Let a vector a⃗ be coplanar with vectors b⃗ = 2î + ĵ + k̂ and c⃗ = î - ĵ + k̂. If a⃗ is perpendicular to d⃗ = 3î + 2ĵ + 6k̂, and | a⃗ | = √10. Then a possible value of [a⃗, b⃗, c⃗ ] + [a⃗, b⃗, d⃗ ] + [a⃗, c⃗, d⃗ ] is equal to :

Solution

Maths sir · 4 steps

Write a⃗ as xb⃗ + yc⃗, let the perpendicular condition fix y, and the modulus fix x.

1. Use coplanarity. Since a⃗ lies in the plane of b⃗ and c⃗, write a⃗ = xb⃗ + yc⃗ = (2x+y)î + (x-y)ĵ + (x+y)k̂. Coplanarity also gives [a⃗ b⃗ c⃗] = 0 straight away.
2. Find the vector. a⃗ · d⃗ = 0 gives 3(2x+y) + 2(x-y) + 6(x+y) = 14x + 7y = 0, so y = -2x and a⃗ = x(3ĵ - k̂). Then |a⃗|² = 10x² = 10, so x = ± 1.
3. Combine two products. Since the triple product is linear in its middle vector, [a⃗ b⃗ d⃗] + [a⃗ c⃗ d⃗] = [a⃗ b⃗ + c⃗ d⃗], with b⃗ + c⃗ = 3î + 2k̂. Take x = 1, so a⃗ = 3ĵ - k̂.

   |0, 3, -1; 3, 0, 2; 3, 2, 6| = 0 - 3(18 - 6) + (-1)(6 - 0) = -36 - 6 = -42

4. So. The sum is 0 + (-42) = -42 for x = 1. It is +42 for x = -1, which is not an option. So the possible value is -42, option (a).

The idea to keep. A vector coplanar with b⃗ and c⃗ is always xb⃗ + yc⃗. Each extra condition fixes one unknown. Then use linearity to merge triple products that share two vectors.

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