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Next: Vector Calculus Up: Vectors Previous: The Scalar Product

The Vector Product

We have discovered how to construct a scalar from the components of two general vectors $\bf a$ and $\bf b$. Can we also construct a vector which is not just a linear combination of $\bf a$ and $\bf b$? Consider the following definition:
{\bf a} \,{\rm x}\, {\bf b} = (a_x \,b_x,\, a_y\, b_y,\, a_z \,b_z).
\end{displaymath} (27)

Is ${\bf a} \,{\rm x}\, {\bf b}$ a proper vector? Suppose that ${\bf a} = (1,\,0,\,0)$ and ${\bf b} = (0,\,1,\,0)$. Clearly, ${\bf a} \,{\rm x}\, {\bf b}= {\bf0}$. However, if we rotate the basis through $45^\circ$ about the $z$-axis then ${\bf a} = (1/\sqrt{2},\, -1/\sqrt{2},\, 0)$, ${\bf b} = (1/\sqrt{2},\,
1/\sqrt{2},\, 0)$, and ${\bf a}\, {\rm x}\, {\bf b} = (1/2,\, -1/2,\,0)$. Thus, ${\bf a} \,{\rm x}\, {\bf b}$ does not transform like a vector, because its magnitude depends on the choice of axes. So, above definition is a bad one.

Consider, now, the cross product or vector product,

{\bf a}\times{\bf b} = (a_y \, b_z-a_z\, b_y,\, a_z\, b_x - a_x\, b_z,\, a_x\, b_y - a_y\, b_x)
={\bf c}.
\end{displaymath} (28)

Does this rather unlikely combination transform like a vector? Let us try rotating the basis through $\theta$ degrees about the $z$-axis using Eqs. (10)-(12). In the new basis,
$\displaystyle c_{x'}$ $\textstyle =$ $\displaystyle (-a_x\, \sin\theta + a_y\,\cos\theta)\,b_z - a_z\,(-b_x\, \sin\theta + b_y\,\cos\theta)$  
  $\textstyle =$ $\displaystyle (a_y\, b_z - a_z\, b_y)\, \cos\theta + (a_z\, b_x-a_x\, b_z)\,\sin\theta$  
  $\textstyle =$ $\displaystyle c_x\,\cos\theta
+c_y\,\sin\theta.$ (29)

Thus, the $x$-component of ${\bf a}\times{\bf b}$ transforms correctly. It can easily be shown that the other components transform correctly as well, and that all components also transform correctly under rotation about the $y$- and $z$-axes. Thus, ${\bf a}\times{\bf b}$ is a proper vector. Incidentally, ${\bf a}\times{\bf b}$ is the only simple combination of the components of two vectors which transforms like a vector (which is non-coplanar with ${\bf a}$ and ${\bf b}$). The cross product is anticommutative,
{\bf a}\times{\bf b} = - {\bf b} \times{\bf a},
\end{displaymath} (30)

{\bf a}\times({\bf b} +{\bf c})= {\bf a} \times{\bf b}+{\bf a}\times{\bf c},
\end{displaymath} (31)

but is not associative:
{\bf a}\times({\bf b} \times{\bf c})\neq ({\bf a}\times{\bf b}) \times{\bf c}.
\end{displaymath} (32)

Note that ${\bf a}\times{\bf b}$ can be written in the convenient, and easy to remember, determinant form
{\bf a}\times {\bf b} = \left\vert\begin{array}{ccc}
{\bf e}...
...a_x& a_y& a_z\\ [0.5ex]
b_x & b_y & b_z\end{array}\right\vert.
\end{displaymath} (33)

The cross product transforms like a vector, which means that it must have a well-defined direction and magnitude. We can show that ${\bf a}\times{\bf b}$ is perpendicular to both ${\bf a}$ and ${\bf b}$. Consider ${\bf a}\cdot {\bf a}\times{\bf b}$. If this is zero then the cross product must be perpendicular to ${\bf a}$. Now

$\displaystyle {\bf a}\cdot {\bf a}\times{\bf b}$ $\textstyle =$ $\displaystyle a_x\,(a_y\, b_z-a_z\, b_y) + a_y\, (a_z\, b_x- a_x \,b_z)
+a_z\,(a_x \,b_y - a_y\, b_x)$  
  $\textstyle =$ $\displaystyle 0.$ (34)

Therefore, ${\bf a}\times{\bf b}$ is perpendicular to ${\bf a}$. Likewise, it can be demonstrated that ${\bf a}\times{\bf b}$ is perpendicular to ${\bf b}$. The vectors $\bf a$, $\bf b$, and ${\bf a}\times{\bf b}$ form a right-handed set, like the unit vectors ${\bf e}_x$, ${\bf e}_y$, and ${\bf e}_z$. In fact, ${\bf e}_x\times
{\bf e}_y={\bf e}_z$. This defines a unique direction for ${\bf a}\times{\bf b}$, which is obtained from a right-hand rule--see Fig. 6.
Figure 6: The right-hand rule for cross products.
\epsfysize =2in

Let us now evaluate the magnitude of ${\bf a}\times{\bf b}$. We have

$\displaystyle ({\bf a}\times{\bf b})^2$ $\textstyle =$ $\displaystyle (a_y \,b_z-a_z\, b_y)^2 +(a_z \,b_x - a_x\, b_z)^2 +(a_x \,b_z
-a_y \,b_x)^2$  
  $\textstyle =$ $\displaystyle (a_x^{~2}+a_y^{~2}+a_z^{~2})\,(b_x^{~2}+b_y^{~2}+b_z^{~2}) -
(a_x\, b_x + a_y \,b_y + a_z\, b_z)^2$  
  $\textstyle =$ $\displaystyle \vert a\vert^2 \,\vert b\vert^2 - ({\bf a}\cdot {\bf b})^2$  
  $\textstyle =$ $\displaystyle \vert a\vert^2 \,\vert b\vert^2 - \vert a\vert^2 \,\vert b\vert^2 \,\cos^2\theta = \vert a\vert^2\,\vert b\vert^2\, \sin^2\theta.$ (35)

\vert{\bf a}\times{\bf b}\vert = \vert a\vert\,\vert b\vert\,\sin\theta.
\end{displaymath} (36)

Clearly, ${\bf a}\times{\bf a} = {\bf0}$ for any vector, since $\theta$ is always zero in this case. Also, if ${\bf a}\times{\bf b} = {\bf0}$ then either $\vert a\vert=0$, $\vert b\vert=0$, or ${\bf b}$ is parallel (or antiparallel) to ${\bf a}$.

Suppose that a force ${\bf F}$ is applied at position ${\bf r}$--see Fig. 7. The moment, or torque, about the origin $O$ is the product of the magnitude of the force and the length of the lever arm $OQ$. Thus, the magnitude of the moment is $\vert F\vert\,\vert r\vert\,\sin\theta$. The direction of the moment is conventionally the direction of the axis through $O$ about which the force tries to rotate objects, in the sense determined by a right-hand grip rule. It follows that the vector moment is given by

{\bf M} = {\bf r}\times{\bf F}.
\end{displaymath} (37)

Figure 7: A torque.
\epsfysize =2.5in

next up previous
Next: Vector Calculus Up: Vectors Previous: The Scalar Product
Richard Fitzpatrick 2007-07-14