Plane, Line, and Point in 3-dimensional Space

 

3D coordinate system

                                       As shown, it is composed by 3 axes X, Y, Z.   The arrow on each

                              axis is to indicate the positive direction on that axis.   There are two

                              kinds of systems for the choice of direction for the axes.  We usually

                              adopt the one shown on the left.  The 3 axes are  perpendicular to

                              each other.

 

                                  

 

 

 

 

Line Equation    There are many ways to describe a line in the 3D space. 

                            For example,  if two points on that line are known,

                            the line equation can be determined as follows:

 

                           Given two points  A(a,b,c), and  B( d,e,f) ,  the line passing

                            the points A and B can be described as  the set containing

                            all the point (x,y,z) that satisfies the following relationship

 

                                             

 

                                 Sometimes,  it is shown as follows :

 

                                                   x = d + (a-d)t

                                                   y = e + (b-e)t

                                                   z = f + (c-f)t               t  R

                                                                       

 

 

 

 

Plane              Plane can be shown as   ax+by+cz = d .

 

Theorem:     If  the equation of a plane is ax+by+cz =d,

                       then  (a,b,c) is a normal vector to this plane.

 

              Proof:

                             Let the point  A(x0, y0, z0) be a point on this plane.  

                             Then it satisfies the plane equation:

 

                                                     ax0 + by0 + cz0 = d

 

                            And any other point (x,y,z) on this plane can determine

                             a vector on this plane :

 

                                                   (x-x0, y-y0, z-z0)

 

                            Consider the inner product

 

                                      (a,b,c)  ( x-x0, y-y0, z-z0 )

                                   = a(x-x0) + b(y-y0) + c(z-z0)

                                   = ax+by+cz – (ax0+by0+cz0)

                                   = d – d = 0

 

                            Thus, (a,b,c) is perpendicular to any vector on this plane.

                            Hence, it is a normal vector to this plane.

 

                                                                       

 

Theorem ( Distance between a point and a plane )

                 The distance between a point  A(x0,y0,z0) and a plane E: ax+by+cz=d  is

                              D(A,E) =    | d – (ax0+by0+cz0) |

 

          Proof:

 

                                        

 

                               Let B(x1,y1,z1) be a point on E.

                               And we know   = (a,b,c) is a normal vector to plane E.

                               Consider the projection of   on   :

 

                                                     =  (    )

 

                                Thus, the distance 

                                                = | (    ) |

                                                     = | (x1-x0, y1-y0, z1-z0)  (a, b, c) | 

                                                     = | ax1+by1+cy1-( ax0+by0+cz0)|

                                                     = | d – (ax0+by0+cz0) |