# Math 241 - Quiz #3

1. Find the parametric equations for the line through the points (3,1,-1) and
(3,2,-6).

2. Find the equation of the plane through the point (6,5,-2) and parallel to
the plane x + y - z + 1 = 0.

3. Change from cylindrical to rectangular (cartesian) coordinates: (1,pi,e).

## Solutions:

1. Let v = P2-P1 = <0,1,-5>. Then using the first point as the starting
point, the parameteric equations are
x = 3

y = 1 + t

z = -1 - 5t

2. The normal for the plane is <1,1,-1>, so the general plane of
this form is x + y - z = d. Plugging in (6,5,-2), we get
d = 6 + 5 - (-2) = 13. Thus the equation is

x + y - z = 13

3. r = 1, theta = pi, z = e, so x = r cos theta = 1 cos pi = -1,
y = r sin theta = 1 sin pi = 0, and z = e. Thus in rectangular
coordinates the point is (-1,0,e).

ccollins@math.utk.edu
*Last Modified: Jan. 29, 1999*