# Homework Help: Double checking vectors

1. Jun 18, 2013

### Jbreezy

1. The problem statement, all variables and given/known data
The points A,B,C, D with position vectors a,b,c,d are coplanar. Given : ∠ BAC = θ
(a) Find an expression for a unit normal of this plane.
(b) Find an expression for the distance of this plane from the origin
(c) Prove that
[abd] + [bac] + [cdb] + [dca] = 0

2. Relevant equations

3. The attempt at a solution

So, I did part a and b but I'm just having a doubt. I'm wondering why I was given ∠ BAC = θ I didn't do part c yet so maybe it comes in there.

(a) Find an expression for a unit normal of this plane.

First I found two vectors on the plane then I crossed them. I chose to use a,b,c.
So I did
(b-a) = u = (<ux, uy,uz>

(c-a) = v = (<vx, vy,vz>

I did u cross v =
< (uyvz-vyuz)-(uxvz- vxuz) + (uxvy-vxuy) >

Lastly I just took u cross v and divided it by its own norm.

< (uyvz-vyuz)-(uxvz- vxuz) + (uxvy-vxuy) > / √(< (uyvz-vyuz)2-(uxvz- vxuz)2 + (uxvy-vxuy)2 >

Call the unit vector that I just got r(hat) then for part b all I did was to do r(hat) dot a

I could of chosen any of the other position vectors to dot with because the question says that they all lie in the same plane and all I need is a vector in that plane right? So once again why
∠ BAC = θ
Thanks

2. Jun 18, 2013

### Jbreezy

Would ∠ BAC = θ be given because if it wasn't I may do a cross product and the angles between might be 0 or 180 resulting in the 0 vector? So this just guarantees I won't get a 0 vector?

3. Jun 18, 2013

### CAF123

Indeed, you are given the condition so that you know vectors a,b,c are not collinear.

4. Jun 18, 2013

### LCKurtz

What does the notation [abd] mean? You need to tell us so we don't have to guess.

Apparently that angle is being named $\theta$. It doesn't tell you anything about the angle.

The cross product is a vector. You just have a single scalar inside the brackets. You need to separate the components of the vector with commas.
Unless there is something you haven't told us, that is just the name of that angle.

5. Jun 18, 2013

### Jbreezy

OK, does the other part look correct? Please and Thanks.

< (uyvz-vyuz)i-(uxvz- vxuz)j + (uxvy-vxuy)k >
This work too no?

And [abd] is just triple scalar product.
Don't worry about that because I have to still do that part. I just want to know if my other parts are correct ...please and thanks

6. Jun 18, 2013

### LCKurtz

No. Either use the bracket notation or the ijk notation, but don't mix them. Also, as others have suggested, when you mean subscripts make them look like subscripts.

7. Jun 18, 2013

### Jbreezy

It was supposed to be subscript like above but it doesn't just copy and paste apparently.

< (uyvz-vyuz),-(uxvz- vxuz), (uxvy-vxuy) >
There we go.

The other part is correct no? a and b?

8. Jun 18, 2013

### LCKurtz

The length of a vector is just a number. What are the < and > brackets in there for?

Yes except for notation. See above.

Also, I would suggest you start using LaTeX. Subscripts are much easier. For example your uyvz-vyuz expression could be typed like this:
$u_yv_z-v_yu_z$. Press the Quote button to see how to do that.

9. Jun 18, 2013

### Jbreezy

OK, I will fix the notation. Also LaTex is quite annoying for me. The other day I tried it and I felt like it took me way to long to write something out and it wasn't even correct I had all sorts of issues. I feel like its more trouble to me.

10. Jun 18, 2013

### Jbreezy

I have a proof but it is ratty but it is right.

(c) Prove that
[abd] + [bac] + [cdb] + [dca] = 0

So if you rearrange it you can see it a little better.

[abd] + [cdb] + [bac] + [dca] = 0

Notice that b cross d = -(d cross b)
So if you are do b cross d + d cross b you get 0

But this is kind of informal. I'm trying to think of a better way to show it.

11. Jun 18, 2013

### Ray Vickson

What you should NOT do is annoy the people whom you want to help you! If you cannot use TeX because it is too much trouble for you, and you cannot use the X2 or X2 buttons because they are too much trouble for you, at least you should use proper ASCII, like this: u_x, etc.

12. Jun 18, 2013

### Jbreezy

What are you even talking about? I copied and pasted something from above that wasn't transferred in the paste so that is what LCKurtz was telling me to use subscripts for. It was an accident all my others have subscripts. Relax dude.

13. Jun 18, 2013

### LCKurtz

The reason the subscripts don't come through is that you are using copy/paste in the first place. You should always use the Quote button instead of copy/pasting. Start your response with the quote button in the post to which you are replying. You can always delete stuff you don't want in the quote.

Did you try pushing the Quote button in my post #8 like I asked you to do? If you haven't yet, do it now. There is no way subscripts and superscripts aren't easier with Latex.

14. Jun 18, 2013

### LCKurtz

It's also wrong. What you have is [abd]-[cbd] for those two terms. They aren't the same so why should you get 0?

15. Jun 18, 2013

### Jbreezy

No No. I have you get the 0 vector. I have [abd] + [cdb] not [abd]-[cbd].
I say you get the 0 vector because [b cross d= -(d cross b)
what [abd] means is ( a dot ( b cross d) It is why I said it was a ratty proof I think it is right though. But I need to think of a better way to show it.

16. Jun 18, 2013

### LCKurtz

Read my post again. The - sign is from switching the b and d.

17. Jun 20, 2013

### LCKurtz

@JBreezy: So have you abandoned this thread even though you are nowhere near a solution of the problem?

18. Jun 20, 2013

### Jbreezy

No I didn't abandoned this thread. I'm thinking. How can triple scalar products be equal to the 0 vector if by definition the operation produces a scalar?

19. Jun 20, 2013

### LCKurtz

Both sides are scalars. Only the letters represent vectors.

20. Jun 20, 2013

### Jbreezy

I mean the left side is a bold 0 so doesn't that mean the zero vector not just 0?

21. Jun 20, 2013

### LCKurtz

So unbold it. You are the one who typed it. Probably too late to change it though. Anyway, both sides are scalars.

22. Jun 20, 2013

### Jbreezy

Thats how the question was given to me! lol.
I didn't just bold it for no reason. How can you say that both sides are scalars then? I agree but why was it given in bold?

23. Jun 20, 2013

### LCKurtz

24. Jun 20, 2013

### Jbreezy

$$\begin{bmatrix} ax & ay & az \\ bx & by & bz\\ dx & dy & dz\end{bmatrix} \ + \begin{bmatrix} cx & cy & cz \\ dx & dy & dz\\ bx & by & bz\end{bmatrix}=$$\begin{bmatrix} ax + cx & ay + cy & az + cz \\ bx + dx & by + dy & bz + dz\\ dx + bx & dy + by& dz + bz\end {bmatrix}

ignore this! sorry

25. Jun 20, 2013

### Jbreezy

$$\begin{bmatrix} ax & ay & az \\ bx & by & bz\\ dx & dy & dz\end{bmatrix} \ + \begin{bmatrix} cx & cy & cz \\ dx & dy & dz\\ bx & by & bz\end{bmatrix}=$$\begin{bmatrix} ax + cx & ay + cy & az + cz \\ bx + dx & by + dy & bz + dz\\ dx + bx & dy + by& dz + bz \end{bmatrix} = 0 because two rows are the same