Let a⃗ = î + ĵ + k̂,b⃗ and c⃗ = ĵ - k̂ be three vectors such that a⃗ × b⃗ = c⃗ and a⃗ . b⃗ = 1. If the length of projection vector of the vector b⃗ on the vector a⃗ × c⃗ is l, then the value of 3l² is equal to _____________.

Solution

Maths sir · 4 steps

Skip finding b⃗: write b⃗ · (a⃗ × c⃗) as a box product.

1. Find a × c. With a⃗ = (1, 1, 1) and c⃗ = (0, 1, -1), the cross product is a⃗ × c⃗ = -2î + ĵ + k̂. So |a⃗ × c⃗| = √(4 + 1 + 1) = √6.
2. Use the box product. Swapping two vectors in a scalar triple product flips its sign, so b⃗ · (a⃗ × c⃗) = [b⃗ a⃗ c⃗] = -[a⃗ b⃗ c⃗] = -(a⃗ × b⃗) · c⃗. Since a⃗ × b⃗ = c⃗, this equals -|c⃗|² = -2.

   l = (|b⃗ · (a⃗ × c⃗)|)/(|a⃗ × c⃗|) = 2/√6

3. Check. Solving the given conditions gives b⃗ = î. Then a⃗ × î = ĵ - k̂ = c⃗ and a⃗ · î = 1. Also î · (-2î + ĵ + k̂) = -2, which matches the box-product result.
4. So. l² = 4/6 = 2/3, so 3l² = 2.

The idea to keep. When a question gives a⃗ × b⃗ and asks for b⃗ dotted with something built from a⃗, try a scalar triple product. Rearrange it until the given cross product appears.

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