# Choose the correct answer in the following : let $$\overrightarrow a$$ and $$\overrightarrow b$$ be two unit vectors and θ is the angle between them. Then $$\overrightarrow a\ + \overrightarrow b$$ is a unit vector if

$\begin{array} (A)\; \theta = \frac{\pi}{4} \quad & (B)\; \theta = \frac{\pi}{3} \quad & (C) \;\theta = \frac{\pi}{2} \quad &(D) \;\theta = \frac{2\pi}{3} \end{array}$

Toolbox:
• $\overrightarrow a.\overrightarrow b=|\overrightarrow a||\overrightarrow b|\cos\theta$
• $\overrightarrow a.\overrightarrow b=\overrightarrow b.\overrightarrow a$
• If $\cos\theta<0$ then $\theta$ lies in the II quadrant (i.e)$(\pi-\theta)$
Step 1:
Given $|\overrightarrow a|=1$ and $|\overrightarrow b|=1$
$|\overrightarrow a+\overrightarrow b|=1$ or $|\overrightarrow a+\overrightarrow b|^2=1$
Let us consider $(\overrightarrow a+\overrightarrow b)^2=1$
(i.e) $(\overrightarrow a+\overrightarrow b)(\overrightarrow a+\overrightarrow b)=|\overrightarrow a|^2+|\overrightarrow b|^2+\overrightarrow a.\overrightarrow b+\overrightarrow b.\overrightarrow a.$
But $\overrightarrow a.\overrightarrow b=\overrightarrow b.\overrightarrow a$
Therefore $(\overrightarrow a+\overrightarrow b)(\overrightarrow a+\overrightarrow b)=|\overrightarrow a|^2+|\overrightarrow b|^2+2\overrightarrow a.\overrightarrow b=1.$
Step 2:
But $|\overrightarrow a|=|\overrightarrow b|=1$
$1^2+1^2+2\times 1\times 1\cos\theta=1$
$2+2\cos\theta=1$
$2(1+\cos\theta)=1$
$(1+\cos\theta)=\large\frac{1}{2}$
Therefore $\cos\theta=-\large\frac{1}{2}$
Step 3:
If $\cos\theta$ is negative,then it lies in the II quadrant.
$\cos(\pi-\large\frac{\pi}{3})=-\large\frac{1}{2}$
$\theta=\large\frac{2\pi}{3}$
Hence D is the correct option.