Let a⃗, b⃗, c⃗ be three mutually perpendicular vectors of the same magnitude and equally inclined at an angle θ, with the vector a⃗ + b⃗ + c⃗. Then 36cos²2θ is equal to ___________.

Solution

Maths sir · 4 steps

Dot a⃗ with the sum: perpendicularity kills two terms, and cos θ drops out at once.

1. Size of the sum. Let each vector have magnitude k. All cross dot products are zero, so |a⃗+b⃗+c⃗|² = 3k² and |a⃗+b⃗+c⃗| = √3 k.
2. Find cos θ. Take the dot product of a⃗ with the sum. Only a⃗ · a⃗ = k² survives, because a⃗ · b⃗ = a⃗ · c⃗ = 0.

   cos θ = (a⃗ · (a⃗+b⃗+c⃗))/(|a⃗| |a⃗+b⃗+c⃗|) = k²/(k × √3 k) = 1/√3

3. Double angle. cos² θ = 1/3, so cos 2θ = 2 cos² θ - 1 = -1/3 and cos² 2θ = 1/9.
4. So. 36 cos² 2θ = 36 × 1/9 = 4.

The common slip: 12. Stopping at 36 cos² θ instead of using the double angle gives 12, not 4.

The idea to keep. Equal, mutually perpendicular vectors behave like the edges of a cube: their sum is the body diagonal, which makes an angle with cos θ = 1/√3 with each edge.

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