Let O(0,0,0),A(1,1,−1) be two points on the cylinder axis, X(x,y,z) be any point on the cylinder surface, P(px,py,pz) be the orthogonal projection of the point X onto the axis. Then the length |XP| is a distance from X to the axis.
The vector OP is a projection of OX , it is equal to OP=OA<OX,OA>/∣OA∣2 .
<OX,OA>=x+y−z
∣OA∣2=12+12+(−1)2=3
PX=OX−OP=(x−px,y−py,z−pz)
x−px=x−(x+y−z)/3=(2x−y+z)/3
y−py=y−(x+y−z)/3=(2y−x+z)/3
z−pz=z+(x+y−z)/3=(x+y+2z)/3
4=∣PX∣2=((2x−y+z)2+(2y−x+z)2+(x+y+2z)2)/9
36=6(x2+y2+z2−xy+yz+zx)
x2+y2+z2−xy+yz+zx=6
This is the equation of the cylinder with the axis OA and radius 2.
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