# Solve the Following Questions

Question:

Let $\overrightarrow{\mathrm{a}}=2 \hat{\mathrm{i}}-\hat{\mathrm{j}}+2 \hat{\mathrm{k}}$ and $\overrightarrow{\mathrm{b}}=\hat{\mathrm{i}}+2 \hat{\mathrm{j}}-\hat{\mathrm{k}}$. Let a vector $\overrightarrow{\mathrm{v}}$ be in the plane containing $\overrightarrow{\mathrm{a}}$ and $\overrightarrow{\mathrm{b}}$. If $\overrightarrow{\mathrm{v}}$ is perpendicular to the vector $3 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}-\hat{\mathrm{k}}$ and its projection on $\vec{a}$ is 19 units, then $|2 \vec{v}|^{2}$ is equal to

Solution:

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

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

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

$\overrightarrow{\mathrm{v}}=\mathrm{x} \overrightarrow{\mathrm{a}}+\mathrm{y} \overrightarrow{\mathrm{b}}$

$\overrightarrow{\mathrm{v}}(3 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}-\mathrm{k})=0$

$\overrightarrow{\mathrm{v}} \cdot \hat{\mathrm{a}}=19$

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

$\overrightarrow{\mathrm{v}}=\lambda[(\overrightarrow{\mathrm{c}} \cdot \overrightarrow{\mathrm{b}}) \overrightarrow{\mathrm{a}}-(\overrightarrow{\mathrm{c}} \cdot \overrightarrow{\mathrm{a}}) \overrightarrow{\mathrm{b}}]$

$=\lambda\left[(3+4+1)(2 \hat{\mathrm{i}}-\hat{\mathrm{j}}+2 \hat{\mathrm{k}})-\left(\frac{6-2-2}{2}\right)(\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+\hat{\mathrm{k}})\right.$

$=\lambda[16 \hat{\mathrm{i}}-8 \hat{\mathrm{j}}+16 \hat{\mathrm{k}}-2 \hat{\mathrm{i}}-4 \hat{\mathrm{j}}+2 \hat{\mathrm{k}}]$

$\lambda[14 \hat{\mathrm{i}}-12 \hat{\mathrm{j}}+18 \hat{\mathrm{k}}] \cdot \frac{(2 \hat{\mathrm{i}}-\hat{\mathrm{j}}+2 \hat{\mathrm{k}})}{\sqrt{4+1+4}}=19$

$\lambda \frac{[28+12+36]}{3}=19$

$\lambda\left(\frac{76}{3}\right)=19$

$4 \lambda=3 \Rightarrow \lambda=\frac{3}{4}$

$\left|2 \mathrm{v}^{2}\right|=\left|2 \times \frac{3}{4}(14 \hat{\mathrm{i}}-12 \hat{\mathrm{j}}+18 \hat{\mathrm{k}})\right|^{2}$

$\frac{9}{4} \times 4(7 \hat{i}-6 \hat{j}+9 \hat{k})^{2}$

$=9(49+36+81)$

$=9(166)$

$=1494$