The torque due to the force (2 î+ĵ+2 k̂) about the origin, acting on a particle whose position vector is (î+ĵ+k̂), would be

Solution

Physics sir · 4 steps

Torque is r⃗ × F⃗, with position first. Expand the determinant and watch the signs.

1. Set up. Torque about the origin is τ⃗ = r⃗ × F⃗, with r⃗ = î+ĵ+k̂ and F⃗ = 2î+ĵ+2k̂. Put r⃗ in the second row and F⃗ in the third.

   τ⃗ = |î, ĵ, k̂; 1, 1, 1; 2, 1, 2|

2. Expand. î part: (1)(2) - (1)(1) = 1. ĵ part: -[(1)(2) - (1)(2)] = 0. k̂ part: (1)(1) - (1)(2) = -1.
3. Quick check. A cross product is perpendicular to both vectors. (î-k̂)·r⃗ = 1 - 1 = 0 and (î-k̂)·F⃗ = 2 - 2 = 0, so the result is consistent.
4. So. τ⃗ = î - k̂, which is option (b).

The idea to keep. Torque is always r⃗ × F⃗, never the other way round. After any cross product, dot the answer with both vectors: you must get zero both times.

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