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Question

Let $$ \vec{a} = x^{2}\hat{i}+2\hat{j}-2\hat{k},\vec{b} = \hat{i}-\hat{j}+\hat{k} $$ and $$ \vec{c} = x^{2}\hat{i}+5\hat{j}-4\hat{k} $$ be three vectors. Find the values of x for which the angle between $$ \vec{a} $$ and $$ \vec{b} $$ is acute and the angle between $$ \vec{b} $$ and $$ \vec{c} $$ is obtuse.


A
(3,2)(2,3)
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B
(3,1)(1,3)
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C
(3,1)(1,3)
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D
(3,2)(2,3)
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Solution

The correct option is A $$ (-3,-2)\cup (2,3) $$
We have 
$$\vec{a}={x}^{2}\hat{i}+2\hat{j}-2\hat{k},\,\vec{b}=\hat{i}-\hat{j}+\hat{k}$$ and $$\vec{c}={x}^{2}\hat{i}+5\hat{j}-4\hat{k}$$
Given that $${\theta}_{1}$$ is acute and $${\theta}_{2}$$ is obtuse.
$$\Rightarrow\,\cos{{\theta}_{1}}>0$$ and $$\cos{{\theta}_{2}}<0$$
$$\Rightarrow\,\dfrac{\vec{a}.\vec{b}}{\left|\vec{a}\right|.\left|\vec{b}\right|}>0$$ and $$\dfrac{\vec{b}.\vec{c}}{\left|\vec{b}\right|.\left|\vec{c}\right|}<0$$
$$\Rightarrow\,\dfrac{{x}^{2}-4}{\sqrt{{x}^{4}+4+4}\sqrt{1+1+1}}>0$$ and $$\dfrac{{x}^{2}-9}{\sqrt{{x}^{4}+25+16}\sqrt{1+1+1}}>0$$
$$\Rightarrow\,{x}^{2}-4>0$$ and $${x}^{2}-9<0$$
$$\Rightarrow\,x\in\left(-\infty,-2\right)\cup\left(2,\infty\right)$$ and $$x\in\left(-3,3\right)$$
$$\therefore\,x\in\left(-3,-2\right)\cup\left(2,3\right)$$

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