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A force $\vec{F}=3 \hat{i}+4 \hat{j}-3 \hat{k}$ is applied at the point $\mathrm{P}$, whose
position vector is $\vec{r}=\widehat{2} i-2 \hat{j}-3 \hat{k}$. What is the magnitude of the moment of the force about the origin?
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position vector is $\vec{r}=\widehat{2} i-2 \hat{j}-3 \hat{k}$. What is the magnitude of the moment of the force about the origin?
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Verified Answer
The correct answer is:
23 units
Moment of force, $\mathrm{m}=\overrightarrow{\mathrm{r}} \times \overrightarrow{\mathrm{F}}$
$\mathrm{m}=\left|\begin{array}{ccc}\hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ 2 & -2 & -3 \\ 3 & 4 & -3\end{array}\right|$
$=\hat{\mathrm{i}}(6+12)-\hat{\mathrm{j}}(-6+9)+\hat{\mathrm{k}}(8+6)$
$=18 \hat{\mathrm{i}}-3 \hat{\mathrm{j}}+14 \hat{\mathrm{k}}$
$=\sqrt{(18)^{2}+(-3)^{2}+(14)^{2}}$
$=\sqrt{529}=23$ units.
$\mathrm{m}=\left|\begin{array}{ccc}\hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ 2 & -2 & -3 \\ 3 & 4 & -3\end{array}\right|$
$=\hat{\mathrm{i}}(6+12)-\hat{\mathrm{j}}(-6+9)+\hat{\mathrm{k}}(8+6)$
$=18 \hat{\mathrm{i}}-3 \hat{\mathrm{j}}+14 \hat{\mathrm{k}}$
$=\sqrt{(18)^{2}+(-3)^{2}+(14)^{2}}$
$=\sqrt{529}=23$ units.
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