• # question_answer 20) A force $\overrightarrow{F}=4\widehat{i}-5\widehat{j}+3\widehat{k}$ is acting a point? ${{\overrightarrow{r}}_{1}}=\widehat{i}+2\widehat{j}+3\widehat{k}$. The torque acting about a point ${{\overrightarrow{r}}_{2}}=3\widehat{i}-2\widehat{j}-3\widehat{k}$ is- A) ZeroB) $42\widehat{i}-30\widehat{j}+6\widehat{k}$C) $42\widehat{i}+30\widehat{j}+6\widehat{k}$D) $42\widehat{i}+30\widehat{j}-6\widehat{k}$

Position vector of the point at which force is acting ${{\overrightarrow{r}}_{1}}=\widehat{i}+2\widehat{j}+3\widehat{k}$ But we have to calculate the torque about another point. So its position vector about that another point. $\overrightarrow{r{{'}_{1}}}=\overrightarrow{{{r}_{1}}}-\overrightarrow{{{r}_{2}}}=(\widehat{i}+2\widehat{j}\,+3\widehat{k})-(3\widehat{i}-2\widehat{j}-3\widehat{k})$ $=-2\widehat{i}+4\widehat{j}+6\widehat{k}$ Now $\overrightarrow{\tau }={{\overrightarrow{r'}}_{1}}\times \overrightarrow{F}=\left( -2\widehat{i}+4\widehat{j}+6\widehat{k} \right)\times (4\widehat{i}-5\widehat{j}+3\widehat{k})$ $\overrightarrow{\tau }\,\,=\,\,\left| \begin{matrix} \overrightarrow{i} & \overrightarrow{j} & \overrightarrow{k} \\ -2 & 4 & 6 \\ 4 & -5 & 3 \\ \end{matrix} \right|\,\,=\,\,\widehat{i}(12+30)-\widehat{j}(-6-24)+\widehat{k}(10-16)$$=\,\,\left( 42\widehat{i}+30\widehat{j}-6\widehat{k} \right)N-m$