# Equation of a plane orthogonal to a vector

1. Oct 3, 2004

### 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

2. Oct 3, 2004

### ComputerGeek

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. Oct 3, 2004

### 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.

4. Oct 3, 2004

### arildno

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

5. Oct 3, 2004

### ComputerGeek

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. Oct 3, 2004

### 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

7. Oct 3, 2004

### arildno

Did you try out my suggestion at all?

8. Oct 3, 2004

### 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.

9. Oct 3, 2004

### 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?

10. Oct 3, 2004

### starbaj12

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

11. Oct 3, 2004

### arildno

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

12. Oct 3, 2004

### starbaj12

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

13. Oct 3, 2004

### arildno

What else?

14. Oct 3, 2004

### 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

15. Oct 3, 2004

### 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,.

16. Oct 4, 2004

### 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.

17. Oct 4, 2004

### 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)?

18. Oct 4, 2004

### 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

19. Oct 4, 2004

### arildno

So, you don't get eq.1?

20. Oct 4, 2004

### 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