Dot Product of Vector Cross Products/Proof 1

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Theorem

Let $\mathbf a, \mathbf b, \mathbf c, \mathbf d$ 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 $\mathbf b$.

Let $\mathbf a \cdot \mathbf b$ denote the dot product of $\mathbf a$ with $\mathbf b$.


Then:

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


Proof

\(\ds \paren {\mathbf a \times \mathbf b} \cdot \paren {\mathbf c \times \mathbf d}\) \(=\) \(\ds \sqbrk {\mathbf a, \mathbf b, \mathbf c \times \mathbf d}\) Definition of Scalar Triple Product
\(\ds \) \(=\) \(\ds \sqbrk {\mathbf b, \mathbf c \times \mathbf d, \mathbf a}\) Equivalent Expressions for Scalar Triple Product
\(\ds \) \(=\) \(\ds \paren {\mathbf b \times \paren {\mathbf c \times \mathbf d} } \cdot \mathbf a\) Definition of Scalar Triple Product
\(\ds \) \(=\) \(\ds \paren {\paren {\mathbf b \cdot \mathbf d} \mathbf c - \paren {\mathbf b \cdot \mathbf c} \mathbf d} \cdot \mathbf a\) Lagrange's Formula on $\mathbf b \times \paren {\mathbf c \times \mathbf d}$
\(\ds \) \(=\) \(\ds \paren {\mathbf b \cdot \mathbf d} \paren {\mathbf c \cdot \mathbf a} - \paren {\mathbf b \cdot \mathbf c} \paren {\mathbf d \cdot \mathbf a}\) Dot Product Distributes over Addition
\(\ds \) \(=\) \(\ds \paren {\mathbf a \cdot \mathbf c} \paren {\mathbf b \cdot \mathbf d} - \paren {\mathbf a \cdot \mathbf d} \paren {\mathbf b \cdot \mathbf c}\) Dot Product Operator is Commutative

Hence the result.

$\blacksquare$


Sources

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