# Derivation of Cross Product

## Main Question or Discussion Point

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 :
$$\left[ \begin{array}{ccc}i & j & k \\n1 & n2 & n3 \\m1 & m2 & m3 \\\end{array}\right]$$

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

Tom Mattson
Staff Emeritus
Gold Member
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 $\vec{A}$ and $\vec{B}$ I would form the dot products $\vec{A}\cdot(\vec{A}\times\vec{B})$ and $\vec{B}\cdot(\vec{A}\times\vec{B})$ and show that they both vanish identically.

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).

Yes I can see that, but what confuses me is why do those two expressions describe/yield the vector perpendicular to n and m.

Tom Mattson
Staff Emeritus
Gold Member
Because the dot product of any vector and its cross product with any other vector vanishes. The "why" is in the proof.

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?

Tom Mattson
Staff Emeritus
Gold Member
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!

es
TD
Homework Helper
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
$$\vec c = \vec a \times \vec b = \left| {\begin{array}{*{20}c} {\vec 1_x } & {\vec 1_y } & {\vec 1_z } \\ {a_1 } & {a_2 } & {a_3 } \\ {b_1 } & {b_2 } & {b_3 } \\ \end{array}} \right|$$
then
$$\left\langle {\vec a,\vec c} \right\rangle = \left| {\begin{array}{*{20}c} {a_1 } & {a_2 } & {a_3 } \\ {a_1 } & {a_2 } & {a_3 } \\ {b_1 } & {b_2 } & {b_3 } \\ \end{array}} \right| = 0 \Rightarrow \vec a \bot \vec c$$

In the same way, $\vec b \bot \vec c$ follows.

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

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).