If $\overrightarrow{\mathrm{F}}=2 \hat{i}+\hat{j}-\hat{k}$ and $\vec{r}=3 \hat{i}+2 \hat{j}-2 \hat{k}$, then the scalar and vector products of $\vec{F}$ and $\vec{r}$ have the magnitudes respectively as:

Question

Aicharya · Accepted Answer

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

If $\overrightarrow{\mathrm{F}}=2 \hat{i}+\hat{j}-\hat{k}$ and $\vec{r}=3 \hat{i}+2 \hat{j}-2 \hat{k}$, then…

Solution