Search any question & find its solution
Question:
Answered & Verified by Expert
Let $\overrightarrow{\mathbf{a}}$ and $\overrightarrow{\mathbf{b}}$ be two unit vectors and $\alpha$ be the angle between
them. If $(\overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{b}})$ is also the unit vectors, then what is the
value of $\alpha ?$
Options:
them. If $(\overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{b}})$ is also the unit vectors, then what is the
value of $\alpha ?$
Solution:
1044 Upvotes
Verified Answer
The correct answer is:
$\frac{2 \pi}{3}$
Let $\vec{a}$ and $\vec{b}$ be two unit vectors.
$\therefore \quad|a|=1$ and $|b|=1$
Since, $\alpha$ is the angle between $\vec{a}$ and $\vec{b}$
$\therefore \quad \cos \alpha=\frac{\vec{a} \cdot \vec{b}}{|a \| b|}$
$\cos \alpha=\frac{\vec{a} \cdot \vec{b}}{1}$
$\cos \alpha=\vec{a} \cdot \vec{b}$
Now, $|\vec{a}+\vec{b}|=1 \quad(\because \vec{a}+\vec{b}$ is unit vector $)$
Squaring both sides
$\Rightarrow|\vec{a}|^{2}+|\vec{b}|^{2}+2 \vec{a} \cdot \vec{b}=1$
$\Rightarrow 1+1+2 \cos \alpha=1$
$\Rightarrow 2 \cos \alpha=-1$
$\Rightarrow \quad \cos \alpha=-\frac{1}{2}=\cos \frac{2 \pi}{3}$
$\Rightarrow \alpha=\frac{2 \pi}{3}$
$\therefore \quad|a|=1$ and $|b|=1$
Since, $\alpha$ is the angle between $\vec{a}$ and $\vec{b}$
$\therefore \quad \cos \alpha=\frac{\vec{a} \cdot \vec{b}}{|a \| b|}$
$\cos \alpha=\frac{\vec{a} \cdot \vec{b}}{1}$
$\cos \alpha=\vec{a} \cdot \vec{b}$
Now, $|\vec{a}+\vec{b}|=1 \quad(\because \vec{a}+\vec{b}$ is unit vector $)$
Squaring both sides
$\Rightarrow|\vec{a}|^{2}+|\vec{b}|^{2}+2 \vec{a} \cdot \vec{b}=1$
$\Rightarrow 1+1+2 \cos \alpha=1$
$\Rightarrow 2 \cos \alpha=-1$
$\Rightarrow \quad \cos \alpha=-\frac{1}{2}=\cos \frac{2 \pi}{3}$
$\Rightarrow \alpha=\frac{2 \pi}{3}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.