Practicing Success
If $\vec a$ and $\vec b$ are unit vectors, then the greatest value of $|\vec a+\vec b|+|\vec a-\vec b|$, is |
2 4 $2\sqrt{2}$ $\sqrt{2}$ |
$2\sqrt{2}$ |
Let θ be the angle between unit vectors $\vec a$ and $\vec b$. Then, $\vec a .\vec b = \cos θ$ Now, $|\vec a+ \vec b|^2$ $=|\vec a|^2+|\vec b|^2+2\vec a .\vec b=2+2\cos θ=4\cos^2\frac{θ}{2}$ and, $|\vec a-\vec b|^2$ $=|\vec a|^2+|\vec b|^2-2\vec a .\vec b=2-2\cos θ=4\sin^2\frac{θ}{2}$ $⇒|\vec a+\vec b|=2\cos\frac{θ}{2},|\vec a-\vec b|=2\sin\frac{θ}{2}$ $⇒|\vec a+\vec b|+|\vec a-\vec b|=2(\cos\frac{θ}{2}+\sin\frac{θ}{2})≤2\sqrt{2}$ |