Vector Algebra
Angle Conditions — Finding Range of Parameter
nta_pyq_2026_jan
Grade 12
Question:
Let a vector $\vec{a}=\sqrt{2}\hat{i}-\hat{j}+\lambda\hat{k}$, $\lambda>0$, 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 $\theta$, $\dfrac{\pi}{6}<\theta<\dfrac{\pi}{2}$, 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 ___.
Step-by-Step Solution
Key Concept: Angle with $z$-axis: $\cos\theta=\frac{\lambda}{\sqrt{3+\lambda^2}}$. Condition $\frac{\pi}{6}<\theta<\frac{\pi}{2}$: $0<\frac{\lambda}{\sqrt{3+\lambda^2}}<\frac{\sqrt{3}}{2}\Rightarrow\lambda>0$ and $4\lambda^2<9+3\lambda^2\Rightarrow\lambda^2<9\Rightarrow\lambda\in(0,3)$.
$\alpha+\beta+\gamma=5$.
Correct Answer: 5