Let a vector overrightarrow{a}=sqrt{2}hat{i}- | vedclass

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Question

(A) Given $\overrightarrow{a}=\sqrt{2}\hat{i}-\hat{j}+\lambda\hat{k}$ and $\overrightarrow{b}=-\lambda^{2}\hat{i}+4\sqrt{2}\hat{j}+4\sqrt{2}\hat{k}$.

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$5$

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(A) Given $\overrightarrow{a}=\sqrt{2}\hat{i}-\hat{j}+\lambda\hat{k}$ and $\overrightarrow{b}=-\lambda^{2}\hat{i}+4\sqrt{2}\hat{j}+4\sqrt{2}\hat{k}$. Since $\overrightarrow{a}$ makes an angle $ heta$ with the positive $z$-axis,$\cos heta = \frac{\overrightarrow{a} \cdot \hat{k}}{|\overrightarrow{a}|} = \frac{\lambda}{\sqrt{(\sqrt{2})^2+(-1)^2+\lambda^2}} = \frac{\lambda}{\sqrt{3+\lambda^2}}$. Given $\frac{\pi}{6} < heta < \frac{\pi}{2}$,so $\cos \frac{\pi}{2} < \cos heta < \cos \frac{\pi}{6}$,which implies $0 < \frac{\lambda}{\sqrt{3+\lambda^2}} < \frac{\sqrt{3}}{2}$. Squaring the inequality,$0 < \frac{\lambda^2}{3+\lambda^2} < \frac{3}{4}$. Since $\lambda > 0$,the left part is always true. For the right part,$4\lambda^2 < 9 + 3\lambda^2 \Rightarrow \lambda^2 < 9 \Rightarrow \lambda < 3$. So $\lambda \in (0, 3)$....$(1)$ Since $\overrightarrow{a}$ makes an obtuse angle with $\overrightarrow{b}$,$\overrightarrow{a} \cdot \overrightarrow{b} < 0$. $\overrightarrow{a} \cdot \overrightarrow{b} = (\sqrt{2})(-\lambda^2) + (-1)(4\sqrt{2}) + (\lambda)(4\sqrt{2}) = -\sqrt{2}(\lambda^2 - 4\lambda + 4) = -\sqrt{2}(\lambda-2)^2 < 0$. Since $\sqrt{2} > 0$,we must have $(\lambda-2)^2 > 0$,which implies $\lambda eq 2$....$(2)$ From $(1)$ and $(2)$,$\lambda \in (0, 3) - \{2\}$. Thus,$\alpha=0, \beta=3, \gamma=2$. Therefore,$\alpha+\beta+\gamma = 0+3+2 = 5$.

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