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Equation of a plane orthogonal to a vector

  1. Oct 3, 2004 #1
    Let vectorB be a vector from the origin to a point D fixed in space. Let vectorW be a vector from the origin to a variable point Q(x,y,z). Show that vectorW (dot) vectorB = B^2 is the equation of a plane perpendicular to vectorB and passing through D.

    Thank you for any help
     
  2. jcsd
  3. Oct 3, 2004 #2
    how do you think you should start? one of the biggest things to remember is that starting a problem down the wrong path is not detrimental to anyone's heath, so guessing will not hurt.
     
  4. Oct 3, 2004 #3
    What do you think that I just looked at the problem and then came here. I have been working on this problem if I typed everything I tried on this forum it would take a lot of time and space.
     
  5. Oct 3, 2004 #4

    arildno

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    Write W=(W-B)+B, and see how your equation simplifies.
     
  6. Oct 3, 2004 #5
    there is no reason to get an attitude. you posted a question with out even giving anyone any idea of the work you have put into i so far. besides that, if we know where you are with this we can help to steer you toward the answer with out just giving it to you and you learn a lot more. you are doing this work to learn it right? if not then why do it?
     
  7. Oct 3, 2004 #6
    Could someone tell me the relationship between vectorW (dot)vectorB = B^2. I'm really baffled with B^2 is this a scalar or the magnitude or what
     
  8. Oct 3, 2004 #7

    arildno

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    Did you try out my suggestion at all?
     
  9. Oct 3, 2004 #8
    Yes I have been working off your suggestion arildno and thank you for it, but I'm coming up short. I do not have the complete understanding of the equation that I just posted.
     
  10. Oct 3, 2004 #9

    arildno

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    We have:
    [tex]\vec{W}\cdot{B}=\vec{B}^{2}[/tex]
    Set
    [tex]\vec{F}=\vec{W}-\vec{B}[/tex]
    Then:
    [tex]\vec{W}=\vec{F}+\vec{B}[/tex]
    (Agreed?)
    Now, put that expression for [tex]\vec{W}[/tex] into
    the first equation; what do you get then?
     
  11. Oct 3, 2004 #10
    arildno the first equation you have did you mean vectorw (dot) vectorb = b^2
    Thank you for all your help
     
  12. Oct 3, 2004 #11

    arildno

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    Yes, that is the equation I meant.
    Remember:
    [tex]\vec{B}^{2}=\vec{B}\cdot\vec{B}=||\vec{B}||^{2}[/tex]
     
  13. Oct 3, 2004 #12
    So your saying that my equation vectorw (dot) vectorb = b^2 is the same as your equation vectorw (dot) b = vectorb^2
     
  14. Oct 3, 2004 #13

    arildno

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    What else?
     
  15. Oct 3, 2004 #14
    * means vector (easier for me to type)
    I did what you said and I end up with *F + *b dot b = *b^2
    Then *f + *b^2 = b^2
    Is this right

    After this the plane needs to be orthogonal to *b so do I just take the dot product of *f*bcos(theta) = 0 so *f = 0
    What about it needing to pass through D

    Thank you for your patience
     
  16. Oct 3, 2004 #15

    arildno

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    [tex](\vec{F}+\vec{B})\cdot\vec{B}=\vec{F}\cdot\vec{B}+\vec{B}^{2}[/tex]
    Hence, using this as your left-hand-side in your original equation, you have:
    [tex]\vec{F}\cdot\vec{B}+\vec{B}^{2}=\vec{B}^{2}[/tex]
    You should be able to figure out the rest,.
     
  17. Oct 4, 2004 #16
    Sorry to push for the help so much arildno but I was up until four in the morning trying to figure the rest out but I seem to be lacking the education to see the relationship between the equations.
     
  18. Oct 4, 2004 #17

    arildno

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    Sorry for keeping you awake!
    Let's finish this in a couple of posts.
    1. Do you accept that your original equation may be rewritten as:
    [tex]\vec{F}\cdot\vec{B}+\vec{B}^{2}=\vec{B}^{2} (eq.1)[/tex]
    where [tex]\vec{F}=\vec{W}-\vec{B} (eq.2)[/tex] ?
    2. Do you also accept that (eq.1) can be rewritten as:
    [tex]\vec{F}\cdot\vec{B}=0 (eq.3)[/tex] ?
    Question:
    What does (eq.3) tell us about the geometrical relationship of [tex]\vec{F},\vec{B}[/tex] ?
    3. Do you agree that, using (eq.2), (eq.3) can be rewritten as:
    [tex](\vec{W}-\vec{B})\cdot\vec{B}=0[/tex] (eq.4)?
     
  19. Oct 4, 2004 #18
    Sorry for the delay reply I had classes, Yes I except eq.1 and eq4. Eq3 tells me that f is orthogonal to b
     
  20. Oct 4, 2004 #19

    arildno

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    So, you don't get eq.1?
     
  21. Oct 4, 2004 #20
    I cannot see what you are saying I thought eq3 was just the dot product and the answer is zero so it is orthogonal
     
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