Begin2.DVI

where λis a scalar parameter. Note that as λvaries from 0 to 1 the position vector

r moves from r 1 to r 2 .An alternative form for the equation of the line is given by

r =r 2 +λ∗(r 1 −r 2 ) −∞ < λ ∗<∞

where λ∗ is some other scalar parameter. This second form for the line has the

position vector r moving from r 2 to r 1 as λ∗varies from 0 to 1. The vector ±(r 2 −r 1 )

is called the direction vector of the line. Equating the coefficients of the unit vectors

in the equation (6.41) there results the scalar parametric equations representing the

line. These parametric equations have the form

x=x 1 +λ(x 2 −x 1 ), y =y 1 +λ(y 2 −y 1 ), z =z 1 +λ(z 1 −z 2 ) −∞ < λ < ∞

If the quantities x 2 −x 1 ,y 2 −y 1 and z 2 −z 1 are different from zero, then the equation

for the line can be represented in the symmetric form

x−x 1 x 2 −x 1 =

y−y 1 y 2 −y 1 =

z−z 1 z 2 −z 1 =λ (6 .42)

Note that the equation of a line can also be represented as the intersection of

two planes
N 1 x+N 2 y+N 3 z+D 1 =0
M 1 x+M 2 y+M 3 z+D 2 =0

or

N ·(r −r 0 ) =0 M ·(r −r 1 ) =0

provided the planes are not parallel or N =kM, for ka nonzero constant.

Example 6-13. Show the perpendicular distance from a point (x 0 , y 0 , z 0 )to a

given line defined by x=x 1 +α 1 t, y =y 1 +α 2 t, z =z 1 +α 3 tis given by

d=

∣∣ ∣∣(r 0 −r 1 )× α |α|

∣∣

∣∣ where α=α 1 ê 1 +α 2 ê 2 +α 3 ê 3

Solution The vector equation of the line is

r =r 1 +α t, where (x 1 , y 1 , z 1 )is a point on the line

described by the position vector r 1 and αis the

direction vector of the line. The vector r 0 −r 1

is a vector pointing from (x 1 , y 1 , z 1 )to the point

(x 0 , y 0 , z 0 ). These vectors are illustrated in the

accompanying figure.