# solve this

Question:

A force $\overrightarrow{\mathrm{F}}=4 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+4 \hat{\mathrm{k}}$ is applied on an

intersection point of $x=2$ plane and $x$-axis. The magnitude of torque of this force about a point $(2,3,4)$ is . (Round off to the Nearest Integer)

Solution:

Ans. $(20)$

$\vec{\tau}=\overrightarrow{\mathrm{r}} \times \overrightarrow{\mathrm{F}}$

$\overrightarrow{\mathrm{r}}=(2 \hat{\mathrm{i}})-(2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+4 \hat{\mathrm{k}})=-3 \hat{\mathrm{j}}-4 \hat{\mathrm{k}}$

$\& \overrightarrow{\mathrm{F}}=4 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+4 \hat{\mathrm{k}}$

$\vec{\tau}=\overrightarrow{\mathrm{r}} \times \overrightarrow{\mathrm{F}}=\left|\begin{array}{ccc}\hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ 0 & -3 & -4 \\ 4 & 3 & 4\end{array}\right|$

$=\hat{\mathrm{i}}(-12+12)-\hat{\mathrm{j}}(0+16)+\hat{\mathrm{k}}(0+12)$

$=-16 \hat{\mathrm{i}}+12 \hat{\mathrm{k}}$

$\therefore \quad|\vec{\tau}|=\sqrt{16^{2}+12^{2}}=20$