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VECTOR ALGEBRA — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsVECTOR ALGEBRA5 Marks
Question
Let $\vec{\text{a}}=\text{x}^2\hat{\text{i}}+2\hat{\text{j}}-2\hat{\text{k}},\vec{\text{b}}=\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}}$ and $\vec{\text{c}}=\text{x}^2\hat{\text{i}}+5\hat{\text{j}}-4\hat{\text{k}}$ be three vectors. Find the valuse of x for which the angle between $\vec{\text{a}}$ and $\vec{\text{b}}$ is acute and the angle between $\vec{\text{b}}$ and $​​\vec{\text{c}}$ is obtuse.
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Answer

We have
$\vec{\text{a}}=\text{x}^2\hat{\text{i}}+2\hat{\text{j}}-2\hat{\text{k}},\vec{\text{b}}=\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}}$ and $\vec{\text{c}}=\text{x}^2\hat{\text{i}}+5\hat{\text{j}}-4\hat{\text{k}}$
Let $\theta_1$ be the angle between $\vec{\text{a}}$ and $\vec{\text{b}}$ and $\theta_2$ be the angle between $\vec{\text{b}}$ and $\vec{\text{c}}.$
Given that $\theta_1$ is acute and $\theta_2$ is obtuse.
$\Rightarrow\cos\theta_1>0$ and $\cos\theta_2<0$
$\Rightarrow\frac{\vec{\text{a}}.\vec{\text{b}}}{|\vec{\text{a}}|.\big|\vec{\text{b}}\big|}>0$ and $\frac{\vec{\text{b}}.\vec{\text{c}}}{\big|\vec{\text{b}}\big|.|\vec{\text{c}}|}<0$
$\Rightarrow\frac{\text{x}^2-4}{\sqrt{\text{x}^2+4+4}\sqrt{\text{1+1+1}}}>0$ and $\frac{\text{x}^2-9}{\sqrt{\text{1+1+1}}\sqrt{\text{x}^4}+25+16}<0$
$\Rightarrow\text{x}^2-4>0$ and $\text{x}^2-9<0$
$\Rightarrow\text{x}\in(-\infty,-2)\cup(2,\infty)$ and $\text{x}\in(-3,3)$
$\Rightarrow\text{x}\in(-3,-2)\cup(2,3)$

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