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By matrix-defined, I mean

$$\left<a,b,c\right>\times\left<d,e,f\right> = \left| \begin{array}{ccc} i & j & k\\ a & b & c\\ d & e & f \end{array} \right|$$

...instead of the definition of the product of the magnitudes multiplied by the sign of their angle, in the direction orthogonal)

If I try cross producting two vectors with no $k$ component, I get one with only $k$, which is expected. But why?

As has been pointed out, I am asking why the algebraic definition lines up with the geometric definition.

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4 Answers 4

up vote 4 down vote accepted

Assuming you know the definition of orthogonal as "a is orthogonal to b iff $a\cdot b=0$ then we could calculate $(a \times b)\cdot a = a_1(a_2b_3-a_3b_2)-a_2(a_1b_3-a_3b_1)-a_3(a_1b_2-a_2b_1)=0$ and $(a \times b)\cdot b-0$, so the cross product is orthogonal to both. As Nold mentioned, if the two vectors a and b lie in the x,y plane, then the orthogonal vectors must be purely in the z direction.

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The obvious but slightly trite answer is "because that's just how the cross-product works as an operation".

If you're looking for an intuitive reason, the cross-product by definition produces a vector that is orthogonal to the two operand (input) vectors. You know that the base vectors i, j, and k are all orthogonal, thus if your two input vectors lie on the (x, y) plane (i.e. only i and j components), you know that any orthogonal vector must have only a component in the z direction (multiple of k).

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As good an answer as any. It works that way because that's how the operation is defined :) –  workmad3 Jul 20 '10 at 19:36
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this is exactly the approach that I specified would not be helpful in my question =/ –  Justin L. Jul 20 '10 at 19:54
    
You did not specify that any approach wouldn't be helpful in your question. Read and learn, please. Also, it's generally not good etiquette to down-vote answers to your own questions. –  Noldorin Jul 20 '10 at 19:56
    
Your non-trite answer is "By definition, they produce mutually orthogonal vectors". I specifically stated that I was aware of the definition of cross product that explicitly stated that the result is mutually orthogonal. This is not the definition I am asking about; I am asking about why the Matrix Determinant definition of the cross product (which has no mention of orthogonality) produces an orthogonal vector. Perhaps I should have clarified more that I was asking specifically about the matrix definition of the cross product, and not the definition that explicitly stated orthogonality –  Justin L. Jul 20 '10 at 20:51
    
@justin-l, Effectively you are asking why the algebraic definition produces the same answer as the geometric definition. –  Neil Mayhew Jul 20 '10 at 21:43

Note that if you replace $i$, $j$, and $k$ with $m$, $n$, and $p$, the determinant becomes the dot-product of the vector $(m, n, p)$ with the cross-product of the two original vectors. If $(m, n, p) = (a, b, c)$ or $(m, n, p) = (d, e, f)$, the determinant is zero (any matrix with two identical rows has determinant zero), so the dot product of $(a, b, c)$ or $(d, e, f)$ with the cross-product is zero. Hence, $(a, b, c)$ and $(d, e, f)$ are orthogonal to their cross-product.

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Here's an explanation in terms of the Hodge dual and the exterior (wedge) product.

Let ${e_1, e_2, e_3}$ be the standard orthonormal basis for $\mathbb{R}^3$. Consider the two vectors $a = a_1 e_1 + a_2 e_2 + a_3 e_3$ and $b = b_1 e_1 + b_2 e_2 + b_3 e_3$. From the matrix computation we obtain the familiar formula

$a\times b = (a_2 b_3 - a_3 b_2) e_1 + (a_3 b_1 - a_1 b_3) e_2 + (a_1 b_2 - a_2 b_1) e_3$.

But (see note at the bottom)

$a \wedge b = (a_1 b_2 - a_2 b_1) e_1 \wedge e_2 + (a_2 b_3 - a_3 b_2) e_2 \wedge e_3 + (a_3 b_1 - a_1 b_3) e_3 \wedge e_1$,

where the wedge $\wedge$ represents the exterior product. One can now compute the dual of this latter expression using that the left contraction of $(e_1 \wedge e_2)$ onto $(e_3 \wedge e_2 \wedge e_1)$ is $e_3$ (and similar relations). The result is that

$a \times b = (a \wedge b)^*$,

that is, the cross product of $a$ and $b$ is the dual of their exterior product.

Geometrically, this is an incredible picture. The exterior product is the plane element spanned by both $a$ and $b$, and the dual is the vector orthogonal to that plane.

This is my favorite interpretation of the cross product, but it's only helpful, of course, if you're familiar with exterior algebra and the Hodge dual.

Note: The wedge product can be found by formally computing

$(a_1 e_1 + a_2 e_2 + a_3 e_3) \wedge (b_1 e_1 + b_2 e_2 + b_3 e_3)$

using the distributivity and anticommutation relations of the exterior product.

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I like this answer because it really highlights why the cross product is a uniquely 3 dimensional phenomenon: you need the hodge dual of a 2-form to be a 1-form. –  Jason DeVito Aug 4 '10 at 0:42

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