1. ## Vector Help

Given:

$\displaystyle \vec{u}\bullet\vec{v}$ = $\displaystyle |\vec{u}||\vec{v}|\cos{\theta}$

Prove that $\displaystyle |\vec{u}\bullet\vec{v}| \leq |\vec{u}||\vec{v}|$

I have:

$\displaystyle \vec{u}\bullet\vec{v}$ = $\displaystyle |\vec{u}||\vec{v}|\cos{\theta}\leq|\vec{u}||\vec{v }| (1)$ since $\displaystyle |\cos{\theta}|\leq1$

How do I jump from that to $\displaystyle |\vec{u}\bullet\vec{v}| \leq |\vec{u}||\vec{v}|$. I realize it's probably very simple. If someone could just help me take the leap I would appreciate it!

I guess I know that the dot product is a scalar so I'm wondering..I don't know...geez louise.

P.S. It's like 1 in the morning, and this is my last quesiton

2. You already got it don't you?

If $\displaystyle \vec{u} \cdot \vec{v} = \left| \vec{u}\right| \left|\vec{v}\right| \cos \theta$, then:

$\displaystyle \left|\vec{u} \cdot \vec{v}\right| = \left| \vec{u}\right| \left|\vec{v}\right| |\cos \theta| \leq \left| \vec{u}\right| \left|\vec{v}\right|(1)$

since $\displaystyle \left| \vec{u}\right| \geq 0$ and $\displaystyle \left| \vec{v}\right| \geq 0$ already. Just a matter of dealing with $\displaystyle \cos \theta$ which you've promptly handled.