When adding vectors in 3D, the steps are basically the same as adding vectors in 2D, except that drawing diagrams for 3D problems is often skipped because it's more complicated to do.

1. Resolve the vectors into components, if necessary.
2. Organize your known information.
3. Add or subtract vectors as required.
4. Convert back to magnitude-angle notation (unless the problem only asks for components).

For step 1, we'll usually use equations of this form to find our components when the components aren't given to us:

$\boxed{A_x=\left|\stackrel{\rightharpoonup }{A}\right|\cos \alpha }$,  $\boxed{A_y=\left|\stackrel{\rightharpoonup }{A}\right|\cos \beta }$, $\boxed{A_z=\left|\stackrel{\rightharpoonup }{A}\right|\cos \gamma }$

Steps 2 and 3 can be done using a table.

For step 4, we'll use the same equations from step 1 solved for α, plus the equation for magnitude:

$\boxed{\left|\stackrel{\rightharpoonup }{A}\right|=\sqrt{{A_x}^2+{A_y}^2+{A_z}^2}}$

This problem looks tricky at first because it's very wordy! But let's tackle it one piece at a time:

Step 1. Notice that the first displacement, let's call it $\stackrel{\rightharpoonup }{A}$, is a quarter-circle that starts with a motion straight upwards. The path is along an arc, but the displacement is from the starting point at the edge of the arc to the final point. If we imagine an origin (0,0) at the center of the arc (or sketch it out that way), it's easy to see that the initial position vector is horizontal, the final position vector is vertical, and both of them have magnitudes equal to the radius of the circle (4.80 cm).

Knowing that, we don't have to do any math to find the components: we can just say

$A_{up}=4.80 \mathrm{cm}$ and $A_{east}=4.80 \mathrm{cm}$.

We can use the same reasoning for the second displacement, $\stackrel{\rightharpoonup }{B}$. We know it's a quarter-circle up and to the north with radius 3.70 cm, so

$B_{up}=3.70 \mathrm{cm}$ and $B_{north}=3.70 \mathrm{cm}$.