Question
If $\vec{\text{a}}$ are $\vec{\text{b}}$ are two unit vectors such that $\vec{\text{a}}+\vec{\text{b}}$ is $\frac{\pi}{6}.$
✓
Answer
Let the angle between $\vec{\text{a}}$ and $\vec{\text{b}}$ be $\theta.$
It is given that $|\vec{\text{a}}|=\big|\vec{\text{b}}\big|=\big|\vec{\text{a}}+\vec{\text{b}}\big|=1.$
$\big|\vec{\text{a}}+\vec{\text{b}}\big|=1$
$\Rightarrow\big|\vec{\text{a}}+\vec{\text{b}}\big|^2=1$
$\Rightarrow|\vec{\text{a}}|^2+2|\vec{\text{a}}|\big|\vec{\text{b}}\big|\cos\theta+\big|\vec{\text{b}}\big|^2=1$
$\Rightarrow1+2\times1\times1\times\cos\theta+1=1$
$\Rightarrow2\cos\theta=-1$
$\Rightarrow\cos\theta=-\frac{1}{2}=\cos\frac{2\pi}{3}$
$\Rightarrow\theta=\frac{2\pi}{3}$
It is given that $|\vec{\text{a}}|=\big|\vec{\text{b}}\big|=\big|\vec{\text{a}}+\vec{\text{b}}\big|=1.$
$\big|\vec{\text{a}}+\vec{\text{b}}\big|=1$
$\Rightarrow\big|\vec{\text{a}}+\vec{\text{b}}\big|^2=1$
$\Rightarrow|\vec{\text{a}}|^2+2|\vec{\text{a}}|\big|\vec{\text{b}}\big|\cos\theta+\big|\vec{\text{b}}\big|^2=1$
$\Rightarrow1+2\times1\times1\times\cos\theta+1=1$
$\Rightarrow2\cos\theta=-1$
$\Rightarrow\cos\theta=-\frac{1}{2}=\cos\frac{2\pi}{3}$
$\Rightarrow\theta=\frac{2\pi}{3}$
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