Question

1) A ball moves with velocity v=(2, 3.4, 6) m/s (meter per second). At time t = 0, it is located at position p=(0, 5, 2).
  
   a) Find the equation of its path in parametric form.

b) At time t = 10 second, what is the ball's position?

2) Consider the plan x + 2y+8z - 90 = 0.
   a) What is the normal vector of this plane?
  

  
   b) What is the distance of the ball in previous question at time t=0 with plane? Which side of the plane is the ball located?
  
  
  
   c) At what time the ball will hit the plane? (hint: find the intersection of the line and the plane. For that, insert x, y, and z from the line into the equation of the plane, and from there solve for t. THis will be the time ball hits the plane.)
  
  

  
   d) What is the position of the collision of the ball with the plane?

Answer 1

s(t) = Qt, 3.44+5, 64+2 Required 6 eam of path at t=10 & ft) = (2x10, 3.4x10+5, 681042) et 2 (20, 39, 62) Scanned with CamScanner