What is the relative velocity of this plane relative to the helicopter? There is a plane moving 100 m/s due east relative to the ground without vertical motion. There is a helicopter facing north moving straight up at 20 m/s. From the perspective of the helicopter, is the plane moving faster or slower than 100 m/s away from the helicopter. The book I am using says the plane's relative velocity is greater than 100 m/s, but I think it is dependent on the initial position of the helicopter relative to the plane.
If the helicopter is 20 meters straight below the plane in one instance, then one second later, the helicopter would be at the same altitude as the plane and the plane would be 100 meters away, so the average relative velocity would then be 80 m/s, no? That would mean that the relative velocity was, at least at some point, less than 100 m/s.
Is my understanding correct, or am I missing something, and the book is correct?
Here is a diagram of what I am trying to express: 
 A: You are missing something. "Velocity" has both magnitude and direction - it is a vector. When you say "the distance changed by x" you can't simply divide x by time taken to get velocity. Imagine you start 10 m to my left, and walk at 2 m/s for ten seconds. Now you are 10 m to my right. So the distance didn't change - I guess your velocity relative to me is zero.  See the point?
You should draw a vector diagram of the motion of the airplane in the frame of reference of the helicopter - this will be a vector with a horizontal component of 100 and a vertical component of 20. The length of this vector is the velocity you want.
Leaving the diagram and math as exercise for the student... this is homework, after all.
A: Further to the correct answer of @floris:
Write the $(x, y, z)$ velocity components of the plane.  Here, conveniently, $x = \text{East}$,$y = \text{North}$, $z = \text{Up}$ 
So, the plane's velocity is $(100,0,0)$
In a similar fashion, the helicopter's velocity is $(0,0,20)$
Now to get the velocity of the plane relative to the helicopter, just subtract the helicopter's three velocity components from the corresponding velocity components of the plane.
So the plane is moving at $(100,0,-20)$ meter/sec relative to the helicopter.  This velocity vector always has a magnitude greater than 100.
Note that the use of the word "away" in the question makes the question unanswerable...
A: You're right - it IS dependent upon the helicopter's initial position.
To begin with, we don't know whether the helicopter is in front of, or behind, the airplane. We don't know if the helicopter is within the plane's line of travel at all - it could be far to either side. It could be far below or above the plane's line of travel, or it could cross that line of travel.
Taken to an extreme. if the helicopter is an infinite distance away from the airplane in any direction, on a line perpendicular to the airplane's line of travel, the airplane will have ZERO (infinitesimal) velocity with respect to the helicopter. If the helicopter crosses the airplane's line of travel, however, then at the moment of crossing the airplane will have 100m/s velocity relative to the helicopter.
Further... since the helicopter is climbing, if its initial position is above the airplane's line of travel, at NO point will the airplane's velocity relative to the helicopter be higher than 100 m/s.
