Q49 / 75 · JEE Main 2026 (22 January Shift 2) · Mathematics
Let a vector $\overrightarrow{\mathrm{a}}=\sqrt{2} \hat{i}-\hat{j}+\lambda \hat{k}, \lambda>0$, make an obtuse angle with the vector $\overrightarrow{\mathrm{b}}=-\lambda^{2} \hat{i}+4 \sqrt{2} \hat{j}+4 \sqrt{2} \hat{k}$ 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 $\_\_\_\_$。
Correct answer: 5
Solution
$\frac{\overrightarrow{\mathrm{a}} . \hat{\mathrm{k}}}{|\overrightarrow{\mathrm{a}}|}=\cos \theta \Rightarrow \frac{\lambda}{\sqrt{3+\lambda^2}}=\cos \theta$
$\Rightarrow 0<\frac{\lambda}{\sqrt{3+\lambda^2}}<\frac{\sqrt{3}}{2}$
$\Rightarrow \lambda>0 \& 4 \lambda^2<9+3 \lambda^2 \Rightarrow \lambda^2<9$
$\Rightarrow \lambda \in(0,3) \quad \quad \ldots .(1)$
$\Rightarrow \overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{~b}}<0 \Rightarrow-\sqrt{2} \lambda^2-4 \sqrt{2}+4 \sqrt{2} \lambda<0$
$\Rightarrow \lambda^2-4 \lambda+4>0 \Rightarrow(\lambda-2)^2>0$
$\Rightarrow \lambda \neq 2$
$\text { from }(1) \&(2)$
$\lambda \in(0,3)-\{2\}$
$\therefore \alpha=0, \beta=3, \gamma=2$
$\Rightarrow \alpha+\beta+\gamma=5$
$\Rightarrow 0<\frac{\lambda}{\sqrt{3+\lambda^2}}<\frac{\sqrt{3}}{2}$
$\Rightarrow \lambda>0 \& 4 \lambda^2<9+3 \lambda^2 \Rightarrow \lambda^2<9$
$\Rightarrow \lambda \in(0,3) \quad \quad \ldots .(1)$
$\Rightarrow \overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{~b}}<0 \Rightarrow-\sqrt{2} \lambda^2-4 \sqrt{2}+4 \sqrt{2} \lambda<0$
$\Rightarrow \lambda^2-4 \lambda+4>0 \Rightarrow(\lambda-2)^2>0$
$\Rightarrow \lambda \neq 2$
$\text { from }(1) \&(2)$
$\lambda \in(0,3)-\{2\}$
$\therefore \alpha=0, \beta=3, \gamma=2$
$\Rightarrow \alpha+\beta+\gamma=5$
Hence, the correct answer is 5.