The torque due to the force vec{F} = (2 hat{i | vedclass

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

Vedclass · Accepted Answer

$\hat{i} - \hat{k}$

Vedclass · Answer

(C) The torque $\vec{	au}$ is given by the cross product of the position vector $\vec{r}$ and the force vector $\vec{F}$.$\vec{	au} = \vec{r} 	imes \vec{F} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & 1 & 1 \ 2 & 1 & 2 \end{vmatrix}$Expanding the determinant:$\vec{	au} = \hat{i}(1 	imes 2 - 1 	imes 1) - \hat{j}(1 	imes 2 - 1 	imes 2) + \hat{k}(1 	imes 1 - 1 	imes 2)$$\vec{	au} = \hat{i}(2 - 1) - \hat{j}(2 - 2) + \hat{k}(1 - 2)$$\vec{	au} = \hat{i} - 0\hat{j} - \hat{k} = \hat{i} - \hat{k}$.

The torque due to the force $\vec{F} = (2 \hat{i} + \hat{j} + 2 \hat{k})$ about the origin,acting on a particle whose position vector is $\vec{r} = (\hat{i} + \hat{j} + \hat{k})$,is:

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