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Math Help - Need help with a vector question.

  1. #1
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    Need help with a vector question.

    Given the following vector:

    a = i + 2j -2k

    Find a possible set of values for y and z such that the unit vector shown below is perpendicular to vector a.

    b = (y)j + (z)k

    I've tried using both of these rules [ (i) |a x b| = |a||b|sin90 (ii)a.b = 0 ]to form an equation but i can't seem to get an answer.

    I suspect it has something to do with matrices also, as this is part of a matrices question.
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  2. #2
    MHF Contributor red_dog's Avatar
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    \mathbf{a}\perp\mathbf{b}\Leftrightarrow \mathbf{a}\cdot\mathbf{b}=0

    \mathbf{a}\cdot\mathbf{b}=2y-2z\Rightarrow y=z
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  3. #3
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    Is that all? that was my answer too. but shouldn't y and z have some values instead of just y=z?
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  4. #4
    MHF Contributor red_dog's Avatar
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    y (or z) can take any value.

    For example:

    y=z=1 or y=z=-10 etc.
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  5. #5
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    i see. Thanks!
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  6. #6
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    Quote Originally Posted by toffeefan View Post
    Given the following vector: a = i + 2j -2k

    Find a possible set of values for y and z such that the unit vector shown below is perpendicular to vector a.
    b = (y)j + (z)k
    Did you notice that vector b should be a unit vector?
    So y^2+z^2=1.
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  7. #7
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    Quote Originally Posted by Plato View Post
    Did you notice that vector b should be a unit vector?
    So y^2+z^2=1.
    ok. but i'm still not sure of how to go about solving x and y. what i've done is:

    |a| = 3
    |b| = 1 (since b is unit vector)

    a x b = (2z + 2y)i - (z)j + (y)k

    |a x b| = square root of (2z + 2y)^2 + z^2 + y^2

    since y^2 + z^2 = 1,

    |a x b| = square root of 1 + 4(1) + 6zy

    Using |a x b| = |a||b|sin90,

    square root of 1 + 4(1) + 6zy = 3
    y = -z/3

    How do i form the other eqn?
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