A force of 3hat{i} + 2hat{j} - hat{k} text{ N | vedclass

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(C) Torque $\vec{	au}$ is given by the cross product of position vector $\vec{r}$ and force vector $\vec{F}$, i.e.,$\vec{	au} = \vec{r} 	imes \vec{

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$\sqrt{6} 	ext{ N m}$

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(C) Torque $\vec{	au}$ is given by the cross product of position vector $\vec{r}$ and force vector $\vec{F}$, i.e.,$\vec{	au} = \vec{r} 	imes \vec{F}$.Given $\vec{r} = \hat{i} + \hat{j} - \hat{k}$ and $\vec{F} = 3\hat{i} + 2\hat{j} - \hat{k}$.$\vec{	au} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & 1 & -1 \ 3 & 2 & -1 \end{vmatrix}$.$\vec{	au} = \hat{i}(-1 - (-2)) - \hat{j}(-1 - (-3)) + \hat{k}(2 - 3)$.$\vec{	au} = \hat{i}(1) - \hat{j}(2) + \hat{k}(-1) = \hat{i} - 2\hat{j} - \hat{k}$.The magnitude of torque is $|\vec{	au}| = \sqrt{(1)^2 + (-2)^2 + (-1)^2} = \sqrt{1 + 4 + 1} = \sqrt{6} 	ext{ N m}$.

$A$ force of $3\hat{i} + 2\hat{j} - \hat{k} \text{ N}$ acts on a particle with position vector $\hat{i} + \hat{j} - \hat{k} \text{ m}$. The magnitude of torque of the given force is

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