1. Question About Parametrizing in 3 Dimensions

Let's say we have a circle of radius 2 centered at (1,2,5) in a plane parallel to the yz-plane.

Are both of these answers correct?
r(t) = (1, 2 + 2 * cos(t), 5 + 2*sin(t))
r(t) = (1, 2 + 2 * sin(t), 5 + 2 * cos(t))

I cannot think of a reason why both would not be correct. The only difference I see is what point t = a represents. I ask, because the back of my book only gives the first.

This is for my calculus III class.

Forget this question. I'm certain in my understanding.

2. Hello, 1005!

That's very perceptive . . . good for you!
You may not need convincing, but I agree with you . . .

We have a circle of radius 2 centered at (1,2,5) in a plane parallel to the yz-plane.

Are both of these answers correct?
. . $\displaystyle \begin{array}{ccc}r(t) &=& (1, \:2 + 2\cos t, \:5 + 2\sin t) \\ r(t) &=& (1, \:2 + 2\sin t, \:5 + 2\cos t)\end{array}$ . . . . Yes!

I cannot think of a reason why both would not be correct.
The only difference I see is what point $\displaystyle t = a$ represents.
I ask because the back of my book only gives the first.
For reference, my coordinate system is set up like this:
Code:
        z
|
|
|
|
+ - - - - - y
/
/
/
/
x

The positive $\displaystyle z$-axis is upward, the positive $\displaystyle y$-axis is to the right,
. . and the positive $\displaystyle x$-axis "comes out of the screen."

The plane of the circle is parallel to the "back wall,"
. . so the circle is facing us.

With your first equation, $\displaystyle t \:=\:0$ starts the circle at 3:00
. . and $\displaystyle t \to 2\pi$ generates the circle counter-clockwise.

With your second equation, $\displaystyle t \:=\:0$ starts the circle at 12:00
. . and $\displaystyle t \to 2\pi$ generates the circle clockwise.

Either way, we get the same circle.