JEE MAIN - Mathematics (2021 - 25th February Evening Shift - No. 23)

Let $$\overrightarrow a = \widehat i + \alpha \widehat j + 3\widehat k$$ and $$\overrightarrow b = 3\widehat i - \alpha \widehat j + \widehat k$$. If the area of the parallelogram whose adjacent sides are represented by the vectors $$\overrightarrow a $$ and $$\overrightarrow b $$ is $$8\sqrt 3 $$ square units, then $$\overrightarrow a $$ . $$\overrightarrow b $$ is equal to __________.
Answer
2

Explanation

$$\overrightarrow a = \widehat i + \alpha \widehat j + 3\widehat k$$

$$\overrightarrow b = 3\widehat i - \alpha \widehat j + \widehat k$$

Area of parallelogram = $$\left| {\overrightarrow a \times \overrightarrow b } \right|$$

$$ = \left| {(\widehat i + \alpha \widehat j + 3\widehat k) \times (3\widehat i - \alpha \widehat j + \widehat k)} \right|$$

$$8\sqrt 3 = \left| {(4\alpha )\widehat i + 8\widehat j - (4\alpha )\widehat k} \right|$$

$$(64)(3) = 16{\alpha ^2} + 64 + 16{\alpha ^2}$$

$$(64)(3) = 32{\alpha ^2} + 64$$

$$6 = {\alpha ^2} + 2$$

$${\alpha ^2} = 4$$

$$ \therefore $$ $$\overrightarrow a = \widehat i + \alpha \widehat j + 3\widehat k$$

$$\overrightarrow b = 3\widehat i - \alpha \widehat j + \widehat k$$

$$\overrightarrow a \,.\,\overrightarrow b = 3 - {\alpha ^2} + 3$$

$$ = 6 - {\alpha ^2}$$

$$ = 6 - 4$$

$$ = 2$$

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