A force vec{F}=(2 hat{i}+3 hat{j}-5 hat{k}) , | vedclass

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(D) The position vector $\vec{r}$ of the point of application relative to the point about which torque is calculated is given by $\vec{r} = \vec{r}_1

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$(13 \hat{j}-22 \hat{i}-\hat{k}) \, N m$

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(D) The position vector $\vec{r}$ of the point of application relative to the point about which torque is calculated is given by $\vec{r} = \vec{r}_1 - \vec{r}_2$.$\vec{r} = (2 \hat{i} + 4 \hat{j} + 7 \hat{k}) - (\hat{i} + 2 \hat{j} + 3 \hat{k}) = \hat{i} + 2 \hat{j} + 4 \hat{k} \, m$.The torque $\vec{	au}$ is defined as $\vec{	au} = \vec{r} 	imes \vec{F}$.$\vec{	au} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & 2 & 4 \ 2 & 3 & -5 \end{vmatrix}$.Calculating the determinant: $\vec{	au} = \hat{i} [ (2)(-5) - (4)(3) ] - \hat{j} [ (1)(-5) - (4)(2) ] + \hat{k} [ (1)(3) - (2)(2) ]$.$\vec{	au} = \hat{i} [ -10 - 12 ] - \hat{j} [ -5 - 8 ] + \hat{k} [ 3 - 4 ]$.$\vec{	au} = -22 \hat{i} + 13 \hat{j} - \hat{k} \, N m$.

$A$ force $\vec{F}=(2 \hat{i}+3 \hat{j}-5 \hat{k}) \, N$ acts at a point $\vec{r}_1=(2 \hat{i}+4 \hat{j}+7 \hat{k}) \, m$. The torque of the force about the point $\vec{r}_2=(\hat{i}+2 \hat{j}+3 \hat{k}) \, m$ is ............. $N m$.

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