Reply To: Question 6 — JEE Main 22 Jan 2026 S Ii
Solution
\[
\begin{aligned}
& \vec{a}^{\wedge} \vec{b} \in\left(\frac{\pi}{2}, \pi\right) ; \vec{a} \cdot \vec{b}<0 \Rightarrow\left(-\lambda^{2} \sqrt{2}-4 \sqrt{2}+4 \sqrt{2} \lambda\right)<0 \\
& \Rightarrow-\sqrt{2}\left(\lambda^{2}+4-4 \lambda\right)<0 \Rightarrow(\lambda-2)^{2}>0 \Rightarrow \lambda \in R-\{2\} \quad \lambda \neq 2 \\
& \theta=\vec{a}^{\wedge} \hat{k} ; \cos \theta=\frac{\vec{a} \cdot \hat{k}}{|\vec{a}||\hat{k}|}=\frac{\lambda}{\sqrt{\lambda^{2}+3}} \\
& \frac{\pi}{6}<\theta<\frac{\pi}{2} \Rightarrow \cos \frac{\pi}{6}>\cos \theta>\cos \frac{\pi}{2}
\end{aligned}
\]
$\Rightarrow \frac{\sqrt{3}}{2}>\frac{\lambda}{\sqrt{\lambda^{2}+3}}>0 ; \lambda>0$
$\Rightarrow \sqrt{3} \sqrt{\lambda^{2}+3}>2 \lambda$
$\Rightarrow 3 \lambda^{2}+9>4 \lambda^{2}=\lambda^{2}-9<0$
$\Rightarrow(\lambda-3)(\lambda+3)<0$
$\Rightarrow \lambda \in(-3,3)$
$(1) \cap(2) \cap(3)$
$\lambda \in(0,3)-\{2\}=(\alpha, \beta)-\{\gamma\}$
$\alpha=0 ; \beta=3 ; \gamma=2 \Rightarrow \alpha+\beta+\gamma=5$ Answer(5)