Question

Find the torque of a force `F=a(hati+2hatj+3hatk)` `N` about a point O. The position vector of point of application of force about `O` is `r=(2hati+3hatj-hatk)` `m`.

Answer 1

Torque `tau=rxxF|{:(hati,hatj,hatk),(2,3,-1),(1,2,-3):}|`
`=hati(-9+2)+hatj(-1+6)+hatk(4-3)`
or `tau=(-7hati+5hatj+hatk)N-m`