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

<p>$\overrightarrow a = \alpha \widehat i + 2\widehat j - \widehat k$ and $\overrightarrow b = - 2\widehat i + \alpha \widehat j + \widehat k$</p> <p>$\therefore$ $$\overrightarrow a \times \overrightarrow b = \left| {\matrix{ {\widehat i} & {\widehat j} & {\widehat k} \cr \alpha & 2 & { - 1} \cr { - 2} & \alpha & 1 \cr } } \right| = (2 + \alpha )\widehat i - (\alpha - 2)\widehat j + ({\alpha ^2} + 4)\widehat k$$</p> <p>Now $$\left| {\overrightarrow a \times \overrightarrow b } \right| = \sqrt {15({\alpha ^2} + 4)} $$</p> <p>$$ \Rightarrow {(2 + \alpha )^2} + {(\alpha - 2)^2} + {({\alpha ^2} + 4)^2} = 15({\alpha ^2} + 4)$$</p> <p>$\Rightarrow {\alpha ^4} - 5{\alpha ^2} - 36 = 0$</p> <p>$\therefore$ $\alpha = \, \pm \,3$</p> <p>Now, $$2{\left| {\overrightarrow a } \right|^2} + \left( {\overrightarrow a - \overrightarrow b } \right){\left| {\overrightarrow b } \right|^{ - 2}} = 2.14 - 14 = 14$$</p>