If a⃗ . b⃗ = 1, b⃗ . c⃗ = 2 and c⃗ . a⃗ = 3, then the value of [ a⃗ × ( b⃗ × c⃗ ), b⃗ × ( c⃗ × a⃗ ), c⃗ × ( b⃗ × a⃗ ) ] is :

Solution

Maths sir · 4 steps

Open each cross product with BAC minus CAB, and the third vector turns out to be the sum of the first two.

1. Expand each one. Use p⃗ × (q⃗ × r⃗) = (p⃗ · r⃗) q⃗ - (p⃗ · q⃗) r⃗ on all three, with a⃗ · b⃗ = 1, b⃗ · c⃗ = 2 and c⃗ · a⃗ = 3.

   a⃗ × (b⃗ × c⃗) = 3b⃗ - c⃗ ; b⃗ × (c⃗ × a⃗) = c⃗ - 2a⃗ ; c⃗ × (b⃗ × a⃗) = 3b⃗ - 2a⃗

2. Spot the relation. Add the first two: (3b⃗ - c⃗) + (c⃗ - 2a⃗) = 3b⃗ - 2a⃗, which is exactly the third vector. So the three vectors are linearly dependent, which means they are coplanar.
3. Check by determinant. In terms of a⃗, b⃗, c⃗ the coefficient rows are (0, 3, -1), (-2, 0, 1) and (-2, 3, 0). Their determinant is 0 - 3(0 + 2) - 1(-6 - 0) = -6 + 6 = 0.
4. So. The scalar triple product of three coplanar vectors is zero. The value is 0, option (a).

The idea to keep. Before expanding a box product, check whether one vector is a combination of the other two. If it is, the vectors are coplanar and the box product is 0. No determinant is needed.

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