Question

The torque of force $ \vec{F}=(2\hat{i}-3\hat{j}+4\hat{k}) $ newton acting at a point $ \vec{r}=(3\hat{i}+2\hat{j}+3\hat{k}) $ metre about origin is:

• A. $ 6\vec{i}-6\vec{j}+12\vec{k}N-m $
• B. $ -6\vec{i}+6\vec{j}-12\vec{k}N-m $
• C. $ 17\vec{i}-6\vec{j}-13\vec{k}N-m $
• D. $ -17\vec{i}+6\vec{j}+13\vec{k}N-m $

Answer 1

Answer: (C)

Here $: \vec{ F }=2 \hat{ i }-3 \hat{ j }+ 4 \mathbf { k } N$
Position vector of a point
$\vec{ r }=3 \hat{ i }+2 \hat{ j }+3 \hat{ k } m$
The torque acting at a point about the origin is given by
$\vec{\tau} =\vec{ r } \times \vec{ F }=( 3 \hat{ i }+2 \hat{ j }+3 \hat{ k }) \times(2 \hat{ i }-3 \hat{ j }+4 \hat{ k }) $
$=\begin{vmatrix} \hat{ i } & \hat{ j } & \hat{ k } \\ 3 & 2 & 3 \\ 2 & -3 & 4\end{vmatrix}$
$=\hat{ i }[8-(-9)]-\hat{ j }(12-6)+\hat{ k }(-9-4) $
$=17 \hat{ i }-6 \hat{ j }-13 \hat{ k } \,N - m$