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Question 6 — JEE Main 22 Jan 2026 S Ii

Question

Let a vector $\overrightarrow{\mathrm{a}}=\sqrt{2} \hat{\mathrm{i}}-\hat{\mathrm{j}}+\lambda \hat{\mathrm{k}}, \lambda>0$, make an obtuse angle with the vector $\vec{b}=-\lambda^{2} \hat{i}+4 \sqrt{2} \hat{j}+4 \sqrt{2} \hat{k}$ and an angle $\theta, \frac{\pi}{6}<\theta<\frac{\pi}{2}$, with the positive $z$-axis. If the set of all possible values of $\lambda$ is $(\alpha, \beta)-\{\gamma\}$, then $\alpha+\beta+\gamma$ is equal to [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)