A) \[6\overrightarrow{i}-6\overrightarrow{j}+12\overrightarrow{k}N-m\]
B) \[-6\overrightarrow{i}+6\overrightarrow{j}-12\overrightarrow{k}N-m\]
C) \[17\overrightarrow{i}-6\overrightarrow{j}-13\overrightarrow{k}N-m\]
D) \[-17\overrightarrow{i}+6\overrightarrow{j}+13\overrightarrow{k}N-m\]
Correct Answer: C
Solution :
Here: \[\overrightarrow{F}=2i-3\hat{j}+4\hat{k}N\] Position vector of a point \[\overrightarrow{r}=3\hat{i}+2\hat{j}+3\hat{k}\,m\] The torque acting at a point about the origin is given by \[\overrightarrow{\tau }=\overrightarrow{r}\times \overrightarrow{F}(3\hat{i}+2\hat{j}+3\hat{k})\times (2\hat{i}-3\hat{j}+4\hat{k})\] \[=\left| \begin{matrix} {\hat{i}} & {\hat{j}} & {\hat{k}} \\ 3 & 2 & 3 \\ 2 & -3 & 4 \\ \end{matrix} \right|\] \[=\hat{i}[8-(-9)-\hat{j}(12-6)+\hat{k}(-9-4)]\] \[=17\hat{i}-6\hat{j}-13\hat{k}N-m\]
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