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

826 Chapter 12 Vectors and the Geometry of Space
In general, the procedure of Example 2 shows that direction numbers of the line L
through the points P 0 sx 0 , y 0 , z 0 d and P 1 sx 1 , y 1 , z 1 d are x 1 2 x 0 , y 1 2 y 0 , and z 1 2 z 0 and so
symmet ric equations of L are
x 2 x 0
− y 2 y 0
− z 2 z 0
x 1 2 x 0 y 1 2 y 0 z 1 2 z 0
Often, we need a description, not of an entire line, but of just a line segment. How, for
instance, could we describe the line segment AB in Example 2? If we put t − 0 in the
parametric equations in Example 2(a), we get the point s2, 4, 23d and if we put t − 1 we
get s3, 21, 1d. So the line segment AB is described by the parametric equations
x − 2 1 t y − 4 2 5t z − 23 1 4t 0 < t < 1
or by the corresponding vector equation
rstd − k2 1 t, 4 2 5t, 23 1 4tl 0 < t < 1
In general, we know from Equation 1 that the vector equation of a line through the (tip
of the) vector r 0 in the direction of a vector v is r − r 0 1 tv. If the line also passes
through (the tip of) r 1 , then we can take v − r 1 2 r 0 and so its vector equation is
r − r 0 1 tsr 1 2 r 0 d − s1 2 tdr 0 1 tr 1
The line segment from r 0 to r 1 is given by the parameter interval 0 < t < 1.
4 The line segment from r 0 to r 1 is given by the vector equation
rstd − s1 2 tdr 0 1 t r 1 0 < t < 1
The lines L 1 and L 2 in Example 3,
shown in Figure 5, are skew lines.
x
5
z
5
L¡ L
_5
FIGURE 5
5
10
y
ExamplE 3 Show that the lines L 1 and L 2 with parametric equations
L 1 : x − 1 1 t y − 22 1 3t z − 4 2 t
L 2 : x − 2s y − 3 1 s z − 23 1 4s
are skew lines; that is, they do not intersect and are not parallel (and therefore do not
lie in the same plane).
SOLUtion The lines are not parallel because the corresponding direction vectors
k1, 3, 21 l and k2, 1, 4 l are not parallel. (Their components are not proportional.) If L 1
and L 2 had a point of intersection, there would be values of t and s such that
1 1 t − 2s
22 1 3t − 3 1 s
4 2 t − 23 1 4s
But if we solve the first two equations, we get t − 11
5 and s − 8 5 , and these values don’t
satisfy the third equation. Therefore there are no values of t and s that satisfy the three
equations, so L 1 and L 2 do not intersect. Thus L 1 and L 2 are skew lines.
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Planes
Although a line in space is determined by a point and a direction, a plane in space is
more difficult to describe. A single vector parallel to a plane is not enough to convey the
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