This section is divided into a number of subsections, links to which are:

Vector products

Tensor products

Cross products

Wedge products

Rotors

Triple Scalar Product

The triple scalar product of three noncoplanar vectors a = (𝑎₁, 𝑎₂, 𝑎₃), b = (b₁, b₂, b₃), and c = (c₁, c₂, c₃) written in Cartesian coordinates is \begin{equation} \label{EqTriple.1} {\bf a} \bullet {\bf b} \times {\bf c} = \det \begin{bmatrix} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{bmatrix} \end{equation}

The scalar triple product of three vectors is the volume of parallelepiped spand on this three vectors:

\[ V = {\bf a} \bullet \left( {\bf b} \times {\bf c} \right) = h \left\vert {\bf b} \times {\bf c} \right\vert , \]
where h is the projection of a onto the direction of b × c.
a = Graphics[{Red, Thickness[0.01], Arrowheads[0.1], Arrow[{{0, 0}, {0.5, 1}}]}]; b = Graphics[{Blue, Thickness[0.01], Arrowheads[0.1], Arrow[{{0, 0}, {1, 0}}]}]; c = Graphics[{Blue, Thickness[0.01], Arrowheads[0.1], Arrow[{{0, 0}, {1.5, 0.5}}]}]; bc = Graphics[{Purple, Thickness[0.01], Arrowheads[0.1], Arrow[{{0, 0}, {0, 1.5}}]}]; line = Plot[1, {x, 0, 1.5}, PlotStyle -> {Thickness[0.01], Dashed, Orange}]\ l1 = Graphics[{Thickness[0.01], Dashed, Orange, Line[{{1, 0}, {1.5, 1}, {2.5, 1.5}, {2, 0.5}, {1, 0}}]}]; l2 = Graphics[{Thickness[0.01], Dashed, Orange, Line[{{1.5, 0.5}, {2, 0.5}}]}]; l3 = Graphics[{Thickness[0.01], Dashed, Orange, Line[{{0.5, 1}, {1.5, 1.5}, {2.5, 1.5}}]}]; th = Graphics[ Text[Style["h", Bold, Black, FontSize -> 24], {-0.15, 1}]]; ta = Graphics[ Text[Style["a", Bold, Black, FontSize -> 24], {0.65, 0.9}]]; tb = Graphics[ Text[Style["b", Bold, Black, FontSize -> 24], {1.2, 0}]]; tc = Graphics[ Text[Style["c", Bold, Black, FontSize -> 24], {1.55, 0.6}]]; tab = Graphics[ Text[Style["a x b", Bold, Black, FontSize -> 24], {0.3, 1.4}]]; Show[bc, a, b, c, line, l1, l2, l3, th, ta, tb, tc, tab]
The scalar product of three vectors
It follows from Eq.(1) and a familiar property of determinants that the scalar triple product is invariant under cyclic permutations of the vectors a, b and c:
\[ {\bf a} \bullet \left( {\bf b} \times {\bf c} \right) = {\bf b} \bullet \left( {\bf c} \times {\bf a} \right) = {\bf c} \bullet \left( {\bf a} \times {\bf b} \right) . \]
Hence, cyclicly permuting the vectors doesn’t change the value of the triple scalar product, which we denote as
\begin{equation} \label{EqTriple.2} \begin{bmatrix} {\bf a} & {\bf b} & {\bf c} \end{bmatrix} = {\bf a} \bullet \left( {\bf b} \times {\bf c} \right) . \end{equation}
Moreover, the scalar triple product vanishes if two of the vectors A, B and C are identical (or parallel) :
\[ {\bf a} \bullet \left( {\bf b} \times {\bf a} \right) = {\bf b} \bullet \left( {\bf a} \times {\bf b} \right) = {\bf } \bullet \left( {\bf a} \times {\bf a} \right) = 0 . \]
Three vectors a, b, and c are coplanar (linearly dependent) if and only if they span a parallelepiped of zero volume, i.e. , if and only if their scalar triple product vanishes. It follows that three vectors a, b, and c form a basis if and only if a• (b × c) ≠ 0. The basis is said to be right-handed if a• (b × c) > 0 and left-handed if a• (b × c) < 0.
Example 1: Let us consider three vectors (that we write as column vectors): \[ {\bf a} = \left( 1, 2, 3 \right) , \quad {\bf b} = \left( -1 2, -4 \right) , \quad {\bf c} = \left( 3, 2, 1 \right) . \] First, we find cross product of vectors b and c: \[ {\bf b} \times {\bf c} = \left( \right). \]
Then the triple scalar product of these three vectors becomes
End of Example 1
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Now we use the triple scalar product to define the reciprocal basis vectors. Suppose we are given non-Euclidean basis vectors

\[ {\bf e}_1 , \qquad {\bf e}_2 , \qquad {\bf e}_3 , \]
where e₁, e₂, e₃ are not necessarily of unit length. The corresponding basis system is sometimes called “covariant”, or “lowered”. The reciprocal (or dual or contracovariant) basis vectors are defined as
\begin{equation} \label{EqTriple.3} {\bf e}^1 = \frac{{\bf e}_2 \times {\bf e}_3}{\left[ {\bf e}_1\,{\bf e}_2\,{\bf e}_3 \right]} , \quad {\bf e}^2 = \frac{{\bf e}_3 \times {\bf e}_1}{\left[ {\bf e}_1\,{\bf e}_2\,{\bf e}_3 \right]} , \quad {\bf e}^3 = \frac{{\bf e}_1 \times {\bf e}_2}{\left[ {\bf e}_1\,{\bf e}_2\,{\bf e}_3 \right]} . \end{equation}
This reciprocal (raised) basis is orthonormal to the covariant (lowered) basis. We can easily verify this by dualing (which is actually dotting) one of the lowered basis vectors into its corresponding reciprocal. We find that
\[ \left\langle {\bf e}^3 , {\bf e}_2 \right\rangle = {\bf e}^3 \bullet {\bf e}_2 = \frac{{\bf e}_1 \times {\bf e}_2}{\left[ {\bf e}_1\,{\bf e}_2\,{\bf e}_3 \right]} \bullet {\bf e}_2 = \frac{\left[ {\bf e}_2\,{\bf e}_1\,{\bf e}_2 \right]}{\left[ {\bf e}_1\,{\bf e}_2\,{\bf e}_3 \right]} = 0 \]
because of the corresponding property above showing that the triple scalar product of parallel vectors is 0.

Using this covariant basis, we can express any real numerical vector either as dual vector (so it becomes bra-vector) or as regular, covariant or ket-vector. We summarize properties of scalar triple products (which is denoted as [x y z]) in the following table.

[u v w] = [v w u] = [w u v] = −[v u w]
[u u w] = [v u v] = 0
[u v w]² = [(u × v) (v × w) (w × u)]
u [v w x] − v [w x u] + w [x u v] − x [u v w] = 0
(u × v) × (w × x) = v [u w x] − u [v w x]
[(u × v) (w × x) (y × z)] = [v y z] [u w x] − [u w z] [v w x]
[(u + v) (v + w) (w + u)] = 2 [u v w]
\( \displaystyle \quad \begin{align*} \left[ ({\bf u} - {\bf x})\ ({\bf v} - {\bf x})\ ({\bf w} - {\bf x}) \right] &= \left[ {\bf u}\ {\bf v}\ {\bf w} \right] - \left[ {\bf u}\ {\bf v}\ {\bf x} \right] - \left[ {\bf u}\ {\bf x}\ {\bf w} \right] - \left[ {\bf x}\ {\bf v}\ {\bf w} \right] \\ &= \left[ ({\bf u} - {\bf x})\ {\bf v} \ {\bf w} \right] - \left[ ({\bf v} - {\bf w}) \ {\bf x} \ {\bf u} \right] \end{align*} \)
\( \displaystyle \quad \left[ {\bf u}\ {\bf v} \ {\bf w} \right] \ \left[ {\bf x}\ {\bf y} \ {\bf z} \right] =\begin{vmatrix} {\bf u} \bullet {\bf x} & {\bf u} \bullet {\bf y} & {\bf u} \bullet {\bf z} \\ {\bf v} \bullet {\bf x} & {\bf v} \bullet {\bf y} & {\bf v} \bullet {\bf z} \\ {\bf w} \bullet {\bf x} & {\bf w} \bullet {\bf y} & {\bf w} \bullet {\bf z} \\ \end{vmatrix}\)

Example 2:
End of Example 2
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Triple Vector Product

By the vector triple product of three vectors a, b, and c we mean the vector a × (b × c). Generally speaking, (a × b) × c ≠ a × (b × c). Clearly, a × (b × c) is perpendicular to a and lies in the plane of b and c.

Suppose the vectors a, b, and c are noncollinear [otherwise a × (b × c)vanishes trivially]. Then a × (b × c) has a unique expansion of the form

\begin{equation} \label{EqTriple.4} {\bf a} \times \left( {\bf b} \times {\bf c} \right) = \alpha\, {\bf b} + \beta\,{\bf c} \end{equation}
for some scalars α and β ∈ ℝ. To determine the scalars α and β, we introduce the vectors
\[ {\bf d} = {\bf b} \times {\bf c} = \left( d_1 , d_2 , d_3 \right) , \quad {\bf e} = {\bf a} \times \left( {\bf b} \times {\bf c} \right) = \left( e_1 , e_2 , e_3 \right) . \]
Using definition of the cross product, we derive
\begin{align*} e_1 &= a_2 d_3 - a_3 d_2 , \\ d_2 &= b_3 c_1 - b_1 c_3 , \\ d_3 &= b_1 c_2 - b_2 c_1 . \end{align*}
It follows that
\begin{align*} e_1 &= a_2 \left( b_1 c_2 - b_2 c_1 \right) - a_3 \left( b_3 c_1 b_1 c_3 \right) \end{align*}
Adding and subtracting 𝑎₁b₁c₁ from the right-hand side and using definition of the dot product, we find that
\begin{align*} e_1 &= b_1 \left( a_1 c_1 + a_2 c_2 + a_3 c_3 \right) - c_1 \left( a_1 b_1 + a_2 b_2 + a_3 b_3 \right) \\ &= b_1 \left( {\bf a} \bullet {\bf c} \right) - c_1 \left( {\bf a} \bullet {\bf c} \right) . \end{align*}
In just the same way, it turns out that
\begin{align*} e_2 &= b_2 \left( {\bf a} \bullet {\bf c} \right) - c_2 \left( {\bf a} \bullet {\bf c} \right) , \\ e_3 &= b_3 \left( {\bf a} \bullet {\bf c} \right) - c_3 \left( {\bf a} \bullet {\bf c} \right) . \end{align*}
Therefore the coefficients α and β in Eq.\eqref{EqTriple.4} are
\[ \alpha = {\bf a} \bullet {\bf c} , \quad \beta = {\bf a} \bullet {\bf b} . \]
Finally, we get
\begin{equation} \label{EqTriple.5} {\bf a} \times \left( {\bf b} \times {\bf c} \right) = {\bf b} \left( {\bf a} \bullet {\bf c} \right) - {\bf c} \left( {\bf a} \bullet {\bf b} \right) . \end{equation}
In a similar way, it can be shown that
\begin{equation} \label{EqTriple.6} \left( {\bf a} \times {\bf b} \right) \times {\bf c} = {\bf b} \left( {\bf a} \bullet {\bf c} \right) - {\bf a} \left( {\bf b} \bullet {\bf c} \right) . \end{equation}
Example 3: Let us consider three vectors \[ {\bf a} = \left( 1, 2, 3 \right) , \quad {\bf b} = \left( 2, -1, 1 \right) , \quad {\bf c} = \left( 3, 2, 1 \right) . \] Calculations show that
Cross[{1, 2, 3}, {2, -1, 1}]
{5, 5, -5}
\[ {\bf a} \times {\bf b} = \left( 5, 5, -5 \right) . \]
Cross[{2, -1, 1}, {3, 2, 1}]
{-3, 1, 7}
\[ {\bf b} \times {\bf c} = \left( -3, 1, 7 \right) . \]
Cross[Cross[{1, 2, 3}, {2, -1, 1}], {3, 2, 1}]
{15, -20, -5}
Cross[{1, 2, 3}, Cross[{2, -1, 1}, {3, 2, 1}]]
{11, -16, 7}
So we see that for these three vectors, we have \[ {\bf a} \times \left( {\bf b} \times {\bf c} \right) = \left( 11, -16, 7 \right) \ne \left( 15, -20, -5 \right) = \left( {\bf a} \times {\bf b} \right) \times {\bf c} . \] Now we verify formulas (3) and (4). First, we calculate dot products: \begin{align*} {\bf a} \bullet {\bf b} &= 3 , \\ {\bf a} \bullet {\bf c} &= 10, \\ {\bf b} \bullet {\bf c} &= 5 . \end{align*}
{1, 2, 3} . {2, -1, 1}
3
{1, 2, 3} . {3, 2, 1}
10
and
{3, 2, 1} . {2, -1, 1}
5
Then we calculate \[ {\bf a} \times \left( {\bf b} \times {\bf c} \right) = {\bf b} \left( {\bf a} \bullet {\bf c} \right) - {\bf c} \left( {\bf a} \bullet {\bf b} \right) = \left( 11, -16, 7 \right) . \]
10*{2, -1, 1} - 3*{3, 2, 1}
{11, -16, 7}
and \[ \left( {\bf a} \times {\bf b} \right) \times {\bf c} = {\bf b} \left( {\bf a} \bullet {\bf c} \right) - {\bf a} \left( {\bf b} \bullet {\bf c} \right) = \left( 15, -20, -5 \right) . \]
10*{2, -1, 1} - 5*{1, 2, 3}
{15, -20, -5}
End of Example 3
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  1. Mecholsky, N., Primer on the Basics of Tensor Analysis and the Laplacian in Generalized Coordinates, 2004.