question_answer

The moment of the force, \[\overrightarrow{\text{F}}\text{=4}\widehat{\text{i}}\text{+5}\widehat{\text{j}}\text{-6}\widehat{\text{k}}\] at (2, 0, -3), about the point (2, -2, -2), is given by [NEET - 2018]

A) \[\text{-7}\widehat{\text{i}}\text{-8}\widehat{\text{j}}\text{-4}\widehat{\text{k}}\]

B) \[\text{-4}\widehat{\text{i}}\text{-}\widehat{\text{j}}\text{-8}\widehat{\text{k}}\]

C) \[\text{-8}\widehat{\text{i}}\text{-4}\widehat{\text{j}}\text{-7}\widehat{\text{k}}\]

D) \[\text{-7}\widehat{\text{i}}\text{-4}\widehat{\text{j}}\text{-8}\widehat{\text{k}}\]

Correct Answer: D

Solution :

[d]

\[\overrightarrow{\tau }=(\overrightarrow{r}-{{\overrightarrow{r}}_{0}})\times \overrightarrow{\text{F}}\]                                  (i)

\[\overrightarrow{r}-{{\overrightarrow{r}}_{_{0}}}=(2\widehat{\text{i}}-0\widehat{\text{j}}-3\widehat{\text{k}})-(2\widehat{i}-2\widehat{\text{j}}-2\widehat{\text{k}})\]

\[=0\widehat{\text{i}}+2\widehat{\text{j}}-\widehat{\text{k}}\]

\[\overrightarrow{\tau }=\left| \begin{matrix}    \widehat{i} & \widehat{j} & \widehat{k}  \\    0 & 2 & -1  \\    4 & 5 & -6  \\ \end{matrix} \right|=-7\widehat{i}-4\widehat{j}-8\widehat{k}\]