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Math Help - Proof of angle between vectors (using dot and cross product)

  1. #1
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    Proof of angle between vectors (using dot and cross product)

    Prove that if \theta is the angle between u and v and u . v =/= 0, then tan \theta = ||u x v||/(u . v)

    where . is dot product

    I am pretty familiar with the properties of both dot and cross product but that tan throws me for a loop, I just can't seem to find a way to do this
    Last edited by Idlewild; January 5th 2011 at 09:26 PM. Reason: math format
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Idlewild View Post
    Prove that if [Theta] is the angle between u and v and u . v =/= 0, then tan [Theta] = ||u x v||/(u . v)

    where . is dot product

    I am pretty familiar with the properties of both dot and cross product but that tan throws me for a loop, I just can't seem to find a way to do this
    Recall that \displaystyle \sin(\theta)=\frac{\|a\times b\|}{\|a\|\|b\|} and \displaystyle \cos(\theta)=\frac{a\cdot b}{\|a\|\|b\|}.
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    Thanks for the speedy reply!

    I'm not sure if this is too basic, or only physics related or something but I guess that means to use SOH CAH TOA, where O = || a x b ||, and A = a . b

    Therefore tan \theta = || a x b || / a . b
    and ah! that turns out to be the given answer!

    now to prove it... I see that u . v =/= 0, which means that u and v are not orthogonal.. That's all I really can come to right now, I haven't had much luck with the geometric interpretation

    Edit: Or am I complicating this, and simply showing the explanation of sin and cos to tan would be sufficient?
    Last edited by Idlewild; January 5th 2011 at 09:35 PM. Reason: Another thought
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    You mean saying \displaystyle\tan\theta = \frac{\sin \theta}{\cos \theta} =\frac{\frac{\|a\times b\|}{\|a\|\|b\|}}{\frac{a\cdot b}{\|a\|\|b\|}} = \frac{\|a\times b\|}{a\cdot b} ?
    Last edited by pickslides; January 5th 2011 at 10:13 PM. Reason: Bad Latex...
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    Yes, I was going to show the SOH CAH TOA thing, but that works.
    Do you think that would be acceptable?
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    Maybe it would be acceptable, depends on your lecturer, you may need to prove

    \displaystyle \sin(\theta)=\frac{\|a\times b\|}{\|a\|\|b\|} and \displaystyle \cos(\theta)=\frac{a\cdot b}{\|a\|\|b\|}

    as well for the answer to be accepted, this brings you back to square one.
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by pickslides View Post
    Maybe it would be acceptable, depends on your lecturer, you may need to prove

    \displaystyle \sin(\theta)=\frac{\|a\times b\|}{\|a\|\|b\|} and \displaystyle \cos(\theta)=\frac{a\cdot b}{\|a\|\|b\|}

    as well for the answer to be accepted, this brings you back to square one.
    You don't really "prove" these though. By definition one takes \displaystyle \cos(\theta)=\frac{a\cdot b}{\|a\|\|b\|} and the other one is gotten from the identity \sin^2(\theta)+\cos^2(\theta)=1.
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  8. #8
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    It is the first part of the first question on this assignment and I do also think that we are given the definition for the cos, so I will just show by identities

    Thanks everyone
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