If the angle theta between the vectors overri | vedclass

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(A) Given, $\overrightarrow{a}=2 x^2 \hat{i}+4 x \hat{j}+\hat{k}$ and $\overrightarrow{b}=7 \hat{i}-2 \hat{j}+x \hat{k}$.We are given that the angle $

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$\left(0, \frac{1}{2}\right)$

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(A) Given, $\overrightarrow{a}=2 x^2 \hat{i}+4 x \hat{j}+\hat{k}$ and $\overrightarrow{b}=7 \hat{i}-2 \hat{j}+x \hat{k}$.We are given that the angle $	heta$ between the vectors satisfies $90^{\circ} This implies that $\cos 	heta We know that $\cos 	heta = \frac{\overrightarrow{a} \cdot \overrightarrow{b}}{|\overrightarrow{a}| |\overrightarrow{b}|}$.Since the magnitudes $|\overrightarrow{a}|$ and $|\overrightarrow{b}|$ are always non-negative, the condition $\cos 	heta Calculating the dot product:$\overrightarrow{a} \cdot \overrightarrow{b} = (2 x^2)(7) + (4 x)(-2) + (1)(x) = 14 x^2 - 8 x + x = 14 x^2 - 7 x$.Setting the dot product to be less than zero:$14 x^2 - 7 x $7 x(2 x - 1) To solve this inequality, we find the critical points $x = 0$ and $x = \frac{1}{2}$.Testing the intervals, we find that the expression $7 x(2 x - 1)$ is negative when $x$ lies between $0$ and $\frac{1}{2}$.Thus, $x \in \left(0, \frac{1}{2}ight)$.

If the angle $\theta$ between the vectors $\overrightarrow{a}=2 x^2 \hat{i}+4 x \hat{j}+\hat{k}$ and $\overrightarrow{b}=7 \hat{i}-2 \hat{j}+x \hat{k}$ is such that $90^{\circ} < \theta < 180^{\circ}$, then $x$ lies in the interval

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