James Stewart-Calculus_ Early Transcendentals-Cengage Learning (2015)

824 Chapter 12 Vectors and the Geometry of Space
z
t=0 t>0
t<0
r¸
x
FIGURE 2
L
y
1 r − r 0 1 tv
which is a vector equation of L. Each value of the parameter t gives the position vector r
of a point on L. In other words, as t varies, the line is traced out by the tip of the vector r. As
Figure 2 indicates, positive values of t correspond to points on L that lie on one side
of P 0 , whereas negative values of t correspond to points that lie on the other side of P 0 .
If the vector v that gives the direction of the line L is written in component form as
v − ka, b, c l, then we have tv − kta, tb, tcl . We can also write r − k x, y, z l and
r 0 − k x 0 , y 0 , z 0 l , so the vector equation (1) becomes
k x, y, z l − k x 0 1 ta, y 0 1 tb, z 0 1 tc l
Two vectors are equal if and only if corresponding components are equal. Therefore we
have the three scalar equations:
x − x 0 1 at y − y 0 1 bt z − z 0 1 ct
where t [ R. These equations are called parametric equations of the line L through the
point P 0 sx 0 , y 0 , z 0 d and parallel to the vector v − ka, b, cl. Each value of the parameter t
gives a point sx, y, zd on L.
2 Parametric equations for a line through the point sx 0 , y 0 , z 0 d and parallel to the
direction vector ka, b, cl are
x − x 0 1 at y − y 0 1 bt z − z 0 1 ct
Figure 3 shows the line L in Exam ple 1
and its relation to the given point and to
the vector that gives its direction.
L
(5, 1, 3)
r¸
z
v=i+4j-2k
y
ExamplE 1
(a) Find a vector equation and parametric equations for the line that passes through the
point s5, 1, 3d and is parallel to the vector i 1 4 j 2 2k.
(b) Find two other points on the line.
SOLUTION
(a) Here r 0 − k5, 1, 3 l − 5i 1 j 1 3k and v − i 1 4 j 2 2k, so the vector equation
(1) becomes
r − s5i 1 j 1 3kd 1 tsi 1 4 j 2 2kd
x
or
r − s5 1 td i 1 s1 1 4td j 1 s3 2 2td k
FIGURE 3
Parametric equations are
x − 5 1 t y − 1 1 4t z − 3 2 2t
(b) Choosing the parameter value t − 1 gives x − 6, y − 5, and z − 1, so s6, 5, 1d is
a point on the line. Similarly, t − 21 gives the point s4, 23, 5d.
■
The vector equation and parametric equations of a line are not unique. If we change
the point or the parameter or choose a different parallel vector, then the equations change.
For instance, if, instead of s5, 1, 3d, we choose the point s6, 5, 1d in Example 1, then the
parametric equations of the line become
x − 6 1 t y − 5 1 4t z − 1 2 2t
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