26
CHAPTER
1V
ECTORS AND VECTORVALUED FUNCTIONS
1.6.
L
INES AND
P
LANES
,P
ART
II
In Section 1.4 we saw the following formula for planes.
Component Equation for a Plane.
The vector equation for the
plane through the point
x
0
,y
0
,z
0
normal to the vector
n
A,B,C
is
A
x
x
0
B
y
y
0
C
z
z
0
0
.
At that time our ability to use this equation was quite limited, because it requires a
normal vector (a vector orthogonal to the plane). Now, however, we have the cross product
to help us. We begin this section with two examples which demonstrate this use of the
cross product. Example 1 is routine, while Example 2 shows a more complex situation.
After that, we discuss how to ±nd the line where two planes intersect and how to measure
distances between points, lines, and planes.
Example 1.
Find the equation of the plane which contains the points
P
3
,
1
,
3
2
,
Q
2
,
2
,
2
, and
R
0
,
0
,
1
.
Solution.
First we write down two vectors on the plane,
PQ
5
,
3
,
1
2
,
PR
3
,
1
,
1
2
.
A normal vector for the plane is then the cross product of these two vectors,
i
j
k
5
3
1
2
3
1
1
2
i
3
1
2
1
1
2
j
5
1
2
3
1
2
k
5
3
3
1
i
3
2
1
2
j
5
2
3
2
k
5
9
2
i
4
j
4
k
.
We therefore get
2
x
3
4
y
1
4
z
3
2
0
,
by choosing
P
3
,
1
,
3
2
as our point on the plane in the plane formula.
Example 2.
Find the equation of the plane which contains the lines
4
t
2
,
3
t,
2
t
1
and
1
2
t,t
2
4
,
3
t
.
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View Full DocumentSECTION 1.6
LINES AND PLANES,PART II
27
Solution.
The two direction vectors are
m
1
4
,
1
,
2
and
m
2
2
,
1
2
,
1
. If we
proceed as in the last example, we run into problems, because
m
1
m
2
i
j
k
4
12
2
1
2
1
i
1
2
1
j
42
2
1
k
4
1
1
2
1
i
1
1
j
4
4
k
2
2
0
.
Of course, this doesn’t mean that the plane’s normal vector is
0
! All it means is that we
took the cross product of two parallel vectors. In order to Fnd the normal vector for the
plane, we need to Fnd some other vector on the plane. Since the point
2
,
3
,
1
lies on the
Frst line and the point
1
,
4
,
3
lies on the second plane, the vector between them,
v
3
,
1
,
2
,
must also lie on the plane. Therefore we can get the normal vector by crossing one of the
direction vectors from before with
v
:
m
1
v
i
j k
4
3
1
2
i
j
32
k
4
1
31
i
2
2
j
8
6
k
4
3
4
i
2
j
7
k
.
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 Fall '08
 Keeran
 Calculus, 2J, 3 1 j

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