QJEEM24Ma00e9: JEE Main 2024 Physics - Class 11 JEEM24Ma00e9 Solution: The least positive integral value of , for which the angle…

$\begin{aligned}& \cos \theta=\frac{(\alpha \hat{\mathrm{i}}-2 \hat{\mathrm{j}}+2 \hat{\mathrm{k}}) \cdot(\alpha \hat{\mathrm{i}}+2 \alpha \hat{\mathrm{j}}-2 \hat{\mathrm{k}})}{\sqrt{\alpha^2+4+4} \sqrt{\alpha^2+4 \alpha^2+4}} \\& \cos \theta=\frac{\alpha^2-4 \alpha-4}{\sqrt{\alpha^2+8} \sqrt{5 \alpha^2+4}} \\& \Rightarrow \alpha^2-4 \alpha-4>0 \quad \Rightarrow(\alpha-2)^2>8 \\& \Rightarrow \alpha^2-4 \alpha+4>8 \quad \alpha-2<-2 \sqrt{2} \\& \Rightarrow \alpha-2>2 \sqrt{2} \text { or } \alpha-2<2 \sqrt{2} \\& \alpha>2+2 \sqrt{2} \text { or } \alpha<2-2 \sqrt{2} \\& \alpha \in(-\infty,-0.82) \cup(4.82, \infty)\end{aligned}$

Least positive integral value of $\alpha \Rightarrow 5$