Vector Cross Product Distributes over Addition

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Theorem

The vector cross product is distributive over addition.

That is, in general:

$\mathbf a \times \paren {\mathbf b + \mathbf c} = \paren {\mathbf a \times \mathbf b} + \paren {\mathbf a \times \mathbf c}$

for $\mathbf a, \mathbf b, \mathbf c \in \R^3$.


Proof 1

Let:

$\mathbf a = \begin {bmatrix} a_x \\ a_y \\a_z \end {bmatrix}$, $\mathbf b = \begin {bmatrix} b_x \\ b_y \\ b_z \end {bmatrix}$, $\mathbf c = \begin {bmatrix} c_x \\ c_y \\ c_z \end {bmatrix}$

be vectors in $\R^3$.


Then:

\(\ds \mathbf a \times \paren {\mathbf b + \mathbf c}\) \(=\) \(\ds \begin {bmatrix} a_x \\ a_y \\a_z \end {bmatrix} \times \paren {\begin {bmatrix} b_x \\ b_y \\ b_z \end {bmatrix} + \begin {bmatrix} c_x \\ c_y \\ c_z \end {bmatrix} }\)
\(\ds \) \(=\) \(\ds \begin {bmatrix} a_x \\ a_y \\a_z \end {bmatrix} \times {\begin {bmatrix} b_x + c_x \\ b_y + c_y \\ b_z + c_z \end {bmatrix} }\) Definition of Vector Sum
\(\ds \) \(=\) \(\ds \begin {bmatrix} a_y \paren {b_z + c_z} - a_z \paren {b_y + c_y} \\ a_z \paren {b_x + c_x} - a_x \paren {b_z + c_z} \\ a_x \paren {b_y + c_y} - a_y \paren {b_x + c_x} \end {bmatrix}\) Definition of Vector Cross Product
\(\ds \) \(=\) \(\ds \begin {bmatrix} a_y b_z + a_y c_z - a_z b_y - a_z c_y \\ a_z b_x + a_z c_x - a_x b_z - a_x c_z \\ a_x b_y + a_x c_y - a_y b_x - a_y c_x \end {bmatrix}\) Real Multiplication Distributes over Addition
\(\ds \) \(=\) \(\ds \begin {bmatrix} a_y b_z - a_z b_y + a_y c_z - a_z c_y \\ a_z b_x - a_x b_z + a_z c_x - a_x c_z \\ a_x b_y - a_y b_x + a_x c_y - a_y c_x \end {bmatrix}\) Real Addition is Commutative
\(\ds \) \(=\) \(\ds \begin {bmatrix} a_y b_z - a_z b_y \\ a_z b_x - a_x b_z \\ a_x b_y - a_y b_x \end {bmatrix} + \begin {bmatrix} a_y c_z - a_z c_y \\ a_z c_x - a_x c_z \\ a_x c_y - a_y c_x \end {bmatrix}\) Definition of Vector Sum
\(\ds \) \(=\) \(\ds \paren {\begin {bmatrix}a_x \\ a_y \\ a_z \end {bmatrix} \times \begin {bmatrix} b_x \\ b_y \\ b_z \end {bmatrix} } + \paren {\begin {bmatrix} a_x \\ a_y \\ a_z \end {bmatrix} \times \begin {bmatrix} c_x \\ c_y \\ c_z \end {bmatrix} }\) Definition of Vector Cross Product
\(\ds \) \(=\) \(\ds \paren {\mathbf a \times \mathbf b} + \paren {\mathbf a \times \mathbf c}\)

$\blacksquare$


Proof 2

We draw a triangular prism whose parallel edges are in the direction of $\mathbf a$ and with its end faces as triangles with sides $\mathbf b$, $\mathbf c$ and $\mathbf b + \mathbf c$.

Cross-product-distributes-over-addition.png

From Magnitude of Vector Cross Product equals Area of Parallelogram Contained by Vectors, the vector areas of these triangular end faces are $\dfrac {\mathbf b \times \mathbf c} 2$ and $\dfrac {\mathbf c \times \mathbf b} 2$.

The remaining vector areas are $\mathbf b \times \mathbf a$, $\mathbf c \times \mathbf a$ and $\mathbf a \times \paren {\mathbf b + \mathbf c}$.

From Total Vector Area of Polyhedron is Zero:

$\paren {\mathbf b \times \mathbf a} + \paren {\mathbf c \times \mathbf a} + \paren {\mathbf a \times \paren {\mathbf b + \mathbf c} } + \dfrac {\mathbf b \times \mathbf c} 2 + \dfrac {\mathbf c \times \mathbf b} 2 = 0$

from which we get:

$\paren {-\mathbf b \times \mathbf a} + \paren {-\mathbf c \times \mathbf a} = \paren {\mathbf a \times \paren {\mathbf b + \mathbf c} }$

The result follows from Vector Cross Product is Anticommutative:

$\mathbf a \times \mathbf b + \mathbf a \times \mathbf c = \mathbf a \times \paren {\mathbf b + \mathbf c}$

$\blacksquare$


Proof 3

Let $\mathbf b'$ and $\mathbf c'$ be the projections of $\mathbf b$ and $\mathbf c$ onto the plane perpendicular to $\mathbf a$.

Then $\mathbf b' + \mathbf c'$ is the projection of $\mathbf b + \mathbf c$ onto that plane.

We have:

\(\ds \mathbf a \times \mathbf b'\) \(=\) \(\ds \mathbf a \times \mathbf b\)
\(\ds \mathbf a \times \mathbf c'\) \(=\) \(\ds \mathbf a \times \mathbf c\)
\(\ds \mathbf a \times \paren {\mathbf b' + \mathbf c'}\) \(=\) \(\ds \mathbf a \times \paren {\mathbf b + \mathbf c}\)

Because:

$\mathbf a$ is perpendicular to $\mathbf b'$ and $\mathbf c'$
$\mathbf a \times \mathbf b'$ lies in the plane perpendicular to $\mathbf a$

it follows that:

$\mathbf a \times \mathbf b'$ is $\norm {\mathbf a}$ times the length of $\mathbf b'$ and is perpendicular to $\mathbf b'$.

Similarly:

$\mathbf a \times \mathbf c'$ is $\norm {\mathbf a}$ times the length of $\mathbf c'$ and is perpendicular to $\mathbf c'$.

Hence:

$\mathbf a \times \mathbf b' + \mathbf a \times \mathbf c'$ also lies in the plane perpendicular to $\mathbf a$.
$\mathbf a \times \mathbf b' + \mathbf a \times \mathbf c'$ is $\norm {\mathbf a}$ times the length of $\mathbf b' + \mathbf c'$ and is perpendicular to $\mathbf b' + \mathbf c'$.

In other words:

$\mathbf a \times \mathbf b + \mathbf a \times \mathbf c = \mathbf a \times \paren {\mathbf b + \mathbf c}$

$\blacksquare$


Sources

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