A force F⃗ = ( î + 2ĵ + 3k̂ ) N acts at a point
( 4î + 3ĵ - k̂ ) m. Then the magnitude of torque
about the point ( î + 2ĵ + k̂ ) m will be √x N m.
The value of x is _______.

Solution

Physics sir · 4 steps

Torque about a point needs the arm from that point, so subtract the positions before taking the cross product.

1. Find the lever arm. The arm runs from the point about which torque is taken to the point where the force acts: r⃗ = (4-1)î + (3-2)ĵ + (-1-1)k̂ = 3î + ĵ - 2k̂ m.

   τ⃗ = r⃗ × F⃗ = |î, ĵ, k̂; 3, 1, -2; 1, 2, 3|

2. Expand the determinant. î: (1)(3) - (-2)(2) = 7. ĵ: -[(3)(3) - (-2)(1)] = -11. k̂: (3)(2) - (1)(1) = 5. So τ⃗ = 7î - 11ĵ + 5k̂ N m.
3. Take the magnitude. |τ⃗| = √(7² + 11² + 5²) = √(49 + 121 + 25) = √195 N m.
4. So. |τ⃗| = √195 N m, so x = 195.

The common slip: 315. Using the position vector 4î + 3ĵ - k̂ from the origin gives torque about the origin, 11î - 13ĵ + 5k̂, and x = 315.

The idea to keep. For torque about any point, first build r⃗ = (point of action) minus (reference point), then cross it with F⃗ in that order, r⃗ × F⃗.

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