It is proved that the surface represented by the equation (x-z) square + (y + z-a) square = a square is a cylinder

It is proved that the surface represented by the equation (x-z) square + (y + z-a) square = a square is a cylinder

The elimination of square expansion energy
We only need to prove that when Z is any value,
X and y are circles with equal radius and the center of the circle is on a straight line

The equation of a straight line passing through point (1,2, - 4) and perpendicular to plane 2x-3y + Z-4 = 0

The line is perpendicular to the plane, that is, the direction vector of the line is parallel to the normal vector of the plane
The normal vector of the plane is {2, - 3,1}, so let the direction vector of the straight line be {2, - 3,1} and the straight line pass through {1,2, - 4}. It is easy to get the equation of the straight line and write it in the form of the ratio of the direction vector
(x-1)/2= (y-2)/-3 =(z+4)/1

Solving the plane equation of the intersection of point m (1, - 2,3) and two planes 2x-3y + Z = 3, x + 3Y + 2Z + 1 = 0