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Section 1.3 Wedge and Cross Product

Definition 1.3.1. Wedge Product.

The wedge product, denoted by \(\wedge\text{,}\) is an associative multiplication of vectors that is linear in each factor and satisfies the alternating property
\begin{equation*} \vu \wedge \vv = -\vv \wedge \vu. \end{equation*}
In particular, \(\vu \wedge \vu = 0\) for every vector \(\vu\text{.}\)
In \(\R^3\) the wedge product of two vectors records an oriented area. Reversing the order reverses the orientation, which changes the sign.
A parallelogram spanned by u and v with boundary orientation matching u then v.
(a) \(\vu \wedge \vv\)
The same parallelogram with the opposite boundary orientation.
(b) \(\vv \wedge \vu = -\vu \wedge \vv\)
Figure 1.3.2. Swapping the order reverses the orientation of the same parallelogram.
Write \(\vu = u_1 \mathbf{e}_1 + u_2 \mathbf{e}_2 + u_3 \mathbf{e}_3\) and \(\vv = v_1 \mathbf{e}_1 + v_2 \mathbf{e}_2 + v_3 \mathbf{e}_3\text{.}\) Using linearity and the alternating property,
\begin{align*} \vu \wedge \vv \amp = (u_1 \mathbf{e}_1 + u_2 \mathbf{e}_2 + u_3 \mathbf{e}_3) \wedge (v_1 \mathbf{e}_1 + v_2 \mathbf{e}_2 + v_3 \mathbf{e}_3)\\ \amp = (u_2 v_3 - u_3 v_2) (\mathbf{e}_2 \wedge \mathbf{e}_3) + (u_3 v_1 - u_1 v_3) (\mathbf{e}_3 \wedge \mathbf{e}_1)\\ \amp\phantom{{}={}} + (u_1 v_2 - u_2 v_1) (\mathbf{e}_1 \wedge \mathbf{e}_2). \end{align*}
We can identify the basic bivectors with the standard basis vectors of \(\R^3\) by
\begin{equation*} \mathbf{e}_2 \wedge \mathbf{e}_3 \leftrightarrow \mathbf{e}_1,\qquad \mathbf{e}_3 \wedge \mathbf{e}_1 \leftrightarrow \mathbf{e}_2,\qquad \mathbf{e}_1 \wedge \mathbf{e}_2 \leftrightarrow \mathbf{e}_3. \end{equation*}
With this identification, the wedge product of two vectors becomes the cross product.
The wedge product of three vectors captures signed volume. The coefficient of \(\mathbf{e}_1 \wedge \mathbf{e}_2 \wedge \mathbf{e}_3\) in \(\vu \wedge \vv \wedge \vw\) is the determinant of the matrix with rows (or columns) \(\vu, \vv, \vw\) (in that order). Thus,
\begin{equation*} \vu \wedge \vv \wedge \vw = \det(\vu,\vv,\vw)\,(\mathbf{e}_1 \wedge \mathbf{e}_2 \wedge \mathbf{e}_3). \end{equation*}
Likewise, for \(\vu\) and \(\vv\) in \(\R^2\text{,}\) the determinant of the matrix
\begin{equation*} \det\begin{bmatrix} u_1 \amp u_2 \\ v_1 \amp v_2 \end{bmatrix} = u_1v_2-u_2v_1 \end{equation*}
is the coefficient of \(\mathbf{e}_1\wedge\mathbf{e}_2\) in \(\vu\wedge\vv\text{.}\)
The explicit formulas for determinants of \(2\times 2\) and \(3\times 3\) matrices are related by expansion by the first row:
\begin{align*} \det\begin{bmatrix} a_{11} \amp a_{12} \amp a_{13} \\ a_{21} \amp a_{22} \amp a_{23} \\ a_{31} \amp a_{32} \amp a_{33} \end{bmatrix} \amp= a_{11}\det\begin{bmatrix}a_{22}\amp a_{23}\\a_{32}\amp a_{33}\end{bmatrix} -a_{12}\det\begin{bmatrix}a_{21}\amp a_{23}\\a_{31}\amp a_{33}\end{bmatrix}\\ \amp\phantom{{}={}} +a_{13}\det\begin{bmatrix}a_{21}\amp a_{22}\\a_{31}\amp a_{32}\end{bmatrix}. \end{align*}
Substituting the \(2\times 2\) formula gives the explicit \(3\times 3\) formula
\begin{align*} \det(A)\amp=a_{11}(a_{22}a_{33}-a_{23}a_{32})\\ \amp\phantom{{}={}}-a_{12}(a_{21}a_{33}-a_{23}a_{31})\\ \amp\phantom{{}={}}+a_{13}(a_{21}a_{32}-a_{22}a_{31}). \end{align*}
More generally, the determinant of an \(n\times n\) matrix \(A\) can be defined recursively. If \(A_{1j}\) is the \((n-1)\times(n-1)\) matrix obtained by deleting row \(1\) and column \(j\) from \(A\text{,}\) then expansion by the first row gives
\begin{equation*} \det(A)=\sum_{j=1}^n(-1)^{1+j}a_{1j}\det(A_{1j}), \end{equation*}
beginning with \(\det[a_{11}]=a_{11}\) for a \(1\times 1\) matrix.

Definition 1.3.3. Right-Hand System.

An ordered triple of vectors \(\vu_1,\vu_2,\vu_3\) in \(\R^3\) forms a right-hand system if the determinant of the matrix whose \(i\)-th row is \(\vu_i\) is positive.

Definition 1.3.4. Cross Product.

Let \(\vu = \gv{u_1, u_2, u_3}\) and \(\vv = \gv{v_1, v_2, v_3}\) be vectors in \(\R^3\text{.}\) Their cross product is
\begin{equation*} \vu \times \vv = \gv{u_2 v_3 - u_3 v_2,\; u_3 v_1 - u_1 v_3,\; u_1 v_2 - u_2 v_1}. \end{equation*}
A convenient way to remember this formula is through the symbolic determinant
\begin{equation*} \vu \times \vv = \det \begin{bmatrix} \ui \amp \uj \amp \uk \\ u_1 \amp u_2 \amp u_3 \\ v_1 \amp v_2 \amp v_3 \end{bmatrix}. \end{equation*}

Example 1.3.5.

Let \(\vu = \gv{1,2,3}\) and \(\vv = \gv{4,5,6}\text{.}\) Compute \(\vu \times \vv\text{.}\)
Solution.
Expanding the determinant gives
\begin{align*} \vu \times \vv \amp = \ui((2)(6)-(3)(5)) - \uj((1)(6)-(3)(4)) + \uk((1)(5)-(2)(4))\\ \amp = -3\ui + 6\uj - 3\uk = \gv{-3,6,-3}. \end{align*}

Proof.
Write \(\vu = \gv{u_1,u_2,u_3}\) and \(\vv = \gv{v_1,v_2,v_3}\text{.}\) A direct dot product computation shows
\begin{equation*} \vu\cdot(\vu\times\vv)=0, \qquad \vv\cdot(\vu\times\vv)=0, \end{equation*}
so \(\vu \times \vv\) is orthogonal to both vectors.
Using the Cauchy–Binet identity, applied to the \(2\times 3\) matrix whose rows are \(\vu\) and \(\vv\) and to its transpose, or by a direct verification from the component formula,
\begin{equation*} \norm{\vu\times\vv}^2=\norm{\vu}^2\norm{\vv}^2-(\vu\cdot\vv)^2. \end{equation*}
For a proof of the general Cauchy–Binet identity, see Theorem 3.96 in [5]. Since \(\vu\cdot\vv=\norm{\vu}\norm{\vv}\cos\theta\text{,}\) we obtain
\begin{equation*} \norm{\vu\times\vv}=\norm{\vu}\norm{\vv}\sin\theta. \end{equation*}
Finally, expansion by the third row gives
\begin{equation*} \det\begin{bmatrix} \vu \\ \vv \\ \vu\times\vv \end{bmatrix} =(\vu\times\vv)\cdot(\vu\times\vv) =\norm{\vu\times\vv}^2. \end{equation*}
Because \(\vu\) and \(\vv\) are linearly independent, \(\vu\times\vv\ne\vz\text{,}\) so this determinant is positive. Hence \(\vu,\vv,\vu\times\vv\) form a right-hand system.
If you curl the fingers of your right hand from \(\vu\) toward \(\vv\text{,}\) your thumb points in the direction of \(\vu \times \vv\text{.}\)
Vectors u and v span a shaded parallelogram. Their cross product points right-handed; its norm equals the area.
Figure 1.3.7. The ordered triple \(\vu,\vv,\vu\times\vv\) has right-hand orientation. Numerically, \(\norm{\vu\times\vv}\) is the area of the shaded parallelogram; the displayed length of the cross-product arrow is schematic.
The cross product gives a practical way to find equations of planes. If a plane passes through a point with position vector \(\vr_0\) and contains two nonparallel vectors \(\vu\) and \(\vv\text{,}\) then \(\vu \times \vv\) is orthogonal to the plane. So the plane equation is obtained by using this normal in the dot product formula:
\begin{equation*} (\vu\times\vv)\cdot(\vr-\vr_0)=0. \end{equation*}
Reversing the order to \(\vv\times\vu\) only changes the sign, so it gives the same plane.

Example 1.3.8.

Find an equation of the plane through the point \((1,-1,2)\) that contains the vectors \(\vu=\gv{1,2,0}\) and \(\vv=\gv{2,1,1}\text{.}\)
Solution.
First compute a normal vector:
\begin{equation*} \vu\times\vv = \gv{(2)(1)-(0)(1),\; (0)(2)-(1)(1),\; (1)(1)-(2)(2)} = \gv{2,-1,-3}. \end{equation*}
Therefore the plane equation is
\begin{equation*} \gv{2,-1,-3}\cdot\gv{x-1,\;y+1,\;z-2}=0, \end{equation*}
which becomes
\begin{equation*} 2(x-1)-(y+1)-3(z-2)=0. \end{equation*}
Simplifying,
\begin{equation*} 2x-y-3z+3=0. \end{equation*}
We can apply the same idea when a plane is given by three non-collinear points with position vectors \(\vp_0\text{,}\) \(\vp_1\text{,}\) and \(\vp_2\text{.}\) The vectors \(\vp_1-\vp_0\) and \(\vp_2-\vp_0\) both lie in the plane, so a normal vector is
\begin{equation*} \vn = (\vp_1-\vp_0)\times(\vp_2-\vp_0). \end{equation*}
Since the points are non-collinear, this cross product is nonzero. The plane equation is then
\begin{equation*} \big((\vp_1-\vp_0)\times(\vp_2-\vp_0)\big)\cdot(\vr-\vp_0)=0. \end{equation*}

Example 1.3.9.

Find an equation of the plane through the points \((1,0,0)\text{,}\) \((0,1,0)\text{,}\) and \((0,0,1)\text{.}\)
Solution.
Take \(\vp_0=\gv{1,0,0}\text{,}\) \(\vp_1=\gv{0,1,0}\text{,}\) and \(\vp_2=\gv{0,0,1}\text{.}\) Then
\begin{equation*} \vp_1-\vp_0=\gv{-1,1,0}, \qquad \vp_2-\vp_0=\gv{-1,0,1}. \end{equation*}
A normal vector is
\begin{equation*} (\vp_1-\vp_0)\times(\vp_2-\vp_0) = \gv{1,1,1}. \end{equation*}
Therefore
\begin{equation*} \gv{1,1,1}\cdot\gv{x-1,\;y,\;z}=0, \end{equation*}
so the plane equation is
\begin{equation*} x+y+z=1. \end{equation*}
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