Question

If the angle between the unit vectors $\overrightarrow{a}$ and $\overrightarrow{b}$ is ${60}^o$, then $|\overrightarrow{a}-\overrightarrow{b}|$ is

• A. 0
• B. 2
• C. 1
• D. 4

Answer 1

The correct option is C 1

As $\overrightarrow{a}$ and $\overrightarrow{b}$ are unit vectors, thus their respective magnitude is $1$.

$\therefore$ $|\overrightarrow{a}|=|\overrightarrow{b}|=1$

Also $\overrightarrow{a}.\overrightarrow{b}=|\overrightarrow{a}||\overrightarrow{b}|cos\theta$ where $\theta ={60}^o$

$\therefore$ $\overrightarrow{a}.\overrightarrow{b}=\left(1\right)\left(1\right)cos{60}^o=0.5$

Now $|\overrightarrow{a}-\overrightarrow{b}{|}^2=(\overrightarrow{a}-\overrightarrow{b}).(\overrightarrow{a}-\overrightarrow{b})$

$|\overrightarrow{a}-\overrightarrow{b}{|}^2=|\overrightarrow{a}{|}^2+|\overrightarrow{b}{|}^2-2(\overrightarrow{a}.\overrightarrow{b})$ $(\because \overrightarrow{a}.\overrightarrow{b}=\overrightarrow{b}.\overrightarrow{a})$

$|\overrightarrow{a}-\overrightarrow{b}{|}^2=1^2+1^2-2\left(0.5\right)=1$

$⟹$ $|\overrightarrow{a}-\overrightarrow{b}|=\sqrt{1}=1$