CALCULUS III SAMPLE FINAL
Problem 1:
(20 points) Find parametric equations for the line of intersection of the planes
z
=
x
+
y,
2
x
−
5
y
−
z
= 1
.
Solution:
The normal vectors to the two planes are
vectorn
1
=
(
1
,
1
,
−
1
)
and
vectorn
2
=
(
2
,
−
5
,
−
1
)
respectively. The direction vector of the intersecting line is
vectorv
=
vectorn
1
×
vectorn
2
=
(
1
,
1
,
−
1
) × (
2
,
−
5
,
−
1
)
=
(−
6
,
−
1
,
−
7
)
To find a point on the line of intersection, try to let
y
= 0, then we get a system
z
=
x
,
2
x
−
z
= 1, and by solving it we get
x
=
z
= 1, so we get a point (1
,
0
,
1) on the line of
intersection. The parametric equation of the line is
(
x, y, z
)
=
(
1
,
0
,
1
)
+
t
(−
6
,
−
1
,
−
7
)
=
(
1
−
6
t,
−
t,
1
−
7
t
)
Problem 2:
(20 points) Find the distance between skew lines
x
=
y
=
z
and
−
x
−
1 =
y/
2 =
z/
3.
Solution:
The point
P
= (0
,
0
,
0) is on the first line, and the point
Q
= (
−
1
,
0
,
0) is on the
second line. The direction vectors of these two lines are
(
1
,
1
,
1
)
and
(−
1
,
2
,
3
)
respectively.
The vector
vectorn
=
(
1
,
1
,
1
) × (−
1
,
2
,
3
)
=
(
1
,
−
4
,
3
)
is orthogonal to both lines. The distance
d
=
vextendsingle
vextendsingle
vextendsingle
comp
vectorn
vector
PQ
vextendsingle
vextendsingle
vextendsingle
=

vector
PQ
·
vectorn


vectorn

=
(−
1
,
0
,
0
) · (
1
,
−
4
,
3
)
√
1
2
+ 4
2
+ 3
3
=
1
√
26
Problem 3.
(20 points)
(a) Find (
−
2
√
3 + 2
i
)
5
using De Moivre’s Theorem:
(b) Find the cube roots of
−
1
−
i
and sketch the roots in the complex plane:
Solution:
(a)
(
−
2
√
3 + 2
i
)
5
=
parenleftBig
4
e
i
5
π/
6
parenrightBig
5
= 4
5
e
i
25
π
6
= 4
5
e
iπ/
6
(b) Cube roots of 1i: write
−
1
−
i
=
√
2
e
i
5
π/
4
, then the three cube roots are
6
√
2
e
i
5
π/
12
,
6
√
2
e
i
(5
π/
12+2
π/
3)
=
6
√
2
e
i
13
π/
12
, and
6
√
2
e
i
(5
π/
12+4
π/
3)
=
6
√
2
e
i
21
π/
12
1