The Cross Product

Given two nonzero vectors \(\mathbf{a} = \langle a_1, a_2, a_3 \rangle\) and \(\mathbf{b} = \langle b_1, b_2, b_3 \rangle\), it is very useful to be able to find a nonzero vector c that is perpendicular to both a and b. If \(\mathbf{c} = \langle c_1, c_2, c_3 \rangle\) is such a vector, then \(\mathbf{a} \cdot \mathbf{c} = 0\) and \(\mathbf{b} \cdot \mathbf{c} = 0\) and so \[ a_1c_1 + a_2c_2 + a_3c_3 = 0 \tag{1} \] \[ b_1c_1 + b_2c_2 + b_3c_3 = 0 \tag{2} \] To eliminate \(c_3\) we multiply (1) by \(b_3\) and (2) by \(a_3\) and subtract: \[ (a_1b_3 - a_3b_1)c_1 + (a_2b_3 - a_3b_2)c_2 = 0 \tag{3} \] Equation 3 has the form \(pc_1 + qc_2 = 0\), for which an obvious solution is \(c_1 = q\) and \(c_2 = -p\). So a solution of (3) is \[ c_1 = a_2b_3 - a_3b_2 \qquad c_2 = a_3b_1 - a_1b_3 \] Substituting these values into (1) and (2), we then get \[ c_3 = a_1b_2 - a_2b_1 \] This means that a vector perpendicular to both a and b is \[ \mathbf{c} = \langle c_1, c_2, c_3 \rangle = \langle a_2b_3 - a_3b_2, a_3b_1 - a_1b_3, a_1b_2 - a_2b_1 \rangle \] The resulting vector is called the cross product of a and b and is denoted by \(\mathbf{a} \times \mathbf{b}\).

Definition 4 If \(\mathbf{a} = \langle a_1, a_2, a_3 \rangle\) and \(\mathbf{b} = \langle b_1, b_2, b_3 \rangle\), then the cross product of a and b is the vector \[ \mathbf{a} \times \mathbf{b} = \langle a_2b_3 - a_3b_2, a_3b_1 - a_1b_3, a_1b_2 - a_2b_1 \rangle \]

Notice that the cross product \(\mathbf{a} \times \mathbf{b}\) of two vectors a and b, unlike the dot product, is a vector. For this reason it is also called the vector product. Note that \(\mathbf{a} \times \mathbf{b}\) is defined only when a and b are three-dimensional vectors.

In order to make Definition 4 easier to remember, we use the notation of determinants. A determinant of order 2 is defined by \[ \begin{vmatrix} a & b \\ c & d \end{vmatrix} = ad - bc \] (Multiply across the diagonals and subtract.) For example, \[ \begin{vmatrix} 2 & 1 \\ -6 & 4 \end{vmatrix} = 2(4) - 1(-6) = 14 \] A determinant of order 3 can be defined in terms of second-order determinants as follows: \[ \begin{vmatrix} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{vmatrix} = a_1 \begin{vmatrix} b_2 & b_3 \\ c_2 & c_3 \end{vmatrix} - a_2 \begin{vmatrix} b_1 & b_3 \\ c_1 & c_3 \end{vmatrix} + a_3 \begin{vmatrix} b_1 & b_2 \\ c_1 & c_2 \end{vmatrix} \tag{5} \] Observe that each term on the right side of Equation 5 involves a number \(a_i\) in the first row of the determinant, and \(a_i\) is multiplied by the second-order determinant obtained from the left side by deleting the row and column in which \(a_i\) appears. Notice also the minus sign in the second term. For example, \[ \begin{vmatrix} 1 & 2 & -1 \\ 3 & 0 & 1 \\ -5 & 4 & 2 \end{vmatrix} = 1 \begin{vmatrix} 0 & 1 \\ 4 & 2 \end{vmatrix} - 2 \begin{vmatrix} 3 & 1 \\ -5 & 2 \end{vmatrix} + (-1) \begin{vmatrix} 3 & 0 \\ -5 & 4 \end{vmatrix} \] \[ = 1(0 - 4) - 2(6 - (-5)) + (-1)(12 - 0) = -4 - 22 - 12 = -38 \] If we now rewrite Definition 4 using second-order determinants and the standard basis vectors i, j, and k, we see that the cross product of the vectors \(\mathbf{a} = a_1\mathbf{i} + a_2\mathbf{j} + a_3\mathbf{k}\) and \(\mathbf{b} = b_1\mathbf{i} + b_2\mathbf{j} + b_3\mathbf{k}\) is \[ \mathbf{a} \times \mathbf{b} = \begin{vmatrix} a_2 & a_3 \\ b_2 & b_3 \end{vmatrix} \mathbf{i} - \begin{vmatrix} a_1 & a_3 \\ b_1 & b_3 \end{vmatrix} \mathbf{j} + \begin{vmatrix} a_1 & a_2 \\ b_1 & b_2 \end{vmatrix} \mathbf{k} \tag{6} \] In view of the similarity between Equations 5 and 6, we often write \[ \mathbf{a} \times \mathbf{b} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \end{vmatrix} \tag{7} \] Although the first row of the symbolic determinant in Equation 7 consists of vectors, if we expand it as if it were an ordinary determinant using the rule in Equation 5, we obtain Equation 6. The symbolic formula in Equation 7 is probably the easiest way of remembering and computing cross products.

EXAMPLE 1 If \(\mathbf{a} = \langle 1, 3, 4 \rangle\) and \(\mathbf{b} = \langle 2, 7, -5 \rangle\), then \[ \mathbf{a} \times \mathbf{b} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 3 & 4 \\ 2 & 7 & -5 \end{vmatrix} = \begin{vmatrix} 3 & 4 \\ 7 & -5 \end{vmatrix} \mathbf{i} - \begin{vmatrix} 1 & 4 \\ 2 & -5 \end{vmatrix} \mathbf{j} + \begin{vmatrix} 1 & 3 \\ 2 & 7 \end{vmatrix} \mathbf{k} \] \[ = (-15 - 28)\mathbf{i} - (-5 - 8)\mathbf{j} + (7 - 6)\mathbf{k} = -43\mathbf{i} + 13\mathbf{j} + \mathbf{k} \]