Let a⃗ = 2î - ĵ + 3k̂, b⃗ = 3î - 5ĵ + k̂ and c⃗ be a vector such that a⃗ × c⃗ = a⃗ × b⃗ = c⃗ × b⃗ and (a⃗ + c⃗) · (b⃗ + c⃗) = 168 . Then the maximum value of |c⃗|² is :

Solution

Maths sir · 5 steps

Turn each cross-product equation into a parallel condition, and c⃗ gets pinned down completely.

1. First equation. a⃗ × c⃗ = a⃗ × b⃗ gives a⃗ × (c⃗ - b⃗) = 0⃗. So c⃗ - b⃗ is parallel to a⃗, which means c⃗ = b⃗ + λ a⃗.
2. Second equation. c⃗ × b⃗ = a⃗ × b⃗ gives (c⃗ - a⃗) × b⃗ = 0⃗. So c⃗ - a⃗ is parallel to b⃗, which means c⃗ = a⃗ + μ b⃗.
3. Compare the two. Equating them gives (λ - 1)a⃗ = (μ - 1)b⃗. The vectors a⃗ and b⃗ are not parallel, so both coefficients must be zero: λ = μ = 1.

   c⃗ = a⃗ + b⃗ = 5î - 6ĵ + 4k̂

4. Check the dot product. a⃗ + c⃗ = 7î - 7ĵ + 7k̂ and b⃗ + c⃗ = 8î - 11ĵ + 5k̂. Their dot product is 56 + 77 + 35 = 168, so the given condition holds.
5. So. There is only one possible c⃗, and |c⃗|² = 25 + 36 + 16 = 77. That single value is also the maximum, so the answer is (a).

The idea to keep. Whenever you see p⃗ × x⃗ = p⃗ × y⃗, rewrite it as x⃗ - y⃗ parallel to p⃗. Two such conditions usually fix the vector completely.

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