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If the angle between the vectors $2 \alpha^2 \hat{\mathbf{i}}+4 \alpha \hat{\mathbf{j}}+\hat{\mathbf{k}}$ and $7 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}+\alpha \hat{\mathbf{k}}$ is obtuse, then
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Verified Answer
The correct answer is:
$0 < \alpha < \frac{1}{2}$
Given vectors
$2 \alpha^2 \hat{\mathbf{i}}+4 \alpha \hat{\mathbf{j}}+\hat{\mathbf{k}} \text { and } 7 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}+\alpha \hat{\mathbf{k}}$
Angle between these vector are obtuse
$\begin{array}{ll}
\therefore \cos \theta < 0 & \\
& \frac{\left(2 \alpha^2 \hat{\mathbf{i}}+4 \alpha \hat{\mathbf{j}}+\hat{\mathbf{k}}\right)(7 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}+\alpha \hat{\mathbf{k}})}{\left|2 \alpha^2 \hat{\mathbf{i}}+4 \alpha \hat{\mathbf{j}}+\hat{\mathbf{k}} \| 7 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}+\alpha \hat{\mathbf{k}}\right|} \\
\Rightarrow & 14 \alpha^2-8 \alpha+\alpha < 0 \Rightarrow 14 \alpha^2-7 \alpha < 0 \\
\Rightarrow 7 \alpha(2 \alpha-1) < 0 \Rightarrow \alpha \in\left(0, \frac{1}{2}\right) \\
\therefore & 0 < \alpha < \frac{1}{2}
\end{array}$
$2 \alpha^2 \hat{\mathbf{i}}+4 \alpha \hat{\mathbf{j}}+\hat{\mathbf{k}} \text { and } 7 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}+\alpha \hat{\mathbf{k}}$
Angle between these vector are obtuse
$\begin{array}{ll}
\therefore \cos \theta < 0 & \\
& \frac{\left(2 \alpha^2 \hat{\mathbf{i}}+4 \alpha \hat{\mathbf{j}}+\hat{\mathbf{k}}\right)(7 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}+\alpha \hat{\mathbf{k}})}{\left|2 \alpha^2 \hat{\mathbf{i}}+4 \alpha \hat{\mathbf{j}}+\hat{\mathbf{k}} \| 7 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}+\alpha \hat{\mathbf{k}}\right|} \\
\Rightarrow & 14 \alpha^2-8 \alpha+\alpha < 0 \Rightarrow 14 \alpha^2-7 \alpha < 0 \\
\Rightarrow 7 \alpha(2 \alpha-1) < 0 \Rightarrow \alpha \in\left(0, \frac{1}{2}\right) \\
\therefore & 0 < \alpha < \frac{1}{2}
\end{array}$
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