Square of Vector Cross Product/Proof 1

Theorem

Let $\mathbf a$ and $\mathbf b$ be vectors in a vector space $\mathbf V$ of $3$ dimensions.

Let $\mathbf a \times \mathbf b$ denote the vector cross product of $\mathbf a$ with $\mathbf b$.

Then:

$\paren {\mathbf a \times \mathbf b}^2 = \mathbf a^2 \mathbf b^2 - \paren {\mathbf a \cdot \mathbf b}^2$

where:

$\paren {\mathbf a \times \mathbf b}^2$ denotes the square of $\mathbf a \times \mathbf b$

$\mathbf a \cdot \mathbf b$ denotes the dot product of $\mathbf a \times \mathbf b$.

Proof

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

\(=\)

\(\ds \paren {\mathbf a \times \mathbf b} \cdot \paren {\mathbf a \times \mathbf b}\)

Definition of Square of Vector Quantity

\(\ds \)

\(=\)

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

Dot Product of Vector Cross Products

\(\ds \)

\(=\)

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

Dot Product Operator is Commutative

\(\ds \)

\(=\)

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

Definition of Square of Vector Quantity

\(\ds \)

\(=\)

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

Definition of Square Function

Sources

• 1957: D.E. Rutherford: Vector Methods (9th ed.) ... (previous) ... (next): Chapter $\text I$: Vector Algebra: $\S 6$: $(14)$