A) \[\text{6}\overset{\hat{\ }}{\mathop{\text{i}}}\,-\text{6}\overset{\hat{\ }}{\mathop{\text{j}}}\,\text{+12}\overset{\hat{\ }}{\mathop{\text{k}}}\,\text{ N}-\text{m}\]
B) \[-\text{6}\overset{\hat{\ }}{\mathop{\text{i}}}\,\text{+6}\overset{\hat{\ }}{\mathop{\text{j}}}\,\text{-12}\overset{\hat{\ }}{\mathop{\text{k}}}\,\text{ N}-\text{m}\]
C) \[17\overset{\hat{\ }}{\mathop{\text{i}}}\,\text{-6}\overset{\hat{\ }}{\mathop{\text{j}}}\,\text{-13}\overset{\hat{\ }}{\mathop{\text{k}}}\,\text{ N}-\text{m}\]
D) \[-17\overset{\hat{\ }}{\mathop{\text{i}}}\,\text{+6}\overset{\hat{\ }}{\mathop{\text{j}}}\,\text{+13}\overset{\hat{\ }}{\mathop{\text{k}}}\,\text{ N}-\text{m}\]
Correct Answer: D
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, \[\vec{F}=2\hat{i}-3\hat{j}+4\,\hat{k},\,r=3\,\hat{i}+2\hat{j}+3\hat{k}\] \[\therefore \] \[\tau =(2\,\hat{i}-3\hat{j}+4\hat{k})\times (3\,\hat{i}+2\hat{j}+3\hat{k})\] \[\tau =\left| \begin{matrix} {\hat{i}} & {\hat{j}} & {\hat{k}} \\ 2 & -3 & 4 \\ 3 & 2 & 3 \\ \end{matrix} \right|\] \[\Rightarrow \] \[\tau =\hat{i}\,(-9-8)-\hat{j}\,(6-12)+\hat{k}\,(4+9)\] \[\Rightarrow \] \[\tau =-17\,\hat{i}+6\hat{j}+13\,\hat{k}\].
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