Plane Passing a Line and a Point

 

Motivation:

                      Given a point P and a line L such that P is not on L,  how many planes we can find to pass the point P and line L ?

 

                       The answer is One – only one plane !

 

                       How do we come up with that solution?  We can think in the following way.  Given a line, there exist many planes containing that line.

 

                                  

 

                      We just start with a plane containing the line L, and use line L as an axis to  rotate the plane slightly.  When the plane hits the point P,  we know that is the plane we are looking for.  And there exists only one such plane.

 

                      After figuring this out, we start to  paraphrase this problem by using equations.

 

Question ( Plane Passing a Point and a Line )

                  Given a point (1,1,1 ) and a line L with the following line equation

 

                                 L : 

                      find the plane passing the point P and the line L.

 

        Sol:

                       

                            We consider the problem in this way:  if a plane is containing a line and a point outside the line,  then the normal vector is perpendicular to the directional vector of the line.

                                     

 

                        Furthermore,  if we can find a point lying on the line, we can create a vector from that point to point P. And the vector we create must be also perpendicular to the normal vector of the plane.  We start finding the normal vector with the observation above.

 

                    From the equation of L, we know that Q(0,1,0 ) is on line L.

                    Thus,

                                    =  -  =  ( 1, 0, 1 )

 

                    Furthermore, the directional vector of line L is  ( 2, 1, 3 ).

 

                    We would like to find a vector that is perpendicular to ( 1, 0, 1)

                    and (2,1,3).  So, we can use it as the normal vector of the plane.

 

                    Naturally, we will consider the outer product of the two vectors:

 

                                  (2,1,3)  (1,0, 1) =  ( 1, 1, -1 )

 

                           2     1       3        2         1

                                 

                           1     0       1         1         0

 

                               Let’s just denote this normal vector as , and set

 

                                    = ( 1, 1, -1 ) = (2,1,3)  (1,0, 1)

 

                       With normal vector and a point on the plane Q,  the plane equation is

                  

                                        ( x-0, y-1, z-0 ) = 0

 

                              x + (y-1) –z =0

                               x + y –z = 1

 

                             That’s the plane equation we are looking for.