Cartesian Components of a Vector

Consider a Cartesian coordinate system $Oxyz$ consisting of an origin, $O$, and three mutually perpendicular coordinate axes, $Ox$, $Oy$, and $Oz$. See Figure A.4. Such a system is said to be right-handed if, when looking along the $Oz$ direction, a $90^\circ$ clockwise rotation about $Oz$ is required to take $Ox$ into $Oy$. Otherwise, it is said to be left-handed. In physics, it is conventional to always use right-handed coordinate systems.

Figure A.4: A right-handed Cartesian coordinate system.

It is convenient to define unit vectors, ${\bf e}_x$, ${\bf e}_y$, and ${\bf e}_z$, parallel to $Ox$, $Oy$, and $Oz$, respectively. Incidentally, a unit vector is a vector whose magnitude is unity. The position vector, ${\bf r}$, of some general point $P$ whose Cartesian coordinates are ($x$, $y$, $z$) is then given by

$${\bf r} = x\,{\bf e}_z + y\,{\bf e}_y+z\,{\bf e}_z.$$ (1.11)

In other words, we can get from $O$ to $P$ by moving a distance $x$ parallel to $Ox$, then a distance $y$ parallel to $Oy$, and then a distance $z$ parallel to $Oz$. Similarly, if ${\bf a}$ is an arbitrary vector then

$${\bf a} = a_x\,{\bf e}_x+ a_y\,{\bf e}_y+a_z\,{\bf e}_z,$$ (1.12)

where $a_x$, $a_y$, and $a_z$ are termed the Cartesian components of ${\bf a}$. It is conventional to write ${\bf a} \equiv (a_x,\,a_y,\,a_z)$. It follows that ${\bf e}_x\equiv (1,\,0,\,0)$, ${\bf e}_y\equiv (0,\,1,\,0)$, and ${\bf e}_z\equiv (0,\,0,\,1)$. Of course, ${\bf0} \equiv (0,\,0,\,0)$.

According to the three-dimensional generalization of the Pythagorean theorem, the distance $OP\equiv \vert{\bf r}\vert=r$ is given by

$$r = \sqrt{x^{\,2} + y^{\,2} + z^{\,2}}.$$ (1.13)

By analogy, the magnitude of a general vector ${\bf a}$ takes the form

$$a = \sqrt{a_x^{\,2} + a_y^{\,2} + a_z^{\,2}}.$$ (1.14)

If ${\bf a} \equiv (a_x,\,a_y,\,a_z)$ and ${\bf b}\equiv (b_x,\,b_y,\,b_z)$ then it is easily demonstrated that

$${\bf a} + {\bf b} \equiv (a_x+b_x,\,a_y+b_y,\,a_z+b_z).$$ (1.15)

Furthermore, if $n$ is a scalar then it is apparent that

$$n\,{\bf a} \equiv (n\,a_x,\,n\,a_y,\,n\,a_z).$$ (1.16)