1
$\begingroup$

Let us consider a fixed reference point P and another point Q in space. Suppose you want to express the position of Q with respect to P in cylindrical coordinate system. Now in the cylindrical coordinate system we imagine a cylinder whose axis is parallel to the z-axis of the Cartesian system and passes through P. Then we imagine the a circle which contains Q and to which the axis is normal. Now the distance between P and Q along the axis, the radius of the circle and the angle between the radius joining up to Q and Cartesian X-axis are the parameters which we need to describe a point completely in Cylindrical coordinate system. Now lets try to express this position of Q w.r.t P in vector form in Cylindrical system. My book says the unit vectors that are to taken are along the directions of Z-axis, the radius mentioned above and the tangent to that radius. However, we can clearly see that by adding up the vertical vector and the radial vector we can completely obtain the position vector of Q with respect to P . So why do we need a unit vector along the direction of the tangent ? Have I gotten anything wrong about the unit vectors to be chosen? Hope the members here would like to help me in sorting this out.

$\endgroup$
1
$\begingroup$

You can certainly express the position of $Q$ with respect to $P$ with just vectors along $z$ and $r$ in cylindrical coordinates. The $r$ vector depends on the angle. Often you need to construct a local coordinate system at $Q$ for whatever problem you are solving. For example, you might be investigating how the forces on $Q$ depend on small changes in its position. For that you need three vectors. It is natural to take two along increasing $z$ and $r$. Then the third is along the tangent, as you say.

$\endgroup$
  • 1
    $\begingroup$ You have some missing latex $\endgroup$ – zodiac Sep 19 '13 at 14:05
  • $\begingroup$ @Ross : Is there anything for which the book doesn't mention about the sufficiency of two unit vectors ? $\endgroup$ – Primeczar Sep 19 '13 at 14:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.