JEE Main 2022 (Online) 28th June Evening ShiftQuestion 1
Let $\vec{a}=\alpha\hat{i}+2\hat{j}-\hat{k}$ and $\vec{b}=-2\hat{i}+\alpha\hat{j}+\hat{k}$, where $\alpha\in\mathbf{R}$. 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 $2|\vec{a}|^{2}+(\vec{a}\cdot\vec{b})|\vec{b}|^{2}$ is equal to :
Vector Algebra
Solution
$\vec a\times\vec b=(2+\alpha,\ 2-\alpha,\ \alpha^2+4)$. $|\vec a\times\vec b|^2=(2+\alpha)^2+(2-\alpha)^2+(\alpha^2+4)^2=2\alpha^2+8+(\alpha^2+4)^2=(\alpha^2+4)(\alpha^2+6)$. Setting this equal to $15(\alpha^2+4)$ gives $\alpha^2=9$, $\alpha=\pm3$. $|\vec a|^2=|\vec b|^2=\alpha^2+5=14$ and $\vec a\cdot\vec b=-2\alpha+2\alpha-1=-1$. $2\cdot14+(-1)\cdot14=14$. Answer: (D).