# Thread: Another Proof for Distributivity of Cross Product

1. ## Another Proof for Distributivity of Cross Product

Greetings,

In the book Field and Wave Electromagnetics by David K. Chen, I am asked whether the distributive law holds for cross product or not. However, I am asked not to resolve the vectors in Cartesian components and I also want to avoid the geometric approach( I found one and understood it). Is it possible to somehow prove the distributive law for cross product without geometric means or use of Cartesian coordinates? I am really confused by this, and I would be glad if you reply if it is not possible. Thanks in advance.

2. ## Re: Another Proof for Distributivity of Cross Product

Look at *exactly* how that book defines the cross product. That's the definition you'll want to use in your proof.

3. ## Re: Another Proof for Distributivity of Cross Product

The books gives the familiar definition that uses the sine of the angle and the magnitude of the vectors without reference to any coordinate system, but I am unable to figure out how can I use this definition to prove the distribution law for cross product.

4. ## Re: Another Proof for Distributivity of Cross Product

I'm sorry - I don't see a clean way to do this. I haven't thought through the methods below, so they might not work, or might not work easily. They're just a ideas for an approach.

I) Define a function $\displaystyle *: \mathbb{R}^3 \times \mathbb{R}^3 \rightarrow \mathbb{R}^3$ that has all the nice algebraic properties as the cross product - in particular, it satisfies the distributive property. (In other words, explicitly state what * is on a basis, then declare that you're extending it linearly in both components to define it for any pair of vectors.) Then show that * satisfies the magnitude and direction required by the definition you're using for the vector cross product. Thus * IS the vector cross product. Thus the vector cross product obeys those nice algebraic properties, since * does, since they're the same thing.

II) Perhaps build it up step by step?

1) Show $\displaystyle \hat{k} \times (a \hat{i} + b \hat{j}) = a(\hat{k} \times \hat{i}) + b(\hat{k} \times \hat{j})$.

2) Show $\displaystyle \hat{k} \times (a \hat{i} + b \hat{j} + c\hat{k}) = \hat{k} \times (a \hat{i} + b \hat{j}) + (\vec{0}). \ (\hat{k} \times c\hat{k} = \vec{0}).$

3) Using #1 and #2, show $\displaystyle \hat{k} \times (\vec{v} + \vec{w}) = (\hat{k} \times \vec{v}) + (\hat{k} \times \vec{w})$.

4) Show $\displaystyle q\hat{k} \times (\vec{v} + \vec{w}) = (q\hat{k} \times \vec{v}) + (q\hat{k} \times \vec{w})$.

5) Argue that a rigid rotation (so preserves angles and "right hand rule") can transform any $\displaystyle \vec{z_0} \times (\vec{z_1} + \vec{z_2})$ into $\displaystyle q\hat{k} \times (\vec{v} + \vec{w})$, then apply #4, then do the reverse of the rotation.

5. ## Re: Another Proof for Distributivity of Cross Product

Thanks. I will think of them, but the second method still uses Cartesian coordinates. In addition to that first method seems cumbersome so I avoided it. Any other suggestions are welcome.

6. ## Re: Another Proof for Distributivity of Cross Product

Originally Posted by Vesnog
Greetings,

In the book Field and Wave Electromagnetics by David K. Chen, I am asked whether the distributive law holds for cross product or not. However, I am asked not to resolve the vectors in Cartesian components and I also want to avoid the geometric approach( I found one and understood it). Is it possible to somehow prove the distributive law for cross product without geometric means or use of Cartesian coordinates? I am really confused by this, and I would be glad if you reply if it is not possible. Thanks in advance.
Yes, but it will require tools from abstract algebra and differential geometry.

The cross product is terms of the wedge product and the hodge dual is

$\displaystyle \vec{v} \times \vec{w} = *( v \wedge w)$

$\displaystyle \vec{v} \times (\vec{w}+\vec{u}) = *( v \wedge (w+u))$

Expanding the wedge product in the exterior algebra gives

$\displaystyle *(v \wedge w +v \wedge u) = *(v \wedge w)+*(v \wedge u)=\vec{v} \times \vec{w} +\vec{v} \times \vec{u}$

7. ## Re: Another Proof for Distributivity of Cross Product

the "standard definition" you are talking about really isn't one.

what we have is: |uxv| = |u||v||sin(arccos((u.v)/(|u||v|))|, which tells us the MAGNITUDE of the cross product, but doesn't tell us the DIRECTION.

however, we can say that w = (uxv)/|(uxv)| is the unit vector in the orthogonal complement to span({u,v}). the trouble is: this space is 1-dimensional (a line), and we have "two choices" for this unit vector (which means we have to pick an orientation for span({u,v,w}) = span({u,v,-w})).

so what you wind up doing is "calling u one axis (a basis vector)" and v "another axis" (another basis vector). then uxv is (one of the two) "perpendicular" vectors to the plane spanned by u and v, with the given magnitude. this IS choosing a coordinate system (that's what a basis IS: a way to "coordinatize" a vector space), but it gives a DIFFERENT coordinate system, for different vectors u and v.

usually, the vector w i have referenced above is called "n" (for "normal"), typically the orientation (the choice whether to use n or -n) is given by "the right-hand rule". if we replace u and v by unit vectors, say u' and v', we get a basis {u',v',n}. this IS a coordinate system, it's just "relative".

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