Question

If $|a|=|b|=|a-b|=1.$ Then angle between $a$ and $b$ is

• A. $\frac{\pi }{6}$
• B. $\frac{\pi }{4}$
• C. $\frac{\pi }{3}$
• D. $\frac{\pi }{2}$
• E. $\frac{\pi }{8}$

Answer 1

Answer: (C)

We know that
$|a-b{|}^2=(a-b)\cdot (a-b)$
$\Rightarrow (l{)}^2=|a{|}^2+|b{|}^2-2a\cdot b[\because |a|-|b|=1]$
$\Rightarrow l=l+l-2|a||b|\cos \theta$
$[\because |a|=|b|=1$ and $\theta$ is angle between $a$ and $b]$
$\Rightarrow 1=2-2\cos \theta$
$\Rightarrow \cos \theta =\frac{1}{2}$
$\Rightarrow \theta =\frac{\pi }{3}\left[\because \cos \frac{\pi }{3}=\frac{1}{2}\right]$