Vector Algebra – Scalar and Vector Products – JEE Main 22 Jan 2026 Shift 2

Solution:

First, we analyze the condition that $\vec{a}$ makes an obtuse angle with $\vec{b}$. This implies their dot product is negative.
$$ \vec{a} \cdot \vec{b} < 0 $$ $$ (\sqrt{2}\hat{i}-\hat{j}+\lambda\hat{k}) \cdot (-\lambda^{2}\hat{i}+4\sqrt{2}\hat{j}+4\sqrt{2}\hat{k}) < 0 $$ $$ -\sqrt{2}\lambda^{2} - 4\sqrt{2} + 4\sqrt{2}\lambda < 0 $$ Dividing by $-\sqrt{2}$ and reversing the inequality: $$ \lambda^{2} - 4\lambda + 4 > 0 $$
$$ (\lambda – 2)^{2} > 0 $$
This inequality holds for all real $\lambda$ except where the square is zero.
$$ \lambda \neq 2 $$

Next, we analyze the angle $\theta$ made by $\vec{a}$ with the positive z-axis.
$$ \cos \theta = \frac{\vec{a} \cdot \hat{k}}{|\vec{a}|} = \frac{\lambda}{\sqrt{(\sqrt{2})^{2} + (-1)^{2} + \lambda^{2}}} = \frac{\lambda}{\sqrt{3+\lambda^{2}}} $$
Given the range $\frac{\pi}{6} < \theta < \frac{\pi}{2}$, the cosine values satisfy: $$ \cos \left(\frac{\pi}{2}\right) < \cos \theta < \cos \left(\frac{\pi}{6}\right) $$ $$ 0 < \frac{\lambda}{\sqrt{3+\lambda^{2}}} < \frac{\sqrt{3}}{2} $$ The left inequality $0 < \frac{\lambda}{\sqrt{3+\lambda^{2}}}$ implies $\lambda > 0$ (which is already given).
Squaring the right inequality:
$$ \frac{\lambda^{2}}{3+\lambda^{2}} < \frac{3}{4} $$ $$ 4\lambda^{2} < 3(3+\lambda^{2}) $$ $$ 4\lambda^{2} < 9 + 3\lambda^{2} $$ $$ \lambda^{2} < 9 $$ $$ \Rightarrow \lambda \in (0, 3) $$
Combining the two conditions:
$\lambda \in (0, 3)$ and $\lambda \neq 2$.
The set of values is $(0, 3) – \{2\}$.
Comparing this with $(\alpha, \beta) – \{\gamma\}$:
$$ \alpha = 0, \quad \beta = 3, \quad \gamma = 2 $$
We need to find $\alpha + \beta + \gamma$:
$$ \alpha + \beta + \gamma = 0 + 3 + 2 = 5 $$

Ans. (5)