Question:
A force $\vec{F}=(\hat{i}+2 \hat{j}+3 \hat{k}) \mathrm{N}$ acts at a point $(4 \hat{i}+3 \hat{j}-\hat{k}) \mathrm{m}$.
Then the magnitude of torque about the point
$(\hat{i}+2 \hat{j}+\hat{k}) \mathrm{m}$ will be $\sqrt{x} \mathrm{~N}-\mathrm{m}$. The value of $x$ is_____
Solution:
$(195)$
Given : $\vec{F}=(\hat{i}+2 \hat{j}+3 \hat{k}) \mathrm{N}$
And, $\vec{r}=[(4 \hat{i}+3 \hat{j}-\hat{k})-(\hat{i}+2 \hat{j}+\hat{k})]=3 \hat{i}+\hat{j}-2 \hat{k}$
Torque, $\tau=\vec{r} \times \vec{F}=(3 \hat{i}+\hat{j}-2 \hat{k}) \times(\hat{i}+2 \hat{j}+3 \hat{k})$
$\tau=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 3 & 1 & -2 \\ 1 & 2 & 3\end{array}\right|=7 \hat{i}-11 \hat{j}+5 \hat{k}$
Magnitude of torque, $|\vec{\tau}|=\sqrt{195}$.