Equivalence of Definitions of Scalar Triple Product

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Theorem

The following definitions of the concept of Scalar Triple Product are equivalent:


Let $\mathbf a$, $\mathbf b$ and $\mathbf c$ be vectors in a Cartesian $3$-space:

\(\ds \mathbf a\) \(=\) \(\ds a_i \mathbf i + a_j \mathbf j + a_k \mathbf k\)
\(\ds \mathbf b\) \(=\) \(\ds b_i \mathbf i + b_j \mathbf j + b_k \mathbf k\)
\(\ds \mathbf c\) \(=\) \(\ds c_i \mathbf i + c_j \mathbf j + c_k \mathbf k\)

where $\tuple {\mathbf i, \mathbf j, \mathbf k}$ is the standard ordered basis.

Definition 1

The scalar triple product of $\mathbf a$, $\mathbf b$ and $\mathbf c$ is defined and denoted as:

$\sqbrk {\mathbf a, \mathbf b, \mathbf c} := \mathbf a \cdot \paren {\mathbf b \times \mathbf c}$

where:

$\cdot$ denotes dot product
$\times$ denotes vector cross product.

Definition 2

The scalar triple product of $\mathbf a$, $\mathbf b$ and $\mathbf c$ is defined and denoted as:

$\sqbrk {\mathbf a, \mathbf b, \mathbf c} := \begin {vmatrix}

a_i & a_j & a_k \\ b_i & b_j & b_k \\ c_i & c_j & c_k \\ \end {vmatrix}$

where $\begin {vmatrix} \ldots \end {vmatrix}$ is interpreted as a determinant.


Proof

\(\ds \mathbf a \cdot \paren {\mathbf b \times \mathbf c}\) \(=\) \(\ds \mathbf a \cdot \paren {\paren {b_j c_k - c_j b_k} \mathbf i + \paren {b_k c_i - c_k b_i} \mathbf j + \paren {b_i c_j - c_i b_j} \mathbf k}\) Definition of Vector Cross Product
\(\ds \) \(=\) \(\ds a_i \paren {b_j c_k - c_j b_k} + a_j \paren {b_k c_i - c_k b_i} + a_k \paren {b_i c_j - c_i b_j}\) Definition of Dot Product


Then:

\(\ds \begin {vmatrix} a_i & a_j & a_k \\ b_i & b_j & b_k \\ c_i & c_j & c_k \end {vmatrix}\) \(=\) \(\ds a_i b_j c_k - a_i b_k c_j - a_j b_i c_k + a_j b_k c_i + a_k b_i c_j - a_k b_j c_i\) Determinant of Order 3
\(\ds \) \(=\) \(\ds a_i \paren {b_j c_k - c_j b_k} + a_j \paren {b_k c_i - c_k b_i} + a_k \paren {b_i c_j - c_i b_j}\) extracting factors

Hence the result.

$\blacksquare$


Sources

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