Question

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

• A. $\sqrt{2}$
• B. $\sqrt{3}$
• C. $\frac{1}{\sqrt{2}}$
• D. $\sqrt{5}$

Answer 1

The correct option is B $\sqrt{3}$

Let $\widehat{n_1}$ and $\widehat{n_2}$ be the two unit vectors. Then their sum is
$\overrightarrow{n_s}=\widehat{n_1}+\widehat{n_2}$ or $n_s^2=n_1^2+n_2^2+2n_1{n}_2\cos \theta$
$=1+1+2\cos \theta$
Since it is given that $\overrightarrow{n_s}$ is also a unit vector,
$1=1+1+2\cos \theta$
$\Rightarrow \cos \theta =-\frac{1}{2}\therefore \theta ={120}^{\circ }$
Now the difference vector is $\overrightarrow{n_d}=\widehat{n_1}-\widehat{n_2}$
or $n_d^2=n_1^2+n_2^2-2n_1{n}_2\cos \theta$
$=1+1-2\cos \left({120}^{\circ }\right)$
$n_d^2=2-2(-1/2)=2+1=3$
$\Rightarrow n_d=\sqrt{3}$