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Vector Calculus An Introduction

Daniel Daners

Section 1.5 The Cross Product

We now introduce another way of multiplying two vectors, the cross product, also sometimes called the vector product because the resulting object is a vector. In contrast to the scalar product it is only defined in \(\mathbb R^3\), not in any other dimension!
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Definition 1.20.

If two vectors \(\vect x=(x_1,x_2,x_3)\) and \(\vect y=(y_1,y_2,y_3)\) in \(\mathbb R^3\) are given then we define
\begin{equation} \vect x\times\vect y = \begin{bmatrix} x_1\\ x_2\\ x_3 \end{bmatrix} \times \begin{bmatrix} y_1\\ y_2\\ y_3 \end{bmatrix} = \begin{bmatrix} x_2y_3-x_3y_2\\ x_3y_1-x_1y_3\\ x_1y_2-x_2y_1 \end{bmatrix}\text{.}\tag{1.8} \end{equation}
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Note that the second and third components are obtained from the first by a cyclic permutation of the indices \(1\to 2\to 3\to 1\text{.}\) There is an easy way to remember the definition of the cross product.
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Observation 1.21. Computation of cross product using determinant.

If we set \(\vect e_1=(1,0,0)\text{,}\) \(\vect e_2=(0,1,0)\) and \(\vect e_3=(0,0,1)\) then, formally,
\begin{equation*} \vect x\times\vect y=\det \begin{bmatrix} \vect e_1\amp \vect e_2\amp \vect e_3\\ x_1\amp x_2\amp x_3\\ y_1\amp y_2\amp y_3 \end{bmatrix} \end{equation*}
expanded along the first row. There is another way to express the components of \(\vect v=(v_1,v_2,v_3)=\vect x\times\vect y\text{,}\) namely
\begin{align} v_1\amp=\det \begin{bmatrix} x_2\amp y_2\\x_3\amp y_3 \end{bmatrix},\notag\\ v_2\amp=-\det \begin{bmatrix} x_1\amp y_1\\x_3\amp y_3 \end{bmatrix},\tag{1.9}\\ v_3\amp=\det \begin{bmatrix} x_1\amp y_1\\x_2\amp y_2 \end{bmatrix}\text{.}\notag \end{align}
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We next state the main algebraic rules for the cross product.
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Algebraic properties of cross product.

If \(\vect x\text{,}\) \(\vect y\) and \(\vect z\) are vectors in \(\mathbb R^3\) and \(\alpha\) a scalar, then it is easily verified from the above definition that
\begin{align*} \vect x\times\vect y\amp =-(\vect y\times\vect x)\\ \vect x\times(\vect y+\vect z) \amp =\vect x\times\vect y+\vect x\times\vect z\\ \alpha(\vect x\times\vect y) \amp =(\alpha\vect x)\times\vect y =\vect x\times(\alpha\vect y)\text{.} \end{align*}
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Note that \(\vect x\times\vect y\) is not the same as \(\vect y\times\vect x\text{,}\) the sign changes! As \(\vect x\times\vect y\) is a vector we can form the scalar product with a third vector. The following lemma shows how to compute such a scalar product.
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Proof.

By (1.9) and the definition of the scalar product (Definition 1.4 ) we have
\begin{align*} (\vect x\times\vect y)\cdot\vect z \amp =z_1\det \begin{bmatrix} x_2\amp y_2\\x_3\amp y_3 \end{bmatrix} -z_2\det \begin{bmatrix} x_1\amp y_1\\x_3\amp y_3 \end{bmatrix} +z_3\det \begin{bmatrix} x_1\amp y_1\\x_2\amp y_2 \end{bmatrix}\\ \amp =\det \begin{bmatrix} x_1\amp y_1\amp z_1\\ x_2\amp y_2\amp z_2\\ x_3\amp y_3\amp z_3 \end{bmatrix} \end{align*}
if we expand the last determinant along the third column.
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The cross product has, as the scalar product, a geometric significance. We will use that geometric property extensively later.
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Given three vectors \(\vect x\text{,}\) \(\vect y\) and \(\vect z\) in \(\mathbb R^3\text{,}\) not all parallel to some plane, recall that the triple \((\vect x, \vect y,\vect z)\) is positively oriented or right handed if
\begin{equation*} \det \begin{bmatrix} \vect x\amp \vect y\amp \vect z \end{bmatrix} =\det \begin{bmatrix} x_1\amp y_1\amp z_1\\ x_2\amp y_2\amp z_2\\ x_3\amp y_3\amp z_3 \end{bmatrix} \gt 0\text{.} \end{equation*}
Geometrically this means that if we take a screw driver, align it with \(\vect z\) and turn it in such a way that \(\vect x\) is moved towards \(\vect y\) through the smaller angle, then the screw moves in the direction of \(\vect z\text{.}\) The situation is depicted in Figure 1.23.
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Figure 1.23. A positively oriented triple \((\vect x,\vect y,\vect z)\)
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Proof.

According to (1.6) the area, \(A\text{,}\) of the parallelogram spanned by \(\vect x\) and \(\vect y\) is given by
\begin{equation*} A=\sqrt{\|\vect x\|^2\|\vect y\|^2-(\vect x\cdot\vect y)^2}\text{.} \end{equation*}
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We then use the definition of the scalar product, the norm and the cross product, and rearrange the terms in the square root:
\begin{align*} A^2\amp =\|\vect x\|^2\|\vect y\|^2-(\vect x\cdot\vect y)^2\\ \amp =(x_1^2+x_2^2+x_3^2)(y_1^2+y_2^2+y_3^2)-(x_1y_1+x_2y_2+x_3y_3)^2\\ \amp =x_1^2y_1^2+x_1^2y_2^2+x_1^2y_3^2 +x_2^2y_1^2+x_2^2y_2^2+x_2^2y_3^2 +x_3^2y_1^2+x_3^2y_2^2+x_3^2y_3^2\\ \amp \qquad -(x_1^2y_1^2+x_2^2y_2^2+x_3^2y_3^2+2x_1x_2y_1y_2+2x_1x_3y_1y_3+2x_2x_3y_2y_3)\\ \amp =x_1^2y_2^2+x_1^2y_3^2 +x_2^2y_1^2+x_2^2y_3^2 +x_3^2y_1^2+x_3^2y_2^2\\ \amp \qquad-2x_1x_2y_1y_2-2x_1x_3y_1y_3-2x_2x_3y_2y_3\\ \amp =(x_2y_3-x_3y_2)^2+(x_3y_1-x_1y_3)^2+(x_1y_2-x_2y_1)^2\\ \amp =\|\vect x\times\vect y\|^2\text{.} \end{align*}
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This shows (1.12). To show that the triple \((\vect x,\vect y,\vect x\times\vect y)\) is positively oriented we apply (1.10) to the matrix with columns \(\vect x\text{,}\) \(\vect y\) and \(\vect z=\vect x\times\vect y\text{.}\) Then using the definition of the norm
\begin{equation*} \det \begin{bmatrix} x_1\amp y_1\amp x_2y_3-x_3y_2\\ x_2\amp y_2\amp x_3y_1-x_1y_3\\ x_3\amp y_3\amp x_1y_2-x_2y_1 \end{bmatrix} =(\vect x\times\vect y)\cdot(\vect x\times\vect y) =\|\vect x\times\vect y\|^2 \geq 0\text{,} \end{equation*}
and thus the triple is positively oriented.
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Figure 1.25. Geometric interpretation of the cross product.
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Remark 1.26.

The above proposition shows that the cross product is independent of the particular coordinate system we choose. Often the cross product is defined by its geometric properties, and then the representation given in Definition 1.20 is derived from them.
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Remark 1.27. Comparison of scalar and cross products.

Let us compare the scalar and cross products. They are both products of two vectors. The result of the former is a scalar, and the result of the latter is a vector.
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Another difference is that the cross product is only defined in \(\mathbb R^3\), whereas the scalar product is defined in \(\mathbb R^N\) for all \(N\text{.}\)
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Given vectors \(\vect x\) and \(\vect y\) it follows from (1.1) and (1.12) that
  • \(\vect x\) and \(\vect y\) are orthogonal (perpendicular) if and only if \(\vect x\cdot\vect y=0\text{;}\)
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  • \(\vect x\) and \(\vect y\) are parallel if and only if \(\vect x\times\vect y=\vect 0\text{.}\)
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