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Math 192, Prelim 1. Solution
September 29, 2005
1) a) In order for the lines to intersect, there must be a point of intersection at which
x
1
=
x
2
,
so 3 + 2
t
=15
−
7
s
,
t
=6
−
7
s/
2. The point of intersection also must have
y
1
=
y
2
,so
3+3
t
=
−
2+
s
. Substituting
t
=6
−
7
t/
2 and solving yields
s
=2
,
t
=
−
1. With these
values of
t
and
s
Fnd
z
1
=
−
1=
z
2
. So there is a point of intersection, and its coordinates
are (1
,
0
,
−
1).
b) Let the plane be given by
A
(
x
−
x
o
)+
B
(
y
−
y
o
)+
C
(
z
−
z
0
) = 0. The point of intersection
is a point in the plane, so can take (
x
o
,y
o
,z
o
)=(1
,
0
,
−
1). The direction of
L
1
is given by
~v
1
=2
~
i
+3
~
j
+2
~
k
,thato
f
L
2
by
~v
2
=
−
7
~
i
+
~
j
+3
~
k
. The normal to the plane is
~v
1
×
~v
2
=
±
±
±
±
±
±
~
i
~
j
~
k
232
−
713
±
±
±
±
±
±
=7
~
i
−
20
~
j
+23
~
k
.So
A
=7
,
B
=
−
20 and
C
= 23. The equation for
the plane is 7(
x
−
1)
−
20(
y
−
0) + 23(
z
+ 1) = 0, or 7
x
−
20
y
+23
z
=
−
16
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This note was uploaded on 06/01/2008 for the course MATH 1920 taught by Professor Pantano during the Spring '06 term at Cornell University (Engineering School).
 Spring '06
 PANTANO
 Math

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