A force $\vec{F}=4 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+4 \hat{\mathrm{k}}$ is applied on an intersection point of $\mathrm{x}=2$ plane and $\mathrm{x}$-axis. The magnitude of torque of this force about a point
$(2,3,4)$ is_____
(Round off to the Nearest Integer)
$(20)$
$\vec{\tau}=\overrightarrow{\mathrm{r}} \times \overrightarrow{\mathrm{F}}$
$\overrightarrow{\mathrm{r}}=(2 \hat{\mathrm{i}})-(2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+4 \hat{\mathrm{k}})=-3 \hat{\mathrm{j}}-4 \hat{\mathrm{k}}$
$\& \overrightarrow{\mathrm{F}}=4 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+4 \hat{\mathrm{k}}$
$\vec{\tau}=\overrightarrow{\mathrm{r}} \times \overrightarrow{\mathrm{F}}=\left|\begin{array}{ccc}\hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ 0 & -3 & -4 \\ 4 & 3 & 4\end{array}\right|$
$=\hat{\mathrm{i}}(-12+12)-\hat{\mathrm{j}}(0+16)+\hat{\mathrm{k}}(0+12)$
$=-16 \hat{\mathrm{i}}+12 \hat{\mathrm{k}}$
$\therefore|\vec{\tau}|=\sqrt{16^{2}+12^{2}}=20$
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