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Babynini

  • one year ago

Projecting Vertex.

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  1. Babynini
    • one year ago
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  2. Babynini
    • one year ago
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    @acxbox22 :)

  3. acxbox22
    • one year ago
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    sorry i dont know this stuff :(

  4. Babynini
    • one year ago
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    @Nnesha

  5. IrishBoy123
    • one year ago
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    \(proj_{\vec{v}}\vec{u} = \frac{\vec{v} \bullet \vec{u}}{|\vec v|^2}\vec{v}\) is the *vector* projection of \(\vec u \) on \(\vec v \). it's just a formula so calculate the various parts, eg \(\vec{v} \bullet \vec{u} = <11,3> \bullet <-6,-4>\)

  6. IrishBoy123
    • one year ago
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    \(| \vec v |^2 = \vec v \bullet \vec v\)

  7. Babynini
    • one year ago
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    so u * v = <-66,-12>

  8. Babynini
    • one year ago
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    yeah?

  9. Babynini
    • one year ago
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    simplified u*v=-78 v*v=52 is it right so far?

  10. Babynini
    • one year ago
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    @IrishBoy123

  11. Babynini
    • one year ago
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    @Loser66 :)

  12. anonymous
    • one year ago
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    can you help me plz

  13. Babynini
    • one year ago
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    Post your question on a new feed and i'll try :)

  14. IrishBoy123
    • one year ago
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    "u • v = <-66,-12>" nah u • v = <11,3).<-6,-4> = (11)(-6) + (3)(-4) = -66 - 12 = - 78

  15. Babynini
    • one year ago
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    yep yep that's what I got for that one :)

  16. IrishBoy123
    • one year ago
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    \(\vec u \bullet \vec v = <u_x,u_y> \bullet <v_x,v_y> = (u_x \times v_x) + (u_y \times v_y) \)

  17. IrishBoy123
    • one year ago
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    sorry, bandwith issues, i am clearly behind times. let me read the thread again

  18. Babynini
    • one year ago
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    It's all good :)

  19. IrishBoy123
    • one year ago
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    v.v = <-6,-4><-6,-4> = 36 + 16 = 52 seems we agree

  20. IrishBoy123
    • one year ago
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    what next?

  21. Babynini
    • one year ago
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    uh..not sure haha

  22. Babynini
    • one year ago
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    plug it into the formula?

  23. IrishBoy123
    • one year ago
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    yes <9,6>?

  24. Babynini
    • one year ago
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    so we have \[\frac{ -78 }{ 52}v\]

  25. Babynini
    • one year ago
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    woah where did you get<9,6> ?

  26. IrishBoy123
    • one year ago
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    v = <-6,-4> and, -78/52 = -3/2 and, -3/2 * <-6,-4> = <9,6> agree?

  27. Babynini
    • one year ago
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    ah I see, thanks :)

  28. Babynini
    • one year ago
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    so, is that the final answer then? o.0

  29. Babynini
    • one year ago
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    or ..is that just what \[u _{1} \] equals?

  30. IrishBoy123
    • one year ago
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    that answers (a). for (b) you have to .....!!! do something v similar

  31. Babynini
    • one year ago
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    let's do it x)

  32. Babynini
    • one year ago
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    we would use (-75/52)v again, yeah?

  33. IrishBoy123
    • one year ago
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    you've already done \(\vec u_1\) that's what you just calculated so you need to do the same on \(\vec u_2\) so what is \(\vec u_2\) ? it is orthogonal to \(\vec v\) ...

  34. Babynini
    • one year ago
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    its \[u - u _{1}\]

  35. Babynini
    • one year ago
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    <2,-3>?

  36. IrishBoy123
    • one year ago
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    <-6,-4>•<x,y> = 0 -6x-4y = 0 try: <-6,-4>•<4,-6> or <-6,-4>•<-4,6> both work we can use either

  37. Babynini
    • one year ago
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    does what I did work? o.o

  38. Babynini
    • one year ago
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    I'm pretty sure that's what the prof did in class but ah well haha

  39. IrishBoy123
    • one year ago
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    let me actually do it

  40. Babynini
    • one year ago
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    k

  41. IrishBoy123
    • one year ago
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    <2,-3> Roger

  42. Babynini
    • one year ago
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    Yay! fantastic :)

  43. Babynini
    • one year ago
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    thanks for taking the time to calculate it all :)

  44. Babynini
    • one year ago
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    so we have \[u _{1}=<9,6> u _{2}=<2,-3>\]

  45. Babynini
    • one year ago
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    what goes into \[proj _{v}u\]?

  46. Babynini
    • one year ago
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    nevermind, they just wanted <9,-6> there too :) thanks!!

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