Dot Product of Vector Cross Products

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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 1

\(\ds \paren {\mathbf a \times \mathbf b} \cdot \paren {\mathbf c \times \mathbf d}\) \(=\) \(\ds \sqbrk {\mathbf a, \mathbf b, \mathbf c \times \mathbf d}\) 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.

$\blacksquare$


Proof 2

Let $\mathbf a, \mathbf b, \mathbf c, \mathbf d$ be embedded in a Cartesian space:

\(\ds \mathbf a\) \(=\) \(\ds a_1 \mathbf e_1 + a_2 \mathbf e_2 + a_3 \mathbf e_3\)
\(\ds \mathbf b\) \(=\) \(\ds b_1 \mathbf e_1 + b_2 \mathbf e_2 + b_3 \mathbf e_3\)
\(\ds \mathbf c\) \(=\) \(\ds c_1 \mathbf e_1 + c_2 \mathbf e_2 + c_3 \mathbf e_3\)
\(\ds \mathbf d\) \(=\) \(\ds d_1 \mathbf e_1 + d_2 \mathbf e_2 + d_3 \mathbf e_3\)

where $\tuple {\mathbf e_1, \mathbf e_2, \mathbf e_3}$ denotes the standard ordered basis of $\mathbf V$.


Then:

\(\ds \paren {\mathbf a \cdot \mathbf c} \paren {\mathbf b \cdot \mathbf d}\) \(=\) \(\ds \paren {\sum_{i \mathop = 1}^3 a_i c_i} \paren {\sum_{j \mathop = 1}^3 b_j d_j}\) Definition of Dot Product
\(\ds \) \(=\) \(\ds \paren {\sum_{i \mathop = 1}^3 a_i d_i} \paren {\sum_{j \mathop = 1}^3 b_j c_j} + \sum_{1 \mathop \le i \mathop < j \mathop \le 3} \paren {a_i b_j - a_j b_i} \paren {c_i d_j - c_j d_i}\) Binet-Cauchy Identity
\(\ds \) \(=\) \(\ds \paren {\mathbf a \cdot \mathbf d} \paren {\mathbf b \cdot \mathbf c} + \paren {a_1 b_2 - a_2 b_1} \paren {c_1 d_2 - c_2 d_1}\) Definition of Dot Product and expanding right hand side
\(\ds \) \(\) \(\, \ds + \, \) \(\ds \paren {a_1 b_3 - a_3 b_1} \paren {c_1 d_3 - c_3 d_1}\)
\(\ds \) \(\) \(\, \ds + \, \) \(\ds \paren {a_2 b_3 - a_3 b_2} \paren {c_2 d_3 - c_3 d_2}\)
\(\ds \) \(=\) \(\ds \paren {\mathbf a \cdot \mathbf d} \paren {\mathbf b \cdot \mathbf c} + \paren {a_1 b_2 - a_2 b_1} \paren {c_1 d_2 - c_2 d_1}\)
\(\ds \) \(\) \(\, \ds + \, \) \(\ds \paren {-\paren {a_1 b_3 - a_3 b_1} } \paren {-\paren {c_1 d_3 - c_3 d_1} }\) two sign changes which cancel each other out
\(\ds \) \(\) \(\, \ds + \, \) \(\ds \paren {a_2 b_3 - a_3 b_2} \paren {c_2 d_3 - c_3 d_2}\)
\(\ds \) \(=\) \(\ds \paren {\mathbf a \cdot \mathbf d} \paren {\mathbf b \cdot \mathbf c} + \paren {a_1 b_2 - a_2 b_1} \paren {c_1 d_2 - c_2 d_1}\)
\(\ds \) \(\) \(\, \ds + \, \) \(\ds \paren {a_3 b_1 - a_1 b_3} \paren {c_3 d_1 - c_1 d_3}\) rearranging
\(\ds \) \(\) \(\, \ds + \, \) \(\ds \paren {a_2 b_3 - a_3 b_2} \paren {c_2 d_3 - c_3 d_2}\)
\(\ds \) \(=\) \(\ds \paren {\mathbf a \cdot \mathbf d} \paren {\mathbf b \cdot \mathbf c} + \paren {\mathbf a \times \mathbf b} \cdot \paren {\mathbf c \times \mathbf d}\) Definition of Vector Cross Product

Hence the result.

$\blacksquare$


Sources

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