If vec{omega} = (3hat{i} + 4hat{j} + 5hat{k}) | vedclass

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(C) The linear velocity $\vec{v}$ is given by the cross product of angular velocity $\vec{\omega}$ and position vector $\vec{r}$:$\vec{v} = \vec{\omeg

Vedclass · Accepted Answer

$(22, -4, -10)$

Vedclass · Answer

(C) The linear velocity $\vec{v}$ is given by the cross product of angular velocity $\vec{\omega}$ and position vector $\vec{r}$:$\vec{v} = \vec{\omega} 	imes \vec{r}$$\vec{v} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 3 & 4 & 5 \ 1 & -2 & 3 \end{vmatrix}$Expanding the determinant:$\vec{v} = \hat{i}((4)(3) - (5)(-2)) - \hat{j}((3)(3) - (5)(1)) + \hat{k}((3)(-2) - (4)(1))$$\vec{v} = \hat{i}(12 + 10) - \hat{j}(9 - 5) + \hat{k}(-6 - 4)$$\vec{v} = 22\hat{i} - 4\hat{j} - 10\hat{k}$Thus,the linear velocity vector is $(22, -4, -10) 	ext{ m/s}$.

If $\vec{\omega} = (3\hat{i} + 4\hat{j} + 5\hat{k}) \text{ rad/s}$ and $\vec{r} = (\hat{i} - 2\hat{j} + 3\hat{k}) \text{ m}$,then the linear velocity vector of the particle is:

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