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C4 - Vectors

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    Find a unit vector ai + bj + ck which is perpendicular to the two vectors 2i + 2j -k and 4i + 2k

    I understand for two vectors to be perpendicular their dot product must be 0, provided they are both non zero vectors.

    So I worked out the dot product for the vector I am trying to find, and the two vectors given.

    2a + 2b - c = 0 and 4a + 2c = 0

    Is this right so far? I don't think there is a way to solve them like this
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    Find the cross product of the two vectors, then divide each coefficient of i, j and k by a constant, such that it's magnitude is 1. If you don't know how to do this yet then I'll explain a little more, but you won't learn anything if I just give you the answer haha.
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    for this you need to remember there wont be a unique solution as if (x,y,z) is perp then so will k(x,y,z)

    so if you could write

    a=kb
    b=b
    c=lb

    for some k,l then your vector would be b(k,1,l) and then choose b so its a unit vector.
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    (Original post by James94)
    Find the cross product of the two vectors, then divide each coefficient of i, j and k by a constant, such that it's magnitude is 1. If you don't know how to do this yet then I'll explain a little more, but you won't learn anything if I just give you the answer haha.
    So you find the point of intersection of the two vectors given, is that what you mean by cross product? I know what a unit vector is, its a vector of magnitude 1. So you divide each component of the vector by the magnitude of the initial vector.
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    Cross product is meant to help you arrive at a vector which is perpendicular to the original constituents.

    ie a cross b produces a vector n; this vector n is perpendicular to both a and b.

    Cross product is not solving for intersections whatsoever.

    Hope this helps. Peace.
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    Just thought I'd add that the cross product is not required for Edexcel C4.


    Also, remember that it isn't the exact values of what you've called a, b and c you are actually interested in - it's their relative values. You can make up any number and use it for a, b or c (it's usually easiest to use 1) and then solve your two simultaneous equations to find the other two.
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    What would be the best way for revising for vectors. I find them the hardest in C4
    Thanks
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    (Original post by bkhan)
    What would be the best way for revising for vectors. I find them the hardest in C4
    Thanks
    Do lots and lots and lots of questions - they don't deviate too much from the norm.
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    You can't find a specific solution as there are more variables than equations,but you can write each in terms of the others: e.g.

    a + b + c = 0
    2a + c = 0

    so c = -2a, and b = a, so you have a vector (a, a, -2a), and then just let a be a constant such that the vector (a, a, -2a) has length is 1

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Updated: May 24, 2012
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