The moment of the force, F → = 4 i ^ + 5 j ^ − 6 k ^ at ( 2 , 0 , − 3 ) , about the point ( 2 , − 2 , − 2 ) …

1. \(-7 \hat{\mathrm{i}}-8 \hat{\mathrm{j}}-4 \widehat{\mathrm{k}}\)
2. \(-4 \hat{\mathrm{i}}-\hat{\mathrm{j}}-8 \widehat{\mathrm{k}}\)
3. \(-8 \hat{\mathrm{i}}-4 \hat{\mathrm{j}}-7 \hat{\mathrm{k}}\)
4. \(-7 \hat{\mathrm{i}}-4 \hat{\mathrm{j}}-8 \widehat{\mathrm{k}}\)

Solution

$\overrightarrow{F}=4\widehat{i}+5\widehat{j}-6\widehat{k}$
$\overrightarrow{r}=\left(2-2\right)\widehat{i}+\left(0-\left(-2\right)\right)\widehat{j}+\left(-3-\left(-2\right)\right)\widehat{k}=2\widehat{j}-1\widehat{k}$
$\overrightarrow{r}\times \overrightarrow{F}=\left(2\widehat{j}-1\widehat{k}\right)\times \left(4\widehat{k}+5\widehat{j}-6\widehat{k}\right)$
$= \left|\begin{array}{ccc}\widehat{i}& \widehat{j}& \widehat{k}\\ 0& 2& -1\\ 4& 5& -6\end{array}\right|=-7\widehat{i}-4\widehat{j}-8\widehat{k}$

Asked in: NEET 2018

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