Question

If the angle between $\overrightarrow{a}=2x^2\hat{i}+4x\hat{j}+\hat{k}$ and $\overrightarrow{b}=7\hat{i}-2\hat{j}+x\hat{k}$ is obtuse and the angle between $\overrightarrow{b}$ and $Z-$ axis is acute and less than $\frac{\pi }{6},$ then

• A. $x\in \varphi$
• B. $x\in (0,\frac{1}{2})$
• C. $x\in (-\sqrt{159},\sqrt{159})$
• D. $x\in (-1,1)$

Answer 1

The correct option is A $x\in \varphi$

Given : Angle between $\overrightarrow{a}$ and $\overrightarrow{b}$ is obtuse.
$\Rightarrow \overrightarrow{a}\cdot \overrightarrow{b}<0$
$\Rightarrow (2x^2\hat{i}+4x\hat{j}+\hat{k})\cdot (7\hat{i}-2\hat{j}+x\hat{k})=14x^2-8x+x<0$
$\Rightarrow 14x^2-7x<0$
$\Rightarrow 7x(2x-1)<0$
$\Rightarrow 0<x<\frac{1}{2}\cdots \left(i\right)$
The angle between $\overrightarrow{b}$ and $Z-$ axis is acute and less than $\frac{\pi }{6}$
$\therefore \frac{\overrightarrow{b}\cdot k}{|\overrightarrow{b}||\overrightarrow{k}|}>\cos \frac{\pi }{6}$
$(\because \theta <\frac{\pi }{6}\Rightarrow \cos \theta >\cos \frac{\pi }{6})$
$\Rightarrow \frac{x}{\sqrt{x^2+49+4}}>\frac{\sqrt{3}}{2}$
$\Rightarrow 4x^2>3x^2+159$
$\Rightarrow x>\sqrt{159}\text{or}x<-\sqrt{159}\cdots \left(ii\right)$
From $\left(i\right)$ and $\left(ii\right),$ we get
$\therefore x\in \varphi$