Is Lagrange's Theorem the Key to Solving This Vector Equation?

In summary, to proof Lagrange's identity, you can use the definitions of the cross product and dot product. Multiply out the cross products and dot products and you should find the equality. This process can be lengthy and requires understanding of matrix operations.
  • #1
rroy81
13
0
Proof using lagrange!

Homework Statement



(A x B) . (C x D) = (A . B) (C . D) - (A . D) (B . C)

Homework Equations



This is all that's given..I am sort of lost on how to proof this. Spent 4hrs +

The Attempt at a Solution



Completely lost and don't know where to start
 
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  • #2


rroy81 said:

Homework Statement



(A x B) . (C x D) = (A . B) (C . D) - (A . D) (B . C)

Homework Equations



This is all that's given..I am sort of lost on how to proof this. Spent 4hrs +

The Attempt at a Solution



Completely lost and don't know where to start

Welcome to PF, rroy! :smile:

What did you try?
Which relevant equations do you have (and are you allowed to use)?

I presume you are allowed to use the definitions of the cross product and dot product.
Did you use those?
Any other identities?
Are you allowed to use identities that you can find with wikipedia?
 
  • #3


So turns out the instructor had typo in the problem. it should be
(A x B) . (C x D) = (A . C) (B . D) - (A . D) (B . C)

The way I started was let A = {a1, a2, a3 } and so on for the remaining (B, C, D). I assume it is a matrice and will proceed as a dot product. Than I am unsure if this right, if so than I stuck.
 
  • #4


What do you think is representing a matrix?

Do you know what the definition for the cross product for vectors is?
 
  • #5


Actually, what you are supposed to proof is Lagrange's identity.
See for instance here.
 
  • #6


I like Serena,

My understanding to cross product is using the matrix, am I right? (ith, jth and kth terms)
and as I have given the terms A=(a1, a2, a3) B=(b1, b2, b3) and so on..than to solve the cross product. I simply took the crossed out the first column and row and than did the cross product of remaining 4 and did the same with 2 column and row and so on
 
  • #7


Ah okay.
Yes, that is the standard way to go.
If you multiply out the cross products and dot products, you should find the equality.
It is a bit of work though.
 

FAQ: Is Lagrange's Theorem the Key to Solving This Vector Equation?

What is proof using Lagrange?

Proof using Lagrange, also known as the Mean Value Theorem, is a mathematical technique used to prove that a function has a critical point or a point of inflection. It is named after the French mathematician Joseph-Louis Lagrange, who first proposed the theorem in the late 18th century.

How does proof using Lagrange work?

The proof using Lagrange involves finding the derivative of a function and setting it equal to zero. This allows us to find the critical points of the function, which are points where the tangent line is horizontal. By analyzing the behavior of the function at these points, we can determine if there is a maximum, minimum, or point of inflection.

What are the assumptions made in proof using Lagrange?

The main assumption in proof using Lagrange is that the function is continuous on a closed interval and differentiable on an open interval within that closed interval. Additionally, the function must have a critical point or a point of inflection within that interval.

What are some real-life applications of proof using Lagrange?

Proof using Lagrange has many practical applications in fields such as economics, physics, and engineering. For example, it can be used to optimize production processes, analyze the motion of objects, and determine the most efficient way to design structures.

Is proof using Lagrange always accurate?

Proof using Lagrange is based on mathematical principles and is considered a very accurate method for analyzing functions. However, it is important to note that the theorem only provides necessary conditions for a maximum, minimum, or point of inflection, and further analysis may be needed to confirm the results.

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