JEE MAIN - Mathematics (2022 - 28th June Evening Shift - No. 14)

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 :
10
7
9
14

Explanation

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

$$\therefore$$ $$\overrightarrow a \times \overrightarrow b = \left| {\matrix{ {\widehat i} & {\widehat j} & {\widehat k} \cr \alpha & 2 & { - 1} \cr { - 2} & \alpha & 1 \cr } } \right| = (2 + \alpha )\widehat i - (\alpha - 2)\widehat j + ({\alpha ^2} + 4)\widehat k$$

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

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