Physics - Vectors Question with Solution | TestHub

$\begin{array}{rl}& \overrightarrow{\mathrm{F}}=2\hat{i}+\hat{j}-\hat{k}\\ & \vec{r}=3\hat{i}+2\hat{j}-2\hat{k}\text{ (given) }\end{array}$ The scalar product of $\overrightarrow{\mathrm{F}}$ and $\vec{r}$ is $\begin{array}{rl}\vec{F}.\vec{r}& =(2)(3)+(1)(2)+(-1)(-2)\\ & =6+2+2=10\end{array}$ The vector product of $\vec{F}$ and $\vec{r}$ is $\begin{array}{rl}\overrightarrow{\mathrm{F}}\times \vec{r}& =\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 2& 1& -1\\ 3& 2& -2\end{array}\right|\\ & =\hat{i}[(1\times -2)-(2\times -1)]-\hat{j}[(2\times -2)\\ & -(-1\times 3)]+\hat{k}[(2\times 2)-(3\times 1)]\\ & =\hat{i}(0)-\hat{j}(-1)+\hat{k}(1)=\hat{j}+\hat{k}\end{array}$ Magnitude $=\sqrt{(1{)}^2+(1{)}^2}=\sqrt{2}$