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Math Help - Vectors ! Please help :)

  1. #1
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    Vectors ! Please help :)

    Hi , not sure if vectors falls under this category , but what are you going to do I just need someone to explain this question for me step by step if you have time . I'm Gavin quite a hard time with it :/

    "The vectors a and b span two space . For what values of m is it true that (m^2+2m-3)a + (m^2+m-6) = 0
    *note that it is a 0 vector. I imagine this'd and they have to cancel each other out some how but I don't know how ....

    Thanks !
    Emma
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  2. #2
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    Re: Vectors ! Please help :)

    Hey elittlewood.

    Did you mean (m^2+2m-3)*a + (m^2+m-6)*b = 0?

    If so try setting up equations involving m and each of the vector components (i.e. x, y, and z) and then solve a set of equations.
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    Re: Vectors ! Please help :)

    Okay... woould that mean you have to expand...? im having a hard tie with that .
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    Re: Vectors ! Please help :)

    Basically you have x,y,z for a and another for b.

    You will get three sets of three equations. For example if we look only at x co-ordinate, we get:

    (m^2 + 2m - 3)*x_a - (m^2 + m - 6)x_b = 0.

    You have two other equations involving the other co-ordinates so if you set x_a and x_b to constants you can solve for m and if m is a common solution for all co-ordinates, then you have a solution.
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  5. #5
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    Re: Vectors ! Please help :)

    Quote Originally Posted by elittlewood View Post
    "The vectors a and b span two space . For what values of m is it true that (m^2+2m-3)a + (m^2+m-6) = 0
    Quote Originally Posted by chiro View Post
    Basically you have x,y,z for a and another for b.
    Please note that there is no second b in the original post. I think there is a typo in the question itself.

    I have seen this sort of question before. I think it should be (m^2+2m-3)\vec{a}+(m^2+m-6)\vec{b}=0.

    Because \vec{a}~\&~\vec{b} span 2-space then \alpha\vec{a}+\beta\vec{b}=\vec{0}\text{ if snd only if }\alpha=\beta=0.

    Thus coordinates really have little to do with it.
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  6. #6
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    Re: Vectors ! Please help :)

    To elaborate on what Plato said,

    set both m^2 + 2m - 3 \space and \space m^2 + m - 6 \space equal to \space 0

    solve both quadratics and then what ever value of m is common to both would be the value that you are looking for
    Last edited by jpritch422; May 27th 2013 at 08:49 AM.
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