# Prove the following

Question:

Let $\overrightarrow{\mathrm{p}}=2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+\hat{\mathrm{k}}$ and $\overrightarrow{\mathrm{q}}=\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+\hat{\mathrm{k}}$ be two vectors.

If a vector $\overrightarrow{\mathrm{r}}=(\alpha \hat{\mathrm{i}}+\beta \hat{\mathrm{j}}+\gamma \hat{\mathrm{k}})$ is perpendicular to each of the vectors $(\overrightarrow{\mathrm{p}}+\overrightarrow{\mathrm{q}})$ and $(\overrightarrow{\mathrm{p}}-\overrightarrow{\mathrm{q}})$, and $|\overrightarrow{\mathrm{r}}|=\sqrt{3}$, then $|\alpha|+|\beta|+|\gamma|$ is equal to

Solution:

$\overrightarrow{\mathrm{p}}=2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+\hat{\mathrm{k}} \quad$ (Given)

$\overrightarrow{\mathrm{q}}=\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+\hat{\mathrm{k}}$

$\operatorname{Now}(\overrightarrow{\mathrm{p}}+\overrightarrow{\mathrm{q}}) \times(\overrightarrow{\mathrm{p}}-\overrightarrow{\mathrm{q}})=\left|\begin{array}{lll}\hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ 3 & 5 & 2 \\ 1 & 1 & 0\end{array}\right|$

$=-2 \hat{i}-2 \hat{j}-2 \hat{k}$

$\Rightarrow \overrightarrow{\mathrm{r}}=\pm \sqrt{3} \frac{((\overrightarrow{\mathrm{p}}+\overrightarrow{\mathrm{q}}) \times(\overrightarrow{\mathrm{p}}-\overrightarrow{\mathrm{q}}))}{|(\overrightarrow{\mathrm{p}}+\overrightarrow{\mathrm{q}}) \times(\overrightarrow{\mathrm{p}}-\overrightarrow{\mathrm{q}})|}=\pm \frac{\sqrt{3}(-2 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}-2 \hat{\mathrm{k}})}{\sqrt{2^{2}+2^{2}+2^{2}}}$

$\overrightarrow{\mathrm{r}}=\pm(-\hat{\mathrm{i}}-\hat{\mathrm{j}}-\hat{\mathrm{k}})$

According to question

$\overrightarrow{\mathrm{r}}=\alpha \hat{\mathrm{i}}+\beta \hat{\mathrm{j}}+\gamma \hat{\mathrm{k}}$

So $|\alpha|=1,|\beta|=1,|\gamma|=1$

$\Rightarrow|\alpha|+|\beta|+|\gamma|=3$