next up previous
Next: Rotation Up: Vectors and Vector Fields Previous: Vector Area

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 that is not just a linear combination of $ \bf a$ and $ \bf b$ ? Consider the following definition:

$\displaystyle {\bf a} \ast{\bf b} \equiv (a_x \,b_x,\, a_y\, b_y,\, a_z \,b_z).$ (A.36)

Is $ {\bf a} \ast {\bf b}$ a proper vector? Suppose that $ {\bf a} =
(0,\,1,\,0)$ , $ {\bf b} = (1,\,0,\,0)$ . In this case, $ {\bf a} \ast {\bf b}= {\bf0}$ . However, if we rotate the coordinate axes through $ 45^\circ$ about $ Oz$ then $ {\bf a} = (1/\!\sqrt{2},\, 1/\!\sqrt{2},\, 0)$ , $ {\bf b} = (1/\!\sqrt{2},\, -1/\!\sqrt{2},\,0)$ , and $ {\bf a}\ast {\bf b} = (1/2,\, -1/2,\,0)$ . Thus, $ {\bf a} \ast {\bf b}$ does not transform like a vector, because its magnitude depends on the choice of axes. So, previous definition is a bad one.

Consider, now, the cross product or vector product:

$\displaystyle {\bf a}\times{\bf b} \equiv (a_y \, b_z-a_z\, b_y,\, a_z\, b_x - a_x\, b_z,\, a_x\, b_y - a_y\, b_x) ={\bf c}.$ (A.37)

Does this rather unlikely combination transform like a vector? Let us try rotating the coordinate axes through an angle $ \theta $ about $ Oz$ using Equations (A.20)-(A.22). In the new coordinate system,

$\displaystyle c_{x'}$ $\displaystyle = (-a_x\, \sin\theta + a_y\,\cos\theta)\,b_z - a_z\,(-b_x\, \sin\theta + b_y\,\cos\theta)$    
  $\displaystyle = (a_y\, b_z - a_z\, b_y)\, \cos\theta + (a_z\, b_x-a_x\, b_z)\,\sin\theta$    
  $\displaystyle = c_x\,\cos\theta +c_y\,\sin\theta.$ (A.38)

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 $ Ox$ and $ Oy$ . 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 that transforms like a vector (which is non-coplanar with $ {\bf a}$ and $ {\bf b}$ ). The cross product is anti-commutative,

$\displaystyle {\bf a}\times{\bf b} = - {\bf b} \times{\bf a},$ (A.39)


$\displaystyle {\bf a}\times({\bf b} +{\bf c})= {\bf a} \times{\bf b}+{\bf a}\times{\bf c},$ (A.40)

but is not associative,

$\displaystyle {\bf a}\times({\bf b} \times{\bf c})\neq ({\bf a}\times{\bf b}) \times{\bf c}.$ (A.41)

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,

$\displaystyle {\bf a}\cdot {\bf a}\times{\bf b}$ $\displaystyle = a_x\,(a_y\, b_z-a_z\, b_y) + a_y\, (a_z\, b_x- a_x \,b_z) +a_z\,(a_x \,b_y - a_y\, b_x)$    
  $\displaystyle =0.$ (A.42)

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 Figure A.8.)

Figure: The right-hand rule for cross products. Here, $ \theta $ is less that $ 180^\circ $ .
\epsfysize =1.75in

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

$\displaystyle ({\bf a}\times{\bf b})^2$ $\displaystyle = (a_y \,b_z-a_z\, b_y)^2 +(a_z \,b_x - a_x\, b_z)^2 +(a_x \,b_y -a_y \,b_x)^2$    
  $\displaystyle = (a_x^{\,2}+a_y^{\,2}+a_z^{\,2})\,(b_x^{\,2}+b_y^{\,2}+b_z^{\,2}) - (a_x\, b_x + a_y \,b_y + a_z\, b_z)^2$    
  $\displaystyle = \vert{\bf a}\vert^{\,2} \,\vert{\bf b}\vert^{\,2} - ({\bf a}\cdot {\bf b})^2$    
  $\displaystyle = \vert{\bf a}\vert^{\,2} \,\vert{\bf b}\vert^{\,2} - \vert{\bf a...
...\cos^2\theta = \vert{\bf a}\vert^{\,2}\,\vert{\bf b}\vert^{\,2}\, \sin^2\theta.$ (A.43)


$\displaystyle \vert{\bf a}\times{\bf b}\vert = \vert{\bf a}\vert\,\vert{\bf b}\vert\,\sin\theta,$ (A.44)

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

Consider the parallelogram defined by the vectors $ {\bf a}$ and $ {\bf b}$ . (See Figure A.9.) The scalar area of the parallelogram is $ a\,b \sin\theta$ . By convention, the vector area has the magnitude of the scalar area, and is normal to the plane of the parallelogram, in the sense obtained from a right-hand circulation rule by rotating $ {\bf a}$ on to $ {\bf b}$ (through an acute angle). In other words, if the fingers of the right-hand circulate in the direction of rotation then the thumb of the right-hand indicates the direction of the vector area. So, the vector area is coming out of the page in Figure A.9. It follows that

$\displaystyle {\bf S} = {\bf a}\times {\bf b},$ (A.45)

Figure A.9: A vector parallelogram.
\epsfysize =1.25in

Suppose that a force $ {\bf F}$ is applied at position $ {\bf r}$ , as illustrated in Figure A.10. The 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 torque is $ \vert{\bf F}\vert\,\vert{\bf r}\vert\,\sin\theta$ . The direction of the torque is conventionally defined as the direction of the axis through $ O$ about which the force tries to rotate objects, in the sense determined by a right-hand circulation rule. Hence, the torque is out of the page in Figure A.10. It follows that the vector torque is given by

$\displaystyle \mbox{\boldmath$\tau$}$$\displaystyle = {\bf r}\times{\bf F}.$ (A.46)

Figure A.10: A torque.
\epsfysize =2.25in

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

$\displaystyle {\bf l}={\bf r}\times {\bf p}.$ (A.47)

next up previous
Next: Rotation Up: Vectors and Vector Fields Previous: Vector Area
Richard Fitzpatrick 2016-03-31