Let a⃗=-5 î+ĵ-3 k̂, b⃗=î+2 ĵ-4 k̂ and
c⃗=(((a⃗ × b⃗) × î) × î) × î. Then c⃗ ·(-î+ĵ+k̂) is equal to :

Solution

Maths sir · 4 steps

Crossing twice with î just flips the sign, so only one cross product needs real work.

1. Find a cross b. With a⃗ = (-5, 1, -3) and b⃗ = (1, 2, -4), the determinant gives a⃗ × b⃗ = 2î - 23ĵ - 11k̂. Call this v⃗.
2. Cross twice with i. By the vector triple product, (v⃗ × î) × î = (v⃗ · î)î - (î · î)v⃗ = 2î - v⃗. Crossing this once more with î removes the 2î part, because î × î = 0⃗.

   c⃗ = (2î - v⃗) × î = -v⃗ × î = 11ĵ - 23k̂

3. Check the cross. For any v⃗ = (x, y, z), v⃗ × î = (0, z, -y). Here that is (0, -11, 23), so c⃗ = (0, 11, -23).
4. So. c⃗ · (-î + ĵ + k̂) = 0 + 11 - 23 = -12, which is option (a).

The idea to keep. For a unit vector u, (v⃗ × u) × u = -v⃗_⊥. Once a vector is perpendicular to u, each pair of crosses with u just flips its sign.

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