A force vec{F} = (2hat{i} - hat{j} + 3hat{k}) | vedclass

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(B) The torque $\vec{	au}$ about the origin is given by the cross product of the position vector $\vec{r}$ and the force vector $\vec{F}$.Given,$\vec

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$3(\hat{i} - \hat{j} - \hat{k})$

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

$A$ force $\vec{F} = (2\hat{i} - \hat{j} + 3\hat{k}) \text{ N}$ is acting at a point $(-1, 2, -3) \text{ m}$. Find its torque about the origin.

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