8/23/2005 The Position Vector.doc 1/7 Jim Stiles The Univ. of Kansas Dept. of EECS The Position Vector Consider a point whose location in space is specified with Cartesian coordinates (e.g., P(x,y,z)). Now consider the directed distance (a vector quantity!) extending from the origin to this point. This particular directed distance—a vector beginning at the origin and extending outward to a point—is a very important and fundamental directed distance known as the position vector r . Using the Cartesian coordinate system, the position vector can be explicitly written as: ˆˆˆ = xyzrxayaza++ x y z P(x,y,z) r8/23/2005 The Position Vector.doc 2/7 Jim Stiles The Univ. of Kansas Dept. of EECS * Note that given the coordinates of some point (e.g., x =1, y =2, z =-3), we can easily determine the corresponding position vector (e.g., 23xyzˆˆˆra a a=+ − ). * Moreover, given some specific position vector (e.g., 42yzˆˆraa=−), we can easily determine the corresponding coordinates of that point (e.g., x =0, y =4, z =-2). In other words, a position vector r is an alternative way to denote the location of a point in space! We can use three coordinate values to specify a point’s location, or we can use a single position vector r. I see! The position vector is essentially a pointer. Look at the end of the vector, and you will find the point specified! P(r) r8/23/2005 The Position Vector.doc 3/7 Jim Stiles The Univ. of Kansas Dept. of EECS The magnitude of r Note the magnitude of any and all position vectors is: 222rrrxyzr=⋅= ++= The magnitude of the position vector is equal to the coordinate value r of the point the position vector is pointing to! A: That’s right! The magnitude of a directed distance vector is equal to the distance between the two points—in this case the distance between the specified point and the origin! Alternative forms of the position vector Be careful! Although the position vector is correctly expressed as: ˆˆˆr = xyzxa ya za++ Q: Hey, this makes perfect sense! Doesn’t the coordinate value r have a physical interpretation as the distance between the point and the origin?8/23/2005 The Position Vector.doc 4/7 Jim Stiles The Univ. of Kansas Dept. of EECS It is NOT CORRECT to express the position vector as: ˆˆˆr pzaazaφρφ++≠ nor ˆˆˆr rra a aθφθφ++≠ NEVER, EVER express the position vector in either of these two ways! It should be readily apparent that the two expression above cannot represent a position vector—because neither is even a directed distance! A: Recall that the magnitude of the position vector r has units of distance. Thus, the scalar components of the position vector must also have units of distance (e.g., meters). The coordinates ,,,xyzρ and r do have units of distance, but coordinates and θφ do not. Thus, the vectors ˆθθa and ˆφφa cannot be vector components of a position vector—or for that matter, any other directed distance! Q: Why sure—it is of course readily apparent to me—but why don’t you go ahead and explain it to those with less insight!8/23/2005 The Position Vector.doc 5/7 Jim Stiles The Univ. of Kansas Dept. of EECS Instead, we can use coordinate transforms to show that: ˆˆˆˆˆˆcos sinˆˆˆsin cos sin sin cosxyzxyzxyzrx y zzrrrρφ ρφθφθφ θ=++=++=++aaaaaaaaa ALWAYS use one of these three expressions of a position vector!! Note that in each of the three expressions above, we use Cartesian base vectors. The scalar components can be expressed using Cartesian, cylindrical, or spherical coordinates, but we must always use Cartesian base vectors. A: The reason we only use Cartesian base vectors for constructing a position vector is that Cartesian base vectors are the only base vectors whose directions are fixed—independent of position in space!Q: Why must we always use Cartesian base vectors? You said that we could express anyvector using spherical or base vectors. Doesn’t this also apply to position vectors?8/23/2005 The Position Vector.doc 6/7 Jim Stiles The Univ. of Kansas Dept. of EECS To see why this is important, let’s go ahead and change the base vectors used to express the position vector from Cartesian to spherical or cylindrical. If we do this, we find: ˆˆˆˆˆˆxyzzrrx y zzrρρ=++=+=aaaaaa Thus, the position vector expressed with the cylindrical coordinate system is ˆˆρzrρz=+aa, while with the spherical coordinate system we get ˆrrr=a. The problem with these two expressions is that the direction of base vectors ˆρa and ˆra are not constant. Instead, they themselves are vector fields—their direction is a function of position! Thus, an expression such as ˆ6rr=a does not explicitly define a point in space, as we do not know in what direction base vector ˆra is pointing! The expression ˆ6rr=a does tell us that the coordinate r =6, but how do we determine what the values of coordinates or θφ are? (answer: we can’t! ) Compare this to the expression: ˆˆˆ 23xyzra aa=+−8/23/2005 The Position Vector.doc 7/7 Jim Stiles The Univ. of Kansas Dept. of EECS Here, the point described by the position vector is clear and unambiguous. This position vector identifies the point P(x =1, y =2, z =-3). Lesson learned: Always express a position vector using Cartesian base vectors (see box on previous