Let $\vec{a}=2 \hat{i}-3 \hat{j}-5 \hat{k}$ and $\vec{b}=3 \hat{i}+2 \hat{j}-5 \hat{k}$ be two vectors and…
Let $\vec{a}=2 \hat{i}-3 \hat{j}-5 \hat{k}$ and $\vec{b}=3 \hat{i}+2 \hat{j}-5 \hat{k}$ be two vectors and $\overrightarrow{\mathrm{r}}$ be a vector in the plane of $\vec{a}$ and $\vec{b}$. If $\vec{r}$ is orthogonal to the vector $5 \hat{i}-2 \hat{j}+3 \hat{k}$ and the magnitude of $\vec{r}$ is $\sqrt{94}$, then $|\vec{r} \cdot \vec{b}|=$
$36$
$38$
$42$
$46$
Solution
Let $\overrightarrow{\mathrm{c}}=\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}=\left|\begin{array}{ccc}\hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ 2 & -3 & -5 \\ 3 & 2 & -5\end{array}\right|=25 \hat{\mathrm{i}}-5 \hat{\mathrm{j}}+13 \hat{\mathrm{k}}$
Since, $\overrightarrow{\mathrm{r}}$ lies on plane containing $\overrightarrow{\mathrm{a}}$ and $\overrightarrow{\mathrm{b}}$ therefore $\overrightarrow{\mathrm{r}}$ is perpendicular to $\overrightarrow{\mathrm{c}}$ and given that $\overrightarrow{\mathrm{r}}$ is also perpendicular to vector $5 \hat{i}-2 \hat{j}+3 \hat{k}$
$\begin{aligned}
& \therefore \overrightarrow{\mathrm{r}}=\lambda\left|\begin{array}{ccc}
\hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\
25 & -5 & 13 \\
5 & -2 & 3
\end{array}\right|=\lambda[11 \hat{\mathrm{i}}-10 \hat{\mathrm{j}}-25 \hat{\mathrm{k}}] \text { and }|\overrightarrow{\mathrm{r}}|=\sqrt{94} \\
& \therefore \overrightarrow{\mathrm{r}}=\frac{1}{3}(11 \hat{\mathrm{i}}-10 \hat{\mathrm{j}}-25 \hat{\mathrm{k}})
\end{aligned}$
Now, $|\vec{r}, \vec{b}|=\frac{1}{3}(33-20+125)=46$.