A triangular plate is shown. A force $\overrightarrow{\mathrm{F}}=4\widehat{\mathrm{i}}-3\widehat{\mathrm{j}}$  is applied at point P. The torque at point P with respect to point ‘O’ and ‘Q’ are:

Torque of force  $\overrightarrow{\mathrm{F}}=4\widehat{\mathrm{i}}-3\widehat{\mathrm{j}}$ w.r.t  O

 $\overrightarrow{{\mathrm{\tau }}_{\mathrm{O}}}=\overrightarrow{{\mathrm{r}}_{\mathrm{OP}}}\times \overrightarrow{\mathrm{F}}=\left(10\cos {60}^0 \widehat{\mathrm{i}}+10\sin {60}^0 \widehat{\mathrm{j}}\right)\times \left(4\widehat{\mathrm{i}}-3\widehat{\mathrm{j}}\right)$
$\Rightarrow \overrightarrow{{\mathrm{\tau }}_{\mathrm{O}}}=\left(5\widehat{\mathrm{i}}+5\sqrt{3}\widehat{\mathrm{j}}\right)\times \left(4\widehat{\mathrm{i}}-3\widehat{\mathrm{j}}\right)=\left|\begin{array}{l}\widehat{\mathrm{i}}\widehat{\mathrm{j}}\widehat{\mathrm{k}}\\ 55\sqrt{3 } 0\\ 4-30\end{array}\right|$

$\Rightarrow \overrightarrow{{\mathrm{\tau }}_{\mathrm{O}}}=\widehat{\mathrm{k}}\left(-15-20\sqrt{3}\right)$       

Torque of force  $\overrightarrow{\mathrm{F}}=4\widehat{\mathrm{i}}-3\widehat{\mathrm{j}}$ w.r.t  Q
  ${\overrightarrow{\mathrm{\tau }}}_{\mathrm{Q}}={\stackrel{ }{\overrightarrow{\mathrm{r}}}}_{\mathrm{PQ}}\times \stackrel{ }{\overrightarrow{\mathrm{F}}}$
$\begin{array}{l}\stackrel{ }{{\overrightarrow{\mathrm{r}}}_{\mathrm{PQ}}}={\stackrel{ }{\overrightarrow{\mathrm{r}}}}_{\mathrm{P}}-{\stackrel{ }{\overrightarrow{\mathrm{r}}}}_{\mathrm{Q}}=(5\widehat{\mathrm{i}}+5\sqrt{3}\widehat{\mathrm{j}})-10\widehat{\mathrm{i}}=(-5\widehat{\mathrm{i}}+5\sqrt{3}\widehat{\mathrm{j}})\\ {\stackrel{ }{\overrightarrow{\mathrm{\tau }}}}_{\mathrm{Q}}=(-5\widehat{\mathrm{i}}+5\sqrt{3}\widehat{\mathrm{j}})\times (4\widehat{\mathrm{i}}-3\widehat{\mathrm{j}})\end{array}$

$=\left|\begin{array}{ccc}\widehat{\mathrm{i}}& \widehat{\mathrm{j}}& \widehat{\mathrm{k}}\\ -5& 5\sqrt{3}& 0\\ 4& -3& 0\end{array}\right|=\left(15-20\sqrt{3}\right)\widehat{\mathrm{k}}$     
     
So the answer for given question is $-15-20\sqrt{3},15-20\sqrt{3}$