Proof that VxU=(determinant) derives from VxU=|V||U|sinαe

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  • #1
PauloE
28
0

Homework Statement


I need to proof that VxU=(determinant) starting from VxU=|V||U|sinαe

Homework Equations


VxU=|V||U|sinαe and what I'm aiming to is VxU=(uy⋅uz - uz⋅vy)i - (ux⋅vz - uz⋅vx)j + (ux⋅vy - uy⋅vx)k

The Attempt at a Solution


U x V = |U||V|Sinαe
(U x V)^2 = |U|^2|V|^2 cos^2α - 1 e
(U x V)^2 = |U|^2|V|^2 cos^2α - |U|^2|V|^2
U x V = (ux + uy + uz) (vx + vy + vz)cos^2α - (ux + uy + uz) (vx + vy + vz) e

This is where I get stuck. Can someone point me on the right direction? that cosine and e are disorienting me. Thanks.
 
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  • #2
PauloE said:

Homework Statement


I need to proof that VxU=(determinant) starting from VxU=|V||U|sinαe

Homework Equations


VxU=|V||U|sinαe and what I'm aiming to is VxU=(uy⋅uz - uz⋅vy)i - (ux⋅vz - uz⋅vx)j + (ux⋅vy - uy⋅vx)k

The Attempt at a Solution


U x V = |U||V|Sinαe
(U x V)^2 = |U|^2|V|^2 cos^2α - 1 e
(U x V)^2 = |U|^2|V|^2 cos^2α - |U|^2|V|^2
U x V = (ux + uy + uz) (vx + vy + vz)cos^2α - (ux + uy + uz) (vx + vy + vz) e

This is where I get stuck. Can someone point me on the right direction? that cosine and e are disorienting me. Thanks.

The magnitude of ##\vec{A} \times \vec{B}## is the area of a certain parallogram with sides ##\vec{A}, \, \vec{B}##. Can you find another formula for that area?
 
  • #3
PauloE said:

Homework Statement


I need to proof that VxU=(determinant) starting from VxU=|V||U|sinαe

Homework Equations


VxU=|V||U|sinαe and what I'm aiming to is VxU=(uy⋅uz - uz⋅vy)i - (ux⋅vz - uz⋅vx)j + (ux⋅vy - uy⋅vx)k
What is e in the formula you show?

Also, your formula is incorrect. V x U is a vector, but what you have on the right side is a scalar, unless e happens to be a vector (which isn't explained).
PauloE said:

The Attempt at a Solution


U x V = |U||V|Sinαe
(U x V)^2 = |U|^2|V|^2 cos^2α - 1 e
What does (U x V)2 mean? With vectors there are three kinds of multiplication: cross product, dot product, multiplication by a scalar.
Where did cos2α = 1e come from? I don't know what this means.
PauloE said:
(U x V)^2 = |U|^2|V|^2 cos^2α - |U|^2|V|^2
U x V = (ux + uy + uz) (vx + vy + vz)cos^2α - (ux + uy + uz) (vx + vy + vz) e

This is where I get stuck. Can someone point me on the right direction? that cosine and e are disorienting me. Thanks.
 
  • #4
Ray Vickson said:
The magnitude of A⃗ ×B⃗ \vec{A} \times \vec{B} is the area of a certain parallelogram with sides A⃗ ,B⃗ \vec{A}, \, \vec{B}. Can you find another formula for that area?
Besides |U||V|Sinα wouldn't that be |U x V|? which confuses me too because it means that |U||V|Sinα = |U x V| = U x V

I'm working on it and post if there is any or not much progress...

 
  • #5
Mark44 said:
What is e in the formula you show?

Also, your formula is incorrect. V x U is a vector, but what you have on the right side is a scalar, unless e happens to be a vector (which isn't explained).
What does (U x V)2 mean? With vectors there are three kinds of multiplication: cross product, dot product, multiplication by a scalar.
Where did cos2α = 1e come from? I don't know what this means.

e is a unit vector.
(U x V)2 I squared both sides
sin2α = cos2α - 1 and here I should have written (cos2α - 1)e
 
  • #6
PauloE said:
it means that |U||V|Sinα = |U x V| = U x V

This is probably wrong anyway
 
  • #7
PauloE said:
e is a unit vector.
In what direction?
PauloE said:
(U x V)2 I squared both sides
How do you square a vector? By that, I mean, which type of multiplication do you mean? I listed three.
PauloE said:
sin2α = cos2α - 1 and here I should have written (cos2α - 1)e
##\sin^2(\alpha) = 1 - \cos^2(\alpha)##. You have ##\cos^2(\alpha) = 1##, which has the opposite sign.
 
  • #8
PauloE said:
Besides |U||V|Sinα wouldn't that be |U x V|? which confuses me too because it means that |U||V|Sinα = |U x V| = U x V

I'm working on it and post if there is any or not much progress...

Google is your friend.
 
  • #9
PauloE said:
##\sin^2(\alpha) = 1 - \cos^2(\alpha)##. You have ##\cos^2(\alpha) = 1##, which has the opposite sign.

Ups..

The unit vector is in the direction of the normal vector I supose.
 
  • #10
Not getting really far on this. I do not understand why this forum makes a point on being so cryptic with the help provided. I have no doubts on the good intentions of its members but after asking a couple of questions related with homework I couldn't help get frustrated for the meager hints (not just in this thread but in others that me and my friends have posted). I'm not trying to slack, I'm not asking you to do my homework, I don't want to copy and paste the answer from google, I need help when I get stuck. I agree you shouldn't give me the answer but c'mon, it is so impractical the way help is shared here. Basically I learned nothing but to copy the thing from google today.

Thank you Ray Vickson and Mark44 anyway.
 
  • #11
PauloE said:
Not getting really far on this. I do not understand why this forum makes a point on being so cryptic with the help provided. I have no doubts on the good intentions of its members but after asking a couple of questions related with homework I couldn't help get frustrated for the meager hints (not just in this thread but in others that me and my friends have posted). I'm not trying to slack, I'm not asking you to do my homework, I don't want to copy and paste the answer from google, I need help when I get stuck. I agree you shouldn't give me the answer but c'mon, it is so impractical the way help is shared here. Basically I learned nothing but to copy the thing from google today.

Thank you Ray Vickson and Mark44 anyway.

Sorry you feel this way, but these are the rules governing HW help here at PF. You knew this going into the process when you signed up. If the HW Helpers don't follow the rules, they get awarded infraction points.

In any event, I think Ray Vickson gave you an excellent hint in Post #2, which you didn't seem to follow up on in the succeeding posts.

FWIW, I've never thought of the determinant form of the cross product as anything more than a handy mnemonic which can be used to find the components of the product. I've never thought that anyone would go to the trouble to construct a full mathematical proof.

http://math.oregonstate.edu/home/pr...estStudyGuides/vcalc/crossprod/crossprod.html
 
  • #12
PauloE said:

Homework Statement


I need to proof that VxU=(determinant) starting from VxU=|V||U|sinαe
The problem is already confusing at this point. The standard definition of the cross product is ##A\times B=(A_2B_3-A_3B_2)i+(A_1B_3-B_1A_3)j+(A_1B_2-B_1A_2)k##, and if you use this, you only need to verify that this is exactly what you get from the determinant formula. This is trivial, so I guess you're not supposed to use the standard definition, and instead take ##V\times U=|V||U|(\sin\alpha) e## as the definition of the cross product. But what then is ##e##?

I think the fact that you haven't properly answered that, is the main reason why there's been so little progress in this thread. You said nothing about it until Mark44 asked you. You gave the incomplete answer that it's a unit vector. So Mark44 had to ask the follow-up question "in what direction?". You gave an incomplete answer to that too. I assume you meant normal to the plane spanned by ##V## and ##U##, but there are two of those. I'm guessing that your answer to the obvious follow-up question would be "the one given by the right-hand rule".

But we have little use for a wordy answer like that. We need a formula for ##e##. The problem is that the obvious formula is the equation ##V\times U=|V||U|(\sin\alpha) e##. This looks more like a definition of ##e## than a definition of ##V\times U##. But if we are to take it as the definition of ##e##, we need another definition of ##V\times U##.

If the equation ##V\times U=|V||U|(\sin\alpha) e## is supposed to be the definition of ##e##, and we are allowed to use the standard definition of the cross product, the problem is trivial. And if that equation is supposed to be the definition of the cross product, the problem doesn't seem solvable at all without a proper definition of ##e##.
 
Last edited:

Related to Proof that VxU=(determinant) derives from VxU=|V||U|sinαe

1. What is the proof for the equation VxU=(determinant)?

The proof for this equation is based on the fact that the cross product of two vectors in three-dimensional space is equal to the determinant of the matrix formed by the vectors and the unit vectors in the x, y, and z directions. This can be mathematically derived using the properties of vector operations and the properties of determinants.

2. How does VxU=|V||U|sinαe relate to the equation VxU=(determinant)?

The equation VxU=|V||U|sinαe is essentially a simplified version of the equation VxU=(determinant), where |V| and |U| represent the magnitudes of the vectors V and U, and α represents the angle between them. This version is often used to calculate the cross product of two vectors in two-dimensional space.

3. What is the significance of the cross product in vector algebra?

The cross product is a fundamental operation in vector algebra that results in a vector perpendicular to the two input vectors. It is useful in finding the direction of a vector perpendicular to a plane or in calculating the area of a parallelogram formed by two vectors.

4. Can the equation VxU=(determinant) be applied to vectors in higher dimensions?

Yes, the equation VxU=(determinant) is applicable to vectors in any number of dimensions. However, it becomes more complex and difficult to visualize in higher dimensions, so the simplified version VxU=|V||U|sinαe is often used instead.

5. Are there any real-world applications of the equation VxU=(determinant)?

Yes, the equation VxU=(determinant) has various applications in physics, engineering, and computer graphics. It is used to calculate torque, force, and angular momentum in physics, and to determine the orientation and rotation of objects in computer graphics. It is also used in the study of fluid dynamics and electromagnetism.

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