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    Hi guys
    I'm struggling to understand part c of the question on the attachment. i know that the scalar multiple of two perpendicular angle is zero bur i don't know how to apple it on part c. any help would be much appreciated
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    a.b = magnitudes of two vectors * cos angle.

    The magnitudes aren't zero but cos can be zero and cos is zero at 90 degrees.
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    (Original post by Alen.m)
    Hi guys
    I'm struggling to understand part c of the question on the attachment. i know that the scalar multiple of two perpendicular angle is zero bur i don't know how to apple it on part c. any help would be much appreciated
    \boldsymbol{p}\cdot \boldsymbol{p} = |\boldsymbol{p}|^2
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    (Original post by notnek)
    \boldsymbol{p}\cdot \boldsymbol{p} = |\boldsymbol{p}|^2
    Just as an FYI - we tend to use \mathbf{p} (\mathbf{p}) for bold vectors in \LaTeX - that should save you some typing time. :yep:
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    this appears to be GCSE.

    Modulus may confuse.
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    (Original post by Zacken)
    Just as an FYI - we tend to use \mathbf{p} (\mathbf{p}) for bold vectors in \LaTeX - that should save you some typing time. :yep:
    Yes I've always used that in mathematical writing but I couldn't remember it so I chose the lazy option!
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    (Original post by Mutleybm1996)
    this appears to be GCSE.

    Modulus may confuse.
    Definitely not GCSE - look at part (b).
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    (Original post by notnek)
    Definitely not GCSE - look at part (b).
    oops.

    It's quite a nice question actually, I remember those days fondly ^_^
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    this is not GCSE

    this is a standard geometric proof involving vectors

    see Q154
    http://www.madasmaths.com/archive/ma...ons_part_a.pdf
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    yeah the question is in c4 , thanks everyone managed to find a great explanation for it
 
 
 
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