Vector Cross Product Distributes over Addition/Proof 3

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

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

• 1957: D.E. Rutherford: Vector Methods (9th ed.) ... (previous) ... (next): Chapter $\text I$: Vector Algebra: $\S 3$