# Let vector e be a vector perpendicular to

Question:

Let $\overrightarrow{\mathrm{c}}$ be a vector perpendicular to the vectors $\overrightarrow{\mathrm{a}}=\hat{\mathrm{i}}+\hat{\mathrm{j}}-\hat{\mathrm{k}}$ and $\overrightarrow{\mathrm{b}}=\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+\hat{\mathrm{k}}$.

If $\overrightarrow{\mathrm{c}} \cdot(\hat{\mathrm{i}}+\hat{\mathrm{j}}+3 \hat{\mathrm{k}})=8$ then the value of $\overrightarrow{\mathrm{c}} .(\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}})$ is equal to________.

Solution:

$\overrightarrow{\mathrm{c}}=\lambda(\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}})$

$\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}=\left|\begin{array}{ccc}\hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ 1 & 1 & -1 \\ 1 & 2 & 1\end{array}\right|$

$(\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}})=3 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}+\hat{\mathrm{k}}$

$\overrightarrow{\mathrm{c}} \cdot(\hat{\mathrm{i}}+\hat{\mathrm{j}}+3 \hat{\mathrm{k}})=\lambda(3 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}+\hat{\mathrm{k}}) \cdot(\hat{\mathrm{i}}+\hat{\mathrm{j}}+3 \hat{\mathrm{k}})$

$\Rightarrow \lambda(4)=8 \Rightarrow \lambda=2$

$\overrightarrow{\mathrm{c}}=2(\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}})$

$\overrightarrow{\mathrm{c}} \cdot(\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}})=2|\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}|^{2}=28$