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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
(a) $\theta=\frac{\pi}{4}$
(b) $\theta=\frac{\pi}{3}$
(c) $\theta=\frac{\pi}{2}$
(d) $\theta=\frac{2 \pi}{3}$
(a) $\theta=\frac{\pi}{4}$
(b) $\theta=\frac{\pi}{3}$
(c) $\theta=\frac{\pi}{2}$
(d) $\theta=\frac{2 \pi}{3}$
Solution:
2252 Upvotes
Verified Answer
We have $|\vec{a}|=1,|\vec{b}|=1 \quad$ and $|\vec{a}+\vec{b}|=1$
or $|\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}|^2=1 \quad$ i.e., $(\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}) \cdot(\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}})=1$
or $\quad|\vec{a}|^2+|\vec{b}|^2+2 \vec{a} \cdot \vec{b}=1 \quad[\because \vec{b} \cdot \vec{a}=\vec{a} \cdot \vec{b}]$
$=1+1+2|\vec{a}||\vec{b}| \cos \theta=1$
$[\because|\vec{a}|=1,|\vec{b}|=1]$ where $\theta$ is the angle between $\vec{a}$ and $\vec{b}$
$\therefore \quad 2+2 \cdot 1 \cos \theta=1 \quad$ or $2 \cos 2 \theta=1$
$$
\Rightarrow \quad \cos \theta=-\frac{1}{2} \quad \Rightarrow \theta=\frac{2 \pi}{3}
$$
or $|\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}|^2=1 \quad$ i.e., $(\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}) \cdot(\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}})=1$
or $\quad|\vec{a}|^2+|\vec{b}|^2+2 \vec{a} \cdot \vec{b}=1 \quad[\because \vec{b} \cdot \vec{a}=\vec{a} \cdot \vec{b}]$
$=1+1+2|\vec{a}||\vec{b}| \cos \theta=1$
$[\because|\vec{a}|=1,|\vec{b}|=1]$ where $\theta$ is the angle between $\vec{a}$ and $\vec{b}$
$\therefore \quad 2+2 \cdot 1 \cos \theta=1 \quad$ or $2 \cos 2 \theta=1$
$$
\Rightarrow \quad \cos \theta=-\frac{1}{2} \quad \Rightarrow \theta=\frac{2 \pi}{3}
$$
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