Question

If $a$ and $b$ are unit vectors, then the greatest value of $|a+b|+|a-b|$ is

• A. $\sqrt{3}$
• B. $\sqrt{2}$
• C. $2\sqrt{2}$
• D. 4
• E. $3\sqrt{3}$

Answer 1

Answer: (C)

Let $\theta$ be an angle between unit vectors a and $b$.
Then, $a.b=\cos \theta$
Now, $|a+b{|}^2=|a{|}^2+|b{|}^2+2a\cdot b$
$=2+2\cos \theta =4{\cos }^2\frac{\theta }{2}$
$|a-b{|}^2=|a{|}^2+|b{|}^2-2a\cdot b$
$=2-2\cos \theta =4{\sin }^2\frac{\theta }{2}$
$\therefore |a+b|=2\cos \frac{\theta }{2},|a-b|=2\sin \frac{\theta }{2}$
$\therefore |a+b|+|a-b|=2\left(\cos \frac{\theta }{2}+\sin \frac{\theta }{2}\right)\le 2\sqrt{2}$