Question

$\text{I}$: If $|\overline{a}.\overline{b}|=|\overline{a}\times \overline{b}|$, then $(\overline{a}.\overline{b})=\frac{\pi }{4}$
$\text{II}$: $|\overline{a}.\overline{b}{|}^2=|\overline{a}\times \overline{b}{|}^2$, then $(\overline{a},\overline{b})=\frac{3\pi }{4}$

• A. Only $\text{I}$ is true
• B. Only $\text{II}$ is true
• C. Both $\text{I}$ and $\text{II}$ true
• D. Neither $\text{I}$ nor $\text{II}$ are true

Answer 1

The correct option is C Both $\text{I}$ and $\text{II}$ true

$\overrightarrow{a}-\overrightarrow{b}=\left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right|cos\theta$ ( where $\theta$ is angle between $\overrightarrow{a}$&$\overrightarrow{b}$ )
$|\overrightarrow{a}\times \overrightarrow{b}|=\left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right|\sin \theta$
$|\overrightarrow{a}||\overrightarrow{b}|cos\theta =\left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right|\sin \theta$
$\tan \theta =1\theta =\frac{\pi }{4}$
$\left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right|cos\theta -\left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right|\sin \theta$
$\tan \theta =-1$
$\theta =\frac{3\pi }{4}$
So both cases are true