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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*}

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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