# Derivation of Cross Product

1. Sep 26, 2005

### thechunk

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'.

2. Sep 26, 2005

### Tom Mattson

Staff Emeritus
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.

3. Sep 26, 2005

### amcavoy

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).

4. Sep 26, 2005

### thechunk

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. Sep 26, 2005

### Tom Mattson

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

6. Sep 26, 2005

### amcavoy

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. Sep 26, 2005

### Tom Mattson

Staff Emeritus
I don't know, but it sure seems easier than doing it the other way around!

8. Sep 27, 2005

### es

9. Sep 28, 2005

### TD

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.

10. Oct 17, 2005

### john fairbanks

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. Oct 17, 2005

### amcavoy

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).