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Vectors and planes watch

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    This should be really simple but I'm getting lost somewhere.
    Please could somebody follow the link:

    http://www.maths.leeds.ac.uk/~sergue...351/index.html

    Then click handouts, then lecture notes and look at the problem at the bottom of Page 20, labelled (1.40) .

    Looking at the bottom line, I understand the |BC| = |BA|cosθ part.
    But then I would say,

    |BC| = |a-b|cosθ

    Can somebody explain why cosθ disappears and where the unit vector n comes from please?
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    You know that BA=a-b. So |BA|=|a-b|=|a-b||n|, since |n|=1 (because it's a unit vector).

    Now, the dot product |P.Q| is defined as |P||Q|cosθ, where θ is the angle between P and Q. So |a-b||n|cosθ=|(a-b).n|.
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    Ah.. I see! Just to clear up though- usually, I'd define the dot product as,

    p.q = |p||q|cos θ [i.e with no modulus], in which case,

    |a-b||n|cosθ = (a-b).n

    So I assume you just added the modulus to the (a-b).n term because it is a length, right?
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    Precisely. Since the dot product gives you a scalar quantity, the modulus signs basically have no bearing at all. What they do is ensure that you don't get a negative distance. (Although in some cases getting negative values might help point out which side of the plane the point is on.)
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    Thanks for clearing that up!
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    oh my gosh i should know this...should i be worried that i don't?!?!? lol its only been what? 3months since iv done this! aaaaaaaaaaaaaaargh thank god im nt doing a degree in maths!!!
 
 
 
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Updated: August 26, 2005

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