Let be the plane which is parallel to and which passes through the intersection of the planes
Find a Cartesian equation for .
First, we find the intersection of the two planes. This is the set of points which simultaneously satisfy
From the first equation we have
Plugging this into the second equation we have
This then gives us . So the line is the set of points where is arbitrary. Thus,
The plane must then contain this line and be parallel to the vector . Since it is parallel to its normal must be perpendicular to , thus,
and we must have
This has the Cartesian equation
Prove that if and are two planes which are not parallel then they intersect in a line.
Proof. Let the Cartesian equations of and be given by
respectively. Then, the intersection is given by the common solutions of these two equations. Since and are not parallel, we know they do not have the same normal vector so that for all . Further, since the normals are nonzero, we know each equation has at least one nonzero coefficient. Without loss of generality, let . Then,
Substituting into the Cartesian equation for we have
is the set of solutions for the points on . But, we know at least one of or is nonzero, otherwise . Hence, we have the equation for a line. Therefore, is a line
Consider a plane given by the equation
Find the Cartesian equation for a plane parallel to this one and the same distance as this plane from the point .
Since the requested plane is parallel to the given plane we know that they must have the same normal vector, . Therefore, the Cartesian equation of the requested plane is of the form
From the previous exercise (Section 13.17, Exercise #19) we know the distance from to a plane is given by the formula
Therefore, the distance from the given plane to the point is
Since the distance from the point to the requested plane must be the same we must have
(Since the solution belongs to the other plane.)
Prove that three planes with normal vectors which are linearly independent intersect in exactly one point.
Proof. Let the normals of the three planes be given by
Then, the Cartesian equations of the three planes are given by
Since the normals are independent we know that they span the zero vector uniquely (by definition of independence). BY Theorem 12.7 (page 463 of Apostol) this implies that they span every vector in . Hence, the vector equation has a unique solution . Therefore, the system of equations has a unique solution. Hence, there is exactly one point on the intersection of three planes with linearly independent normal vectors
Consider three planes given by the Cartesian equations
Find all of the points which are on the intersections of these three planes.
The points on the intersection must satisfy the system of equations
To find the points satisfying these equations we row-reduce the coefficient matrix of the system (if you don’t have any linear algebra, then you can use Gaussian elimination the long way),
Therefore, we have , , and . So, the only point on the intersection is the point .