Question

If $\vec{a}$ and $\vec{b}$ are unit vectors, then the greatest value of $|\vec{a}+\vec{b}|+|\bar{a}-\vec{b}|$ is

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

Answer 1

Answer: (A)

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