Question

Consider the points
P1 = (1, -1, 1) , P2 = (2, 1, 0) , P3 = (-2, 2, 3)

Find an equation for the plane through P1, P2, P3 (of the form ax + by + cz = d), and determine if it meets the line segment with endpoints Q1 = (0, -1, 0) and Q2 = (2, 0, -1)

Answer 1

To find the equation of a plane we need a vector in the plane (r-r0) and a vector normal to the plane (n), such that:\[n*(r-r _{0})=0\]We have three points within the plane from which we can derive two vectors in the plane. The vectors
P1P2=<1,2,-1> and P1P3=<-3,3,2> must lie within our plane. Now if we only knew a vector normal to our plane we would have our answer. How do we get a normal vector? Take the cross product of our vectors P1P2 and P1P3!
P1P2XP1P3=<7,1,9>. Now we have all we need to derive our plane equation.
\[n*(r-r _{0})=0\]Taking n=<7,1,9>, r=<x,y,z> and r0=P1=<1,-1,1> we get,
<7,1,9>*<x-1,y+1,z-1>=0
7x-7+y+1+9z-9=0
7x+y+9z=15

Answer 2

Now what is the relationship of our plane given by 7x+y+9z=15 with the line segment given by Q1=(0,-1,0) and Q2=(2,0,-1)? Okay, lets enter point Q1 into our plane equation and see where it is relative to the plane:
-1<15
So Q1 is not on the plane but it lies below the plane (since -1 is less than 15; think of the plane as splitting space in two parts, those points (x,y,z) with 7x+y+9z>15 and those points (x,y,z) with 7x+y+9z<15). Now try Q2:
5<15
Again Q2 is not on the plane and in fact is beneath the plane. So we have two points beneath the plane. Connecting them to form a line-segment gives is a line segment beneath the plane. Our plane does not meet the line-segment given by Q1, Q2.