A) Zero
B) \[42\widehat{i}-30\widehat{j}+6\widehat{k}\]
C) \[42\widehat{i}+30\widehat{j}+6\widehat{k}\]
D) \[42\widehat{i}+30\widehat{j}-6\widehat{k}\]
Correct Answer: D
Solution :
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 \[\overrightarrow{\tau }={{\overrightarrow{r'}}_{1}}\times \overrightarrow{F}=\left( -2\widehat{i}+4\widehat{j}+6\widehat{k} \right)\times (4\widehat{i}-5\widehat{j}+3\widehat{k})\] \[\overrightarrow{\tau }\,\,=\,\,\left| \begin{matrix} \overrightarrow{i} & \overrightarrow{j} & \overrightarrow{k} \\ -2 & 4 & 6 \\ 4 & -5 & 3 \\ \end{matrix} \right|\,\,=\,\,\widehat{i}(12+30)-\widehat{j}(-6-24)+\widehat{k}(10-16)\]\[=\,\,\left( 42\widehat{i}+30\widehat{j}-6\widehat{k} \right)N-m\]
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