Question

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

• A. 1
• B.
• C.
• D. None of these

Answer 1

The correct option is D

None of these

Let $\widehat{n_1}$ and $\widehat{n_2}$ are the two unit vectors, then their sum is $\overrightarrow{n_s}$ = $\widehat{n_1}$ + $\widehat{n_2}\Rightarrow |\overrightarrow{n_s}|=|\widehat{n_1}+\widehat{n_2}|$

Squaring both sides,

$\Rightarrow |\overrightarrow{n_s}.\overrightarrow{n_s}|=|(\widehat{n_1}+\widehat{n_2}).(\widehat{n_1}+\widehat{n_2})|$

$\Rightarrow n_s{}^2=n_1{}^2+n_2{}^2+2n_1{n}_{2}cos\theta =1+1+2cos\theta$

Since it is given that $\overrightarrow{n_s}$ is a unit vector,so $n_s=1$.

Therefore $1=1+1+2cos\theta \Rightarrow cos\theta =-\frac{1}{2}\Rightarrow \theta ={120}^{\circ }$

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

$\Rightarrow n_d{}^2=n_1{}^2+n_2{}^2-2n_1{n}_{2}cos\theta =1+1-2cos\left({120}^{\circ }\right)=3$

$\Rightarrow n_d=\sqrt{3}$