The plane: traces and classification

Plane α has normal vector n(1, 3, 3) and passes through A(15, 7, 8). Find its traces and classify it.

Step-by-step solution

Plane equation: x + 3y + 3z − 60 = 0.
Horizontal trace α₁: set z = 0 and draw the resulting line on the PH. If no variable remains, the plane is parallel to the PH.
Vertical trace α₂: set y = 0 and draw the resulting line on the PV. If no variable remains, the plane is parallel to the PV.
Classify from the normal: a zero component means a projecting plane or a plane parallel to the ground line.

Result: oblique plane (general). α₁: x + 3y − 60 = 0; α₂: x + 3z − 60 = 0.

If the normal has a zero component, the plane contains that direction: that is where projecting and ground-line-parallel planes come from.