The torque of the force vec{F} = (2hat{i} - 3 | vedclass

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(B) 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

$17\hat{i} - 6\hat{j} - 13\hat{k}$

Vedclass · Answer

(B) 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}$$\vec{	au} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 3 & 2 & 3 \ 2 & -3 & 4 \end{vmatrix}$Expanding the determinant:$\vec{	au} = \hat{i} [(2)(4) - (3)(-3)] - \hat{j} [(3)(4) - (3)(2)] + \hat{k} [(3)(-3) - (2)(2)]$$\vec{	au} = \hat{i} [8 + 9] - \hat{j} [12 - 6] + \hat{k} [-9 - 4]$$\vec{	au} = 17\hat{i} - 6\hat{j} - 13\hat{k} 	ext{ N m}$

The torque of the force $\vec{F} = (2\hat{i} - 3\hat{j} + 4\hat{k}) \text{ N}$ acting at the point $\vec{r} = (3\hat{i} + 2\hat{j} + 3\hat{k}) \text{ m}$ about the origin is:

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