Solve Double Cross Product Problem in $\mathbb{R}^3$

In summary: Some detailed calculations are just unavoidable. However, if you pay attention to what I said in #3, that will lead to just about the shortest and most straightforward proof available. Besides, you have NOT reached the point where ## (a,b) \in \{\pm (u.w, - u.v) \} ##, because you had one relation between ##a## and ##b##, and you said that your particular choice "works". However, in principle, there are infinitely many other choices of ##a,b## that also work for that single condition. Somehow, you need more conditions in order to pin down ##a,b## convincingly and uniquely.Oops you're right, I
  • #1
geoffrey159
535
72

Homework Statement


If ##u,v,w\in\mathbb{R}^3##, show that ## u\times(v\times w) = (u.w) v - (u.v) w ##.

Homework Equations



The Attempt at a Solution


Since ## u\times(v\times w)##, ##v## and ##w## are orthogonal to ##v\times w##, these vectors are coplanar. Therefore, there must be reals ## a,b## such that ## u\times(v\times w) = a v + b w ##.
Then, ## 0 = u.(u\times(v\times w)) = a u.v + b u.w ## and ## a = u.w## and ## b = - u.v## work. But why not ## a = -u.w ## and ## b = u.v ## ? Thanks
 
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  • #2
geoffrey159 said:

Homework Statement


If ##u,v,w\in\mathbb{R}^3##, show that ## u\times(v\times w) = (u.w) v - (u.v) w ##.

Homework Equations



The Attempt at a Solution


Since ## u\times(v\times w)##, ##v## and ##w## are orthogonal to ##v\times w##, these vectors are coplanar. Therefore, there must be reals ## a,b## such that ## u\times(v\times w) = a v + b w ##.
Then, ## 0 = u.(u\times(v\times w)) = a u.v + b u.w ## and ## a = u.w## and ## b = - u.v## work. But why not ## a = -u.w ## and ## b = u.v ## ? Thanks
Write the vectors in Cartesian components and expand the double product.
 
  • #3
geoffrey159 said:

Homework Statement


If ##u,v,w\in\mathbb{R}^3##, show that ## u\times(v\times w) = (u.w) v - (u.v) w ##.

Homework Equations



The Attempt at a Solution


Since ## u\times(v\times w)##, ##v## and ##w## are orthogonal to ##v\times w##, these vectors are coplanar. Therefore, there must be reals ## a,b## such that ## u\times(v\times w) = a v + b w ##.
Then, ## 0 = u.(u\times(v\times w)) = a u.v + b u.w ## and ## a = u.w## and ## b = - u.v## work. But why not ## a = -u.w ## and ## b = u.v ## ? Thanks

An alternative method is to write vectors as ##/vec{v} = (v_1,v_2,v_3)## for example, and use the so-called Levi-Civita symbol to express the cross-product:
[tex] (\vec{u} \times \vec{v}) _i = \sum_{j,k} \epsilon_{ijk} u_j v_k [/tex]
Here the Levi-Civita symbol ##\epsilon## is
[tex] \epsilon_{ijk} = \begin{cases} +1 & \text{if} \:\; ijk \; \text{is an even permutation of} \: \; 123 \\
-1 & \text{if} \:\; ijk \; \text{is an odd permutation of} \:\; 123 \\
0 & \text{otherwise} .
\end{cases}
[/tex]
There are identities available for sums of products like ##\sum_k \epsilon_{ijk} \epsilon_{klm}## which would be the thing needed in the triple product. See, eg., http://www.ucl.ac.uk/~ucappgu/seminars/levi-civita.pdf or https://en.wikipedia.org/wiki/Levi-Civita_symbol .
 
  • #4
Thank you for your answers, but is it possible to avoid lengthy calculations ?
In the OP, I am to the point where ## (a,b) \in \{\pm (u.w, - u.v) \} ##.
How should I decide for the sign ?
 
  • #5
geoffrey159 said:
Thank you for your answers, but is it possible to avoid lengthy calculations ?
In the OP, I am to the point where ## (a,b) \in \{\pm (u.w, - u.v) \} ##.
How should I decide for the sign ?

Some detailed calculations are just unavoidable. However, if you pay attention to what I said in #3, that will lead to just about the shortest and most straightforward proof available. Besides, you have NOT reached the point where ## (a,b) \in \{\pm (u.w, - u.v) \} ##, because you had one relation between ##a## and ##b##, and you said that your particular choice "works". However, in principle, there are infinitely many other choices of ##a,b## that also work for that single condition. Somehow, you need more conditions in order to pin down ##a,b## convincingly and uniquely.
 
Last edited:
  • #6
Oops you're right, I'm just at ## \begin{pmatrix}a\\b\end{pmatrix} \in \mathbb{R} \begin{pmatrix}u.w \\ -u.v \end{pmatrix} ##.
Thanks again, I will try your proof.
 

FAQ: Solve Double Cross Product Problem in $\mathbb{R}^3$

1. What is a double cross product in $\mathbb{R}^3$?

A double cross product is a mathematical operation that involves two vectors in three-dimensional space, resulting in a new vector that is perpendicular to both of the original vectors.

2. How do you solve a double cross product problem in $\mathbb{R}^3$?

To solve a double cross product problem in $\mathbb{R}^3$, you can use the formula (A x B) x C = (A · C)B - (B · C)A, where A, B, and C are the three vectors involved. First, perform the individual cross products (A x B) and (A x C), then take the dot product of each result with the remaining vector and subtract them from each other to get the final solution.

3. What are some real-world applications of double cross products in $\mathbb{R}^3$?

Double cross products are commonly used in physics and engineering, particularly in mechanics and electromagnetism. They can be used to calculate torque, angular momentum, and magnetic fields, among other things.

4. Are there any special properties of double cross products in $\mathbb{R}^3$?

One special property of double cross products is that they are not commutative, meaning that the order in which you take the cross products matters. (A x B) x C is not the same as (C x B) x A. They are also distributive, meaning (A + B) x C = (A x C) + (B x C).

5. Can a double cross product problem have more than three vectors in $\mathbb{R}^3$?

Yes, a double cross product problem can involve more than three vectors in $\mathbb{R}^3$. However, the resulting vector will still be perpendicular to all of the original vectors. To solve for this, you can use the same formula (A x B) x C = (A · C)B - (B · C)A and continue to expand it for each additional vector.

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