JEE MAIN - Mathematics (2022 - 28th July Evening Shift - No. 13)
Let S be the set of all a $$\in R$$ for which the angle between the vectors $$
\vec{u}=a\left(\log _{e} b\right) \hat{i}-6 \hat{j}+3 \hat{k}$$ and $$\vec{v}=\left(\log _{e} b\right) \hat{i}+2 \hat{j}+2 a\left(\log _{e} b\right) \hat{k}$$, $$(b>1)$$ is acute. Then S is equal to :
$$\left(-\infty,-\frac{4}{3}\right)$$
$$\Phi $$
$$\left(-\frac{4}{3}, 0\right)$$
$$\left(\frac{12}{7}, \infty\right)$$
Explanation
$$\overrightarrow u = a({\log _e}b)\widehat i - 6\widehat j + 3\widehat k$$
$$\overrightarrow v = ({\log _e}b)\widehat i + 2\widehat j + 2a({\log _e}b)\widehat k$$
For acute angle $$\overrightarrow u \,.\,\overrightarrow v > 0$$
$$ \Rightarrow a{({\log _e}b)^2} - 12 + 6a({\log _e}b) > 0$$
$$\because$$ $$b > 1$$
Let $${\log _e}b = t \Rightarrow t > 0$$ as $$b > 1$$
$$a{t^2} + 6at - 12 > 0\,\,\,\,\,\,\,\forall t > 0$$
$$ \Rightarrow a \in \phi $$
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