In (east,north,up) coordinate system, the distance from the
airplane to the radar tower is (3,0,-4) miles　and the velocity
of the airplane is (a) (450,0,5) mph, (b) (450/√2,450/√2,5) mph.
The rate the airplane is approaching the tower is the component
of the velocity vector in the direction of the distance vector.
So, the rate is calculated as a innerproduct of the vectors above.
(a) (450,0,5)・(3,0,-4)/|(3,0,-4)| = 266 mph
(b) (450/√2,450/√2,5)・(3,0,-4)/|(3,0,-4)| = 135√2 - 4 mph
英語が合っとるかは、知らん。