Let λ ∈ ℤ, a⃗=λ î+ĵ-k̂ and b⃗=3 î-ĵ+2 k̂. Let c⃗ be a vector such that (a⃗+b⃗+c⃗) × c⃗=0⃗, a⃗ · c⃗=-17 and b⃗ · c⃗=-20. Then |c⃗ ×(λ î+ĵ+k̂)|² is equal to :

Solution

Maths sir · 4 steps

A zero cross product means parallel, so write c⃗ as a multiple of a⃗+b⃗ first.

1. Use the cross product. Since c⃗ × c⃗ = 0⃗, the condition becomes (a⃗+b⃗) × c⃗ = 0⃗. So c⃗ = μ(a⃗+b⃗) for some scalar μ, where a⃗+b⃗ = (λ+3)î + k̂.
2. Two dot products. Dot a⃗+b⃗ with each vector: a⃗·(a⃗+b⃗) = λ²+3λ-1 and b⃗·(a⃗+b⃗) = 3λ+11. So μ(λ²+3λ-1) = -17 and μ(3λ+11) = -20.
3. Find the scalars. Dividing the two equations removes μ and gives 20λ² + 9λ - 207 = 0, whose roots are λ = 3 and λ = -69/20. Since λ is an integer, λ = 3. Then 20μ = -20, so μ = -1 and c⃗ = -6î - k̂.

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

4. So. The required value is 1² + 3² + 6² = 1 + 9 + 36 = 46, which is option (d).

The idea to keep. Whenever (u⃗+c⃗) × c⃗ = 0⃗, drop the c⃗ × c⃗ term and write c⃗ as a scalar multiple of u⃗. One scalar is easier than three components.

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