How to prove the distributive property of cross product

Question

That is, how to prove the following identity: $$a \times (b+c) = a \times b + a \times c$$ where the $\times$ represents cross product of two vectors in 3-dimensional Euclidean space.

Answer 1

To see it geometrically:

Recall that $\lVert x\times y\rVert$ represents the area of the parallelogram with sides $x,y$. If you then "glue" together the parallelograms with sides $a,b$ and $a,c$ along the side $a$, you get a hexagon (not regular) with the same area of the parallelogram with sides $a,b+c$.

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Edit: As another user correctly noted, the above proof works only when $a$ lies in the plane spanned by $b$ and $c$. If not, then we can reduce to this case by considering the triangle $T$ with base $b+c$ and sides $b,c$, as suggested by this image¹:

Indeed, without loss of generality we may assume that $a$ is orthogonal to $b$ and $c$. Then the angle and proportion between $a \times b$ and $a \times c$ are the same as those between $b$ and $c$, because in this case the cross product with $a$ is equivalent to the composition of a rotation by $90$ degrees and a dilation by $\lVert a \rVert$. Therefore if $P$ is the prism with base $T$ and height $a$, then the area of the projection of the face $[a,b]$ on the face $[a,b+c]$ is the same as the length of the cross product between $a$ and the projection of $b$ on $b+c$, and similarly for the face $[a,c]$.

^{1. Courtesy of WikiMedia.}

Answer 2

Just to enrich the post for future readers, I would like to add another derivation that I found on the internet. This proof uses the distributivity of the dot product (which is easier to prove), and the property that the circular commutation of vectors doesn't change the triple product of the vectors (which is quite obvious, since the triple product is just the volume of the parallelepiped formed by the vectors).

Let $d = a \times (b + c) - a \times b - a \times c$

so it is required to prove that $d = 0$:

$d^2$ = $d \cdot d$

$= d \cdot (a \times (b + c) - a \times b - a \times c)$

$= d \cdot (a \times (b + c)) - d \cdot (a \times b) - d \cdot (a \times c)$

$= (d \times a) \cdot (b + c) - (d \times a) \cdot b - (d \times a) \cdot c$

$= (d \times a) \cdot (b + c) - (d \times a) \cdot (b + c)$

$= 0$

Therefore $d = 0$, so $a \times (b + c) = a \times b + a \times c$.

Answer 3

Let $a=(a_1,a_2,a_3),~~b=(b_1,b_2,b_3),~~c=(c_1,c_2,c_3)\in\mathbb R^3$ so $$a\times(b+c)=\begin{vmatrix} i\;\;\;j\;\;\;k \\ a_1\;\;\;a_2\;\;\;a_3 \\ b_1+c_1\;\;\;b_2+c_2\;\;\;b_3+c_3 \end{vmatrix}=i\left(a_2b_3+a_2c_3-a_3b_2-a_3c_2\right)-j(...)+k(...)$$ Now try to rearrange the above terms to find the result. See that in the first term we have $i\left(a_2b_3+a_2c_3-a_3b_2-a_3c_2\right)=i(a_2b_3-a_3b_2)+i(a_2c_3-a_3c_2)$.

Answer 4

Thought I'd share my geometric proof. Even though it in essence doesn't differ much from the one linked to by @Srivatsa, I've tried to add some more clarity through descriptive diagrams.

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The geometric definition of the cross product of two vectors $\mathbf{A}$ and $\mathbf{B}$ is given by $$\mathbf{A} \times \mathbf{B} \overset{\operatorname{def}}{=} |\mathbf{A}||\mathbf{B}|\sin \theta \,\hat{\mathbf{N}},$$ where $\theta \in [0,\pi]$ is the angle between the two vectors when placed tail to tail, and $\hat{\mathbf{N}}$ is a unit vector whose orientation is determined by the right-hand rule. We are to prove that $$\mathbf{C} \times (\mathbf{A} + \mathbf{B}) = \mathbf{C} \times \mathbf{A} + \mathbf{C} \times \mathbf{B}.$$ To do this, first consider some arbitrary plane in $\mathbb{R}^3$ with unit normal vector $\hat{\mathbf{n}}$, as seen below.

We can always place the tails of two vectors $\mathbf{A}$ and $\mathbf{B}$ onto this plane as vectors (their magnitude and orientation) stay the same no matter where in space they reside. Now let’s establish two important facts.

Define $\mathbf{P}(\mathbf{A})$ as the orthogonal vector projection of $\mathbf{A}$ onto the plane. From the figure above and the right-hand rule we see that that $\hat{\mathbf{n}} \times \mathbf{A}$ and $\hat{\mathbf{n}} \times \mathbf{P}(\mathbf{A})$ have the same orientation. Moreover, they also have the same magnitude; $$ |\hat{\mathbf{n}} \times\mathbf{A}| = |\hat{\mathbf{n}}||\mathbf{A}|\sin \alpha = |\hat{\mathbf{n}}||\mathbf{P}(\mathbf{A})| \sin \frac{\pi}{2} = |\hat{\mathbf{n}} \times \mathbf{P}(\mathbf{A})|. $$ Thus, we have that $$ \hat{\mathbf{n}} \times \mathbf{A} = \hat{\mathbf{n}} \times \mathbf{P}(\mathbf{A}) \tag{1} $$ and a similar argument of course applies to $\mathbf{B}$ (or any other vector). Furthermore, from the figure above we can recognize that the projection operation is linear; $$ \mathbf{P}(\mathbf{A} + \mathbf{B}) = \mathbf{P}(\mathbf{A}) + \mathbf{P}(\mathbf{B}) \tag{2}. $$ These two properties allows us to work with projections rather than the vectors themselves, which simplifies things considerably.

In this last figure we can study the projections of all relevant vectors and see that $$ \hat{\mathbf{n}} \times \big(\mathbf{P}(\mathbf{A}) + \mathbf{P}(\mathbf{B})\big) = \hat{\mathbf{n}} \times \mathbf{P}(\mathbf{A}) + \hat{\mathbf{n}} \times \mathbf{P}(\mathbf{B}) $$ since taking the cross product with $\hat{\mathbf{n}}$ and a vector projection effectively rotates the projection $\pi/2$ radians counterclockwise about $\hat{\mathbf{n}}$ without changing its magnitude. We now use the linearity of projections $(2)$ to rewrite the left-hand side $$ \hat{\mathbf{n}} \times\mathbf{P}(\mathbf{A}+ \mathbf{B}) = \hat{\mathbf{n}} \times \mathbf{P}(\mathbf{A}) + \hat{\mathbf{n}} \times \mathbf{P}(\mathbf{B}) $$ and property $(1)$ gives us $$ \hat{\mathbf{n}} \times (\mathbf{A}+ \mathbf{B}) = \hat{\mathbf{n}} \times \mathbf{A} + \hat{\mathbf{n}} \times \mathbf{B}. $$ We can muliply both sides by a scalar $C$ and set $\mathbf{C} = C\hat{\mathbf{n}}$ (remember, the orientation of the plane and thereby that of $\hat{\mathbf{n}}$ and $\mathbf{C}$ is arbitrary) to finally arrive at $$ \mathbf{C} \times (\mathbf{A} + \mathbf{B}) = \mathbf{C} \times \mathbf{A} + \mathbf{C} \times \mathbf{B}, $$ Q.E.D.

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We can from here easily use the antisymmetric property of the cross product to also prove that $$ (\mathbf{A} + \mathbf{B}) \times \mathbf{C} = \mathbf{A} \times \mathbf{C} + \mathbf{B} \times \mathbf{C}. $$

Answer 5

Found a geometric proof in https://en.wikiversity.org/wiki/Cross_product. Refer to section: Equivalence of the two Definitions

Answer 6

Here's a geometric proof in the same vein as those by Srivatsa and Engelmark.

Notation: in the context of a cross product $\vec a\times\vec b$ we write $\vec a_\perp$ for the component of $\vec a$ perpendicular to $\vec b$.

From the definition of the cross product it is clear that $\vec a\times\vec b = \vec a_\perp\times\vec b$ (for example if $\vec a\times\vec b$ is defined with the area of a parallelogram, that area is base $\|\vec b\|$ times height $\|\vec a_\perp\|$ which discards the component of $\vec a$ parallel to $\vec b$). This means that we only need to prove

$$ (\vec a+\vec b)_{\perp}\times\vec c=\vec a_{\perp}\times\vec c+\vec b_{\perp}\times\vec c. $$

The perpendicular components $\vec a_\perp$, $\vec b_\perp$ and $(\vec a+\vec b)_\perp$ are obtained by projecting $\vec a$, $\vec b$ and $\vec a+\vec b$ onto a plane orthogonal to $\vec c$:

Since $\vec a_\perp$ is perpendicular to $\vec c$, the cross product $\vec a_\perp\times\vec c$ amounts to scaling of $\vec a_\perp$ by a factor $\|\vec c\|$, then rotating the result clockwise in the plane by an angle $\frac\pi2$. The same goes for $\vec b_\perp\times\vec c$ and $(\vec a+\vec b)_\perp\times\vec c$.

The vectors $\vec a_\perp$ and $\vec b_\perp$ build a parallelogram of which $(\vec a+\vec b)_\perp$ is a diagonal. This will still be the case after a scaling and rotation, therefore $(\vec a+\vec b)_{\perp}\times\vec c=\vec a_{\perp}\times\vec c+\vec b_{\perp}\times\vec c$.