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$
$2|\vec{a}|^{2}+(\vec{a} \cdot \vec{b})|\vec{b}|^{2}$
$|\vec{a}|^{2}=\alpha^{2}+4+1=\alpha^{2}+5$
$|\vec{b}|^{2}=4+\alpha^{2}+1=\alpha^{2}+5$
$\vec{a} \cdot \vec{b}=-2 \alpha+2 \alpha-1=-1$
$\therefore 2|\vec{a}|^{2}+(\vec{a} \cdot \vec{b})|\vec{b}|^{2}$
$2\left(\alpha^{2}+5\right)-1\left(\alpha^{2}+5\right)=\alpha^{2}+5=14$