Let a = √2 i - j + k, 0 and b = -^2 i + 4√2 j + 4√2 k.

Question

Let $\vec a = \sqrt2\hat i - \hat j + \lambda\hat k,\ \lambda>0$ and $\vec b = -\lambda^2\hat i + 4\sqrt2\hat j + 4\sqrt2\hat k$. Let $	heta$ be the angle between $\vec a$ and $\hat k$, with $\dfrac{\pi}{6} < 	heta < \dfrac{\pi}{2}$, and let $\vec a$ make an obtuse angle with $\vec b$. If the set of all possible values of $\lambda$ is $(\alpha,\beta) - \{\gamma\}$, then $\alpha+\beta+\gamma$ is equal to \_\_\_\_

Answer: 5

Accepted Answer

$\cos\theta = \dfrac{\vec a \cdot \hat k}{|\vec a|} = \dfrac{\lambda}{\sqrt{3+\lambda^2}}$. The condition $\dfrac{\pi}{6}<\theta<\dfrac{\pi}{2}$ gives $0 < \dfrac{\lambda}{\sqrt{3+\lambda^2}} < \dfrac{\sqrt3}{2}$, and with $\lambda>0$ this yields $\lambda \in (0,3)$ \ldots (1) $\vec a$ obtuse with $\vec b$ means $\vec a\cdot\vec b < 0$: $$-\sqrt2\lambda^2 - 4\sqrt2 + 4\sqrt2\lambda < 0 \implies \lambda^2 - 4\lambda + 4 > 0 \implies (\lambda-2)^2>0 \implies \lambda \ne 2 \quad (2)$$ Combining (1) and (2): $\lambda \in (0,3) - \{2\}$, so $\alpha=0,\ \beta=3,\ \gamma=2$, giving $\alpha+\beta+\gamma = 5$.

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