# Solve this following

Question:

Let three vectors $\overrightarrow{\mathrm{a}}, \overrightarrow{\mathrm{b}}$ and $\overrightarrow{\mathrm{c}}$ be such that $\overrightarrow{\mathrm{c}}$ is coplanar with $\vec{a}$ and $\vec{b}, \vec{a} \cdot \vec{c}=7$ and $\vec{b}$ is perpendicular to $\overrightarrow{\mathrm{c}}$, where $\overrightarrow{\mathrm{a}}=-\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}$ and

$\overrightarrow{\mathrm{b}}=2 \hat{\mathrm{i}}+\hat{\mathrm{k}}$, then the value of $2|\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}}|^{2}$

is

Solution:

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

$=\lambda((\overrightarrow{\mathrm{b}} \cdot \overrightarrow{\mathrm{b}}) \overrightarrow{\mathrm{a}}-(\overrightarrow{\mathrm{b}} \cdot \overrightarrow{\mathrm{a}}) \overrightarrow{\mathrm{b}})$

$=\lambda(5(-\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}})+2 \hat{\mathrm{i}}+\hat{\mathrm{k}})$

$=\lambda(-3 \hat{\mathrm{i}}+5 \hat{\mathrm{j}}+6 \hat{\mathrm{k}})$

$\vec{c} \cdot \vec{a}=7 \Rightarrow 3 \lambda+5 \lambda+6 \lambda=7$

$\lambda=\frac{1}{2}$

$\therefore 2\left|\left(\frac{-3}{2}-1+2\right) \hat{i}+\left(\frac{5}{2}+1\right) \hat{j}+(3+1+1) \hat{k}\right|^{2}$

$=2\left(\frac{1}{4}+\frac{49}{4}+25\right)=25+50=75$