Why was ∠ BAC = θ given in this problem?

[Jbreezy]

Homework Statement

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

Homework Equations

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 = (<u_x, u_y,u_z>

(c-a) = v = (<v_x, v_y,v_z>

I did u cross v =
< (u_yv_z-v_yu_z)-(u_xv_z- v_xu_z) + (u_xv_y-v_xu_y) >

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

< (u_yv_z-v_yu_z)-(u_xv_z- v_xu_z) + (u_xv_y-v_xu_y) > / √(< (u_yv_z-v_yu_z)²-(u_xv_z- v_xu_z)² + (u_xv_y-v_xu_y)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

 

[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?

 

[CAF123]

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?

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

 

[LCKurtz]

Homework Statement

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

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

Homework Equations

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

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

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 = (<u_x, u_y,u_z>

(c-a) = v = (<v_x, v_y,v_z>

I did u cross v =
< (u_yv_z-v_yu_z)-(u_xv_z- v_xu_z) + (u_xv_y-v_xu_y) >

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.

So once again why
∠ BAC = θ
Thanks

Unless there is something you haven't told us, that is just the name of that angle.

 

[Jbreezy]

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

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

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.

< (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

 

[LCKurtz]

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

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.

 

[Jbreezy]

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

< (u_yv_z-v_yu_z),-(u_xv_z- v_xu_z), (u_xv_y-v_xu_y) >
There we go.

The other part is correct no? a and b?

 

[LCKurtz]

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

< (u_yv_z-v_yu_z)-(u_xv_z- v_xu_z) + (u_xv_y-v_xu_y) > / √(< (u_yv_z-v_yu_z)²-(u_xv_z- v_xu_z)² + (u_xv_y-v_xu_y)² >

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

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

< (u_yv_z-v_yu_z),-(u_xv_z- v_xu_z), (u_xv_y-v_xu_y) >
There we go.

The other part is correct no? a and b?

Yes except for notation. See above.

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

 

[Jbreezy]

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

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.

 

[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.

 

[Ray Vickson]

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.

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.

 

[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.

 

[LCKurtz]

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.

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.

 

[LCKurtz]

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.

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?

 

[Jbreezy]

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?

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.

 

[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?

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.

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

 

[LCKurtz]

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

 

[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?

 

[LCKurtz]

Both sides are scalars. Only the letters represent vectors.

 

[Jbreezy]

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

 

[LCKurtz]

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

 

[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?

 

[LCKurtz]

If you want more discussion about this problem, you need to address the issue of what is wrong with your "proof" discussed in posts 14-16.

 

[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

 

[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

What about this dude?

 

[LCKurtz]

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

What about this dude?

Are those matrices or determinants? I'm guessing determinants since you claim the result is 0. But you have added them like they are matrices. Altogether it is just nonsense. The determinant of a sum of two matrices is not the sum of the determinants of them.

Instead of trying to defend your incorrect argument you might be better off trying to understand why it is wrong.

I have to leave now. Think about that while I am gone.

 

[Jbreezy]

I didn't know there was a difference. Oh well.

 

[HallsofIvy]

If you do not know what matrices and determinants are, then it is not a good idea to try to use them!

 

[Jbreezy]

If you do not know what matrices and determinants are, then it is not a good idea to try to use them!
__________________
"Euclid alone has looked on beauty bare"
Edna St. Vincent Millay

Steller advice amigo.

Are those matrices or determinants? I'm guessing determinants since you claim the result is 0. But you have added them like they are matrices. Altogether it is just nonsense. The determinant of a sum of two matrices is not the sum of the determinants of them.

Instead of trying to defend your incorrect argument you might be better off trying to understand why it is wrong.

I have to leave now. Think about that while I am gone.
Y 09:39 PM

Well the problem is the dot product with different vectors.[abd]-[cbd]
I wish a = c then I would be good. But it doesn't.

 

[LCKurtz]

Well the problem is the dot product with different vectors.[abd]-[cbd]
I wish a = c then I would be good. But it doesn't.

Yes, that's why your argument doesn't work. You could have guessed you didn't have a solution to the original problem because you never used the fact that the four points are given to be coplanar. Before you will be able to actually solve the original problem, you have to know some things about the triple scalar product [abc] (sometimes called the box product) of three vectors in general. So here are some preliminary questions about that:

1. Does [abc] mean $a\cdot (b \times c)$ or $(a\times b) \cdot c$, or does it matter?
2. What happens to the value of [abc] if you swap two of the vectors?
3. What happens to the value of [abc] it two of the vectors are the same?
4. What happens if you have a constant k times one of the vectors: [(ka)bc]?
5. What happens to its value if one of the vectors is a sum: [(a+d)bc]?
6. What does it mean geometrically about the vectors if [abc]=0?

Start by answering those questions. You need to know them to work out the hint I will give you next.

 

[Jbreezy]

1. Does [abc] mean a⋅(b×c) or (a×b)⋅c , or does it matter?
2. What happens to the value of [abc] if you swap two of the vectors?
3. What happens to the value of [abc] it two of the vectors are the same?
4. What happens if you have a constant k times one of the vectors: [(ka)bc]?
5. What happens to its value if one of the vectors is a sum: [(a+d)bc]?
6. What does it mean geometrically about the vectors if [abc]=0?

1. It doesn't matter you can exchange the dot and cross.
2. It depends if it cyclicic or not. Cyclic rotations leave it unchanged. If it is not cyclicic I would say you would get the negative.
3. Evaluates to 0.
4. The constant just changes the value of cross procuct by k times.
5.[(a+d)bc]?
Isn't the cross product distributive over addition so you would have a cross b + d cross c
6. If [abc] = 0 the angle between the vectors is 0 or 180 and the vectors are coplanar.

 

[LCKurtz]

1. Does [abc] mean $a\cdot (b \times c)$ or $(a\times b) \cdot c$, or does it matter?
2. What happens to the value of [abc] if you swap two of the vectors?
3. What happens to the value of [abc] it two of the vectors are the same?
4. What happens if you have a constant k times one of the vectors: [(ka)bc]?
5. What happens to its value if one of the vectors is a sum: [(a+d)bc]?
6. What does it mean geometrically about the vectors if [abc]=0?

Start by answering those questions. You need to know them to work out the hint I will give you next.

1. It doesn't matter you can exchange the dot and cross.

Yes. That's why they can both be left out of the symbol [abc].

2. It depends if it cyclicic or not. Cyclic rotations leave it unchanged. If it is not cyclicic I would say you would get the negative.

I didn't ask about rotations. If you interchange two of the vectors, what does that always do to the value?

3. Evaluates to 0.

Yes.

4. The constant just changes the value of cross procuct by k times.

So you would write [(ka)bc] = k[abc].

5.[(a+d)bc]?
Isn't the cross product distributive over addition so you would have a cross b + d cross c

No. You are just guessing. This isn't a cross product, it is a triple scalar product. Work it out and see what happens.

6. If [abc] = 0 the angle between the vectors is 0 or 180 and the vectors are coplanar.

The angle between what vectors is 0 or 180 and why? Just because a box product comes out zero doesn't mean the internal cross product is zero. Do you know what the box product [abc] represents geometrically, and why it's called a box product in the first place?

 

[Jbreezy]

I didn't ask about rotations. If you interchange two of the vectors, what does that always do to the value?

3. Evaluates to 0.
Yes.

4. The constant just changes the value of cross procuct by k times.
So you would write [(ka)bc] = k[abc].

Yes!

5.[(a+d)bc]?
Isn't the cross product distributive over addition so you would have a cross b + d cross c
No. You are just guessing. This isn't a cross product, it is a triple scalar product. Work it out and see what happens.

(a+d) dot b cross (a+d) dot c
?


6. If [abc] = 0 the angle between the vectors is 0 or 180 and the vectors are coplanar.
The angle between what vectors is 0 or 180 and why? Just because a box product comes out zero doesn't mean the internal cross product is zero. Do you know what the box product [abc] represents geometrically, and why it's called a box product in the first place?

It is the volume of the parallelepiped OK yes it doesn't mean that the cross product is 0. So it just means that the volume of the parallelepiped is 0. COPLANAR!

 

[Jbreezy]

2 is it negates the value

 

[LCKurtz]

What about #5? It is important. Try looking here:
http://www.vitutor.com/alg/determinants/properties_determinants.html