Part 1: Lines in 3-dimensional Space

In this section, we use vectors, the cross product, and the dot product to explore lines and planes in 3-dimensional space. A key idea in this exploration is that if a vector P1 is fixed at the origin and if it terminates at P1( x1,y1,z1) , then P1 is a position vector for the point P1(x1,y1,z1) .  As a result, each point in R3 corresponds to a 3-dimensional position vector. 

If v is a vector and  P1( x1, y1, z1) = P1 is a point (i.e., position vector), then any point  P2 on the line through  P1  parallel to v corresponds to P1 + vt  for some t.   Thus, the position vectors (i.e., points) of the form
L( t) = P1 + vt
(1)
are on the line through P1 parallel to v. We say that L( t) is the parametric equation of the line in the parameter t.
For the line L(t) through points P1(x1,y1,z1) and P2( x2,y2,z2) , the vector v is

v =  

P1P2
 

 

EXAMPLE 1    Find the equation of the line which passes through the points P1( 1,0,1) and P2( 4,3,2) .        

Solution: The vector v is given by

v =

P1P2

  =  á4-1,3-0,2-1 ñ = á 3,3,1 ñ
 
As a result, the equation of the line is
L(t) = P1+tv = á1,0,1 ñ +t á 3,3,1 ñ
which reduces to L( t) = á3t+1,3t,t+1 ñ . For example, t = 0 and t = 1 yields
L( 0) = á 1,0,1 ñ = P1  and  L( 1) = á 3( 1) +1,3(1) ,1+1 ñ = á 4,3,2 ñ = P2
Other points on the line follow from other choices of t. For example, when t = 2, we obtain the position vector L( 2) = á 3·2+1,3·2,2+1 ñ = á7,6,3 ñ , which yields the point P3( 7,6,3) .  Likewise, when t = -1, we obtain the position vector L(-1) = á -2,-3,0 ñ , which corresponds to the point P0( -2,-3,0) .

Maple/Javaview Figure       

Alternatively, since v = P2-P1, we have L( t) = P1+t( P2-P1) , which reduces to
L( t) = ( 1-t) P1+tP2
(2)
Clearly, L( 0) = P1 and L(1) = P2, so that if 0 £ t £ 1, then L( t) is the line segment with endpoints P1 and P2. Moreover, (2) shows that the order of the points is not important in the equation of a line.

Let's look at an application. If L( t) and K( s) are the equations of two lines, then equating their components leads to 3 equations in the 2 unknowns s and t.  If a solution exists, then the two lines intersect.  Otherwise, they do not.       

EXAMPLE 2    Find the equation of the line through P1(0,1,3) and P2( 2,1,6) .  Then find the equation of the line through P3( -1,1,1) and P4( 2,1,-2) and determine if the two lines intersect.       

Solution:  The line through P1( 0,1,3) and P2( 2,0,6) is
L( t) = ( 1-t) á 0,1,3 ñ+t á 2,0,6 ñ = á 2t,1-t,3t+3 ñ
while the line through P3( -1,1,1) and P4(2,2,-2) is
K( s) = ( 1-s) á-1,1,1 ñ +s á 2,2,-2 ñ = á3s-1,s+1,-3s+1 ñ
The lines intersect when L( t) = K(s) , which is when
3s-1
=
2t,   s+1 = 1-t,   -3s+1 = 3t+3
 3
2
 s-  1
2
=
t,    s = -t,     s = -t-  2
3
Combining the first two equations leads to
-  3
2
 t-  1
2
 = t   or    t =  -1
5
Thus, s = -t implies that s = 1/5.  However, substituting into the last equation leads to
 1
5
 =  1
5
 -  2
3
      or   0 =  -2
3
  
Since s = 1/5 and t = -1/5 are not solutions to the last equation, the 3 equations together have no solution.  Thus, the lines do not intersect.

Maple/Javaview Figure

Check your Reading: What is the parameter for the parametric equation K( s) ?