Vector Algebra Question 5
Question 5 - 2024 (27 Jan Shift 1)
The least positive integral value of $\alpha$, for which the angle between the vectors $\alpha \hat{i}-2 \hat{j}+2 k$ and $\alpha \hat{i}+2 \alpha \hat{j}-2 k$ is acute, is
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Answer (5)
Solution
$\cos \theta=\frac{(\alpha \hat{i}-2 \hat{j}+2 \hat{k}) \cdot(\alpha \hat{i}+2 \alpha \hat{j}-2 \hat{k})}{\sqrt{\alpha^{2}+4+4} \sqrt{\alpha^{2}+4 \alpha^{2}+4}}$
$\cos \theta=\frac{\alpha^{2}-4 \alpha-4}{\sqrt{\alpha^{2}+8} \sqrt{5 \alpha^{2}+4}}$
$\Rightarrow \alpha^{2}-4 \alpha-4>0$
$\Rightarrow \alpha^{2}-4 \alpha+4>8 \quad \Rightarrow(\alpha-2)^{2}>8$
$\Rightarrow \alpha-2>2 \sqrt{2}$ or $\alpha-2<-2 \sqrt{2}$
$\alpha>2+2 \sqrt{2}$ or $\alpha<2-2 \sqrt{2}$
$\alpha \in(-\infty,-0.82) \cup(4.82, \infty)$
Least positive integral value of $\alpha \Rightarrow 5$
