If the sum of two unit vectors is a unit vector, then find the magnitude of difference of these two vectors

Let ${\hat{\mathrm{n}}}_1 \mathrm{and} {\hat{\mathrm{n}}}_2$ are the two unit vectors, then the sum and difference of these two vectors be represented by
$\begin{array}{l}{\hat{\mathrm{n}}}_{\mathrm{s}}={\hat{\mathrm{n}}}_1+{\hat{\mathrm{n}}}_2\text{ and }{\hat{\mathrm{n}}}_d={\hat{\mathrm{n}}}_1-{\hat{\mathrm{n}}}_2\\ \Rightarrow {\mathrm{n}}_{\mathrm{s}}^2={\mathrm{n}}_1^2+{\mathrm{n}}_2^2+2{\mathrm{n}}_1{\mathrm{n}}_2\cos \mathrm{\theta }=1+1+2\cos \mathrm{\theta }\end{array}$
Since it is given that n_s is also a unit vector, therefore we have
1 = 1+1+ 2 cos$\mathrm{\theta }$
$\begin{array}{l}\Rightarrow \cos \mathrm{\theta }=-\frac{1}{2}\\ \Rightarrow \mathrm{\theta }=120\end{array}$
The difference of the two vectors is given by
$\begin{array}{l}{\hat{\mathrm{n}}}_{\mathrm{d}}={\hat{\mathrm{n}}}_1-{\widehat{\mathrm{n}}}_2\\ \Rightarrow {\mathrm{n}}_d^2={\mathrm{n}}_1^2+{\mathrm{n}}_2^2-2{\mathrm{n}}_1{\mathrm{n}}_2\cos \mathrm{\theta }=1+1-2\cos \left({120}^{\mathrm{o}}\right)\\ \Rightarrow {\mathrm{n}}_{\mathrm{d}}^2=2-2\left(-\frac{1}{2}\right)=2+1=3\\ \Rightarrow {\mathrm{n}}_{\mathrm{d}}=\sqrt{3}=1.73\end{array}$