Thank you for reporting, we will resolve it shortly
Q.
The torque of force $\vec{ F }=(2 \hat{ i }-3 \hat{ j }+4 \hat{ k }) N$ acting at a point $\vec{ r }=(3 \hat{ i }+2 \hat{ j }+3 \hat{ k }) m$ about origin is
AMUAMU 2005
Solution:
The moment of force or torque about an axis is equal to the vector product of force $(F)$ and perpendicular distance of line of action of force from the axis of rotation $(r)$
$ \therefore $ $ \tau =F\times r $
Given, $ \vec{F}=2\hat{i}-3\hat{j}+4\,\hat{k},\,$
$r=3\,\hat{i}+2\hat{j}+3\hat{k} $
$ \therefore $ $ \tau =(2\,\hat{i}-3\hat{j}+4\hat{k})\times (3\,\hat{i}+2\hat{j}+3\hat{k}) $
$ \tau =\left| \begin{matrix} {\hat{i}} & {\hat{j}} & {\hat{k}} \\ 2 & -3 & 4 \\ 3 & 2 & 3 \\ \end{matrix} \right| $
$ \Rightarrow $ $ \tau =\hat{i}\,(-9-8)-\hat{j}\,(6-12)+\hat{k}\,(4+9) $
$ \Rightarrow $ $ \tau =-17\,\hat{i}+6\hat{j}+13\,\hat{k} $ .