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Equation of a plane orthogonal to a vector

  • Thread starter starbaj12
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  • #1
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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
 

Answers and Replies

  • #2
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starbaj12 said:
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
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.
 
  • #3
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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.
 
  • #4
arildno
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Write W=(W-B)+B, and see how your equation simplifies.
 
  • #5
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starbaj12 said:
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.
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?
 
  • #6
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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
 
  • #7
arildno
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Did you try out my suggestion at all?
 
  • #8
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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.
 
  • #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?
 
  • #10
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arildno the first equation you have did you mean vectorw (dot) vectorb = b^2
Thank you for all your help
 
  • #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]
 
  • #12
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So your saying that my equation vectorw (dot) vectorb = b^2 is the same as your equation vectorw (dot) b = vectorb^2
 
  • #13
arildno
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What else?
 
  • #14
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* 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
 
  • #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,.
 
  • #16
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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.
 
  • #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)?
 
  • #18
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Sorry for the delay reply I had classes, Yes I except eq.1 and eq4. Eq3 tells me that f is orthogonal to b
 
  • #19
arildno
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So, you don't get eq.1?
 
  • #20
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I cannot see what you are saying I thought eq3 was just the dot product and the answer is zero so it is orthogonal
 
  • #21
arildno
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You're ABSOLUTELY RIGHT about eq.3!
However, I asked you:
Do you understand how eq.1 is really the same equation as:
[tex]\vec{W}\cdot\vec{B}=\vec{B}^{2}[/tex]
(Given relationship eq.2 between F and W)
 
  • #22
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no is it where you plug f in to the first equation and after using one of the dot product rules you end up with a negative b^2 and cancels the positive vectorb^2
 
  • #23
arildno
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Let's start with YOUR equation:
[tex]\vec{W}\cdot\vec{B}=\vec{B}^{2} (eq.0)[/tex]
Now, clearly, from eq.2, we have:
[tex]\vec{W}=\vec{F}+\vec{B}[/tex]
Agreed?
Hence, let's substitute this expression for [tex]\vec{W}[/tex] into (eq.0)
(You agree that I am allowed to do that?)
[tex](\vec{F}+\vec{B})\cdot\vec{B}=\vec{B}^{2} (eq.5)[/tex]
Now, let's rewrite the left-hand side of (eq.5):
[tex](\vec{F}+\vec{B})\cdot\vec{B}=\vec{F}\cdot\vec{B}+\vec{B}\cdot\vec{B} (eq.6)[/tex]
Do you agree that the right-hand side of (eq.6) is equal to the left-hand side of (eq.6)?
Now, we also have:
[tex]\vec{B}\cdot\vec{B}=\vec{B}^{2}[/tex]
(That's what we MEAN with [tex]\vec{B}^{2}[/tex])
But this means, that we may rewrite (eq.6) as:
[tex](\vec{F}+\vec{B})\cdot\vec{B}=\vec{F}\cdot\vec{B}+\vec{B}^{2} (eq.7)[/tex]
But this means, that the right-hand side of (eq.5) must equal the right-hand side of (eq.7)! (Since their left-hand sides are identical!)
Hence,
[tex]\vec{F}\cdot\vec{B}+\vec{B}^{2}=\vec{B}^{2} (eq.1)[/tex]

Agreed?
 
  • #24
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Could you clear something up for me how is it vectorb^2 is the same as just b^2 (on the right hand side). which is the first equation
 
  • #25
arildno
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Just a question:
Given "vectorb" do you then regard "b" as the magnitude of "vectorb"?
 

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