A vector force with components $(1,2,3)$acts at the point$(3,2,1)$. Find the vector torque about the origin due to this force and find the torque about each of the coordinate axes.

The torque, $\mathcal{t}$is the cross product of the position role="math" localid="1653047985022" $\text{r}$and force role="math" localid="1653047993987" $\text{F}$, mathematically$\mathcal{t}=\text{r×F}$.

The given components of force is $(1,2,3)$and of point is $(3,2,1)$

The torque by the force about all the coordinate axes is to be found.

Find the position vector about the origin.

$$\begin{gathered} {\text{r}}_0\text{=r=3i+2j+k-(0i+0j+0k)} \\ =\text{3i+2j+k} \end{gathered}$$

About the origin the position vector is ${\text{r}}_0\text{=3i+2j+k}$.

From the given components of force, the force vector is $\text{F=}\mathbf{\text{i}}\text{+2}\mathbf{\text{j}}\text{+3}\mathbf{\text{k}}$.

Take the cross product of the force vector and the position vector about origin to obtain torque about origin.

role="math" localid="1653048748688" $$\begin{gathered} \mathcal{t}=\left|\begin{array}{ccc}\mathbf{\text{i}}& \mathbf{\text{j}}& \mathbf{\text{k}}\\ 3& 2& 1\\ 1& 2& 3\end{array}\right| \\ =\mathbf{\text{i}}\text{(2.3-1.2)-}\mathbf{\text{j}}\text{(3.3-1.1)+}\mathbf{\text{k}}\text{(3.2-2.1)} \\ =\mathbf{\text{i}}\text{(6-2)-}\mathbf{\text{j}}\text{(9-1)+}\mathbf{\text{k}}\text{(6-2)} \\ =4\mathbf{\text{i}}\text{-8}\mathbf{\text{j}}\text{+4}\mathbf{\text{k}} \end{gathered}$$

Therefore, the torque about origin is$=4\mathbf{\text{i}}\text{-8}\mathbf{\text{j}}\text{+4}\mathbf{\text{k}}$

Substitute $\hat{n}=1$and the obtained value of $\mathbf{\text{r}}\text{×}\mathbf{\text{F}}$into $\mathbf{\text{n}}\text{.}\mathbf{\text{(r×F)}}$to find the torque about the localid="1653049259253" $x$-axis,

localid="1653278700627" $$\begin{gathered} {\mathcal{t}}_x=\mathbf{\text{i}}\text{.(4}\mathbf{\text{i}}\text{-8}\mathbf{\text{j}}\text{+4}\mathbf{\text{k)}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}=4\text{N} \end{gathered}$$

Substitute $\hat{\text{n}}=\mathbf{\text{j}}$and the obtained value of $\mathbf{\text{r}}\text{×}\mathbf{\text{F}}$into $\mathbf{\text{n}}\text{.}\mathbf{\text{(r}}\text{×}\mathbf{\text{F)}}$to find the torque about the $y$-axis,

$$\begin{gathered} {\mathcal{\text{t}}}_y=\text{j.(4}\mathbf{\text{i}}\text{+8}\mathbf{\text{j}}\text{+4}\mathbf{\text{k}}\text{)} \\ =-8\text{N} \end{gathered}$$

Substitute $\hat{\text{n}}=\mathbf{\text{k}}$and the obtained value of $\mathbf{\text{r}}\times \mathbf{\text{F}}$into $\mathbf{\text{n.}}\text{(}\mathbf{\text{r}}\times \mathbf{\text{F)}}$to find the torque about the $y$-axis,

$$\begin{gathered} {\mathcal{t}}_z=\text{k.(4}\mathbf{\text{i}}\text{-8}\mathbf{\text{j}}\text{+4}\mathbf{\text{k}}\text{)} \\ =4\text{N} \end{gathered}$$

Therefore, torque about the origin, $x$-axis, $y$-axis and $z$-axis is 0 N, $4$N,localid="1653049699231" $-8$N and$4$N respectively.