QJEEM22Mada89: JEE Main 2022 Physics - Class 11 JEEM22Mada89 Solution: Let and , where . If the area of the parallelogram whose…

$\overrightarrow a = \alpha \widehat i + 2\widehat j - \widehat k$ and $\overrightarrow b = - 2\widehat i + \alpha \widehat j + \widehat k$

$\therefore$ $$\overrightarrow{a}\times \overrightarrow{b}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ \alpha & 2& -1\\ -2& \alpha & 1\end{array}\right|=(2+\alpha )\hat{i}-(\alpha -2)\hat{j}+({\alpha }^2+4)\hat{k}$$

Now $\left| {\overrightarrow a \times \overrightarrow b } \right| = \sqrt {15({\alpha ^2} + 4)}$

$\Rightarrow {(2 + \alpha )^2} + {(\alpha - 2)^2} + {({\alpha ^2} + 4)^2} = 15({\alpha ^2} + 4)$

$\Rightarrow {\alpha ^4} - 5{\alpha ^2} - 36 = 0$

$\therefore$ $\alpha = \, \pm \,3$

Now, $2{\left| {\overrightarrow a } \right|^2} + \left( {\overrightarrow a - \overrightarrow b } \right){\left| {\overrightarrow b } \right|^{ - 2}} = 2.14 - 14 = 14$