JEE MAIN - Mathematics (2021 - 27th August Morning Shift - No. 12)
Let $$\overrightarrow a = \widehat i + 5\widehat j + \alpha \widehat k$$, $$\overrightarrow b = \widehat i + 3\widehat j + \beta \widehat k$$ and $$\overrightarrow c = - \widehat i + 2\widehat j - 3\widehat k$$ be three vectors such that, $$\left| {\overrightarrow b \times \overrightarrow c } \right| = 5\sqrt 3 $$ and $${\overrightarrow a }$$ is perpendicular to $${\overrightarrow b }$$. Then the greatest amongst the values of $${\left| {\overrightarrow a } \right|^2}$$ is _____________.
Answer
90
Explanation
Since, $$\overrightarrow a .\,\overrightarrow b = 0$$
$$1 + 15 + \alpha \beta = 0 \Rightarrow \alpha \beta = - 16$$ .... (1)
Also,
$${\left| {\overrightarrow b \, \times \overrightarrow c } \right|^2} = 75 \Rightarrow (10 + {\beta ^2})14 - {(5 - 3\beta )^2} = 75$$
$$\Rightarrow$$ 5$$\beta$$2 + 30$$\beta$$ + 40 = 0
$$\Rightarrow$$ $$\beta$$ = $$-$$4, $$-$$2
$$\Rightarrow$$ $$\alpha$$ = 4, 8
$$ \Rightarrow \left| {\overrightarrow a } \right|_{\max }^2 = {(26 + {\alpha ^2})_{\max }} = 90$$
$$1 + 15 + \alpha \beta = 0 \Rightarrow \alpha \beta = - 16$$ .... (1)
Also,
$${\left| {\overrightarrow b \, \times \overrightarrow c } \right|^2} = 75 \Rightarrow (10 + {\beta ^2})14 - {(5 - 3\beta )^2} = 75$$
$$\Rightarrow$$ 5$$\beta$$2 + 30$$\beta$$ + 40 = 0
$$\Rightarrow$$ $$\beta$$ = $$-$$4, $$-$$2
$$\Rightarrow$$ $$\alpha$$ = 4, 8
$$ \Rightarrow \left| {\overrightarrow a } \right|_{\max }^2 = {(26 + {\alpha ^2})_{\max }} = 90$$
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