|Feb22-09, 01:31 PM||#1|
Find the equation of the plane that satisfies the stated conditions.
The plane that contains the line x = -2 + 3t, y = 4 + 2t, z = 3 - t and is perpendicular to the plane x - 2y + z = 5.
|Feb22-09, 01:35 PM||#2|
Nevermind, I kinda just solved it myself.
|Feb22-09, 06:49 PM||#3|
Find the equation of the plane (if you dare) that is tangent to the following spheres:
sphere 1: r=2 center P(2,2,2)
sphere 2: r=3 center Q(3,4,5)
Ok, I know that the plane will be parallel to the vector PQ = <1,2,3>
For the equation of a plane I need a normal vector, and a point on the plane. I know that one of the points on the plane will be one of the points on the sphere (either sphere will do, right?).
The equations for the spheres are:
(x-2)^2 + (y-2)^2 + (z-2)^2=4
(x-3)^2 + (y-4)^2 + (z-5)^2=9
What i am thinking now is that i should take the midpoint between the spheres' centers. Use that to find another vector parallel to the plane in question, and then generate a normal vector from that - is this correct thinking?
And from there, i am unsure how to find a point common to the plane and the sphere (either sphere).
I have drawn this out on paper and understand the geometry...
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