A force vec{F} = 4hat{i} - 5hat{j} + 3hat{k} | vedclass

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(D) The position vector of the point where the force is applied is $\vec{r_1} = \hat{i} + 2\hat{j} + 3\hat{k}$.The position vector of the point about

Vedclass · Accepted Answer

$42\hat{i} + 30\hat{j} - 6\hat{k}$

Vedclass · Answer

(D) The position vector of the point where the force is applied is $\vec{r_1} = \hat{i} + 2\hat{j} + 3\hat{k}$.The position vector of the point about which the torque is to be calculated is $\vec{r_2} = 3\hat{i} - 2\hat{j} - 3\hat{k}$.The relative position vector $\vec{r}$ is given by $\vec{r} = \vec{r_1} - \vec{r_2} = (\hat{i} + 2\hat{j} + 3\hat{k}) - (3\hat{i} - 2\hat{j} - 3\hat{k}) = -2\hat{i} + 4\hat{j} + 6\hat{k}$.The torque $\vec{	au}$ is given by the cross product $\vec{	au} = \vec{r} 	imes \vec{F}$.$\vec{	au} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ -2 & 4 & 6 \ 4 & -5 & 3 \end{vmatrix}$.$\vec{	au} = \hat{i}(4 	imes 3 - 6 	imes (-5)) - \hat{j}((-2) 	imes 3 - 6 	imes 4) + \hat{k}((-2) 	imes (-5) - 4 	imes 4)$.$\vec{	au} = \hat{i}(12 + 30) - \hat{j}(-6 - 24) + \hat{k}(10 - 16)$.$\vec{	au} = 42\hat{i} + 30\hat{j} - 6\hat{k} 	ext{ Nm}$.

$A$ force $\vec{F} = 4\hat{i} - 5\hat{j} + 3\hat{k}$ is applied at a point with position vector $\vec{r_1} = \hat{i} + 2\hat{j} + 3\hat{k}$. The torque about the point with position vector $\vec{r_2} = 3\hat{i} - 2\hat{j} - 3\hat{k}$ is:

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