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 $90^{\circ}

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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 $90^{\circ} We know that $\cos 	heta = \frac{\overrightarrow{a} \cdot \overrightarrow{b}}{|\overrightarrow{a}| |\overrightarrow{b}|}$.Since $|\overrightarrow{a}|$ and $|\overrightarrow{b}|$ are magnitudes,they are always positive. Thus,$\cos 	heta Calculating the dot product:$\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:For $x  0$.For $0 For $x > \frac{1}{2}$,$7 x(2 x - 1) > 0$.Thus,the interval for which the expression is negative is $x \in \left(0, \frac{1}{2}ight)$.

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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