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Math Help - [SOLVED] Vector Proof

  1. #1
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    Exclamation [SOLVED] Vector Proof

    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


    Last edited by the1u2001; October 13th 2008 at 12:38 PM.
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by the1u2001 View Post
    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

    Thanks Muqtasid

    recall, \bold{u} \cdot \bold{v} = |\bold{u}||\bold{v}| \cos \theta

    where \theta is the angle between the vectors

    what does \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
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    Quote Originally Posted by Jhevon View Post
    recall, \bold{u} \cdot \bold{v} = |\bold{u}||\bold{v}| \cos \theta

    where \theta is the angle between the vectors

    what does \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

    u is parallel to v if and only if the angle between u and v is 0 or (pie/2).


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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by the1u2001 View Post
    u is parallel to v if and only if the angle between u and v is 0 or (pie).


    yes, and what happens in those two case, in particular, what is \cos \theta? can you answer the problem now?
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    Quote Originally Posted by Jhevon View Post
    yes, and what happens in those two case, in particular, what is \cos \theta? can you answer the problem now?

    I understand what happens but i don't know where to start with the proof
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by the1u2001 View Post
    I understand what happens but i don't know where to start with the proof
    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, 0 \le \theta \le 2 \pi
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    Quote Originally Posted by Jhevon View Post
    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, 0 \le \theta \le 2 \pi
    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
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  8. #8
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    Since the two vectors are not parallel, i.e. \theta \neq 0, 2\pi, then that must mean \cos \theta < 1. Can you see how to get inequality \bold{u} \cdot \bold{v} < |\bold{u}||\bold{v}| from \bold{u} \cdot \bold{v} = |\bold{u}||\bold{v}|\cos \theta now?
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  9. #9
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    Quote Originally Posted by o_O View Post
    Since the two vectors are not parallel, i.e. \theta \neq 0, 2\pi, then that must mean \cos \theta < 1. Can you see how to get inequality \bold{u} \cdot \bold{v} < |\bold{u}||\bold{v}| from \bold{u} \cdot \bold{v} = |\bold{u}||\bold{v}|\cos \theta now?

    Yes know i get it, thanks for the Help. I really appriate it.
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