The torque of force vec F = -2hat i + 2hat j | 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 $\vec{F}$:$\vec{	au}

Vedclass · Accepted Answer

$-8\hat i - 5\hat j - 2\hat k$

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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 $\vec{F}$:$\vec{	au} = \vec{r} 	imes \vec{F}$Given $\vec{r} = \hat{i} - 2\hat{j} + \hat{k}$ and $\vec{F} = -2\hat{i} + 2\hat{j} + 3\hat{k}$.Calculating the cross product using the determinant method:$\vec{	au} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & -2 & 1 \ -2 & 2 & 3 \end{vmatrix}$Expanding the determinant:$\vec{	au} = \hat{i}[(-2)(3) - (1)(2)] - \hat{j}[(1)(3) - (1)(-2)] + \hat{k}[(1)(2) - (-2)(-2)]$$\vec{	au} = \hat{i}[-6 - 2] - \hat{j}[3 + 2] + \hat{k}[2 - 4]$$\vec{	au} = -8\hat{i} - 5\hat{j} - 2\hat{k}$Thus,the torque is $-8\hat{i} - 5\hat{j} - 2\hat{k}$.

The torque of force $\vec F = -2\hat i + 2\hat j + 3\hat k$ acting on a point $\vec r = \hat i - 2\hat j + \hat k$ about the origin will be:

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