If the sum of two unit vectors is a unit v

Let ${\hat n_1}$ and ${\hat n_2}$ are the two unit vectors, then the sum is

${\overrightarrow n _s} = {\hat n_1} + {\hat n_2}$ or

$n_s^2 = n_1^2 + n_2^2 + 2{n_1}{n_2}\cos \theta $

$ = 1 + 1 + 2 \cos \theta $

Since it is given that $n_s$ is also a unit vector, therefore $1 = 1 + 1 + 2\cos \theta $

$⇒$ $\cos \theta = - \frac{1}{2}$

$\therefore \theta = 120^\circ $

Now the difference vector is ${\hat n_d} = {\hat n_1} - {\hat n_2}$ or

$n_d^2 = n_1^2 + n_2^2 - 2{n_1}{n_2}\cos \theta $

$ = 1 + 1 - 2\cos (120^\circ )$

$n_d^2 = 2 - 2( - 1/2) = 2 + 1 = 3$

$ \Rightarrow \,\,{n_d} = \sqrt 3 $