A) \[-21\mathbf{\hat{i}}+3\mathbf{\hat{j}}+5\mathbf{\hat{k}}\]
B) \[-14\mathbf{\hat{i}}+3\mathbf{\hat{j}}+16\mathbf{\hat{k}}\]
C) \[4\mathbf{\hat{i}}+4\mathbf{\hat{j}}+6\mathbf{\hat{k}}\]
D) \[14\mathbf{\hat{i}}+38\mathbf{\hat{j}}+16\mathbf{\hat{k}}\]
Correct Answer: D
Solution :
Key Idea: The torque of a force is the cross product of \[\overrightarrow{\mathbf{r}}\] and \[\overrightarrow{\mathbf{F}}\] in the same order. Given: \[\overrightarrow{\mathbf{r}}=7\widehat{\mathbf{i}}+3\mathbf{\hat{j}}+\mathbf{\hat{k}},\,\overrightarrow{\mathbf{F}}=-3\mathbf{\hat{i}}+\mathbf{\hat{j}}+5\mathbf{\hat{k}}\] \[\overrightarrow{\tau }=\overrightarrow{\mathbf{r}}\times \overrightarrow{\mathbf{F}}=(7\mathbf{\hat{i}}+3\mathbf{\hat{j}}+\mathbf{\hat{k}})\,\times (-3\mathbf{\hat{i}}+\mathbf{j}+5\mathbf{\hat{k}})\] \[=\left| \begin{matrix} {\mathbf{\hat{i}}} & {\mathbf{\hat{j}}} & {\mathbf{\hat{k}}} \\ 7 & 3 & 1 \\ -3 & 1 & 5 \\ \end{matrix} \right|\] \[=\mathbf{\hat{i}}(15-1)\mathbf{\hat{j}}\,(35+3)+\mathbf{\hat{k}}(7+9)\] \[=14\mathbf{\hat{i}}-38\mathbf{\hat{j}}+16\mathbf{\hat{k}}\] Alternative: \[\overrightarrow{\tau }=\overrightarrow{\mathbf{r}}\times \overrightarrow{\mathbf{F}}=(7\mathbf{\hat{i}}+3\mathbf{\hat{j}}+\mathbf{\hat{k}})\times (-3\mathbf{\hat{i}}+\mathbf{\hat{j}}+5\mathbf{\hat{k}})\] \[=-21(\mathbf{\hat{i}}\times \mathbf{\hat{i}})+7(\mathbf{\hat{i}}\times \mathbf{\hat{j}})+35(\mathbf{\hat{i}}\times \mathbf{\hat{k}})\] \[-9(\mathbf{\hat{j}}\times \mathbf{\hat{i}})+3(\mathbf{\hat{j}}\times \mathbf{\hat{j}})+15(\mathbf{\hat{j}}\times \mathbf{\hat{k}})\] \[-3(\mathbf{\hat{k}}\times \mathbf{\hat{i}})+(\mathbf{\hat{k}}\times \mathbf{\hat{j}})+5(\mathbf{\hat{k}}\times \mathbf{\hat{k}})\] \[=0+7\mathbf{\hat{k}}-35\mathbf{\hat{j}}+9\mathbf{\hat{k}}+0+15\mathbf{\hat{i}}-3\mathbf{\hat{j}}-\mathbf{\hat{i}}=0\] \[=14\mathbf{\hat{i}}-38\mathbf{\hat{j}}+16\mathbf{\hat{k}}\]
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