Prove that if u is parallel to v then |u · v| = |u||v|.

Prove that if u is not parallel to v then |u · v| < |u||v|.

Can anyone help me prove this please

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- Oct 13th 2008, 12:27 PMthe1u2001[SOLVED] Vector ProofProve that if u is parallel to v then |u · v| = |u||v|.

Prove that if u is not parallel to v then |u · v| < |u||v|.

Can anyone help me prove this please

- Oct 13th 2008, 12:31 PMJhevon
recall, $\displaystyle \bold{u} \cdot \bold{v} = |\bold{u}||\bold{v}| \cos \theta$

where $\displaystyle \theta$ is the angle between the vectors

what does $\displaystyle \theta$ have to be for the vectors to be parallel? what does it have to be for them not to be parallel? how does this affect the dot product - Oct 13th 2008, 12:33 PMthe1u2001
- Oct 13th 2008, 12:35 PMJhevon
- Oct 13th 2008, 12:46 PMthe1u2001
- Oct 13th 2008, 12:49 PMJhevon
start with the formula i gave you. say what theta has to be if the vectors are parallel and plug the values in and show what happens.

then say what theta has to be for the vectors not to be parallel and show what happens.

pretty straight forward stuff. don't fall into the trap of making this more complicated than it already is just because the word "prove" is there.

you may also want to say, $\displaystyle 0 \le \theta \le 2 \pi$ - Oct 13th 2008, 01:29 PMthe1u2001
K thank for the help and i have proved that if u is parallel that theta=0

but i am find it difficlut for if U is not parallel. is

theta http://www.mathhelpforum.com/math-he...b7f5ef8f-1.gif - Oct 13th 2008, 01:34 PMo_O
Since the two vectors are not parallel, i.e. $\displaystyle \theta \neq 0, 2\pi$, then that must mean $\displaystyle \cos \theta < 1$. Can you see how to get inequality $\displaystyle \bold{u} \cdot \bold{v} < |\bold{u}||\bold{v}|$ from $\displaystyle \bold{u} \cdot \bold{v} = |\bold{u}||\bold{v}|\cos \theta$ now?

- Oct 13th 2008, 01:38 PMthe1u2001