The torque of a force overrightarrow{F} = -3h | vedclass

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Question

(A) The torque $\overrightarrow{	au}$ is given by the cross product of the position vector $\overrightarrow{r}$ and the force vector $\overrightarrow

Vedclass · Accepted Answer

$14\hat{i} - 38\hat{j} + 16\hat{k}$

Vedclass · Answer

(A) The torque $\overrightarrow{	au}$ is given by the cross product of the position vector $\overrightarrow{r}$ and the force vector $\overrightarrow{F}$.$\overrightarrow{	au} = \overrightarrow{r} 	imes \overrightarrow{F}$$\overrightarrow{	au} = (7\hat{i} + 3\hat{j} + \hat{k}) 	imes (-3\hat{i} + \hat{j} + 5\hat{k})$Using the determinant method for the cross product:$\overrightarrow{	au} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 7 & 3 & 1 \ -3 & 1 & 5 \end{vmatrix}$$\overrightarrow{	au} = \hat{i}(3 	imes 5 - 1 	imes 1) - \hat{j}(7 	imes 5 - 1 	imes (-3)) + \hat{k}(7 	imes 1 - 3 	imes (-3))$$\overrightarrow{	au} = \hat{i}(15 - 1) - \hat{j}(35 + 3) + \hat{k}(7 + 9)$$\overrightarrow{	au} = 14\hat{i} - 38\hat{j} + 16\hat{k}$

The torque of a force $\overrightarrow{F} = -3\hat{i} + \hat{j} + 5\hat{k}$ acting at the point $\overrightarrow{r} = 7\hat{i} + 3\hat{j} + \hat{k}$ is:

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