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

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