# Equation of a plane orthogonal to a vector

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

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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.

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.

arildno
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Write W=(W-B)+B, and see how your equation simplifies.

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?

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

arildno
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Did you try out my suggestion at all?

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.

arildno
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We have:
$$\vec{W}\cdot{B}=\vec{B}^{2}$$
Set
$$\vec{F}=\vec{W}-\vec{B}$$
Then:
$$\vec{W}=\vec{F}+\vec{B}$$
(Agreed?)
Now, put that expression for $$\vec{W}$$ into
the first equation; what do you get then?

arildno the first equation you have did you mean vectorw (dot) vectorb = b^2
Thank you for all your help

arildno
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Yes, that is the equation I meant.
Remember:
$$\vec{B}^{2}=\vec{B}\cdot\vec{B}=||\vec{B}||^{2}$$

So your saying that my equation vectorw (dot) vectorb = b^2 is the same as your equation vectorw (dot) b = vectorb^2

arildno
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What else?

* 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

arildno
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$$(\vec{F}+\vec{B})\cdot\vec{B}=\vec{F}\cdot\vec{B}+\vec{B}^{2}$$
Hence, using this as your left-hand-side in your original equation, you have:
$$\vec{F}\cdot\vec{B}+\vec{B}^{2}=\vec{B}^{2}$$
You should be able to figure out the rest,.

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.

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:
$$\vec{F}\cdot\vec{B}+\vec{B}^{2}=\vec{B}^{2} (eq.1)$$
where $$\vec{F}=\vec{W}-\vec{B} (eq.2)$$ ?
2. Do you also accept that (eq.1) can be rewritten as:
$$\vec{F}\cdot\vec{B}=0 (eq.3)$$ ?
Question:
What does (eq.3) tell us about the geometrical relationship of $$\vec{F},\vec{B}$$ ?
3. Do you agree that, using (eq.2), (eq.3) can be rewritten as:
$$(\vec{W}-\vec{B})\cdot\vec{B}=0$$ (eq.4)?

Sorry for the delay reply I had classes, Yes I except eq.1 and eq4. Eq3 tells me that f is orthogonal to b

arildno
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So, you don't get eq.1?

I cannot see what you are saying I thought eq3 was just the dot product and the answer is zero so it is orthogonal

arildno
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Do you understand how eq.1 is really the same equation as:
$$\vec{W}\cdot\vec{B}=\vec{B}^{2}$$
(Given relationship eq.2 between F and W)

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

arildno
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$$\vec{W}\cdot\vec{B}=\vec{B}^{2} (eq.0)$$
Now, clearly, from eq.2, we have:
$$\vec{W}=\vec{F}+\vec{B}$$
Agreed?
Hence, let's substitute this expression for $$\vec{W}$$ into (eq.0)
(You agree that I am allowed to do that?)
$$(\vec{F}+\vec{B})\cdot\vec{B}=\vec{B}^{2} (eq.5)$$
Now, let's rewrite the left-hand side of (eq.5):
$$(\vec{F}+\vec{B})\cdot\vec{B}=\vec{F}\cdot\vec{B}+\vec{B}\cdot\vec{B} (eq.6)$$
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:
$$\vec{B}\cdot\vec{B}=\vec{B}^{2}$$
(That's what we MEAN with $$\vec{B}^{2}$$)
But this means, that we may rewrite (eq.6) as:
$$(\vec{F}+\vec{B})\cdot\vec{B}=\vec{F}\cdot\vec{B}+\vec{B}^{2} (eq.7)$$
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,
$$\vec{F}\cdot\vec{B}+\vec{B}^{2}=\vec{B}^{2} (eq.1)$$

Agreed?

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

arildno