Let P be a plane passing through the points (1, 0, 1), (1, -2, 1) and (0, 1, -2). Let a vector a⃗ = α î + β ĵ + γ k̂ be such that a⃗ is parallel to the plane P, perpendicular to (î + 2ĵ + 3k̂) and a⃗ . (î + ĵ + 2k̂) = 2, then (α - β + γ )² equals ____________.

Solution

Maths sir · 4 steps

Two perpendicularity conditions fix the direction of a; the dot product condition then fixes its length.

1. Normal to the plane. With A(1,0,1), B(1,-2,1), C(0,1,-2), take AB⃗ = (0,-2,0) and AC⃗ = (-1,1,-3). Their cross product is (6, 0, -2), so the normal is n⃗ = (3, 0, -1).
2. Direction of a. Parallel to the plane means a⃗ ⊥ n⃗. The question also needs a⃗ ⊥ (1,2,3). A vector perpendicular to both is their cross product.

   n⃗ × (î + 2ĵ + 3k̂) = |î, ĵ, k̂; 3, 0, -1; 1, 2, 3| = 2î - 10ĵ + 6k̂

3. Fix the length. So a⃗ = λ(1, -5, 3). Then a⃗ · (1,1,2) = λ(1 - 5 + 6) = 2λ = 2, which gives λ = 1 and a⃗ = î - 5ĵ + 3k̂.
4. So. α - β + γ = 1 + 5 + 3 = 9, and (α - β + γ)² = 81.

The common slip: 324. Using a⃗ = (2, -10, 6) straight from the cross product, without applying a⃗ · (î + ĵ + 2k̂) = 2, gives 18² = 324 instead of 81.

The idea to keep. A vector perpendicular to two given vectors lies along their cross product. Find that direction first, then use the extra condition to fix the scalar multiple.

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