Converting Polar coordinates to Cartesian coordinates

[uchicago2012]

Homework Statement

Write the vectors B,D, and F in the figure in Cartesian form, with unit vectors.
(See attachments)

Homework Equations

a_x = a cos theta
a_y = a sin theta
where a = magnitude of vector a, and theta = the angle vector a makes with the positive direction of the x axis

The Attempt at a Solution

I'm having trouble with this "the angle vector a makes with the positive direction of the x axis" business. What does that mean, exactly? The smallest angle that the vector can make with the positive direction of the x axis? Because there are two angles it could make, technically. My book is not specific and I'm confused.

In one part of my book it has a note: "Angles that are measured relative to the positive direction of the x-axis are positive if they are measured in the counter clockwise direction and negative if measured clockwise. For example, 210 degrees and -150 degrees are the same angle"

I thought maybe it's just talking about a protractor but then I wondered if it had anything to do with the angle theta business.

So how exactly does one find the angle a vector makes with the positive direction of the x axis?

The thetas I found for the vectors in the images are:
Vector B: 53 degrees
Vector D: 143 degrees
Vector F: I haven't any idea

For Vector B, I arbitrarily decided to use the smaller angle.
For Vector D, I stuck with the smaller angle idea and found theta to be 143 degrees.
For Vector F, I'm puzzled by the addition of the z axis, as our book doesn't have a formula for resolving the vector of a three dimensional vector, or at least we haven't covered it and I haven't found it. I also wonder how I'm supposed to interpret the "the angle the vector makes with the positive direction of the x axis" business on a three dimensional coordinate system.

 

[Delphi51]

You did the B and D vectors correctly. For B, you could say either 53 degrees or -307 degrees (going clockwise from the x-axis around to the vector) and both give the same answer when you do the sine or cosine of them. That's all there is to it. The F vector example is 3D; save it for next year.

 

[uchicago2012]

But I have to know theta for vector F. This is an actual problem. Or I think I have to know it to put vector F in Cartesian form.

I couldn't figure out theta by looking at the diagram (is it even possible to figure out the angle by looking at the diagram?) so I tried this equation:

u dot v = |u| |v| cos theta

where u and v are vectors. If I let u = vector f and v = unit vector xhat, then theta should be the angle between them, correct?

Only I can't quite figure it out, because I need to know what F dot xhat is and I don't know. I was hoping it was something that didn't require the formula, so I could sub it in and then solve for theta. Or perhaps this is wishful thinking.

So any help for F dot xhat or another method of finding theta for vector F?

 

[Delphi51]

It isn't clear to me what the question is for F. Usually in 3D work, you give two angles, one in the xy plane plus the elevation so you would give the direction for F as 45 degrees from x plus a 30 degree elevation.

If, however, you want the angle between x and F you could find it by working with the triangle formed by the origin, point (1,0,0) on the x-axis and a point on the F vector 1 unit from the origin. I think its coords would be
[cos(30)*cos(45), cos(30)*sin(45), sin(30)].
You could find the distance from (1,0,0) to that point and knowing all three sides of the triangle figure out the angle you want with the law of cosines.

 

[uchicago2012]

For F, I'm assuming he wants us to just put it in regular Cartesian coordinates like the others. Maybe the z component will come out in the Cartesian coordinates, i.e. for vector D my answer was D = -16.0 xhat + 12.0 yhat, but I'm assuming for vector F it will be something like F = 1 xhat + 2 yhat + 3 zhat.

I tried to use the coordinates to find theta, but I got to one side of the right triangle is 1 unit and the hypotenuse is .88 units (from the distance formula, using the coordinates suggested) then I used the Pythagorean theorem to find the other side of the triangle,
1² + b² = (.88)²
b² = -.22

I can't take the root of a negative, so I'm at square one again. So any ideas on how to get theta (the angle between vector F and the x axis?)

On a side note, how did you get the coordinates you suggested, Delphi51? It seems useful.

 

[Delphi51]

Okay, I see the question now. And I even noticed on the diagram that the magnitude of F is 10. That makes it easy to find its cartesian coordinates. Focus on the triangle formed by F and the two dashed lines, right angle at their intersection, angle 30 degrees at the origin (90 - 60). The dashed line in the horizontal plane has length 10*cos(30) = 8.66. You can find the vertical dashed line easily and it is the z cartesian coordinate of F. To get the x coordinate, use the triangle in the x-y plane with side 8.66, a line parallel to the y-axis extending from the end of the 8.66 to the x-axis, and the 45 degree angle. You should see that the side on the x-axis has length 8.66*cos(45) = 6.12. So the final answer is
(6.12, 6.12, the length of the vertical dashed line)

 

[uchicago2012]

I don't exactly understand what you're doing. I got similar numbers, but I didn't apply them to the figure in the same way:

To find the distance between the origin and the point at which the dotted line intersects the y axis, I took (cos 30 degrees)(10) = 8.66. Shouldn't 8.66 be the y component of F, then, or am I not thinking 3-D?

To find the z component, I used a similar method. I found the distance between the point at which the dotted line intersects the y-axis and the vector F and I thought that might be the z component. I got (sin 30 degrees)(10) = 5.

I'm not sure how to find the x component though, I didn't really understand which triangle you were talking about using.

And wouldn't the answer be (x, 8.66, 5)? I don't understand where you got your numbers, or perhaps which triangles you are manipulating and how.

 

[Delphi51]

To find the distance between the origin and the point at which the dotted line intersects the y axis, I took (cos 30 degrees)(10) = 8.66. Shouldn't 8.66 be the y component of F, then, or am I not thinking 3-D?

According to my eye, the dotted line does not intersect the y axis. It falls well in front of the y-axis, hitting the xy plane at an angle of 45 degrees away from the y-axis. 10*cos(30) is the length of the dashed line in the xy plane that comes out from the origin 45 degrees away from the y-axis. This 10*cos(30) is not the x or the y coordinate. Multiply by sin(45) to get the x part and by cos(45) to get the y part.

 

[uchicago2012]

Ah, I see the 3-D triangle. I was seeing it two dimensionally. One last question: how do you know "the triangle in the x-y plane with side 8.66, a line parallel to the y-axis extending from the end of the 8.66 to the x-axis, and the 45 degree angle" has a 90 degree angle? I don't see it. Or I may not know enough geometry, or a combination of the above.

Thanks for your help, it makes sense now.

 

[Delphi51]

a line parallel to the y-axis extending from the end of the 8.66 to the x-axis

A line parallel to the y-axis meets the x-axis at 90 degrees.