Derivation of Cross Product

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Does anyone know where I can find the derivation of the cross product. I know how to use it and the like but I do not understand why the norm of the matrix :
[tex]
\left[ \begin{array}{ccc}i & j & k \\n1 & n2 & n3 \\m1 & m2 & m3 \\\end{array}\right] [/tex]

yields the vector perpendicular to 'n' and 'm'.
 

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  • #2
Tom Mattson
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I wouldn't prove that the cross product can be written as a determinant, I would simply define it that way. Then I would prove that the result is perpendicular to both of the vectors in the cross product. How would I do that? For vectors [itex]\vec{A}[/itex] and [itex]\vec{B}[/itex] I would form the dot products [itex]\vec{A}\cdot(\vec{A}\times\vec{B})[/itex] and [itex]\vec{B}\cdot(\vec{A}\times\vec{B})[/itex] and show that they both vanish identically.
 
  • #3
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Also, if you want to show that your definition above is equal to ABsinθ, you can find the determinant, square it, and then rearrange and use the definition of a dot product (you will see something similar after squaring).
 
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Yes I can see that, but what confuses me is why do those two expressions describe/yield the vector perpendicular to n and m.
 
  • #5
Tom Mattson
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Because the dot product of any vector and its cross product with any other vector vanishes. The "why" is in the proof.
 
  • #6
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Just out of curiosity: was this stumbled upon just like you said? (defining the cross product as a determinant) and then later shown that the resulting vector orthogonal to both? Or was that the goal and then later shown that the determinant did the trick?
 
  • #7
Tom Mattson
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apmcavoy said:
Just out of curiosity: was this stumbled upon just like you said? (defining the cross product as a determinant) and then later shown that the resulting vector orthogonal to both?
I don't know, but it sure seems easier than doing it the other way around!
 
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  • #9
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thechunk said:
Yes I can see that, but what confuses me is why do those two expressions describe/yield the vector perpendicular to n and m.
If
[tex]\vec c = \vec a \times \vec b = \left| {\begin{array}{*{20}c}
{\vec 1_x } & {\vec 1_y } & {\vec 1_z } \\
{a_1 } & {a_2 } & {a_3 } \\
{b_1 } & {b_2 } & {b_3 } \\
\end{array}} \right|[/tex]
then
[tex]\left\langle {\vec a,\vec c} \right\rangle = \left| {\begin{array}{*{20}c}
{a_1 } & {a_2 } & {a_3 } \\
{a_1 } & {a_2 } & {a_3 } \\
{b_1 } & {b_2 } & {b_3 } \\
\end{array}} \right| = 0 \Rightarrow \vec a \bot \vec c[/tex]

In the same way, [itex]\vec b \bot \vec c[/itex] follows.
 
  • #10
derivation of Vector Cross Product

Using determinants to describe cross products I think was discovered by a mathamatician named William Rowan Hamiliton. He came up with the algebraic forms in dot and cross products, but I have no idea how he did it. Does anybody know. I just memorize the cross product formula but don't know where it comes from. Thanks
 
  • #11
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john fairbanks said:
Using determinants to describe cross products I think was discovered by a mathamatician named William Rowan Hamiliton. He came up with the algebraic forms in dot and cross products, but I have no idea how he did it. Does anybody know. I just memorize the cross product formula but don't know where it comes from. Thanks
Have you taken linear algebra? You learn a lot about how determinants equal the volume of the parallelopiped made by the three vectors (or other dimensions).
 

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