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Math Help - Projection of Vectors

  1. #1
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    Projection of Vectors

    The rectangular prism is measured 1x2x3 as shown, find the the projection of vector u onto vector v.

    i'm having trouble finding the points of u and v

    my guess:
    u(1,0,3)
    v(1,2,3)
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  2. #2
    MHF Contributor Calculus26's Avatar
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    The formula for the projection of u on v is


    proju on v = (u*v/||v||^2) v


    what you are using for u and v is incorrect

    u = (3,0,1) v = (3,2,-1)
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  3. #3
    Junior Member enjam's Avatar
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    The projection of u along v is simply:
    u . v/|v|
    So you have:
    (3i + 0j + k) . (3i + 2j - k) / (3^2 + 2^2 + (-1)^2)^0.5
    = (3x3) + (0x2) + (1x-1) / (14)^0.5
    = 8 / sqrt 14
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  4. #4
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    How did you get the x value to be 3? isn't the x value the width/depth of the rectangular prism, thus 1? and the height 3 = z value and the length 2=y value?
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  5. #5
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    Quote Originally Posted by skeske1234 View Post
    How did you get the x value to be 3? isn't the x value the width/depth of the rectangular prism, thus 1? and the height 3 = z value and the length 2=y value?
    You just learn the rule for projections.

    \text{proj}_v u = \frac{{u \cdot v}}<br />
{{\left\| v \right\|^2 }}v = \frac{8}<br />
{{14}}\left\langle {3,2, - 1} \right\rangle
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  6. #6
    MHF Contributor Calculus26's Avatar
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    The projection of u along v is simply:
    u . v/|v|
    So you have:
    (3i + 0j + k) . (3i + 2j - k) / (3^2 + 2^2 + (-1)^2)^0.5
    = (3x3) + (0x2) + (1x-1) / (14)^0.5
    = 8 / sqrt 14

    The projection is a vector --what you have here is the magnitude of the projection
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  7. #7
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    Plato, I mean how did you get the point (3,2,-1)? like the vector points of u and v, in order to plug in formula
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  8. #8
    MHF Contributor Calculus26's Avatar
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    At firs tglance it appears as though u goes from(0,0,1) to (3,0,1)

    so u should b3 3i not 3i + k as I suggested

    and v goes from (0,0,1) to (3,2,0) v = 3 i + 2j -k
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