If overrightarrow{F}=(4 hat{i}-10 hat{j}) tex | vedclass

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(B) The cross product of two vectors $\overrightarrow{r}$ and $\overrightarrow{F}$ is given by the determinant method:$\overrightarrow{r} 	imes \over

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$62 \hat{k} 	ext{ Nm}$

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(B) The cross product of two vectors $\overrightarrow{r}$ and $\overrightarrow{F}$ is given by the determinant method:$\overrightarrow{r} 	imes \overrightarrow{F} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ r_x & r_y & r_z \ F_x & F_y & F_z \end{vmatrix}$Given $\overrightarrow{r} = (-5 \hat{i} - 3 \hat{j} + 0 \hat{k}) 	ext{ m}$ and $\overrightarrow{F} = (4 \hat{i} - 10 \hat{j} + 0 \hat{k}) 	ext{ N}$.Substituting the values:$\overrightarrow{r} 	imes \overrightarrow{F} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ -5 & -3 & 0 \ 4 & -10 & 0 \end{vmatrix}$$= \hat{i}((-3)(0) - (0)(-10)) - \hat{j}((-5)(0) - (0)(4)) + \hat{k}((-5)(-10) - (-3)(4))$$= \hat{i}(0) - \hat{j}(0) + \hat{k}(50 - (-12))$$= \hat{k}(50 + 12) = 62 \hat{k} 	ext{ Nm}$.

If $\overrightarrow{F}=(4 \hat{i}-10 \hat{j}) \text{ N}$ and $\overrightarrow{r}=(-5 \hat{i}-3 \hat{j}) \text{ m}$,then $(\overrightarrow{r} \times \overrightarrow{F})$ is

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