Norm of Vector Cross Product

Theorem

Let $\mathbf a$ and $\mathbf b$ be vectors in the real Euclidean space $\R^3$.

Let $\times$ denote the vector cross product.

Then:

$\norm {\mathbf a \times \mathbf b} = \norm {\mathbf a} \norm {\mathbf b} \size {\sin \theta}$

where $\theta$ is the angle between $\mathbf a$ and $\mathbf b$, or an arbitrary number if $\mathbf a$ or $\mathbf b$ is the zero vector.

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Proof

Suppose either $\mathbf a$ or $\mathbf b$ is the zero vector.

Then by Norm Axiom $\text N 1$: Positive Definiteness:

$\norm {\mathbf a} = 0$

or:

$\norm {\mathbf b} = 0$

By calculation, it follows that $\mathbf a \times \mathbf b$ is also the zero vector.

Hence:

$\norm {\mathbf a \times \mathbf b} = 0$

and equality holds.

Now suppose that both $\mathbf a$ or $\mathbf b$ are non-zero vectors.

We have:

\(\ds \norm {\mathbf a \times \mathbf b}^2\)

\(=\)

\(\ds \norm {\mathbf a}^2 \norm {\mathbf b}^2 - \paren {\mathbf a \cdot \mathbf b}^2\)

Square of Norm of Vector Cross Product

\(\ds \)

\(=\)

\(\ds \norm {\mathbf a}^2 \norm {\mathbf b}^2 - \norm {\mathbf a}^2 \norm {\mathbf b}^2 \cos^2 \theta\)

Cosine Formula for Dot Product

\(\ds \)

\(=\)

\(\ds \norm {\mathbf a}^2 \norm {\mathbf b}^2 \paren{1 - \cos^2 \theta}\)

\(\ds \)

\(=\)

\(\ds \norm {\mathbf a}^2 \norm {\mathbf b}^2 \sin^2 \theta\)

Sum of Squares of Sine and Cosine

\(\ds \leadstoandfrom \ \ \)

\(\ds \norm {\mathbf a \times \mathbf b}\)

\(=\)

\(\ds \norm {\mathbf a} \norm {\mathbf b} \size{\sin \theta}\)

taking the square root of both sides of the equality

$\blacksquare$

Sources

• 1994: Robert Messer: Linear Algebra: Gateway to Mathematics: $\S 7.5$