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burhan101

  • 3 years ago

Cross Product

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  1. burhan101
    • 3 years ago
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    |dw:1362205701992:dw|

  2. arjont
    • 3 years ago
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    \[u*v=|u||v|\cos\theta\]

  3. arjont
    • 3 years ago
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    that would give you u dot v... but you need to find the components of each vector so that you can take the cross product. is it all thats given? is plotted on a accurate graph? maybe you can just read off the components from the graph

  4. burhan101
    • 3 years ago
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    Thats all that's given !

  5. arjont
    • 3 years ago
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    oh i see, you are basically given the hypotenuse of two triangles, i think you just take the components of each, where z component is 0 for both... but im not sure about your picture, since you can't tell the angles they with the horizontal

  6. arjont
    • 3 years ago
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    *the angles they make with the horizontal

  7. burhan101
    • 3 years ago
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    it has something to do with the right hand rule but im so confused with it !

  8. harsimran_hs4
    • 3 years ago
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    magnitude of u cross v = |u||v| sin(angle between them) as regards the direction is is always perpendicular to the plane containing these vectors

  9. harsimran_hs4
    • 3 years ago
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    there are two perpendicular directions for a plane so: to determine which side follow this suppose you are to see the direction of a cross b keep your right hand`s base on a vector and curl to towards b and the direction in which the thumb points is the required direction

  10. burhan101
    • 3 years ago
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    how do you know which one to curl it towards tho ?

  11. harsimran_hs4
    • 3 years ago
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    a cross b you are to move from a to b

  12. harsimran_hs4
    • 3 years ago
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    clear?

  13. burhan101
    • 3 years ago
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    yup, thanks !

  14. harsimran_hs4
    • 3 years ago
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    :)

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