Question

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

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

Answer 1

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

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 $n_s^2=n_s^2=n_1^2+n_2^2+2n_1{n}_2\cos \theta$
$=1+1+2\cos \theta$
Since it is given that $n_s$ is also a unit vector, therefore
$1=1+1+2\cos \theta \Rightarrow \cos \theta =-\frac{1}{2}\therefore \theta ={120}^o$
Now the difference vector is ${\hat{n}}_d={\hat{n}}_1-{\hat{n}}_2$
$n_d^2=n_1^2+n_2^2-2n_1{n}_2\cos \theta =1+1-2\cos \left({120}^o\right)$
$\therefore n_d^2=2-2(-1/2)=2+1=3\Rightarrow n_d=\sqrt{3}$