Part 1: Equations of Lines

In this section, we use vectors, the cross product, and the dot product to explore lines and planes in 3-dimensional space. We begin by exploring the vector equation of a line in the xy-plane.

To begin with, a position vector is a vector P = á x, y ñ whose initial point is fixed at the origin, so that each point P(x,y) in R2 corresponds to a position vector P = á x, y ñ.
(note: because a position vector cannot be translated, it is not really a vector but should instead should be considered as a means of using the arithmetic of vectors with points in the plane). Thus, a line corresponds to the endpoints of a set of 2-dimensional position vectors. 

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Indeed, if we define L(t) to be a vector-valued function, which is a function that maps inputs t to output vectors L(t), then a line in 2-dimensions is a vector-valued function of the form
L( t) =  v t + b
 
where b is a fixed point ( = position vector) on the line and v is a constant "slope" vector for the line.
The variable t is called a parameter, and it can be thought of as the "label" or "index" of each point on the line. We use t for the parameter because in applications a point is often labeled (i.e., indexed) by the time at which an object is located at that point.    

EXAMPLE 1    Find the vector form of the line which passes through the points P1( 1,3) and P2( 4,0), and then determine the slope-intercept form of the line from the vector form.        

Solution: To begin with, the vector v is given by

v =

P1P2

  =  á4 - 1, 0 - 3 ñ = á 3,-3 ñ
 
As a result, the equation of the line is
L(t) = vt + b = á3,-3 ñt + á 1,3 ñ = á 3t+1, -3t+3 ñ  
To find the slope-intercept form of the line, let us notice that the x-coordinates are given by x = 3t + 1 and the y-coordinates are given by y = -3t+3.  Solving for t in the first equation yields
x- 1 = 3t   and  t = x - 1
3
Substituting into y = − 3t + 3 yields
y = -3
  x -1
3		  + 3 = − ( x − 1) + 3 = − 3 x + 4
That is,  y = 4 − 3x is the slope-intercept form of the line.

The advantage of a vector equation of a line over point-slope and slope-intercept forms of a line is that a vector equation generalizes to 3 or more dimensions, whereas slope-based forms of a line do not.  Indeed, if we notice that each point P( x,y,z) in R3 corresponds to a 3-dimensional position vector P =  á x, y, z ñ. 

then a line in 3-dimensional space through two points ( = position vectors ) P1 and  P2 is of the form 
L( t) = P1 + vt
(1)
where v is the vector from P1 to P2 and b is the position vector of a point on the line (such as b = P1 ).
A physical interpretation of (1) is that if an object moves along the line L(t) at a constant speed equal to the magnitude of v, then t would be the time at which the object is located at point L(t) on the line.  

EXAMPLE 2    Find the equation of the line which passes through the points P1( 1,0,1) and P2( 4,3,2) , and interpret the result given that an object is located at point P1 at time t = 0 sec and at point P2 at time t = 1 sec.        

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 = á 7,6,3 ñ
 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       

Finally, we can obtain an especially useful form of a line if we notice that v = P2-P1. Substituting into (1) leads to 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 L(t) with 0 £ t £ 1 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.

Check your Reading: What is the parameter for a line with vector equation K( s) = m s ?