A force `vec(F)=4hati-5hatj+3hatk` is acting a point `vec(r_(1))=hati+2hatj+3hatk`. The torque acting about a point `vec(r_(2))=3hati-2hatj-3hatk` is

Options

A. Zero
B. `42hat(i)-30hat(j)+6hat(k)`
C. `42hat(i)+30hat(j)+6hat(k)`
D. `42hat(i)+30hat(j)-6hat(k)`

Answer

D

Explanation

`vec(F)=4hat(i)-5hat(j)+3hat(k)`
`vec(r)=-vec(r)_(2)+vec(r)_(1)=-2hat(i)+4hat(j)+6hat(k)`
`vec(tau)=vec(r)xxvec(F)`
`=|{:(hati,hatj,hatk),(-2,4,6),(4,-5,+3):}|`
= `(42hat(i)+30hat(j)-6hat(k))N-m`