Question

If $\overline{a}$ and $\overline{b}$ are unit vectors such that $|\overline{a}\times \overline{b}|=\overline{a}.\overline{b}$, then $|a-b{|}^2=$

• A. $2$
• B. $2+\sqrt{2}$
• C. $2-\sqrt{2}$
• D. $4\sqrt{2}$

Answer 1

Answer: (C)

$2-\sqrt{2}$

$|\overrightarrow{a}\times \overrightarrow{b}=|\overrightarrow{a}||\overrightarrow{b}|\sin \theta$ ( $\theta$ is angle between $\overrightarrow{a}+\overrightarrow{b})$
$\overrightarrow{a}\overrightarrow{b}=|\overrightarrow{a}|\overrightarrow{b}|\cos \theta$
$\tan \theta =1$
$\theta =\frac{\pi }{\lambda }$
$|\overrightarrow{a}-\overrightarrow{b}{|}^2={\left(\sqrt{a^2+b^2-2\overrightarrow{a}-\overrightarrow{b}}\right)}^2$
$|\overrightarrow{a}-\overrightarrow{b}|=(1+1-2\cos \frac{\pi }{\lambda })$
$=2\frac{-2}{\sqrt{2}}$
$=2-\sqrt{2}$