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Section 1.4 The Cross Product

Definition 1.4.1. Cross product of two vectors in \(\R^3\).

Given two vectors in \(\R^3\text{,}\) \(\vec{u} = \left[u_1, u_2, u_3 \right]^{\T}\) and \(\vec{v} = \left[v_1, v_2, v_3 \right]^{\T}\text{,}\) their cross product denoted \(\vec{u} \times \vec{v} \) is the \(\R^3\) vector found as
\begin{equation*} \vec{u} \times \vec{v} = \left[ \begin{array}{c} 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{array} \right]. \end{equation*}

Insight 1.4.2. Using determinants to find a cross product.

The cross product can also be found as the determinant of the \(3\times 3\) matrix whose columns are given by the \(x, y\) and \(z\) coordinates of the standard unit vectors (\(\hat{\mathbf{i}}, \hat{\mathbf{j}}, \hat{\mathbf{k}}\)) and the two vectors in the cross product.
\begin{equation*} \text{Let } \vec{u} = \left[\begin{array}{c} u_1\\u_2\\u_3 \end{array}\right] \text{ and } \vec{v} = \left[\begin{array}{c}v_1\\v_2\\v_3 \end{array}\right], \text{ then, } \end{equation*}
\begin{equation*} \begin{array}{ccl} \vec{u} \times \vec{v} \amp = \amp \left| \begin{array}{ccc} \hat{\mathbf{i}} \amp \hat{\mathbf{j}} \amp \hat{\mathbf{k}}\\ u_1 \amp u_2 \amp u_3\\ v_1 \amp v_2 \amp v_3\\ \end{array} \right| \\ \amp = \amp \hat{\mathbf{i}} \left| \begin{array}{cc} u_2 \amp u_3\\ v_2 \amp v_3\\ \end{array} \right| - \hat{\mathbf{j}}\left| \begin{array}{cc} u_1 \amp u_3\\ v_1\amp v_3\\ \end{array} \right| + \hat{\mathbf{k}}\left| \begin{array}{cc} u_1 \amp u_2 \\ v_1 \amp v_2 \\ \end{array} \right| \\ \amp = \amp \hat{\mathbf{i}} \left( u_2 \, v_3 - u_3 \, v_2 \right) - \hat{\mathbf{j}} \left( u_1 \, v_3 - u_3 \, v_1 \right) + \hat{\mathbf{k}} \left( u_1 \, v_2 - u_2 \, v_1 \right) \\ \amp \amp \\ \amp = \amp \left[\begin{array}{c} 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{array}\right] \end{array} \end{equation*}

Subsection 1.4.1 Properties of the Cross Product

  • \(\begin{array}{l} \text{Let } \vec{u}, \vec{v} \text{ and } \vec{w} \text{ be vectors in } \R^3 \text{ and } c \text{ a scalar, }\amp\\ \end{array}\)
    \begin{equation*} \begin{array}{clcl} \bullet \amp \vec{u} \times \vec{v} = - \left(\vec{v} \times \vec{u} \right) \amp \hspace{0.4cm} \amp \text{(Anticommutative)}\\ \bullet \amp \vec{u} \times \left(\vec{v}+ \vec{w} \right) = \vec{u} \times \vec{u} + \vec{u} \times \vec{w} \amp \hspace{0.4cm} \amp \text{(Distributive)}\\ \bullet \amp c\, \left(\vec{u} \times \vec{v}\right) = \left(c\,\vec{u}\right) \times \vec{v} = \vec{u} \times \left(c\,\vec{v}\right)\amp \hspace{0.4cm} \amp \text{(Multiplication by a scalar)}\\ \bullet \amp \vec{u} \times \vec{0} = \vec{0} \times \vec{u} = \vec{0} \amp\hspace{0.4cm} \amp \text{(Cross product of the zero vector)}\\ \bullet \amp \vec{u} \times \vec{u} = \vec{0} \amp\hspace{0.4cm} \amp \text{(Cross product of a vector with itself)}\\ \bullet \amp \vec{u} \cdot \left(\vec{v} \times \vec{w}\right) = \left(\vec{u} \times \vec{v}\right) \cdot \vec{w} \amp\hspace{0.4cm} \amp \text{(Scalar triple product)} \end{array} \end{equation*}
  • The result of a cross product of two vectors is another vector; unlike the dot product which is a scalar.
  • Let \(\vec{w} = \vec{u} \times \vec{v}\text{,}\) then
    • The vector \(\vec{w}\) is orthogonal (perpendicular) to both \(\vec{u} \) and \(\vec{v}.\)
      \begin{equation*} \vec{w} \cdot \vec{u} = 0 = \vec{w} \cdot \vec{v}. \end{equation*}
    • The magnitude of \(\vec{w}\) corresponds to the area of the parallelogram having \(\vec{u}\) and \(\vec{v} \) as sides.
      \begin{equation*} \Vert \vec{w} \Vert = \Vert \vec{u} \times \vec{v} \Vert = \Vert \vec{u} \Vert \Vert \vec{v} \Vert |\sin \theta|, \end{equation*}
      where \(\theta\) is the angle between \(\vec{u}\) and \(\vec{v} \text{.}\)
  • Volume of a parallelepiped.
    The volume of a parallelepiped with adjacent edges given by the vectors \(\vec{u}\text{,}\) \(\vec{v}\text{,}\) and \(\vec{w}\) is the absolute value of the triple scalar product.
    \begin{equation*} V = \left| \vec{u} \cdot \left( \vec{v} \times \vec{w}\right)\right|. \end{equation*}
Let \(\vec{u} = -\hat{i} - 2\hat{j} + \hat{k}\text{,}\) \(\vec{v} = 4\hat{i} + 3\hat{j} + 2\hat{k}\text{,}\) and \(\vec{w} = - 4\hat{i} - 2\hat{k}\text{.}\) Find the volume of the parallelepiped with adjacent edges \(\vec{u}, \vec{v} \) and \(\vec{w}\text{.}\)
\begin{equation*} \begin{array}{ccc} \vec{u} = \left[\begin{array}{c} -1\\-2\\1 \end{array}\right], \hspace{0.2cm} \vec{v} = \left[\begin{array}{c} 4\\3\\2 \end{array}\right], \hspace{0.2cm} \vec{w} = \left[\begin{array}{c} -4\\0\\-2 \end{array}\right] \end{array} \end{equation*}
\begin{equation*} \begin{array}{ccc} \vec{v} \times \vec{w} \amp = \amp \left[ \begin{array}{c} 3 \cdot (-2) - (2) \cdot 0\\ 2 \cdot (-4) - (4) \cdot (-2)\\ 4 \cdot 0 - (3) \cdot (-4)\\ \end{array}\right] = \left[ \begin{array}{c} -6\\ 0\\ 12 \end{array}\right]. \\ \vec{u} \cdot \left(\vec{v} \times \vec{w} \right) \amp = \amp \left[\begin{array}{c} -1\\-2\\1 \end{array}\right]\cdot \left[\begin{array}{c} -6\\ 0\\ 12 \end{array}\right] = (-1)(-6) + (-2)(0) + (1)(12) = 18. \end{array} \end{equation*}
\begin{equation*} V = 18 \text{ cubic units}. \end{equation*}
Find the area of the parallelogram with sides given by the vectors \(\vec{u} = \left[ {3}, {2},{2}\right]^{\T}\) and \(\vec{v} = \left[ {-3}, {3},{-1}\right]^{\T}\text{.}\)
Area = .
Answer.
\(17.2627\)
Solution.
\begin{equation*} \begin{array}{ccc} \vec{u} \times \vec{v} \amp = \amp \left[ \begin{array}{c} {3}\\{2}\\{2} \end{array}\right] \times \left[ \begin{array}{c} {-3}\\{3}\\{-1} \end{array}\right]\\ \amp = \amp \left[ \begin{array}{c} ({2} \cdot {-1}) - ({2} \cdot {3})\\ ({2} \cdot {-3}) - ({3} \cdot {-1})\\ ({3} \cdot {3}) - ({2} \cdot {-3})\\ \end{array}\right]\\ \amp = \amp \left[ \begin{array}{c} ({-2})- ( {6})\\ ({-6})- ({-3})\\({9})-({-6}) \end{array}\right]\\ \amp = \amp \left[ \begin{array}{c} {-8}\\{-3}\\{15}\end{array}\right]\\ \end{array} \end{equation*}
\begin{equation*} \text{Area = } \Vert \vec{u} \times \vec{v} \Vert = \sqrt{({-8})^2 + ({-3})^2 + ({15})^2} = {17.2627}. \end{equation*}
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