MCQ
If $|\vec{a}-\vec{b}|=|\vec{a}|=|\vec{b}|=1$, then the angle between $\vec{a}$ and $\vec{b}$ is
  • $\frac{\pi}{3}$
  • B
    $\frac{3 \pi}{4}$
  • C
    $\frac{\pi}{2}$
  • D
    $\frac{\pi}{6}$

Answer

Correct option: A.
$\frac{\pi}{3}$
(a) : Given, $|\vec{a}-\vec{b}|=|\vec{a}|=|\vec{b}|=1$
$
\Rightarrow|\vec{a}-\vec{b}|^2=|\vec{a}|^2+|\vec{b}|^2-2 \vec{a} \cdot \vec{b} \Rightarrow 1=1+1-2|\vec{a}||\vec{b}| \cos \theta
$
(Here $\theta$ is angle between $\vec{a}$ and $\vec{b}$ )
$
\Rightarrow \cos \theta=\frac{1}{2} \Rightarrow \theta=\frac{\pi}{3}
$

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