The torque of a force vec{F} = 5 hat{i} + 3 h | 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

$-17 \hat{i} + 19 \hat{j} - 4 \hat{k}$

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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 $\vec{F}$:$\vec{	au} = \vec{r} 	imes \vec{F}$Using the determinant form for the cross product:$\vec{	au} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 2 & 2 & 1 \ 5 & 3 & -7 \end{vmatrix}$Expanding the determinant:$\vec{	au} = \hat{i} [(2)(-7) - (1)(3)] - \hat{j} [(2)(-7) - (1)(5)] + \hat{k} [(2)(3) - (2)(5)]$$\vec{	au} = \hat{i} [-14 - 3] - \hat{j} [-14 - 5] + \hat{k} [6 - 10]$$\vec{	au} = -17 \hat{i} + 19 \hat{j} - 4 \hat{k}$

The torque of a force $\vec{F} = 5 \hat{i} + 3 \hat{j} - 7 \hat{k}$ about the origin is $\vec{\tau}$. If the force acts on a particle whose position vector is $\vec{r} = 2 \hat{i} + 2 \hat{j} + \hat{k}$,then the value of $\vec{\tau}$ will be:

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