Square of Vector Cross Product/Proof 1
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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: $\S 6$: $(14)$