Question

If $a$ and $b$ are two unit vectors inclined at an angle $\frac{\pi }{3}$, then the value of $|a+b|$ is

• A. greater than 1
• B. less than 1
• C. equal to 1
• D. equal to 0

Answer 1

Answer: (A)

Given, $|a|=|b|=1$ and $\theta =\pi /3$
Now, $|a+b{|}^2=|a{|}^2+|b{|}^2+2|a||b|\cos \theta$
$=1^2+1^2+2\times 1\times 1\times \cos \frac{\pi }{3}$
$=1+1+2\times \frac{1}{2}=1+1+1=3$
$\therefore |a+b|=\sqrt{3}$
$\therefore |a+b|>1$