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Vector Product (AVANCED)

 

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:

\begin{displaymath} {\bf a} \,{\rm x}\, {\bf b} = (a_x \,b_x,\, a_y\, b_y,\, a_z \,b_z). \end{displaymath} (1)

Is ${\bf a} \,{\rm x}\, {\bf b}$ a proper vector? Suppose that ${\bf a} = (1,\,0,\,0)$, ${\bf b} = (0,\,1,\,0)$. 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:

\begin{displaymath} {\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} (2)

Does this rather unlikely combination transform like a vector? Let us try rotating the basis through $\theta$ degrees about the $z$-axis. In the new basis,
$\displaystyle c_{x'}$ $\textstyle =$ $\displaystyle (-a_x\, \sin\theta + a_y\,\cos\theta)\,b_z - a_z\,(-b_x\, \sin\theta + b_y\,\cos\theta)$  
  $\textstyle =$ $\displaystyle (a_y\, b_z - a_z\, b_y)\, \cos\theta + (a_z\, b_x-a_x\, b_z)\,\sin\theta$  
  $\textstyle =$ $\displaystyle c_x\,\cos\theta +c_y\,\sin\theta.$ (3)

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. 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,
\begin{displaymath} {\bf a}\times{\bf b} = - {\bf b} \times{\bf a}, \end{displaymath} (4)

distributive,
\begin{displaymath} {\bf a}\times({\bf b} +{\bf c})= {\bf a} \times{\bf b}+{\bf a}\times{\bf c}, \end{displaymath} (5)

but is not associative:
\begin{displaymath} {\bf a}\times({\bf b} \times{\bf c})\neq ({\bf a}\times{\bf b}) \times{\bf c}. \end{displaymath} (6)

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,

$\displaystyle {\bf a}\cdot {\bf a}\times{\bf b}$ $\textstyle =$ $\displaystyle a_x\,(a_y\, b_z-a_z\, b_y) + a_y\, (a_z\, b_x- a_x \,b_z) +a_z\,(a_x \,b_y - a_y\, b_x)$  
  $\textstyle =$ $\displaystyle 0.$ (7)

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 the right-hand rule (see Fig. 1).
Figure 1: The right-hand rule.

\begin{figure} \epsfysize =2in \centerline{\epsffile{fig6.eps}} \end{figure}

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

$\displaystyle ({\bf a}\times{\bf b})^2$ $\textstyle =$ $\displaystyle (a_y \,b_z-a_z\, b_y)^2 +(a_z \,b_x - a_x\, b_z)^2 +(a_x \,b_z -a_y \,b_x)^2$  
  $\textstyle =$ $\displaystyle (a_x^{~2}+a_y^{~2}+a_z^{~2})\,(b_x^{~2}+b_y^{~2}+b_z^{~2}) - (a_x\, b_x + a_y \,b_y + a_z\, b_z)^2$  
  $\textstyle =$ $\displaystyle \vert a\vert^2 \,\vert b\vert^2 - ({\bf a}\cdot {\bf b})^2$  
  $\textstyle =$ $\displaystyle \vert a\vert^2 \,\vert b\vert^2 - \vert a\vert^2 \,\vert b\vert^2 \,\cos^2\theta = \vert a\vert^2\,\vert b\vert^2\, \sin^2\theta.$ (8)

Thus,
\begin{displaymath} \vert{\bf a}\times{\bf b}\vert = \vert a\vert\,\vert b\vert\,\sin\theta. \end{displaymath} (9)

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 ${\bf a}$.

Figure 2: A vector area.

\begin{figure} \epsfysize =1.5in \centerline{\epsffile{fig7.eps}} \end{figure}
Consider the parallelogram defined by vectors ${\bf a}$ and ${\bf b}$ (see Fig. 2). The scalar area is $a\,b \sin\theta$. By definition, the vector area has the magnitude of the scalar area, and is normal to the plane of the parallelogram, which means that it is perpendicular to both ${\bf a}$ and ${\bf b}$. Clearly, the vector area is given by
\begin{displaymath} {\bf S} = {\bf a}\times {\bf b}, \end{displaymath} (10)

with the sense obtained from the right-hand grip rule by rotating ${\bf a}$ on to ${\bf b}$.

Suppose that a force ${\bf F}$ is applied at position ${\bf r}$ (see Fig. 3). 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 the right-hand grip rule. It follows that the vector moment is given by

\begin{displaymath} {\bf M} = {\bf r}\times{\bf F}. \end{displaymath} (11)

Figure 3: Torque.

\begin{figure} \epsfysize =2.5in \centerline{\epsffile{fig8.eps}} \end{figure}

The angular momentum, ${\bf l}$, of a particle of linear momentum ${\bf p}$ and position vector ${\bf r}$ about the origin is simply defined as the moment of its momentum about the origin. Hence,

\begin{displaymath} {\bf l} = {\bf r}\times{\bf p}. \end{displaymath}

 

 

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