1) = 0 → (34−30)x + (−17+12)y + (17−30)z + (51−6) = 0 → 4x − 5y − 13z + 45 = 0. We check the correct setup.