Question

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

Answer 1

Normal of plane containing $\overrightarrow{a}$ and $\overrightarrow{b}$ is
$\overrightarrow{n}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 2& -1& 2\\ 1& 2& -1\end{array}\right|=-3\hat{i}+4\hat{j}+5\hat{k}$
Also $\overrightarrow{v}$ is perpendicular to $(3\hat{i}+2\hat{j}-\hat{k})\text{and}\overrightarrow{n}$
$\overrightarrow{v}=\lambda \left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 3& 2& -1\\ -3& 4& 5\end{array}\right|=\lambda [14\hat{i}-12\hat{j}+18\hat{k}]$
Given
$\frac{\overrightarrow{a}\cdot \overrightarrow{v}}{|\overrightarrow{a}|}=19\Rightarrow \frac{\lambda \left(\right(2\left)\right(14)+(-12\left)\right(-1)+(18\left)\right(2\left)\right)}{3}=19$
$\Rightarrow \lambda =\frac{3}{4}$
$\therefore \overrightarrow{v}=\frac{3}{4}(14\hat{i}-12\hat{j}+18\hat{k})\Rightarrow 2\overrightarrow{v}=3(7\hat{i}-6\hat{j}+9\hat{k})$
$\Rightarrow |2\overrightarrow{v}{|}^2=1494$