Question

Let $\vec{a}=\alpha \hat{i}+2\hat{j}-\hat{k}$ and $\vec{b}=-2\hat{i}+\alpha \hat{j}+\hat{k}$, where $\alpha \in R$. If the area of the parallelogram whose adjacent sides are represented by the vectors $\vec{a}$ and $\vec{b}$ is $\sqrt{15\left({\alpha }^2+4\right)}$, then the value of $2|\overrightarrow{a}{|}^2+(\overrightarrow{a}\cdot \overrightarrow{b})|\overrightarrow{b}{|}^2$ is equal to

• A. 10
• B. 7
• C. 9
• D. 14

Answer 1

Answer: (D)

$\vec{a}=\alpha \hat{i}+2\hat{j}-\hat{k},\vec{b}=-2\hat{i}+\alpha \hat{j}+\hat{k}$
area of parallelogram $=|\hat{a}\times \hat{b}|$
$|\hat{a}\times \hat{b}|=\sqrt{(\alpha +2{)}^2+(\alpha -2{)}^2+{\left({\alpha }^2+4\right)}^2}$
Given $|\hat{a}\times \hat{b}|=\sqrt{15\left({\alpha }^2+4\right)}$
$2\left({\alpha }^2+4\right)+{\left({\alpha }^2+4\right)}^2=15\left({\alpha }^2+4\right)$
${\left({\alpha }^2+4\right)}^2=13\left({\alpha }^2+4\right)$
$\Rightarrow {\alpha }^2+4=13$
$\therefore {\alpha }^2=9$
$2|\vec{a}{|}^2+(\vec{a}\cdot \vec{b})|\vec{b}{|}^2$
$|\vec{a}{|}^2={\alpha }^2+4+1={\alpha }^2+5$
$|\vec{b}{|}^2=4+{\alpha }^2+1={\alpha }^2+5$
$\vec{a}\cdot \vec{b}=-2\alpha +2\alpha -1=-1$
$\therefore 2|\vec{a}{|}^2+(\vec{a}\cdot \vec{b})|\vec{b}{|}^2$
$2\left({\alpha }^2+5\right)-1\left({\alpha }^2+5\right)={\alpha }^2+5=14$