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 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 ${\bf v} \ = \ \pm \Bigl( \frac{2}{7} {\bf i} + \frac{6}{7} {\bf j} - \frac{3}{7} {\bf k}\Bigl) \, .$

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:

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