# Geometric proof cross product distributes over addition

1. Dec 26, 2015

### Odious Suspect

If the cross product in ℝ3 is defined as the area of the parallelogram determined by the constituent vectors joined at the tail, how does one go about proving this product to distribute over vector addition?

I've attached a drawing showing cyan x yellow, cyan x magenta, and cyan x (magenta + yellow). If the parallelogram cyan x yellow is parallel transported along the magenta vector, an open-ended prism is formed, with the these cross products as faces.

If the open ends are capped by congruent triangles and the vector sum of all faces, treated as surface elements with outward-facing normals is taken, the result is the null vector. Since the end-caps cancel, this tells me the sum of any two of the remaining faces is equal to the other.

I believe this result, somehow demonstrates the proposition, but I don't know if it would persuade an independent observer.

My biggest concern is that the general proof that the sum of outward facing surface elements is the null vector relies on the distributive property of the cross product.

2. Dec 26, 2015

### Staff: Mentor

The cross-product is not the area, it is a vector with a length given by that area. This length is not additive, only the vector itself is, and I don't see how you want to add that geometrically.

3. Dec 27, 2015

### Svein

The standard representation of the cross product of two vectors $\vec{a}= (a_{i}, a_{j}, a_{k})$ and $\vec{b}= (b_{i}, b_{j}, b_{k})$ is given by $\begin{vmatrix} \vec{i} &\vec{j} & \vec{k}\\ a_{i} &a_{j} & a_{k} \\ b_{i} &b_{j} &b_{k} \\ \end{vmatrix}$. From this representation you can easily derive the distributive law.

Last edited: Jan 6, 2016
4. Dec 27, 2015

### Odious Suspect

"The vector product of two vectors $\vec{A}$ and $\vec{B}$ is defined as the vector $\vec{P}$ which is perpendicular to the plane determined by $\vec{A}$ and $\vec{B}$ and whose magnitude is equal to the parallelogram formed by $\vec{A}$ and $\vec{B}$, i.e. equal to AB sin($\vec{A}\vec{B}$)."

He also states that distribution of the product over vector addition is not difficult to prove, but is cumbersome, and so omits the demonstration.

5. Dec 27, 2015

### Odious Suspect

That is a matter of preference. One man's definition is another man's theorem. There are several ways to approach the topic of vectors and tensors. Joos begins with naive geometry in "real" space. He was certainly aware of the concise development presented by the likes of Dirac. https://books.google.com/books?id=Jtu-n8AzJhMC&printsec=frontcover#v=onepage&q&f=false Apparently Joos believed the didactic value of the naive approach justified its use, despite its limitations.

6. Dec 29, 2015

### Odious Suspect

I stand by the legitimacy of my question. The way Joos defines things, the distribution of the vector product follows closely from the geometric definition of the vector product. This is an essential first step in arriving at the more common expression (usual defintion) of the vector product. One merit to Joos's approach is that it is coordinate system agnostic.

I will return to this matter in my next review cycle. Perhaps then I will be able to formulate my question better, and even give a better answer.

7. Dec 29, 2015

### geoffrey159

The way I've learned about the cross product is this:
In a euclidean space $E$ of dimension 3 fitted with a direct orthonormal basis ${\cal B}$ (in your case $E = \mathbb{R}^3$ fitted with the canonical basis), the vector product of $\vec u$ and $\vec v$ is the unique vector written $\vec u \times \vec v$ such that for any $\vec w$ in this vector space: $\text{det}_{\cal B}(\vec u,\vec v,\vec w) = (\vec u \times \vec v).\vec w$.

Antisymetry of the vector product:
$\text{det}_{\cal B}(\vec u,\vec v,\vec w) = - \text{det}_{\cal B}(\vec v,\vec u,\vec w) \iff (\vec u \times \vec v + \vec v\times \vec u ) .\vec w = 0 \iff \vec u \times \vec v + \vec v\times \vec u \in E^\perp = \{0 \} \iff \vec u\times \vec v = - \vec v\times \vec u$

Bilinearity of the vector product:
1. $\text{det}_{\cal B}(\lambda \vec u_1 + \vec u_2 ,\vec v,\vec w) = \lambda \text{det}_{\cal B}(\vec u_1,\vec v,\vec w) + \text{det}_{\cal B}(\vec u_2,\vec v,\vec w) \iff ((\lambda \vec u_1 + \vec u_2) \times \vec v -\lambda \vec u_1 \times \vec v - \vec u_2 \times \vec v ).\vec w = 0 \iff (\lambda \vec u_1 + \vec u_2) \times \vec v = \lambda \vec u_1 \times \vec v + \vec u_2 \times \vec v$.
2. $\vec u \times (\lambda \vec v_1 + \vec v_2) = - (\lambda \vec v_1 + \vec v_2) \times \vec u = -\lambda \vec v_1 \times \vec u - \vec v_2 \times \vec u = \lambda \vec u\times \vec v_1 + \vec u \times \vec v_2$

etc...

8. Dec 30, 2015

### Odious Suspect

Very nice. I believe I will incorporate this into my review notes. Though it vindicates my original question, it doesn't directly address it. Nonetheless, the new insight is appreciated.

9. Jan 5, 2016

### Hawkeye18

In fact, post # 7 essentially answered your question, it only need to be interpreted geometrically.

To do that we need to introduce the notion of signed volume. For a triple of vectors $\mathbf u$, $\mathbf v$, $\mathbf w$ as the volume of the corresponding parallelepiped, taken with sign "+" if the triple $\mathbf u$, $\mathbf v$, $\mathbf w$ is positively oriented, and with sign "-" otherwise. (Note that the sign depends on the order of vectors.)

Using the notion of signed volume we can define the cross product $\mathbf u \times \mathbf v$ as follows: among all unit vectors $\mathbf e$ consider one that maximizes the signed volume of the triple $(\mathbf u, \mathbf v, \mathbf e)$. Then the cross product $\mathbf u \times \mathbf v$ is defined as the vector in the direction of this maximizing unit vector $\mathbf e$ and of length equal to this maximal signed volume.

It is easy to see that the maximizing vector $\mathbf e$ is orthogonal to the vectors $\mathbf u$ and $\mathbf v$, and that the maximal volume is exactly $\|\mathbf u\| \|\mathbf v\| \sin \alpha$, where $\alpha$ is the angle between $\mathbf u$ and $\mathbf v$; thus the definition via signed volume coincides with the classical geometric definition of the cross product.

From this definition of the cross product (or from the classical geometric one) we can see that the signed volume of the triple $\mathbf u$, $\mathbf v$, $\mathbf w$ is exactly the dot product $(\mathbf u\times \mathbf v)\cdot \mathbf w$ (which is also known as the triple product). So we can see that the cross product $\mathbf u\times \mathbf v$ is the unique vector $\mathbf z$ such that for all vectors $\mathbf w$ the signed volume of the triple $\mathbf u$, $\mathbf v$, $\mathbf w$ equals $\mathbf z \cdot \mathbf w$.

Now we can use the fact that the signed volume (triple product) can be computed as the determinant, where at each row we have the coordinates of the corresponding vector; the coordinates here must be with respect to any positively oriented orthonormal basis. Using this fact we arrive to the definition of the cross product presented by geoffrey159 in post # 7. From here, using cofactor expansion of determinant you can get the standard definition of the cross product via determinant, and the distributive law for which follows trivially from the properties of determinants.

If you do not want to use linear algebra, and determinants in particular, it is also possible to present a more "geometric" proof of the distributive law for the cross product, without using the determinants, although the determinants will be lurking in the background. Namely, we first note that the signed volume of a triple $\mathbf u$, $\mathbf v$, $\mathbf w$ (let me use the notation $[\mathbf u, \mathbf v, \mathbf w]$ for it) is linear in each argument $\mathbf u$, $\mathbf v$, $\mathbf w$, and in particular satisfies the distributive law in each of the argument (with the other 2 being fixed). So, for a vector $\mathbf w$ and vectors $\mathbf u$, $\mathbf v_1$, $\mathbf v_2$ we have $$[\mathbf u, (\mathbf v_1+\mathbf v_2), \mathbf w ]= [\mathbf u, \mathbf v_1 , \mathbf e ] + [\mathbf u, \mathbf v_2 ,\mathbf w ] .$$ Let $\mathbf z= \mathbf u\times ( \mathbf v_1+\mathbf v_2)$, $\mathbf z_1= \mathbf u\times \mathbf v_1$, $\mathbf z_2= \mathbf u\times \mathbf v_2$. Then using my second definition of the cross product (the geometric version of the definition from the post # 7) we can rewrite the above identity as $$\mathbf z \cdot \mathbf w = \mathbf z_1 \cdot \mathbf w +\mathbf z_2 \cdot \mathbf w.$$ Using the distributive law for the dot product we can rewrite it as $$\mathbf z \cdot \mathbf w = (\mathbf z_1 +\mathbf z_2 )\cdot \mathbf w.$$ Since it is true for all vectors $\mathbf w$ we can conclude that $\mathbf z=\mathbf z_1+\mathbf z_2$, which is exactly the distributive law for the cross product.

10. Jan 7, 2016

### Staff: Mentor

I agree with mfg -- the cross product is not an area (a scalar). It is a vector.
The definition above is a bit sloppy. An improved version would include, "... and whose magnitude is equal to the area of the paralellogram formed by ..."

11. Jan 8, 2016

### suremarc

It appears to be a typo by OP--the author says "[...] whose magnitude is equal to the area of the parallelogram" etc. I think his confusion stems from misreading.

Regarding geoffrey's post--one need not associate an orthonormal basis to $\mathbb{R}^3$ as the determinant is invariant under a change of basis. So his presentation is therefore "coordinate system agnostic" as desired.

12. Jan 14, 2016

### Odious Suspect

OK, that is a new (to me) and interesting way to define the cross product.

What Joos does is introduces the geometric definition of the cross product; asserts without demonstration that the cross product is distributive over vector addition; demonstrates the anti-symmetric nature of the cross products of ortho-normal basis vectors; and then combines these results to produce the familiar form of the cross product as a determinate with the basis vectors forming the first row. From there it's easy to see that the triple product can be expressed as a determinant. I suspect was suggesting a purely geometric argument to prove the distributive property.

I'm not exactly what it means to say the triple product is linear in each argument. After I have a way to resolve things into components and can write the triple product out component-wise, I can see that it is linear in each component. If, however, I consider the triple product to be defined as a signed volume, and only consider the constituent vectors and volume as geometric objects, the notion of linearity in a vector is nebulous to me.

Certainly, if we change the magnitude of any of the vector arguments, the volume will change proportionally. But a change in the direction of one of the arguments will also change the volume. Without resorting to a demonstration in terms of components, I'm not sure how to show linearity.

13. Jan 14, 2016

### Hawkeye18

It is quite simple. Let me outline the proof of linearity in the 3rd argument $\mathbf w$. As we know, the volume of a parallelepiped is the "area of the base" $\times$ "height", where in our case the "area of the base" is the area of the parallelogram defined by vectors $\mathbf u$ and $\mathbf v$, and the "height" is the length of the projection of the vector $\mathbf w$ onto the axis $l$ orthogonal to the plane spanned by $\mathbf u$ and $\mathbf v$.

The signed volume can then be computed as "area of the base" $\times$ "signed height", where the "signed height" is the length of the above projection, taken with sign $+$ or $-$, depending on its direction. In other words, if you pick a unit vector of appropriate direction as a basis vector in the axis $l$, the "signed height" is exactly the coordinate of this projection. An alternative point of view is to consider an orthonormal basis such that one of the basis vectors is orthogonal to the plane spanned by $\mathbf u$ and $\mathbf v$ and is looking in the right direction: then the "signed height" is just the corresponding coordinate in this basis. The linearity of the "signed height" then should be obvious, and so the linearity of the signed volume.

Joos says about the distributive law that "The proof of this formula is somewhat cumbersome but is not difficult...", and I really do not know what did he mean. I doubt that there is an "elementary" proof (more "elementary" than I presented), at least I am not aware of it. Probably he was just cheating.

14. Jan 15, 2016

### Odious Suspect

As I understand things, projecting onto orthonormal basis vectors is the resolution of the vector into components. That's why I prefer to speak of the signed volume being linear in each of the components of its arguments, rather than in the arguments themselves. Now, we might argue that the statement that the signed volume is linear in each component of its arguments (provided the arguments are parameterized by arc length) is a statement about the geometric object, and not about any specific coordinate system.

I suspect Joos intended a proof similar to his 2D demonstration of the distributive property of the scalar product over vector addition. I don't believe he actually provides a formal generalization of that proof to 3 dimensions. Nonetheless, it is conceptually easy to grasp. https://books.google.com/books?id=btrCAgAAQBAJ&lpg=PP1&pg=PA12#v=onepage&q&f=false

This is why I attempted to appeal to the diagram in the head message of this thread. It communicates in terms of the geometric spirit of the definition of the vector product as a directed plane area. If I'm following Joos correctly, he places the proof of the distribution of vector multiplication over vector addition prior to its resolution into components, which, in his development, relies upon that distributive property. Unfortunately, my vague argument seems to rely on the results presented in fig. 8 and related text. https://books.google.com/books?id=btrCAgAAQBAJ&lpg=PP1&pg=PA12#v=onepage&q&f=false

As for the suggestion that Joos was cheating; considering the cultural station of that masterpiece, I find the possibility vanishingly minute. He wanted me to grasp something fundamental and significant.

I'm also somewhat uneasy about defining the vector product as an artifact of the signed volume element, in this context. My intuition tells me that the vector product is intrinsic to the surface, and describing it as a vector normal to that surface is a convenience which uses the dimension normal to the surface for dual purposes. That is, both as a native geometric dimension, and as an embedding space.

15. Jan 15, 2016

### Svein

16. Jan 15, 2016

### Hawkeye18

I was a bit hasty last time, there is an elementary proof of distributive property for the cross product, one you can present to high school students.

Namely, we first note that the cross product $\mathbf u\times \mathbf v$ does not change if we replace $\mathbf v$ by its projection $\widetilde{\mathbf v}$ onto the orthogonal complement $\mathbf u^\perp$ of $\mathbf u$. You can see that because $\widetilde{\mathbf v}$ belongs to the plane spanned by $\mathbf u$ and $\mathbf v$, the areas of parallelograms defined by $\mathbf u$, $\mathbf v$ and $\mathbf u$, $\widetilde{\mathbf v}$ coincide, and the direction of the shortest rotation from $\mathbf u$ to $\mathbf v$ is the same as he direction of the shortest rotation from $\mathbf u$ to $\widetilde{\mathbf v}$.

So $\mathbf u\times \mathbf v= \mathbf u\times \widetilde{\mathbf v}$. Second, the projection onto $\mathbf u^\perp$ (in fact any projection) is a linear transformation, so $\widetilde{\mathbf v_1+\mathbf v_2} = \widetilde{\mathbf v}_1 + \widetilde{\mathbf v}_2$.

So, to prove the distributive property we need to show that $$\mathbf u\times (\widetilde{\mathbf v}_1 + \widetilde{\mathbf v}_2) = \mathbf u\times \widetilde{\mathbf v}_1 + \mathbf u\times \widetilde{\mathbf v}_2. \tag{1}$$ But what is the cross product $\mathbf u\times \widetilde{\mathbf v}$? We can see from the definition of the cross product, that it is obtained by rotating the vector $\widetilde{\mathbf v}$ in the plane $\mathbf u^\perp$ by $90^\circ$ (clockwise if we are looking at this plane in the direction of the vector $\mathbf u$) and then multiplying it by the length of $\mathbf u$. Rotation and multiplication by a scalar are linear transformations, so (1) should be obvious.

17. Jan 15, 2016

### Hawkeye18

The proof presented there on fig. 5 on p. 11 works in 3 dimensions: you just need to interpret the picture as 3-dimensional one, with the third axis looking toward you. The vertical dashed lines should be interpreted as planes orthogonal to the vector $A$.