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experimentX Group TitleBest ResponseYou've already chosen the best response.0
dw:1362585994965:dw
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.0
dw:1362586038538:dw
 one year ago

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

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

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

experimentX Group TitleBest ResponseYou've already chosen the best response.0
dw:1362586404101:dw
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.0
dw:1362586480133:dw
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.0
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.
 one year ago

Callisto Group TitleBest ResponseYou've already chosen the best response.1
dw:1362586938881:dw
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.0
dw:1362587300516:dw
 one year ago

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

experimentX Group TitleBest ResponseYou've already chosen the best response.0
yes .. .lol, that will be equal to u
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.0
use brackets to separate it ... and don't do arithmetic.
 one year ago
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