M408M Learning Module Pages
Main page

Chapter 10: Parametric Equations and Polar Coordinates

Chapter 12: Vectors and the Geometry of Space

Learning module LM 12.1: 3-dimensional rectangular coordinates:

Learning module LM 12.2: Vectors:

Learning module LM 12.3: Dot products:

Learning module LM 12.4: Cross products:

      Areas in the plane
      Definition
      Computing cross products
      Applications

Learning module LM 12.5: Equations of Lines and Planes:

Learning module LM 12.6: Surfaces:


Chapter 13: Vector Functions


Chapter 14: Partial Derivatives


Chapter 15: Multiple Integrals



Computing cross products

Cross Products p3

Computing Cross Products: On the previous page, we saw the definition of a cross product:

Definition: The Cross Product, ${\bf u} \times {\bf v}$, of vectors $\bf u$ and $\bf v$ is the vector defined by $${\bf u} \times {\bf v}\ = \ (\|{\bf u}\|\|{\bf v}\| \sin \theta){\bf n}$$ where $ \theta $ is the angle between $\bf u$ and $\bf v$, and ${\bf n}$ is the unit vector perpendicular to ${\bf u}$ and ${\bf v}$ such that $\{{\bf u},\, {\bf v},\, {\bf n}\}$ forms a right-handed system. Since $\sin 0\,=\, \sin \pi \,=\, 0$, vectors ${\bf u}, \, {\bf v}$ are parallel when ${\bf u}\times {\bf v} \ = \ 0$ and ${\bf u},\, {\bf v} \, \ne\,0$.


A number of properties follow immediately from this definition and the fact that $\{{\bf i},\, {\bf j},\, {\bf k}\}$ is a right-handed system:


Algebraic Formula: The previously properties provide good algebraic ways of computing the cross product of $${\bf a}\,=\,\langle\,a_1,\, a_2,\, a_3\,\rangle\,=\, a_1\,{\bf i}+ a_2\,{\bf j}+a_3\,{\bf k}\,,\qquad \qquad {\bf b}\,=\, \langle\,b_1,\, b_2,\, b_3\,\rangle\,=\, b_1\, {\bf i}+b_2 \,{\bf j}+b_3\, {\bf k}\,.$$
By Properties 1, 2, and 3, \begin{eqnarray*} {\bf a}\times{\bf b} &=& a_1b_1 {\bf i} \times {\bf i} + a_1 b_2 {\bf i} \times {\bf j} + a_1 b_3 {\bf i} \times {\bf k} + \hbox{(6 more terms like these)} \cr\cr &=& (a_2 b_3 - a_3 b_2)\, {\bf i}- (a_1 b_3 - a_3 b_1)\,{\bf j}+(a_1 b_2 - a_2b_1)\, {\bf k}\,.\end{eqnarray*} By properties of determinants, therefore, $${\bf a}\times{\bf b}\ = \ \left|\begin{array}{cc} a_2 & a_3 \\ b_2 & b_3 \end{array}\right|{\bf i}- \left| \begin{array}{cc} a_1 & a_3 \\ b_1 & b_3 \end{array}\right|{\bf j}+\left| \begin{array}{cc} a_1 & a_2 \\ b_1 & b_2 \end{array}\right|{\bf k} \ = \ \left|\begin{array}{ccc} {\bf i} & {\bf j} & {\bf k} \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \end{array}\right|\,.$$ The two products of vectors can be combined: the (scalar) triple product of vectors $\bf a$, $\bf b$ and $\bf c$ is $${\bf a}\, \cdot \,({\bf b} \times {\bf c}) \ = \ a_1 \left|\begin{array}{cc} b_2 & b_3 \\ c_2 & c_3 \end{array}\right|- a_2\left| \begin{array}{cc} b_1 & b_3 \\ c_1 & c_3 \end{array}\right|+a_3\left| \begin{array}{cc} b_1 & b_2 \\ c_1 & c_2 \end{array}\right| \ = \ \left|\begin{array}{ccc} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{array}\right|\,.$$ There's also a vector triple product ${\bf a} \times ({\bf b} \times {\bf c})$, but we won't need it.

  Example: Determine all unit vectors $\bf v$ orthogonal to $${\bf a} \ = \ 3\, {\bf i} +{\bf j} + 4\, {\bf k}\,, \quad {\bf b} \ = \ 3\, {\bf i} +2\, {\bf j} +6\, {\bf k}\,.$$

Solution: The non-zero vectors orthogonal to ${\bf a},\, {\bf b}$ are all of the form $${\bf v} \ = \ \lambda({\bf a} \times {\bf b}), \quad \lambda \ne 0$$ with $\lambda $ a scalar. This means the only unit vectors orthogonal to ${\bf a},\, {\bf b}$ are $${\bf v} \ = \ \pm \,\frac{{\bf a} \times {\bf b}}{\|{\bf a} \times {\bf b}\|}\,.$$
Now $${\bf a}\times{\bf b} \ = \ \left|\begin{array}{ccc} {\bf i} & {\bf j} & {\bf k} \\ 3 & 1 & 4 \\ 3 & 2 & 6 \end{array}\right|\qquad \qquad$$ $$= \ \left|\begin{array}{cc} 1 & 4 \\ 2 & 6 \end{array}\right|{\bf i}- \left|\begin{array}{cc} 3 & 4 \\ 3 & 6 \end{array}\right|{\bf j}+\left|\begin{array}{cc} 3 & 1 \\ 3 & 2 \end{array}\right|{\bf k}\,.$$ So $${\bf a}\times{\bf b} \ = \ -2\, {\bf i} -6\, {\bf j} +3\, {\bf k} , \quad \|{\bf a}\times{\bf b}\| \ = \ 7\,.$$ Thus the unit vectors orthogonal to ${\bf a},\, {\bf b}$ are $${\bf v} \ = \ \pm \Bigl( \frac{2}{7} {\bf i} + \frac{6}{7} {\bf j} - \frac{3}{7} {\bf k}\Bigl) \, .$$