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experimentX
 3 years ago
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experimentX
 3 years ago
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experimentX
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1362585994965:dw

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1362586038538:dw

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1362586072732:dw \[ \cos \theta = \frac{<u, v>}{u\; v}\]

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.0so you have the component of u along v is \[ \large \frac{u\;v}{v} \times \frac{<u, v>}{u\; v}= \frac{v <u,v>}{v^2} = \frac{v <u,v>}{<v,v>}\]

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.0using vector law of addition we get .. the component perpendicular to V is \[V  \frac{V <U,V>}{<V,V>} \]

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1362586404101:dw

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1362586480133:dw

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.0dw: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.

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1362587300516:dw

Callisto
 3 years ago
Best ResponseYou've already chosen the best response.1Once we find u along v, we can express u as a linear combination of v and v⊥?!

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.0yes .. .lol, that will be equal to u

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.0use brackets to separate it ... and don't do arithmetic.
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