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

• $\theta = \cfrac{\pi }{4}$

• $\theta = \cfrac{\pi }{3}$

• $\theta = \cfrac{\pi }{2}$

• $\theta = \cfrac{{2\pi }}{3}$

[NCERT,Page 459,Misc,Q.No 17]

[Correct Option is d]

We have: $\left| {\vec a} \right| = 1,\left| {\vec b} \right| = 1$

…(i)
Now, $\left| {\vec a + \vec b} \right| = 1 \Rightarrow {\left| {\vec a + \vec b} \right|^2} = 1$

$\Rightarrow$ $\left( {\vec a + \vec b} \right) \cdot \left( {\vec a + \vec b} \right) = 1 \Rightarrow \vec a \cdot \vec a + \vec a \cdot \vec b + \vec b \cdot \vec a + \vec b \cdot \vec b = 1$

$\Rightarrow$ ${\left| {\vec a} \right|^2} + 2\vec a \cdot \vec b + {\left| {\vec b} \right|^2} = 1$

$\Rightarrow$ ${1^2} + 2\vec a \cdot \vec b + {1^2} = 1$

[Using (i)]
$\Rightarrow$ $1 + 2\left| {\vec a} \right|\left| {\vec b} \right|\cos \theta + 1 = 1$

$\Rightarrow$ $1 + 2\left( 1 \right)\left( 1 \right)\cos \theta + 1 = 1$

[Using (i)]
$\Rightarrow$ $2\cos \theta = - 1 \Rightarrow \cos \theta = - \cfrac{1}{2}$
$\Rightarrow$ $\theta = \cfrac{{2\pi }}{3}$