5: Equations of Lines and Planes
Tải bản đầy đủ  0trang 97817_12_ch12_p838847.qk_97817_12_ch12_p838847 11/8/10 8:54 AM Page 841
SECTION 12.5
z
P(x, y, z)
O
841
representations OP
A0 and OP
A). If a is the vector with representation A,
P0 P as in Figure 1, then
the Triangle Law for vector addition gives r r0 ϩ a. But, since a and v are parallel vectors, there is a scalar t such that a tv. Thus
P¸(x¸, y¸, z¸)
a
L
EQUATIONS OF LINES AND PLANES
r¸ r
v
r r0 ϩ t v
1
x
y
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 P0 , whereas negative values of t correspond to points that lie on the other side of P0 .
If the vector v that gives the direction of the line L is written in component form as
v ͗a, b, c͘ , then we have t v ͗ ta, tb, tc͘ . We can also write r ͗x, y, z͘ and
r0 ͗ x 0 , y0 , z0 ͘ , so the vector equation 1 becomes
FIGURE 1
z
t>0
t=0
L
t<0
r¸
͗x, y, z͘ ͗x 0 ϩ ta, y0 ϩ tb, z0 ϩ tc ͘
x
Two vectors are equal if and only if corresponding components are equal. Therefore we
have the three scalar equations:
y
FIGURE 2
2
x x 0 ϩ at
y y0 ϩ bt
z z0 ϩ ct
where t ʦ ޒ. These equations are called parametric equations of the line L through the
point P0͑x 0 , y0 , z0͒ and parallel to the vector v ͗ a, b, c͘ . Each value of the parameter t
gives a point ͑x, y, z͒ on L.
Figure 3 shows the line L in Example 1 and its
relation to the given point and to the vector that
gives its direction.
z
SOLUTION
L
(5, 1, 3)
(a) Here r0 ͗ 5, 1, 3͘ 5i ϩ j ϩ 3k and v i ϩ 4 j Ϫ 2 k, so the vector equation 1 becomes
r¸
v=i+4j2k
x
EXAMPLE 1
(a) Find a vector equation and parametric equations for the line that passes through the
point ͑5, 1, 3͒ and is parallel to the vector i ϩ 4 j Ϫ 2k.
(b) Find two other points on the line.
r ͑5 i ϩ j ϩ 3k͒ ϩ t͑i ϩ 4 j Ϫ 2 k͒
y
or
r ͑5 ϩ t͒ i ϩ ͑1 ϩ 4t͒ j ϩ ͑3 Ϫ 2t͒ k
Parametric equations are
FIGURE 3
x5ϩt
y 1 ϩ 4t
z 3 Ϫ 2t
(b) Choosing the parameter value t 1 gives x 6, y 5, and z 1, so ͑6, 5, 1͒ is a
point on the line. Similarly, t Ϫ1 gives the point ͑4, Ϫ3, 5͒.
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 ͑5, 1, 3͒, we choose the point ͑6, 5, 1͒ in Example 1, then the parametric equations of the line become
x6ϩt
y 5 ϩ 4t
z 1 Ϫ 2t
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
97817_12_ch12_p838847.qk_97817_12_ch12_p838847 11/8/10 8:54 AM Page 842
842
CHAPTER 12
VECTORS AND THE GEOMETRY OF SPACE
Or, if we stay with the point ͑5, 1, 3͒ but choose the parallel vector 2i ϩ 8j Ϫ 4 k, we arrive
at the equations
x 5 ϩ 2t
y 1 ϩ 8t
z 3 Ϫ 4t
In general, if a vector v ͗a, b, c͘ is used to describe the direction of a line L, then the
numbers a, b, and c are called direction numbers of L. Since any vector parallel to v could
also be used, we see that any three numbers proportional to a, b, and c could also be used
as a set of direction numbers for L.
Another way of describing a line L is to eliminate the parameter t from Equations 2. If
none of a, b, or c is 0, we can solve each of these equations for t, equate the results, and
obtain
3
x Ϫ x0
y Ϫ y0
z Ϫ z0
a
b
c
These equations are called symmetric equations of L. Notice that the numbers a, b, and
c that appear in the denominators of Equations 3 are direction numbers of L, that is, components of a vector parallel to L. If one of a, b, or c is 0, we can still eliminate t. For
instance, if a 0, we could write the equations of L as
x x0
y Ϫ y0
z Ϫ z0
b
c
This means that L lies in the vertical plane x x 0.
Figure 4 shows the line L in Example 2 and the
point P where it intersects the xyplane.
z
1
B
x
2
1
P
_1
(a) We are not explicitly given a vector parallel to the line, but observe that the vector v
l
with representation AB is parallel to the line and
v ͗3 Ϫ 2, Ϫ1 Ϫ 4, 1 Ϫ ͑Ϫ3͒͘ ͗1, Ϫ5, 4͘
A
FIGURE 4
SOLUTION
4
y
L
EXAMPLE 2
(a) Find parametric equations and symmetric equations of the line that passes through
the points A͑2, 4, Ϫ3͒ and B͑3, Ϫ1, 1͒.
(b) At what point does this line intersect the xyplane?
Thus direction numbers are a 1, b Ϫ5, and c 4. Taking the point ͑2, 4, Ϫ3͒ as
P0, we see that parametric equations 2 are
x2ϩt
y 4 Ϫ 5t
z Ϫ3 ϩ 4t
and symmetric equations 3 are
xϪ2
yϪ4
zϩ3
1
Ϫ5
4
(b) The line intersects the xyplane when z 0, so we put z 0 in the symmetric equations and obtain
yϪ4
3
xϪ2
1
Ϫ5
4
11
1
11 1
This gives x 4 and y 4 , so the line intersects the xyplane at the point ( 4 , 4 , 0).
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
97817_12_ch12_p838847.qk_97817_12_ch12_p838847 11/8/10 8:54 AM Page 843
SECTION 12.5
EQUATIONS OF LINES AND PLANES
843
In general, the procedure of Example 2 shows that direction numbers of the line L through
the points P0͑x 0 , y0 , z0 ͒ and P1͑x 1, y1, z1͒ are x 1 Ϫ x 0 , y1 Ϫ y0 , and z1 Ϫ z0 and so symmetric equations of L are
x Ϫ x0
y Ϫ y0
z Ϫ z0
x1 Ϫ x0
y1 Ϫ y0
z1 Ϫ z0
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 ͑2, 4, Ϫ3͒ and if we put t 1 we get
͑3, Ϫ1, 1͒. So the line segment AB is described by the parametric equations
x2ϩt
y 4 Ϫ 5t
z Ϫ3 ϩ 4t
0ഛtഛ1
or by the corresponding vector equation
r͑t͒ ͗2 ϩ t, 4 Ϫ 5t, Ϫ3 ϩ 4t͘
0ഛtഛ1
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 ϩ t v. If the line also passes through
(the tip of ) r1, then we can take v r1 Ϫ r 0 and so its vector equation is
r r 0 ϩ t ͑r1 Ϫ r 0͒ ͑1 Ϫ t͒r 0 ϩ tr1
The line segment from r 0 to r1 is given by the parameter interval 0 ഛ t ഛ 1.
4
The line segment from r 0 to r1 is given by the vector equation
r͑t͒ ͑1 Ϫ t͒r 0 ϩ t r1
v
The lines L 1 and L 2 in Example 3, shown in
Figure 5, are skew lines.
EXAMPLE 3 Show that the lines L 1 and L 2 with parametric equations
z
L¡
0ഛtഛ1
x1ϩt
y Ϫ2 ϩ 3t
z4Ϫt
x 2s
y3ϩs
z Ϫ3 ϩ 4s
5
are skew lines; that is, they do not intersect and are not parallel (and therefore do not lie
in the same plane).
L™
SOLUTION The lines are not parallel because the corresponding vectors ͗1, 3, Ϫ1͘ and
5
10
5
x
y
͗2, 1, 4͘ 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 ϩ t 2s
_5
Ϫ2 ϩ 3t 3 ϩ s
4 Ϫ t Ϫ3 ϩ 4s
FIGURE 5
11
8
But if we solve the first two equations, we get t 5 and s 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.
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
“direction” of the plane, but a vector perpendicular to the plane does completely specify
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
97817_12_ch12_p838847.qk_97817_12_ch12_p838847 11/8/10 8:54 AM Page 844
844
CHAPTER 12
VECTORS AND THE GEOMETRY OF SPACE
its direction. Thus a plane in space is determined by a point P0͑x 0 , y0 , z0͒ in the plane and
a vector n that is orthogonal to the plane. This orthogonal vector n is called a normal
vector. Let P͑x, y, z͒ be an arbitrary point in the plane, and let r0 and r be the position vectors of P0 and P. Then the vector r Ϫ r0 is represented by P
A.
0 P (See Figure 6.) The normal
vector n is orthogonal to every vector in the given plane. In particular, n is orthogonal to
r Ϫ r0 and so we have
z
n
P(x, y, z)
r
rr¸
r¸
0
P¸(x¸, y¸, z¸)
n ؒ ͑r Ϫ r0 ͒ 0
5
x
y
FIGURE 6
which can be rewritten as
n ؒ r n ؒ r0
6
Either Equation 5 or Equation 6 is called a vector equation of the plane.
To obtain a scalar equation for the plane, we write n ͗ a, b, c͘ , r ͗x, y, z͘ , and
r0 ͗x 0 , y0 , z0 ͘ . Then the vector equation 5 becomes
͗a, b, c͘ ؒ ͗x Ϫ x 0 , y Ϫ y0 , z Ϫ z0 ͘ 0
or
7
a͑x Ϫ x 0 ͒ ϩ b͑y Ϫ y0 ͒ ϩ c͑z Ϫ z0 ͒ 0
Equation 7 is the scalar equation of the plane through P0͑x 0 , y0 , z0 ͒ with normal vector
n ͗a, b, c͘ .
v EXAMPLE 4 Find an equation of the plane through the point ͑2, 4, Ϫ1͒ with normal
vector n ͗2, 3, 4 ͘ . Find the intercepts and sketch the plane.
SOLUTION Putting a 2, b 3, c 4, x 0 2, y0 4, and z0 Ϫ1 in Equation 7, we
z
see that an equation of the plane is
(0, 0, 3)
2͑x Ϫ 2͒ ϩ 3͑y Ϫ 4͒ ϩ 4͑z ϩ 1͒ 0
(0, 4, 0)
(6, 0, 0)
x
FIGURE 7
2x ϩ 3y ϩ 4z 12
or
y
To find the xintercept we set y z 0 in this equation and obtain x 6. Similarly, the
yintercept is 4 and the zintercept is 3. This enables us to sketch the portion of the plane
that lies in the first octant (see Figure 7).
By collecting terms in Equation 7 as we did in Example 4, we can rewrite the equation
of a plane as
8
ax ϩ by ϩ cz ϩ d 0
where d Ϫ͑ax 0 ϩ by0 ϩ cz0 ͒. Equation 8 is called a linear equation in x, y, and z. Conversely, it can be shown that if a, b, and c are not all 0, then the linear equation 8 represents a plane with normal vector ͗a, b, c͘ . (See Exercise 81.)
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
97817_12_ch12_p838847.qk_97817_12_ch12_p838847 11/8/10 8:54 AM Page 845
SECTION 12.5
Figure 8 shows the portion of the plane in
Example 5 that is enclosed by triangle PQR.
z
EQUATIONS OF LINES AND PLANES
845
EXAMPLE 5 Find an equation of the plane that passes through the points P͑1, 3, 2͒,
Q͑3, Ϫ1, 6͒, and R͑5, 2, 0͒.
l
l
SOLUTION The vectors a and b corresponding to PQ and PR are
Q(3, _1, 6)
a ͗2, Ϫ4, 4͘
b ͗4, Ϫ1, Ϫ2͘
Since both a and b lie in the plane, their cross product a ϫ b is orthogonal to the plane
and can be taken as the normal vector. Thus
P(1, 3, 2)
Խ Խ
y
i
naϫb 2
4
x
R(5, 2, 0)
FIGURE 8
j
Ϫ4
Ϫ1
k
4 12 i ϩ 20 j ϩ 14 k
Ϫ2
With the point P͑1, 3, 2͒ and the normal vector n, an equation of the plane is
12͑x Ϫ 1͒ ϩ 20͑y Ϫ 3͒ ϩ 14͑z Ϫ 2͒ 0
6x ϩ 10y ϩ 7z 50
or
EXAMPLE 6 Find the point at which the line with parametric equations x 2 ϩ 3t,
y Ϫ4t, z 5 ϩ t intersects the plane 4x ϩ 5y Ϫ 2z 18.
SOLUTION We substitute the expressions for x, y, and z from the parametric equations
into the equation of the plane:
4͑2 ϩ 3t͒ ϩ 5͑Ϫ4t͒ Ϫ 2͑5 ϩ t͒ 18
This simplifies to Ϫ10t 20, so t Ϫ2. Therefore the point of intersection occurs
when the parameter value is t Ϫ2. Then x 2 ϩ 3͑Ϫ2͒ Ϫ4, y Ϫ4͑Ϫ2͒ 8,
z 5 Ϫ 2 3 and so the point of intersection is 4, 8, 3.
n ă nĂ
Two planes are parallel if their normal vectors are parallel. For instance, the planes
x ϩ 2y Ϫ 3z 4 and 2x ϩ 4y Ϫ 6z 3 are parallel because their normal vectors are
n1 ͗ 1, 2, Ϫ3͘ and n 2 ͗ 2, 4, Ϫ6͘ and n 2 2n1 . If two planes are not parallel, then
they intersect in a straight line and the angle between the two planes is defined as the acute
angle between their normal vectors (see angle in Figure 9).
ă
FIGURE 9
v
Figure 10 shows the planes in Example 7 and
their line of intersection L.
x2y+3z=1
x+y+z=1
EXAMPLE 7
(a) Find the angle between the planes x ϩ y ϩ z 1 and x Ϫ 2y ϩ 3z 1.
(b) Find symmetric equations for the line of intersection L of these two planes.
SOLUTION
(a) The normal vectors of these planes are
6
4
2
z 0
_2
_4
n1 ͗ 1, 1, 1͘
L
n 2 ͗ 1, Ϫ2, 3͘
and so, if is the angle between the planes, Corollary 12.3.6 gives
cos
_2
FIGURE 10
0
y
2
2
0
x
_2
n1 ؒ n 2
1͑1͒ ϩ 1͑Ϫ2͒ ϩ 1͑3͒
2
n1 n 2
s1 ϩ 1 ϩ 1 s1 ϩ 4 ϩ 9
s42
Խ ԽԽ Խ
ͩ ͪ
cosϪ1
2
s42
Ϸ 72Њ
(b) We first need to find a point on L. For instance, we can find the point where the line
intersects the xyplane by setting z 0 in the equations of both planes. This gives the
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
97817_12_ch12_p838847.qk_97817_12_ch12_p838847 11/8/10 8:54 AM Page 846
846
VECTORS AND THE GEOMETRY OF SPACE
CHAPTER 12
equations x ϩ y 1 and x Ϫ 2y 1, whose solution is x 1, y 0. So the point
͑1, 0, 0͒ lies on L.
Now we observe that, since L lies in both planes, it is perpendicular to both of the
normal vectors. Thus a vector v parallel to L is given by the cross product
Խ Խ
i
j
v n1 ϫ n 2 1
1
1 Ϫ2
Another way to find the line of intersection is
to solve the equations of the planes for two of
the variables in terms of the third, which can
be taken as the parameter.
k
1 5i Ϫ 2 j Ϫ 3 k
3
and so the symmetric equations of L can be written as
xϪ1
y
z
5
Ϫ2
Ϫ3
y
x1
= _2
5
2
L
1
z 0 y
_1
2
z
=3
_2
_1
y
0
1
2
NOTE Since a linear equation in x, y, and z represents a plane and two nonparallel
planes intersect in a line, it follows that two linear equations can represent a line. The
points ͑x, y, z͒ that satisfy both a 1 x ϩ b1 y ϩ c1 z ϩ d1 0 and a 2 x ϩ b2 y ϩ c2 z ϩ d2 0
lie on both of these planes, and so the pair of linear equations represents the line of intersection of the planes (if they are not parallel). For instance, in Example 7 the line L was
given as the line of intersection of the planes x ϩ y ϩ z 1 and x Ϫ 2y ϩ 3z 1. The
symmetric equations that we found for L could be written as
_2
0 _1
x
1
FIGURE 11
Figure 11 shows how the line L in Example 7
can also be regarded as the line of intersection
of planes derived from its symmetric equations.
y
xϪ1
5
Ϫ2
and
y
z
Ϫ2
Ϫ3
which is again a pair of linear equations. They exhibit L as the line of intersection of the
planes ͑x Ϫ 1͒͞5 y͑͞Ϫ2͒ and y͑͞Ϫ2͒ z͑͞Ϫ3͒. (See Figure 11.)
In general, when we write the equations of a line in the symmetric form
x Ϫ x0
y Ϫ y0
z Ϫ z0
a
b
c
we can regard the line as the line of intersection of the two planes
x Ϫ x0
y Ϫ y0
a
b
and
y Ϫ y0
z Ϫ z0
b
c
EXAMPLE 8 Find a formula for the distance D from a point P1͑x 1, y1, z1͒ to the
plane ax ϩ by ϩ cz ϩ d 0.
SOLUTION Let P0͑x 0 , y0 , z0 ͒ be any point in the given plane and let b be the vector
corresponding to PA.
0 P1 Then
b ͗ x 1 Ϫ x 0 , y1 y0 , z1 z0
PĂ
ă
b
Pá
FIGURE 12
D
n
From Figure 12 you can see that the distance D from P1 to the plane is equal to the
absolute value of the scalar projection of b onto the normal vector n ͗a, b, c͘ . (See
Section 12.3.) Thus
nؒb
D compn b
n
Խ ԽԽ ԽԽ
Խ
Խ a͑x
Ϫ x0 ͒ ϩ b͑y1 Ϫ y0 ͒ ϩ c͑z1 Ϫ z0 ͒
sa 2 ϩ b 2 ϩ c 2
Խ ͑ax
ϩ by1 ϩ cz1 ͒ Ϫ ͑ax0 ϩ by0 ϩ cz0 ͒
sa 2 ϩ b 2 ϩ c 2
1
1
Խ
Խ
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
97817_12_ch12_p838847.qk_97817_12_ch12_p838847 11/8/10 8:54 AM Page 847
EQUATIONS OF LINES AND PLANES
SECTION 12.5
847
Since P0 lies in the plane, its coordinates satisfy the equation of the plane and so we
have ax 0 ϩ by0 ϩ cz0 ϩ d 0. Thus the formula for D can be written as
D
9
Խ ax
ϩ by1 ϩ cz1 ϩ d
sa 2 ϩ b 2 ϩ c 2
1
Խ
EXAMPLE 9 Find the distance between the parallel planes 10x ϩ 2y Ϫ 2z 5 and
5x ϩ y Ϫ z 1.
SOLUTION First we note that the planes are parallel because their normal vectors
͗10, 2, Ϫ2͘ and ͗5, 1, Ϫ1͘ are parallel. To find the distance D between the planes, we
choose any point on one plane and calculate its distance to the other plane. In particular,
if we put y z 0 in the equation of the first plane, we get 10x 5 and so ( 12 , 0, 0)
is a point in this plane. By Formula 9, the distance between ( 12 , 0, 0) and the plane
5x ϩ y Ϫ z Ϫ 1 0 is
D
Խ 5( ) ϩ 1͑0͒ Ϫ 1͑0͒ Ϫ 1 Խ
1
2
s5 ϩ 1 ϩ ͑Ϫ1͒
2
2
2
3
2
3s3
s3
6
So the distance between the planes is s3͞6.
EXAMPLE 10 In Example 3 we showed that the lines
L1: x 1 ϩ t
y Ϫ2 ϩ 3t
z4Ϫt
L 2 : x 2s
y3ϩs
z Ϫ3 ϩ 4s
are skew. Find the distance between them.
SOLUTION Since the two lines L 1 and L 2 are skew, they can be viewed as lying on two
parallel planes P1 and P2 . The distance between L 1 and L 2 is the same as the distance
between P1 and P2 , which can be computed as in Example 9. The common normal vector to both planes must be orthogonal to both v1 ͗ 1, 3, Ϫ1͘ (the direction of L 1 ) and
v2 ͗ 2, 1, 4͘ (the direction of L 2 ). So a normal vector is
n v1 ϫ v2
Խ Խ
i j
1 3
2 1
k
Ϫ1 13i Ϫ 6 j Ϫ 5k
4
If we put s 0 in the equations of L 2 , we get the point ͑0, 3, Ϫ3͒ on L 2 and so an equation for P2 is
13͑x Ϫ 0͒ Ϫ 6͑ y Ϫ 3͒ Ϫ 5͑z ϩ 3͒ 0
or
13x Ϫ 6y Ϫ 5z ϩ 3 0
If we now set t 0 in the equations for L 1 , we get the point ͑1, Ϫ2, 4͒ on P1 . So
the distance between L 1 and L 2 is the same as the distance from ͑1, Ϫ2, 4͒ to
13x Ϫ 6y Ϫ 5z ϩ 3 0. By Formula 9, this distance is
D
Խ 13͑1͒ Ϫ 6͑Ϫ2͒ Ϫ 5͑4͒ ϩ 3 Խ
s13 ϩ ͑Ϫ6͒ ϩ ͑Ϫ5͒
2
2
2
8
Ϸ 0.53
s230
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
97817_12_ch12_p848857.qk_97817_12_ch12_p848857 11/8/10 8:56 AM Page 848
848
CHAPTER 12
12.5
VECTORS AND THE GEOMETRY OF SPACE
Exercises
1. Determine whether each statement is true or false.
(a)
(b)
(c)
(d)
(e)
(f )
(g)
(h)
(i)
( j)
(k)
Two lines parallel to a third line are parallel.
Two lines perpendicular to a third line are parallel.
Two planes parallel to a third plane are parallel.
Two planes perpendicular to a third plane are parallel.
Two lines parallel to a plane are parallel.
Two lines perpendicular to a plane are parallel.
Two planes parallel to a line are parallel.
Two planes perpendicular to a line are parallel.
Two planes either intersect or are parallel.
Two lines either intersect or are parallel.
A plane and a line either intersect or are parallel.
2–5 Find a vector equation and parametric equations for the line.
2. The line through the point ͑6, Ϫ5, 2͒ and parallel to the
vector ͗ 1, 3, Ϫ 23 ͘
3. The line through the point ͑2, 2.4, 3.5͒ and parallel to the
vector 3 i ϩ 2 j Ϫ k
4. The line through the point ͑0, 14, Ϫ10͒ and parallel to the line
x Ϫ1 ϩ 2t, y 6 Ϫ 3t, z 3 ϩ 9t
5. The line through the point (1, 0, 6) and perpendicular to the
plane x ϩ 3y ϩ z 5
6–12 Find parametric equations and symmetric equations for the
line.
6. The line through the origin and the point ͑4, 3, Ϫ1͒
7. The line through the points (0, 2 , 1) and ͑2, 1, Ϫ3͒
16. (a) Find parametric equations for the line through ͑2, 4, 6͒ that
is perpendicular to the plane x Ϫ y ϩ 3z 7.
(b) In what points does this line intersect the coordinate
planes?
17. Find a vector equation for the line segment from ͑2, Ϫ1, 4͒
to ͑4, 6, 1͒.
18. Find parametric equations for the line segment from ͑10, 3, 1͒
to ͑5, 6, Ϫ3͒.
19–22 Determine whether the lines L 1 and L 2 are parallel, skew, or
intersecting. If they intersect, find the point of intersection.
19. L 1: x 3 ϩ 2t,
y 4 Ϫ t, z 1 ϩ 3t
L 2: x 1 ϩ 4s, y 3 Ϫ 2s,
20. L 1: x 5 Ϫ 12t,
L 2: x 3 ϩ 8s,
z 4 ϩ 5s
y 3 ϩ 9t,
y Ϫ6s,
21. L 1:
yϪ3
zϪ1
xϪ2
1
Ϫ2
Ϫ3
L 2:
xϪ3
yϩ4
zϪ2
1
3
Ϫ7
22. L 1:
x
yϪ1
zϪ2
1
Ϫ1
3
L 2:
yϪ3
z
xϪ2
2
Ϫ2
7
z 1 Ϫ 3t
z 7 ϩ 2s
23– 40 Find an equation of the plane.
1
8. The line through the points ͑1.0, 2.4, 4.6͒ and ͑2.6, 1.2, 0.3͒
9. The line through the points ͑Ϫ8, 1, 4͒ and ͑3, Ϫ2, 4͒
10. The line through ͑2, 1, 0͒ and perpendicular to both i ϩ j
and j ϩ k
11. The line through ͑1, Ϫ1, 1͒ and parallel to the line
xϩ2 yzϪ3
1
2
12. The line of intersection of the planes x ϩ 2y ϩ 3z 1
and x Ϫ y ϩ z 1
13. Is the line through ͑Ϫ4, Ϫ6, 1͒ and ͑Ϫ2, 0, Ϫ3͒ parallel to the
line through ͑10, 18, 4͒ and ͑5, 3, 14͒ ?
14. Is the line through ͑Ϫ2, 4, 0͒ and ͑1, 1, 1͒ perpendicular to the
line through ͑2, 3, 4͒ and ͑3, Ϫ1, Ϫ8͒ ?
15. (a) Find symmetric equations for the line that passes
through the point ͑1, Ϫ5, 6͒ and is parallel to the vector
͗ Ϫ1, 2, Ϫ3 ͘ .
(b) Find the points in which the required line in part (a) intersects the coordinate planes.
23. The plane through the origin and perpendicular to the
vector ͗1, Ϫ2, 5 ͘
24. The plane through the point ͑5, 3, 5͒ and with normal
vector 2 i ϩ j Ϫ k
25. The plane through the point (Ϫ1, 2 , 3) and with normal
1
vector i ϩ 4 j ϩ k
26. The plane through the point ͑2, 0, 1͒ and perpendicular to the
line x 3t, y 2 Ϫ t, z 3 ϩ 4t
27. The plane through the point ͑1, Ϫ1, Ϫ1͒ and parallel to the
plane 5x Ϫ y Ϫ z 6
28. The plane through the point ͑2, 4, 6͒ and parallel to the plane
zxϩy
29. The plane through the point (1, 2 , 3 ) and parallel to the plane
1 1
xϩyϩz0
30. The plane that contains the line x 1 ϩ t, y 2 Ϫ t,
z 4 Ϫ 3t and is parallel to the plane 5x ϩ 2y ϩ z 1
31. The plane through the points ͑0, 1, 1͒, ͑1, 0, 1͒, and ͑1, 1, 0͒
32. The plane through the origin and the points ͑2, Ϫ4, 6͒
and ͑5, 1, 3͒
1. Homework Hints available at stewartcalculus.com
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
97817_12_ch12_p848857.qk_97817_12_ch12_p848857 11/8/10 8:56 AM Page 849
EQUATIONS OF LINES AND PLANES
SECTION 12.5
849
33. The plane through the points ͑3, Ϫ1, 2͒, ͑8, 2, 4͒, and
57–58 (a) Find parametric equations for the line of intersection of
the planes and (b) find the angle between the planes.
34. The plane that passes through the point ͑1, 2, 3͒ and contains
57. x ϩ y ϩ z 1,
͑Ϫ1, Ϫ2, Ϫ3͒
the line x 3t, y 1 ϩ t, z 2 Ϫ t
35. The plane that passes through the point ͑6, 0, Ϫ2͒ and contains
the line x 4 Ϫ 2t, y 3 ϩ 5t, z 7 ϩ 4 t
36. The plane that passes through the point ͑1, Ϫ1, 1͒ and
contains the line with symmetric equations x 2y 3z
37. The plane that passes through the point ͑Ϫ1, 2, 1͒ and contains
the line of intersection of the planes x ϩ y Ϫ z 2 and
2 x Ϫ y ϩ 3z 1
x ϩ 2y ϩ 2z 1
58. 3x Ϫ 2y ϩ z 1,
2x ϩ y Ϫ 3z 3
59–60 Find symmetric equations for the line of intersection of the
planes.
59. 5x Ϫ 2y Ϫ 2z 1,
60. z 2x Ϫ y Ϫ 5,
4x ϩ y ϩ z 6
z 4x ϩ 3y Ϫ 5
38. The plane that passes through the points ͑0, Ϫ2, 5͒ and
61. Find an equation for the plane consisting of all points that are
39. The plane that passes through the point ͑1, 5, 1͒ and is perpen
62. Find an equation for the plane consisting of all points that are
40. The plane that passes through the line of intersection of the
63. Find an equation of the plane with xintercept a, yintercept b,
͑Ϫ1, 3, 1͒ and is perpendicular to the plane 2z 5x ϩ 4y
dicular to the planes 2x ϩ y Ϫ 2z 2 and x ϩ 3z 4
planes x Ϫ z 1 and y ϩ 2z 3 and is perpendicular to the
plane x ϩ y Ϫ 2z 1
equidistant from the points ͑1, 0, Ϫ2͒ and ͑3, 4, 0͒.
equidistant from the points ͑2, 5, 5͒ and ͑Ϫ6, 3, 1͒.
and zintercept c.
64. (a) Find the point at which the given lines intersect:
41– 44 Use intercepts to help sketch the plane.
r ͗1, 1, 0͘ ϩ t ͗1, Ϫ1, 2 ͘
41. 2x ϩ 5y ϩ z 10
42. 3x ϩ y ϩ 2z 6
r ͗2, 0, 2͘ ϩ s ͗Ϫ1, 1, 0͘
43. 6x Ϫ 3y ϩ 4z 6
44. 6x ϩ 5y Ϫ 3z 15
(b) Find an equation of the plane that contains these lines.
65. Find parametric equations for the line through the point
45– 47 Find the point at which the line intersects the given plane.
45. x 3 Ϫ t, y 2 ϩ t, z 5t ;
x Ϫ y ϩ 2z 9
46. x 1 ϩ 2t, y 4t, z 2 Ϫ 3t ;
47. x y Ϫ 1 2z ;
x ϩ 2y Ϫ z ϩ 1 0
4x Ϫ y ϩ 3z 8
48. Where does the line through ͑1, 0, 1͒ and ͑4, Ϫ2, 2͒ intersect
the plane x ϩ y ϩ z 6 ?
49. Find direction numbers for the line of intersection of the planes
x ϩ y ϩ z 1 and x ϩ z 0.
50. Find the cosine of the angle between the planes x ϩ y ϩ z 0
and x ϩ 2y ϩ 3z 1.
͑0, 1, 2͒ that is parallel to the plane x ϩ y ϩ z 2 and
perpendicular to the line x 1 ϩ t, y 1 Ϫ t, z 2t.
66. Find parametric equations for the line through the point
͑0, 1, 2͒ that is perpendicular to the line x 1 ϩ t,
y 1 Ϫ t, z 2t and intersects this line.
67. Which of the following four planes are parallel? Are any of
them identical?
P1 : 3x ϩ 6y Ϫ 3z 6
P2 : 4x Ϫ 12y ϩ 8z 5
P3 : 9y 1 ϩ 3x ϩ 6z
P4 : z x ϩ 2y Ϫ 2
68. Which of the following four lines are parallel? Are any of them
identical?
51–56 Determine whether the planes are parallel, perpendicular, or
L 1 : x 1 ϩ 6t,
y 1 Ϫ 3t,
neither. If neither, find the angle between them.
L 2 : x 1 ϩ 2t,
y t,
51. x ϩ 4y Ϫ 3z 1,
L 3 : 2x Ϫ 2 4 Ϫ 4y z ϩ 1
52. 2z 4y Ϫ x,
53. x ϩ y ϩ z 1,
Ϫ3x ϩ 6y ϩ 7z 0
3x Ϫ 12y ϩ 6z 1
xϪyϩz1
54. 2 x Ϫ 3y ϩ 4z 5,
55. x 4y Ϫ 2z,
x ϩ 6y ϩ 4z 3
8y 1 ϩ 2 x ϩ 4z
56. x ϩ 2y ϩ 2z 1,
2x Ϫ y ϩ 2z 1
z 12t ϩ 5
z 1 ϩ 4t
L 4 : r ͗3, 1, 5͘ ϩ t ͗4, 2, 8 ͘
69–70 Use the formula in Exercise 45 in Section 12.4 to find the
distance from the point to the given line.
69. ͑4, 1, Ϫ2͒;
70. ͑0, 1, 3͒;
x 1 ϩ t, y 3 Ϫ 2t, z 4 Ϫ 3t
x 2t, y 6 Ϫ 2t, z 3 ϩ t
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
97817_12_ch12_p848857.qk_97817_12_ch12_p848857 11/8/10 8:56 AM Page 850
850
VECTORS AND THE GEOMETRY OF SPACE
CHAPTER 12
71–72 Find the distance from the point to the given plane.
71. ͑1, Ϫ2, 4͒,
3x ϩ 2y ϩ 6z 5
72. ͑Ϫ6, 3, 5͒,
x Ϫ 2y Ϫ 4z 8
equations x 1 ϩ t, y 1 ϩ 6t, z 2t, and x 1 ϩ 2s,
y 5 ϩ 15s, z Ϫ2 ϩ 6s.
79. Let L1 be the line through the origin and the point ͑2, 0, Ϫ1͒.
73–74 Find the distance between the given parallel planes.
73. 2x Ϫ 3y ϩ z 4,
74. 6z 4y Ϫ 2x,
4x Ϫ 6y ϩ 2z 3
75. Show that the distance between the parallel planes
ax ϩ by ϩ cz ϩ d1 0 and ax ϩ by ϩ cz ϩ d2 0 is
Խ
Let L 2 be the line through the points ͑1, Ϫ1, 1͒ and ͑4, 1, 3͒.
Find the distance between L1 and L 2.
80. Let L1 be the line through the points ͑1, 2, 6͒ and ͑2, 4, 8͒.
Let L 2 be the line of intersection of the planes 1 and 2,
where 1 is the plane x Ϫ y ϩ 2z ϩ 1 0 and 2 is the plane
through the points ͑3, 2, Ϫ1͒, ͑0, 0, 1͒, and ͑1, 2, 1͒. Calculate
the distance between L1 and L 2.
9z 1 Ϫ 3x ϩ 6y
D
78. Find the distance between the skew lines with parametric
Խ
d1 Ϫ d2
sa 2 ϩ b 2 ϩ c 2
81. If a, b, and c are not all 0, show that the equation
ax ϩ by ϩ cz ϩ d 0 represents a plane and ͗a, b, c ͘ is
a normal vector to the plane.
Hint: Suppose a 0 and rewrite the equation in the form
ͩ ͪ
a xϩ
76. Find equations of the planes that are parallel to the plane
x ϩ 2y Ϫ 2z 1 and two units away from it.
77. Show that the lines with symmetric equations x y z and
x ϩ 1 y͞2 z͞3 are skew, and find the distance between
these lines.
d
a
ϩ b͑ y Ϫ 0͒ ϩ c͑z Ϫ 0͒ 0
82. Give a geometric description of each family of planes.
(a) x ϩ y ϩ z c
(c) y cos ϩ z sin 1
(b) x ϩ y ϩ cz 1
L A B O R AT O R Y P R O J E C T PUTTING 3D IN PERSPECTIVE
Computer graphics programmers face the same challenge as the great painters of the past: how
to represent a threedimensional scene as a flat image on a twodimensional plane (a screen or a
canvas). To create the illusion of perspective, in which closer objects appear larger than those
farther away, threedimensional objects in the computer’s memory are projected onto a rectangular screen window from a viewpoint where the eye, or camera, is located. The viewing
volume––the portion of space that will be visible––is the region contained by the four planes that
pass through the viewpoint and an edge of the screen window. If objects in the scene extend
beyond these four planes, they must be truncated before pixel data are sent to the screen. These
planes are therefore called clipping planes.
1. Suppose the screen is represented by a rectangle in the yzplane with vertices ͑0, Ϯ400, 0͒
and ͑0, Ϯ400, 600͒, and the camera is placed at ͑1000, 0, 0͒. A line L in the scene passes
through the points ͑230, Ϫ285, 102͒ and ͑860, 105, 264͒. At what points should L be clipped
by the clipping planes?
2. If the clipped line segment is projected on the screen window, identify the resulting line
segment.
3. Use parametric equations to plot the edges of the screen window, the clipped line segment,
and its projection on the screen window. Then add sight lines connecting the viewpoint to
each end of the clipped segments to verify that the projection is correct.
4. A rectangle with vertices ͑621, Ϫ147, 206͒, ͑563, 31, 242͒, ͑657, Ϫ111, 86͒, and
͑599, 67, 122͒ is added to the scene. The line L intersects this rectangle. To make the rectangle appear opaque, a programmer can use hidden line rendering, which removes portions
of objects that are behind other objects. Identify the portion of L that should be removed.
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
97817_12_ch12_p848857.qk_97817_12_ch12_p848857 11/8/10 8:56 AM Page 851
SECTION 12.6
12.6
CYLINDERS AND QUADRIC SURFACES
851
Cylinders and Quadric Surfaces
We have already looked at two special types of surfaces : planes (in Section 12.5) and
spheres (in Section 12.1). Here we investigate two other types of surfaces: cylinders and
quadric surfaces.
In order to sketch the graph of a surface, it is useful to determine the curves of intersection of the surface with planes parallel to the coordinate planes. These curves are called
traces (or crosssections) of the surface.
Cylinders
z
A cylinder is a surface that consists of all lines (called rulings) that are parallel to a given
line and pass through a given plane curve.
v
EXAMPLE 1 Sketch the graph of the surface z x 2.
SOLUTION Notice that the equation of the graph, z x 2, doesn’t involve y. This means
that any vertical plane with equation y k (parallel to the xzplane) intersects the graph
in a curve with equation z x 2. So these vertical traces are parabolas. Figure 1 shows
how the graph is formed by taking the parabola z x 2 in the xzplane and moving it in
the direction of the yaxis. The graph is a surface, called a parabolic cylinder, made up
of infinitely many shifted copies of the same parabola. Here the rulings of the cylinder are
parallel to the yaxis.
0
y
x
FIGURE 1
The surface z=≈ is a
parabolic cylinder.
We noticed that the variable y is missing from the equation of the cylinder in Example 1.
This is typical of a surface whose rulings are parallel to one of the coordinate axes. If one
of the variables x, y, or z is missing from the equation of a surface, then the surface is a
cylinder.
z
EXAMPLE 2 Identify and sketch the surfaces.
(a) x 2 ϩ y 2 1
(b) y 2 ϩ z 2 1
0
SOLUTION
(a) Since z is missing and the equations x 2 ϩ y 2 1, z k represent a circle with
radius 1 in the plane z k, the surface x 2 ϩ y 2 1 is a circular cylinder whose axis is
the zaxis. (See Figure 2.) Here the rulings are vertical lines.
(b) In this case x is missing and the surface is a circular cylinder whose axis is the
xaxis. (See Figure 3.) It is obtained by taking the circle y 2 ϩ z 2 1, x 0 in the
yzplane and moving it parallel to the xaxis.
y
x
FIGURE 2 ≈+¥=1
z

y
x
FIGURE 3 ¥+z@=1
NOTE When you are dealing with surfaces, it is important to recognize that an equation
like x 2 ϩ y 2 1 represents a cylinder and not a circle. The trace of the cylinder
x 2 ϩ y 2 1 in the xyplane is the circle with equations x 2 ϩ y 2 1, z 0.
Quadric Surfaces
A quadric surface is the graph of a seconddegree equation in three variables x, y, and z.
The most general such equation is
Ax 2 ϩ By 2 ϩ Cz 2 ϩ Dxy ϩ Eyz ϩ Fxz ϩ Gx ϩ Hy ϩ Iz ϩ J 0
where A, B, C, . . . , J are constants, but by translation and rotation it can be brought into one
of the two standard forms
Ax 2 ϩ By 2 ϩ Cz 2 ϩ J 0
or
Ax 2 ϩ By 2 ϩ Iz 0
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.