Let three vectors

Question:

Let three vectors $\vec{a}, \vec{b}$ and $\vec{c}$ be such that $\vec{c}$ is coplanar with $\vec{a}$ and $\vec{b}, \vec{a} \cdot \vec{c}=7$ and $\vec{b}$ is perpendicular to $\vec{c}$, where $\vec{a}=-\hat{i}+\hat{j}+\hat{k}$ and $\vec{b}=2 \hat{i}+\hat{k}$, then the value of $2|\vec{a}+\vec{b}+\vec{c}|^{2}$ is

Solution:

$\vec{C}=\lambda(\vec{b} \times(\vec{a} \times \vec{b}))$

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

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

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

$\vec{C} \cdot \vec{a}=7 \rightarrow 3 \lambda+5 \lambda+6 \lambda=7$

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

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

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