What is the torque of the force vec{F} = (2ha | vedclass

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(C) The torque $\vec{	au}$ of a force $\vec{F}$ acting at a position vector $\vec{r}$ relative to the origin is given by the cross product: $\vec{	a

Vedclass · Accepted Answer

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

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(C) The torque $\vec{	au}$ of a force $\vec{F}$ acting at a position vector $\vec{r}$ relative to the origin is given by the cross product: $\vec{	au} = \vec{r} 	imes \vec{F}$.Given:$\vec{r} = (3\hat{i} + 2\hat{j} + 3\hat{k}) 	ext{ m}$$\vec{F} = (2\hat{i} - 3\hat{j} + 4\hat{k}) 	ext{ N}$Calculating the cross product using the determinant method:$\vec{	au} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 3 & 2 & 3 \ 2 & -3 & 4 \end{vmatrix}$Expanding the determinant:$\vec{	au} = \hat{i}((2)(4) - (-3)(3)) - \hat{j}((3)(4) - (2)(3)) + \hat{k}((3)(-3) - (2)(2))$$\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}$.

What is 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?

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