# Vectors in Cartesian Cylindrical Spherical

• ha9981
In summary, at point (x, y, z) = (1, 0, 0), if vectors A and B are perpendicular, then theta is 0. If vectors A and B are not perpendicular, then theta is undefined.

#### ha9981

I do not understand when we are given a vector at point P(x,y,z) or in different forms cylindrical and spherical. What does it mean at point?? I mean aren't vectors supposed to start at origin, even if they don't how will that make a difference in their magnitude or angle between them.

For example we are given two vectors A(2,-2, 1) and B(1,2,-2). They are in different systems A is in spherical and B in in cylindrical and I don't have a problem handling that. Now what does it mean when it says: if they are perpendicular at spherical coordinate point P(2, theta, 60 deg) find theta. Now I know dot product is zero when they are perpendicular. I have no idea how how this has anything to do with theta as what does it mean when it says at point P?

ha9981 said:
I do not understand when we are given a vector at point P(x,y,z) or in different forms cylindrical and spherical. What does it mean at point?? I mean aren't vectors supposed to start at origin, even if they don't how will that make a difference in their magnitude or angle between them.

For example we are given two vectors A(2,-2, 1) and B(1,2,-2). They are in different systems A is in spherical and B in in cylindrical and I don't have a problem handling that. Now what does it mean when it says: if they are perpendicular at spherical coordinate point P(2, theta, 60 deg) find theta. Now I know dot product is zero when they are perpendicular. I have no idea how how this has anything to do with theta as what does it mean when it says at point P?
Hello ha9981.

Position vectors start at the origin. In general vectors are associated with points quite removed from the origin. Perhaps it's the way that vector addition is presented that makes it seem that vectors may be moved around at will.

The direction of unit vectors, $$\hat{i},\,\hat{j},\,\hat{k}$$ for Cartesian coordinates, (x, y, z) is invariant under change of position.

The direction of the unit vectors for spherical coordinates, $$\hat{r},\,\hat{\theta},\,\hat{\phi}$$ depends on the coordinate of the point at which they are applied.

Translate your vectors, $$\vec{A},\,\text{ and }\vec{B}$$ into rectilinear coordinates at different positions, and you will get different results.

Look at this link to http://mathworld.wolfram.com/SphericalCoordinates.html" [Broken].

At point (x, y, z) = (1, 0, 0):

$$\hat{r}=\hat{i},\ \hat{\theta}=\hat{j},\ \hat{\phi}=-\hat{k}\,.$$

At point (x, y, z) = (–1, 0, 0):

$$\hat{r}=-\hat{i},\ \hat{\theta}=\hat{j},\ \hat{\phi}=-\hat{k}\,.$$

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ha9981 said:
I do not understand when we are given a vector at point P(x,y,z) or in different forms cylindrical and spherical. What does it mean at point?? I mean aren't vectors supposed to start at origin, even if they don't how will that make a difference in their magnitude or angle between them.

For example we are given two vectors A(2,-2, 1) and B(1,2,-2). They are in different systems A is in spherical and B in in cylindrical and I don't have a problem handling that. Now what does it mean when it says: if they are perpendicular at spherical coordinate point P(2, theta, 60 deg) find theta.
This doesn't make sense. Two vectors are either perpendicular or they're not. Coordinates don't have anything to do with it. However, if you're dealing with the tangent vectors of a curve, this is of course a bunch of different vectors, one at each point on the curve:

If C:[a,b]→ℝ3 is a curve, then C'(t)=(C'1(t),C'2(t),C'3(t)) is the tangent vector of the curve at C(t). In particular, if C(t) is the position at time t, then C'(t) is the velocity at time t.

So is it possible that your book is talking about the tangent vectors of a curve?

SammyS said:
Look at this link to http://mathworld.wolfram.com/SphericalCoordinates.html" [Broken].

At point (x, y, z) = (1, 0, 0):

$$\hat{r}=\hat{i},\ \hat{\theta}=\hat{j},\ \hat{\phi}=-\hat{k}\,.$$

At point (x, y, z) = (–1, 0, 0):

$$\hat{r}=-\hat{i},\ \hat{\theta}=\hat{j},\ \hat{\phi}=-\hat{k}\,.$$
These results are found by choosing a specific point in ℝ3 (i.e. a specific vector in ℝ3), and treating one of the three spherical coordinates as a variable while holding the others constant. This defines a curve through the point. The tangent of that curve is found by differentiation (as I described above), and the result is then normalized to length 1.

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