Question

Find the value of $x$ for which the angle between the vectors $\hat{a}=2x^2\hat{i}+4x\hat{j}+\hat{k}$ and $\hat{b}=7\hat{i}-2\hat{j}+x\hat{k}$ is obtuse.

Answer 1

Angle $\theta$ between vector $\overrightarrow{a}$ and $\overrightarrow{b}$

$\cos \theta =\frac{\overrightarrow{a}.\overrightarrow{b}}{|\overrightarrow{a}||\overrightarrow{b}|}$

for obtuse angle,

$\cos \theta <0$

$\frac{\overrightarrow{a}.\overrightarrow{b}}{|\overrightarrow{a}||\overrightarrow{b}|}<0$

$\overrightarrow{a}.\overrightarrow{b}<0$

$14x^2-8x+x<0$

$\left(7x\right)(2x-1)<0$

$0<x<\frac{1}{2}$

Hence anlge between given vectors are obtuse if $x\in (0,\frac{1}{2})$