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Scalar Triple Product

Level 1 - Math II (Physics) topic page in Vectors.

Principle

The scalar triple product combines three vectors in \(\mathbb R^3\) and outputs one scalar:

Scalar triple product
\[\mathbf a\cdot(\mathbf b\times\mathbf c)\]

The cross product \(\mathbf b\times\mathbf c\) is perpendicular to the base spanned by \(\mathbf b\) and \(\mathbf c\). Dotting that normal vector with \(\mathbf a\) measures how much \(\mathbf a\) points through the base. Orientation records whether the ordered triple follows the positive right-handed orientation or the opposite orientation. The result is an oriented volume: its magnitude is the physical volume, while its sign records orientation.

A parallelepiped is the three-dimensional solid formed by translating a parallelogram through a third edge direction. If its three edge vectors from one vertex are \(\mathbf a\), \(\mathbf b\), and \(\mathbf c\), then its physical volume is

Parallelepiped volume
\[V=|\mathbf a\cdot(\mathbf b\times\mathbf c)|\]

Notation

\((\mathbf a,\mathbf b,\mathbf c)\)
an ordered triple of vectors; the order is part of the data
\([\mathbf a,\mathbf b,\mathbf c]\)
bracket notation for the scalar triple product \mathbf a\cdot(\mathbf b\times\mathbf c)
\(\mathbf b\times\mathbf c\)
the cross product of \mathbf b and \mathbf c, perpendicular to their spanned base
\(\theta\)
the angle between \mathbf a and the vector \mathbf b\times\mathbf c
\(|\mathbf b\times\mathbf c|\)
the base area of the parallelogram spanned by \mathbf b and \mathbf c
\(|\mathbf a|\cos\theta\)
the signed height of \mathbf a in the direction of \mathbf b\times\mathbf c
\(V\)
physical volume, which is non-negative
\(\det(\mathbf a,\mathbf b,\mathbf c)\)
compact determinant notation for [\mathbf a,\mathbf b,\mathbf c] when the vectors are columns in the same Cartesian basis

An ordered triple \((\mathbf a,\mathbf b,\mathbf c)\) keeps track of first, second, and third vector positions. Changing the order can change the sign of the scalar triple product.

The base area is the area of the parallelogram spanned by two chosen base vectors. The height is the perpendicular distance from the opposite face to the base plane.

Method

Step 1: Compute the base normal

Use the base vectors \(\mathbf b\) and \(\mathbf c\). If \(\mathbf b=(b_1,b_2,b_3)\) and \(\mathbf c=(c_1,c_2,c_3)\), then

Base normal
\[\mathbf b\times\mathbf c=(b_2c_3-b_3c_2,\;b_3c_1-b_1c_3,\;b_1c_2-b_2c_1)\]

The magnitude \(|\mathbf b\times\mathbf c|\) is the base area, and the direction gives the oriented normal to the base.

Step 2: Dot with the remaining edge

If \(\mathbf a=(a_1,a_2,a_3)\) and \(\mathbf b\times\mathbf c=(n_1,n_2,n_3)\), then

Name the base normal
\[\mathbf n=\mathbf b\times\mathbf c=(n_1,n_2,n_3)\]
Dot with the first edge
\[\mathbf a\cdot(\mathbf b\times\mathbf c)=\mathbf a\cdot\mathbf n\]
Use the component dot product
\[\mathbf a\cdot\mathbf n=a_1n_1+a_2n_2+a_3n_3\]

This scalar is positive, negative, or zero depending on the orientation of \((\mathbf a,\mathbf b,\mathbf c)\).

Step 3: Take absolute value for physical volume

The oriented volume is \([\mathbf a,\mathbf b,\mathbf c]=\mathbf a\cdot(\mathbf b\times\mathbf c)\). The physical volume ignores orientation:

Physical volume
\[V=|[\mathbf a,\mathbf b,\mathbf c]|=|\mathbf a\cdot(\mathbf b\times\mathbf c)|\]

Rules

Bracket notation
\[[\mathbf a,\mathbf b,\mathbf c]=\mathbf a\cdot(\mathbf b\times\mathbf c)\]
Cyclic equality
\[[\mathbf a,\mathbf b,\mathbf c]=[\mathbf b,\mathbf c,\mathbf a]=[\mathbf c,\mathbf a,\mathbf b]\]
Swap reverses sign
\[[\mathbf b,\mathbf a,\mathbf c]=-[\mathbf a,\mathbf b,\mathbf c]\]
Parallelepiped volume
\[V=|\mathbf a\cdot(\mathbf b\times\mathbf c)|\]
Tetrahedron volume
\[V=\frac{|\mathbf a\cdot(\mathbf b\times\mathbf c)|}{6}\]

The tetrahedron formula applies when \(\mathbf a\), \(\mathbf b\), and \(\mathbf c\) are the three edge vectors sharing one vertex of the tetrahedron.

When Cartesian components are written as columns in the same basis, determinant notation is a compact optional form:

Determinant form
\[[\mathbf a,\mathbf b,\mathbf c]=\det\begin{pmatrix}a_1&b_1&c_1\\a_2&b_2&c_2\\a_3&b_3&c_3\end{pmatrix}\]

Volume from base area and height

Let \(\theta\) be the angle between \(\mathbf a\) and \(\mathbf b\times\mathbf c\). The base area is \(|\mathbf b\times\mathbf c|\). The signed height is the scalar projection of \(\mathbf a\) onto the oriented normal direction, so it is \(|\mathbf a|\cos\theta\).

Start with volume as base area times signed height
\[\text{oriented volume}=|\mathbf b\times\mathbf c|\,|\mathbf a|\cos\theta\]
Reorder scalar factors
\[\text{oriented volume}=|\mathbf a|\,|\mathbf b\times\mathbf c|\cos\theta\]
Use the geometric dot product
\[\mathbf a\cdot(\mathbf b\times\mathbf c)=|\mathbf a|\,|\mathbf b\times\mathbf c|\cos\theta\]
Identify the scalar triple product
\[\text{oriented volume}=\mathbf a\cdot(\mathbf b\times\mathbf c)\]
Remove orientation for physical volume
\[V=|\mathbf a\cdot(\mathbf b\times\mathbf c)|\]

Examples

Question
Find the volume of the parallelepiped with edge vectors
\[-3\mathbf i+2\mathbf j+2\mathbf k\]
\[\mathbf i+2\mathbf j+\mathbf k\]
and
\[-\mathbf i+\mathbf j+3\mathbf k\]
Answer
Let
\[\mathbf a=(-3,2,2)\]
\[\mathbf b=(1,2,1)\]
and
\[\mathbf c=(-1,1,3)\]
First compute
\[\mathbf b\times\mathbf c\]
The first component is
\[b_2c_3-b_3c_2=2\cdot3-1\cdot1=6-1=5\]
The second component is
\[b_3c_1-b_1c_3=1\cdot(-1)-1\cdot3=-1-3=-4\]
The third component is
\[b_1c_2-b_2c_1=1\cdot1-2\cdot(-1)=1+2=3\]
Thus
\[\mathbf b\times\mathbf c=(5,-4,3)\]
Now dot with
\[\mathbf a\]
\[\mathbf a\cdot(\mathbf b\times\mathbf c)=(-3)\cdot5+2\cdot(-4)+2\cdot3=-15-8+6=-17\]
The physical volume is
\[V=|-17|=17\]

Checks

  • The scalar triple product can be negative because it is an oriented volume.
  • Physical volume is non-negative: use \(|\mathbf a\cdot(\mathbf b\times\mathbf c)|\), not the signed value alone.
  • The order of vectors matters for orientation: swapping two vectors reverses the sign.

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