MCQ
Let $\vec{a}$ and $\vec{b}$ be two unit vectors and $\theta$ is the angle between them. Then $\vec{a}+\vec{b}$ is a unit vector if $\theta =$
- A$\frac{ \pi}{2}$
- B$\frac{ \pi}{3}$
- C$\frac{ \pi}{4}$
- ✓$\frac{2 \pi}{3}$
Then, $|\vec{a}|=|\vec{b}|=1$
Now, $\vec{a}+\vec{b}$ is a unit vector if $|\vec{a}+\vec{b}|=1$
$|\vec{a}+\vec{b}|=1$
$\Rightarrow(\vec{a}+\vec{b})^{2}=1$
$\Rightarrow(\vec{a}+\vec{b}) \cdot(\vec{a}+\vec{b})=1$
$\Rightarrow \vec{a} \cdot \vec{a}+\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{a}+\vec{b}\cdot \vec{b}=1$
$\Rightarrow|\vec{a}|^{2}+2 \vec{a} \cdot \vec{b}+|\vec{b}|^{2}=1$
$\Rightarrow 1^{2}+2|\vec{a}||\vec{b}| \cos \theta+1^{2}=1$
$\Rightarrow 1+2.1 .1 \cos \theta+1=1$
$\Rightarrow \cos \theta=-\frac{1}{2}$
$\Rightarrow \theta=\frac{2 \pi}{3}$
Hence, $\vec{a}+\vec{b}$ is a unit vector if $\theta=\frac{2 \pi}{3}$
The correct answer is $D$.
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