Chapter Review Exercises
443
We want to find the value of
t
for which the point
(
x
,
y
,
z
)
lies on the plane. We substitute the parametric equations in
the equation of the plane and solve for
t
:
2
(
3
t
+
2
)
−
3
·
1
+
(
−
7
t
)
=
5
6
t
+
4
−
3
−
7
t
=
5
−
t
=
4
⇒
t
= −
4
The point
P
of intersection of the line and the plane has the coordinates:
x
=
3
·
(
−
4
)
+
2
= −
10
,
y
=
1
,
z
= −
7
·
(
−
4
)
=
28
and thus,
P
=
(
−
10
,
1
,
28
) .
47.
Find the trace of the plane 3
x
−
2
y
+
5
z
=
4 in the
xy
-plane.
SOLUTION
The
xy
-plane has equation
z
=
0, therefore the intersection of the plane 3
x
−
2
y
+
5
z
=
4 with the
xy
-plane must satisfy both
z
=
0 and the equation of the plane. Therefore the trace has the following equation:
3
x
−
2
y
+
5
·
0
=
4
⇒
3
x
−
2
y
=
4
We conclude that the trace of the plane in the
xy
-plane is the line 3
x
−
2
y
=
4 in the
xy
-plane.
48.
Find the intersection of the planes
x
+
y
+
z
=
1 and 3
x
−
2
y
+
z
=
5.
SOLUTION
The line of intersection of the planes
x
+
y
+
z
=
1 and 3
x
−
2
y
+
z
=
5 consists of all points that satisfy
the equations of the two planes. Therefore, we must solve the following equations:
x
+
y
+
z
=
1
3
x
−
2
y
+
z
=
5
The first equation implies
y
=
1
−
x
−
z
. Substituting in the second equation and solving for
x
in terms of
z
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- Winter '08
- GANGliu
- Equations, Parametric Equations, Tan, Hyperboloid, quadric surface
-
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