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Zale101

  • one year ago

Show that if v1, v2, and v3 are mutually orthogonal nonzero vectors in 3-space, and if a vector v in 3-space is expressed as: v=c1v1+c1v2+c3v3

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  1. Zale101
    • one year ago
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    Then the scalars c1, c2, and c3 are given by the formula: \(\Large c_i=\LARGE \frac{(v*v_i)}{||V_i||^2}\)

  2. beginnersmind
    • one year ago
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    v=c1v1+c1v2+c3v3 What happens if you take the scalar product with v1 on both sides?

  3. phi
    • one year ago
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    i.e. what is the dot product of two orthogonal vectors?

  4. Jhannybean
    • one year ago
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    |dw:1441576756278:dw| oo this drawing tool is so cool!!

  5. Zale101
    • one year ago
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    Hi! Sorry for the late reply! The dot product of the two orthogonal vectors is zero?

  6. Zarkon
    • one year ago
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    yes

  7. Zale101
    • one year ago
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    So v1*v2=0?

  8. beginnersmind
    • one year ago
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    yes. So take the scalar product of v=c1v1+c1v2+c3v3 and simplify.

  9. beginnersmind
    • one year ago
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    So take the scalar product of v=c1v1+c1v2+c3v3 *with v1* and simplify.

  10. beginnersmind
    • one year ago
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    You mean c1?

  11. beginnersmind
    • one year ago
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    Zale are you there?

  12. Zale101
    • one year ago
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    Yes, i'm sorry i dont understand what you meant :(

  13. Jhannybean
    • one year ago
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    yeah that got a bit confusing @beginnersmind

  14. beginnersmind
    • one year ago
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    Ok. I'll try to exlain: We were given v = c1v1+c2v2+c3v3. Both sides of the equation represent the same vector (v) written in different ways. So if we take the scalar product of v with v1 and c1v1+c2v2+c3v3 with v1 we get the same number. I.e v*v1 = ( c1v1+c2v2+c3v3)*v1 Clear so far?

  15. Jhannybean
    • one year ago
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    Yep!

  16. beginnersmind
    • one year ago
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    If not, can you tell me exactly which point is confusing?

  17. Zale101
    • one year ago
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    Makes sense now, i got confused when you said "take the scale product" and i couldn't know which vectors to deal with.

  18. Zale101
    • one year ago
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    so then the dot product would be v*v1?

  19. Zale101
    • one year ago
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    Also, we can choose to simplify with either v2, v3 or v1?

  20. beginnersmind
    • one year ago
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    v*v1 on the left hand side. right hand side is (c1v1 + c2v2 +c3v3)*v1 = c1(v1*v1) + c2(v2*v1) + c3(v3*v1)

  21. Zale101
    • one year ago
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    Okay. I'm following along.

  22. Zale101
    • one year ago
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    When it say i=1,2,3 what does that mean?

  23. Zale101
    • one year ago
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    i'm sorry for asking too many questions lol

  24. beginnersmind
    • one year ago
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    It means that \[\Large c_i=\LARGE \frac{(v*v_i)}{||v_i||^2}\] is actually 3 equations. \[\Large c_1=\LARGE \frac{(v*v_1)}{||v_1||^2}\] \[\Large c_2=\LARGE \frac{(v*v_2)}{||v_2||^2}\] \[\Large c_3=\LARGE \frac{(v*v_3)}{||v_3||^2}\]

  25. Zale101
    • one year ago
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    OH! That makes sense!! Thank you

  26. beginnersmind
    • one year ago
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    Nah, it's cool. I'd rather you ask if you don't understand something :)

  27. Zale101
    • one year ago
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    Now the question makes a lot sense now.

  28. Zale101
    • one year ago
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    So the simplifications you did above when you used v*v1 can also be done with the rest: v*v2, v*v3 I though i'd have to do something different to get i.

  29. Zale101
    • one year ago
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    Thanks @beginnersmind :)

  30. beginnersmind
    • one year ago
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    No problem. Just to clarify though, do you understand how we get \[\Large c_1=\LARGE \frac{(v*v_1)}{||v_1||^2}\] from \[v*v1= (c_1v_1+c_2v_2+c_3v_3)*v1\]

  31. Zale101
    • one year ago
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    No, because i dont understand how you got v2, v3? when it only had v and vc in c1=(v*v1)/(||v||)^2

  32. Zale101
    • one year ago
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    when it only had v and v1*

  33. beginnersmind
    • one year ago
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    I started with \[v*v1= (c_1v_1+c_2v_2+c_3v_3)*v1 \] and got \[\Large c_1=\LARGE \frac{(v*v_1)}{||v_1||^2}\]

  34. Zale101
    • one year ago
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    c1||v||^2=v*v1

  35. Zale101
    • one year ago
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    Oh okay.

  36. Zale101
    • one year ago
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    Yeah it makes sense now.

  37. beginnersmind
    • one year ago
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    So the question shouldn't be how I got v2,v3, but where did they go? :)

  38. Zale101
    • one year ago
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    Yes

  39. Zale101
    • one year ago
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    lol

  40. Zale101
    • one year ago
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    I dont know if i can do this because it's different from your approach. Where it says C2=(v*v2)/(||v||^2) is it possible for me to use v=c1v1+c2v2+c3v3 and for v2, i solve for v2 by using v=c1v1+c2v2+c3v3 ?

  41. beginnersmind
    • one year ago
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    Ok, I'll write it out properly then: We start with \(v= (c_1v_1+c_2v_2+c_3v_3)\) (this was given) \(v*v1= (c_1v_1+c_2v_2+c_3v_3)*v1\) \(v*v1= c_1(v_1*v_1)+c_2(v_2*v_1)+c_3(v_3*v_1)\) now we note that \((v_1*v_2) = 0\) and \((v_1*v_3) = 0\) because v1 is orthogonal to v2 and v3. This was the key step. So \(v*v1= c_1(v_1*v_1)\) remembering that a vector's scalar product with itself is the square of its length (v1*v1 = ||v1||^2): \[\Large v*v_1=c_1||v_1||^2\] or \[\Large c_1=\LARGE \frac{(v*v_1)}{||v_1||^2}\]

  42. beginnersmind
    • one year ago
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    For c2 you do the same thing, except now you take the scalar product with v2. This gets rid of the terms with v1 and v3.

  43. Zale101
    • one year ago
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    I had to keep in my that it states that the orthogonal of v1*v2 is zero. Alright, thanks for all the beautiful and detailed explanation.

  44. beginnersmind
    • one year ago
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    No problem, thanks for sticking with it :)

  45. Zarkon
    • one year ago
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    you should use \cdot. it looks nicer and also use \| v_1\| for \(\|v_1\|\) \[\Huge c_1=\LARGE \frac{v\cdot v_1}{\|v_1\|^2}\] vs \[\Huge c_1=\LARGE \frac{v*v_1}{||v_1||^2}\]

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