Question

suppose P=(2,-4,3) Q=(0,-1,2) and R=(1,1,1). Find the plane PQR

Answer 1

First, find two vectors in the plane PQR:
PQ=<0-2,-1-(-4),2-3>=<-2,3,-1>
PR=<1-2,1-(-4),1-3>=<-1,5,-2>
PQxPR is the normal to the plane PQR:
i j k
-2 3 -1
-1 5 -2
=<-1,-3,-7>

The plane is defined by the normal passing through one of the three points, we can choose (1,1,1) for convenience:
-1(x-1) -3(y-1) -7(z-1)=0
which simplifies to
x+3y+7z-11=0
as the equation of the required plane.
Check that all points PQR lie in the plane.

Answer 2

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