Answer to Question #152945 in Analytic Geometry for Sourav Mondal

Question

Find the equation of the cylinder whose
axis is x = y = - z and radius is 2.

Answer 1

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, $P(p_x, p_y, p_z)$ 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 $OP = OA<OX,OA>/|OA|^2$ .

$<OX,OA> = x + y - z$

$|OA|^2 = 1^2 + 1^ 2 + (-1)^2 = 3$

$PX =OX - OP = (x-p_x,y-p_y,z-p_z)$

$x-p_x = x - (x+y-z)/3 = (2x-y+z)/3$

$y-p_y = y - (x+y-z)/3 = (2y-x+z)/3$

$z-p_z = z + (x+y-z)/3 = (x+y+2z)/3$

$4 =|PX|^2=((2x-y+z)^2 + (2y-x+z)^2+(x+y+2z)^2)/9$

$36 = 6(x^2+y^2+z^2-xy+yz+zx)$

$x^2+y^2+z^2-xy+yz+zx=6$

This is the equation of the cylinder with the axis $OA$ and radius 2.