Equation of a plane orthogonal to a vector

[starbaj12]

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

[ComputerGeek]

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.

[starbaj12]

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]

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

[ComputerGeek]

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?

[starbaj12]

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]

Did you try out my suggestion at all?

[starbaj12]

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]

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?

[starbaj12]

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

[arildno]

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

[starbaj12]

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

[arildno]

What else?

[starbaj12]

* 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

[arildno]

(\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,.

[starbaj12]

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]

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)?

[starbaj12]

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]

So, you don't get eq.1?

[starbaj12]

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]

You're ABSOLUTELY RIGHT about eq.3!
However, I asked you:
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)

[starbaj12]

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]

Let's start with YOUR equation:
\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?

[starbaj12]

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]

Just a question:
Given "vectorb" do you then regard "b" as the magnitude of "vectorb"?

[starbaj12]

it is square root vectorb^2

[arildno]

Yes, so b^2=(square root vectorb^2)^2=vectorb^2
Will you agree to this?
(

[starbaj12]

yes I agree

[arildno]

All right, then!
Go back to post "23" and see if you understand it now!

[starbaj12]

Yes! yes! I follow now and I went back to post 17 and followed all the way through to equation 4. (I hope this was the way to go)

[arildno]

That's great!
Now, try to describe where you are RIGHT NOW (that is,up to the point you understand so far), so that we can proceed further!

Am I right in saying that you now understand all steps up to (eq.4)?

[starbaj12]

yes I'm up to equation 4

[arildno]

Let's look at eq.3 again:
\vec{F}\cdot\vec{B}=0
Now, we agree that this means that F is orthogonal to B.
Now, B is a fixed vector.
Do you agree then, that we can say that every vector which is orthogonal to B is a solution to (eq.3)?
(That is, "can play the role of F"?)

[starbaj12]

yes I agree due to the dot product rule

[arildno]

Let us now look a bit closer at the "solution set" of (eq.3)
(that is, the set of those vectors which are orthogonal to vector B).
Do you agree that:
a) One solution of (eq.3) is the choice of \vec{F}=\vec{0}?
(That is, putting the zero vector into F's place in (eq.3) gives us 0 at the left-hand-side of (eq.3), and hence, that (eq.3) is fulfilled?
In other words, that the zero vector is a SOLUTION of (eq.3))

b) A plane is a geometric surface characterized by, that at all points on the plane, the vector normal is the same?
c) Hence, from b), that the solution set of (eq.3) is, in fact the plane containing the origin and with vector B as its vector normal?

[starbaj12]

yes I agree, and I tried to go further to see if I could work something out but still no lightbulb yet

[arildno]

Ok, so rewriting (eq.2) a bit, we have:
\vec{W}=\vec{F}+\vec{B}
(You've seen this one before..)

We have already established, that all acceptable choices of \vec{F} together constitutes a plane, with \vec{B} as the normal vector.

Now then, if you combine this insight with the above equation, what sort of geometric surface must all acceptable choices for \vec{W} taken together constitute?

[starbaj12]

A plane i think

[starbaj12]

a plane parellel to the normal vector b

[arildno]

It sure is!
Let us argue like this:
1) Take any acceptable choice of \vec{F}
(Such a point lies in the plane containing the origin, with \vec{B} as normal vector.
2) To find the corresponding \vec{W} we go straight up along \vec{B}
(Adding that is, \vec{B} to our \vec{F})
3). Now, the DISTANCE that \vec{W} is removed from the plane in which \vec{F} lies, is how far along the vector normal to the plane \vec{W} is.
4) But this distance is simply the magnitude of \vec{B}, since \vec{B} IS normal to the plane in which \vec{F} lies.
5) Clearly, therefore, EVERY ACCEPTABLE CHOICE OF \vec{W} LIES AN EQUAL DISTANCE FROM OUR PLANE (that is given by the length of \vec{B}
6) But this means, that the set of acceptable \vec{W} is A PLANE PARALLELL TO THE PLANE OF ACCEPTABLE \vec{F}!!

Do you agree with this reasoning?

[starbaj12]

I'm not understanding #5

[arildno]

1.Each acceptable choice of \vec{W} corresponds to an acceptable choice of \vec{F}
2. The distance of a point to a plane is found by measuring the length of the perpendicular (normal) joining that point to the plane.
3. In our case, the perpendicular to the plane in which our \vec{F} lies, is given by \vec{B}

Do you agree with this?

[starbaj12]

yes I'm following now

[arildno]

So, you accept that the set of acceptable \vec{W} represents a plane with normal vector given by \vec{B} ?

[starbaj12]

yes, but this is due in another twenty minutes if you do not mind could you walk me through the steps. Another thing this is a physics assingment, but I placed it in the math due to it being math oriented. I took intro to linear algebra and I realize that in linear algebra it has to be more in dept, but for physics not as much. But I really appreciate your help, nothing like understanding a problem fully through.

[arildno]

But now we're really done!
Because:
1) We've shown that the set of acceptable \vec{W} is a plane.
2) In addition, since \vec{F}=\vec{0} was a solution of (eq.3), we know that one acceptable \vec{W}=\vec{0}+\vec{B}=\vec{B}
But the physical point corresponding to \vec{B} is D..

This was what you had to show:
That your original equation represents a plane with normal vector given by \vec{B} containing D.

[starbaj12]

I had to leave to fast I just want to say thank you for all your help arildno