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If the angle between the vectors $\bar{a}=2 \lambda^2 \hat{i}+4 \lambda \hat{j}+\hat{k}$ and $\overline{\mathrm{b}}=7 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}+\lambda \hat{\mathrm{k}}$ is obtuse, then $\lambda \in$
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The correct answer is:
$\left(0, \frac{1}{2}\right)$
We have $\bar{a}=2 \lambda^2 \hat{i}+4 \lambda \hat{j}+\hat{k}$ and $\bar{b}=7 \hat{i}-2 \hat{j}+\lambda \hat{k}$ $\bar{a} \cdot \bar{b}=|\bar{a}| \cdot|\bar{b}| \cdot \cos \theta$, where $\theta$ is angle between $\bar{a}$ and $\bar{b}$.
$\therefore \quad \cos \theta=\frac{\overline{\mathrm{a}} \cdot \overline{\mathrm{b}}}{|\overline{\mathrm{a}}| \cdot|\overline{\mathrm{b}}|} < 0$, as $\theta$ is an obtuse angle.
$\therefore \quad|\overline{\mathrm{a}}| \cdot|\overline{\mathrm{b}}| < 0$
$\therefore \quad\left(2 \lambda^2\right)(7)+(4 \lambda)(-2)+(1)(\lambda) < 0$
$\therefore \quad 14 \lambda^2-7 \lambda < 0 \Rightarrow 7 \lambda(2 \lambda-1) < 0 \Rightarrow \lambda(2 \lambda-1) < 0$
$\therefore \quad 0 < \lambda < \frac{1}{2}$ i.e. $\lambda \in\left(0, \frac{1}{2}\right)$
$\therefore \quad \cos \theta=\frac{\overline{\mathrm{a}} \cdot \overline{\mathrm{b}}}{|\overline{\mathrm{a}}| \cdot|\overline{\mathrm{b}}|} < 0$, as $\theta$ is an obtuse angle.
$\therefore \quad|\overline{\mathrm{a}}| \cdot|\overline{\mathrm{b}}| < 0$
$\therefore \quad\left(2 \lambda^2\right)(7)+(4 \lambda)(-2)+(1)(\lambda) < 0$
$\therefore \quad 14 \lambda^2-7 \lambda < 0 \Rightarrow 7 \lambda(2 \lambda-1) < 0 \Rightarrow \lambda(2 \lambda-1) < 0$
$\therefore \quad 0 < \lambda < \frac{1}{2}$ i.e. $\lambda \in\left(0, \frac{1}{2}\right)$
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