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experimentX

  • 3 years ago

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  1. experimentX
    • 3 years ago
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    |dw:1362585994965:dw|

  2. experimentX
    • 3 years ago
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    |dw:1362586038538:dw|

  3. experimentX
    • 3 years ago
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    |dw:1362586072732:dw| \[ \cos \theta = \frac{<u, v>}{||u||\; ||v||}\]

  4. experimentX
    • 3 years ago
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    so you have the component of u along v is \[ \large \frac{||u||\;v}{||v||} \times \frac{<u, v>}{||u||\; ||v||}= \frac{v <u,v>}{||v||^2} = \frac{v <u,v>}{<v,v>}\]

  5. experimentX
    • 3 years ago
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    using vector law of addition we get .. the component perpendicular to V is \[V - \frac{V <U,V>}{<V,V>} \]

  6. experimentX
    • 3 years ago
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    |dw:1362586404101:dw|

  7. experimentX
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    |dw:1362586480133:dw|

  8. experimentX
    • 3 years ago
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    |dw:1362586557753:dw| you can check for dot product ... and show be able to verify which should be perpendicular to which and ...same goes for parallel.

  9. Callisto
    • 3 years ago
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    |dw:1362586938881:dw|

  10. experimentX
    • 3 years ago
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    |dw:1362587300516:dw|

  11. Callisto
    • 3 years ago
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    Once we find u along v, we can express u as a linear combination of v and v⊥?!

  12. experimentX
    • 3 years ago
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    yes .. .lol, that will be equal to u

  13. experimentX
    • 3 years ago
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    use brackets to separate it ... and don't do arithmetic.

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