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burhan101

  • one year ago

Area of a parallelogram a= [1,-4], = [3,-5]

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  1. tomo
    • one year ago
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    is there supposed to be a b = [3,-5]

  2. burhan101
    • one year ago
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    yup, typo *

  3. amistre64
    • one year ago
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    magnitude of the cross product i think is a way to determine it

  4. amistre64
    • one year ago
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    |a||b| sin(t); where t is the angle between a and b

  5. amistre64
    • one year ago
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    \[sin(t)=\frac{|axb|}{|a||b|}\] so yeah, the magnitude of the cross product

  6. burhan101
    • one year ago
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    @amistre64 how can i do the cross product when only x and y are given ?

  7. amistre64
    • one year ago
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    plug in z=0 :)

  8. burhan101
    • one year ago
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    ohhh! .. facepalm

  9. amistre64
    • one year ago
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    a= [1,-4], = [3,-5] x 1 3 y -4 -5 z 0 0 x = 0-0 -y = 0-0 z = -5+12 the magnitude of [0, 0, 7 ] = 7

  10. amistre64
    • one year ago
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    of course the cross of R^2 simplifies as the determinant

  11. burhan101
    • one year ago
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    Okay thanks ! :)

  12. burhan101
    • one year ago
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    wouldnt axb = 0 ? :S

  13. burhan101
    • one year ago
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    oh nvm

  14. amistre64
    • one year ago
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    my teacher did an R^2 cross a few semesters ago, and I was like, aint it just the determinant? and she was like or just plug in z=0 :)

  15. burhan101
    • one year ago
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    yeah i think its easier to plug in the zero :P

  16. amistre64
    • one year ago
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    have fun ;)

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