Question

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

Answer 1

Answer: 1494

$\vec{a}=2\hat{i}-\hat{j}+2\hat{k}$
$\vec{b}=\hat{i}+2\hat{j}-\hat{k}$
$\vec{c}=3\hat{i}+2\hat{j}-\hat{k}$
$\vec{v}=x\vec{a}+yb$
$\vec{v}(3\hat{i}+2\hat{j}-k)=0$
$\vec{v}\cdot \hat{a}=19$
$\vec{v}=\lambda \vec{c}\times (\vec{a}\times \vec{b})$
$\vec{v}=\lambda [(\vec{c}\cdot \vec{b})\vec{a}-(\vec{c}\cdot \vec{a})\vec{b}]$
$=\lambda [(3+4+1)(2\hat{i}-\hat{j}+2\hat{k})-\left(\frac{6-2-2}{2}\right)(\hat{i}+2\hat{j}+\hat{k})$
$=\lambda [16\hat{i}-8\hat{j}+16\hat{k}-2\hat{i}-4\hat{j}+2\hat{k}]$
$\vec{v}=\lambda [14\hat{i}-12\hat{j}+18\hat{k}]$
$\lambda [14\hat{i}-12\hat{j}+18\hat{k}]\cdot \frac{(2\hat{i}-\hat{j}+2\hat{k})}{\sqrt{4+1+4}}=19$
$\lambda \frac{[28+12+36]}{3}=19$
$\lambda \left(\frac{76}{3}\right)=19$
$4\lambda =3\Rightarrow \lambda =\frac{3}{4}$
$\left|2v^2\right|={\left|2\times \frac{3}{4}(14\hat{i}-12\hat{j}+18\hat{k})\right|}^2$
$\frac{9}{4}\times 4(7\hat{i}-6\hat{j}+9\hat{k}{)}^2$
$=9(49+36+81)$
$=9(166)$
$=1494$