What Is the Displacement Vector of a Clock's Minute Hand at Different Times?

[-Dragoon-]

Homework Statement

The minute hand on a watch is 2.0 cm in length. What is the displacement vector of the tip of the minute hand:
a) From 8:00 to 8:20 A.M.?
b) From 8:00 to 9:00 A.M.?

Homework Equations

N/A.

The Attempt at a Solution

So, first I drew a diagram of the clock and its initial position. Assuming 3:00 and 9:00 to be the x-axis, and 12:00 and 6:00 to be the Y-axis, and that 6:00 makes a 90° with 3:00, then when the minute hand points to 4:00 (denoting 8:20), it's angle, theta, would then make an angle of 30° with the negative x-axis. So, the displacement of the minute hand would then be:
$\vec{d} = 4 * 2.0 cm = 8 cm$ [30° below the horizontal]
Is this correct?

As for b, the displacement is equal to 0, since the minute hand ends up exactly where it started after a change in distance from 8:00 to 9:00 A.M., correct?

 

[NascentOxygen]

So, the displacement of the minute hand would then be:
$\vec{d} = 4 * 2.0 cm = 8 cm$
Is this correct?

So you have a triangle with two sides of 2 cm, and the third side of 8 cm?

Can you draw that?

 

[-Dragoon-]

So you have a triangle with two sides of 2 cm, and the third side of 8 cm?

Can you draw that?

So, now I would need break the 2 cm side that makes an angle of 30 degrees with the negative x-axis into components, correct?

 

[NascentOxygen]

So, now I would need break the 2 cm side that makes an angle of 30 degrees with the negative x-axis into components, correct?

At 20 mins past, the hand makes an angle with the positive x-axis.

Any method that gets the right answer is okay. I broke the triangle up into two equal triangles.