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:

$${\bf a} \ast{\bf b} \equiv (a_x \,b_x,\, a_y\, b_y,\, a_z \,b_z).$$ (1.36)

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, the previous definition is a bad one.

Consider, now, the cross product or vector product:

$${\bf a}\times{\bf b} \equiv (a_y \, b_z-a_z\, b_y,\, a_z\, b_x - a_x\, b_z,\, a_x\, b_y - a_y\, b_x) ={\bf c}.$$ (1.37)

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,

$$c_{x'} = (-a_x\, \sin\theta + a_y\,\cos\theta)\,b_z - a_z\,(-b_x\, \sin\theta + b_y\,\cos\theta)$$

$$= (a_y\, b_z - a_z\, b_y)\, \cos\theta + (a_z\, b_x-a_x\, b_z)\,\sin\theta$$

$$= c_x\,\cos\theta +c_y\,\sin\theta.$$ (1.38)

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 (and is non-coplanar with ${\bf a}$ and ${\bf b}$). The cross product is anticommutative,

$${\bf a}\times{\bf b} = - {\bf b} \times{\bf a},$$ (1.39)

distributive,

$${\bf a}\times({\bf b} +{\bf c})= {\bf a} \times{\bf b}+{\bf a}\times{\bf c},$$ (1.40)

but is not associative,

$${\bf a}\times({\bf b} \times{\bf c})\neq ({\bf a}\times{\bf b}) \times{\bf c}.$$ (1.41)

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,

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

$$=0.$$ (1.42)

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: A.8 The right-hand rule for cross products. Here, $\theta$ is less that $180^\circ$.

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

$$({\bf a}\times{\bf b})^2 = (a_y \,b_z-a_z\, b_y)^2 +(a_z \,b_x - a_x\, b_z)^2 +(a_x \,b_y -a_y \,b_x)^2$$

$$= (a_x^{~2}+a_y^{~2}+a_z^{~2})\,(b_x^{~2}+b_y^{~2}+b_z^{~2}) - (a_x\, b_x + a_y \,b_y + a_z\, b_z)^2$$

$$= \vert{\bf a}\vert^{\,2} \,\vert{\bf b}\vert^{\,2} - ({\bf a}\cdot {\bf b})^2$$

$$= \vert{\bf a}\vert^{\,2} \,\vert{\bf b}\vert^{\,2} - \vert{\bf a... ...{\,2} \,\cos^2\theta = \vert{\bf a}\vert^2\,\vert{\bf b}\vert^2\, \sin^2\theta.$$ (1.43)

Thus,

$$\vert{\bf a}\times{\bf b}\vert = \vert{\bf a}\vert\,\vert{\bf b}\vert\,\sin\theta,$$ (1.44)

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); that is, 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

$${\bf S} = {\bf a}\times {\bf b},$$ (1.45)

Figure A.9: A vector parallelogram.

Suppose that a force ${\bf F}$ is applied at position ${\bf r}$. See 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

$$\mbox{\boldmath$\tau$} = {\bf r}\times{\bf F}.$$ (1.46)

Figure A.10: A torque.

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,

$${\bf l}={\bf r}\times {\bf p}.$$ (1.47)