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, $|\vec{a}-\vec{b}|=2 \sin \frac{\theta}{2}$
$\therefore |\vec{a}+\vec{b}|+|\vec{a}-\vec{b}|$
$=2\left(\cos \frac{\theta}{2}+\sin \frac{\theta}{2}\right) \leq 2 \sqrt{2}$