Question

If the magnitude of difference of two unit vectors is $\sqrt{3}$, then the magnitude of sum of the two vectors is

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

Answer 1

The correct option is B 1 unit

Let $\overrightarrow{a}$ and $\overrightarrow{b}$ are two unit vectors.

$|\overrightarrow{a}-\overrightarrow{b}|=\sqrt{3}$

or, $\sqrt{a^2+b^2-2ab\cos \theta }=\sqrt{3}$ where, $\theta$ is the angle between the vectors

or, $a^2+b^2-2ab\cos \theta =3$

or, $1+1-2\cos \theta =3$ $(|\overrightarrow{a}|=|\overrightarrow{b}|=1)$

or, $\cos \theta =-\frac{1}{2}$

Now,

$|\overrightarrow{a}+\overrightarrow{b}|=$$\sqrt{a^2+b^2+2ab\cos \theta }$

$=\sqrt{1+1+2\times (-\frac{1}{2})}$

$=1$ unit

Hence, B is the correct answer.