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
±n
1
=
±
1
,
1
,

1
²
and
2
=
±
2
,

5
,

1
²
respectively. The direction vector of the intersecting line is
±v
=
1
×
2
=
±
1
,
1
,

1
² × ±
2
,

5
,

1
²
=
±
6
,

1
,

7
²
To ±nd 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
²
=
±
1
,
0
,
1
²
+
t
±
6
,

1
,

7
²
=
±
1

6
t,

t,
1

7
t
²
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 ±rst line, and the point
Q
=(

1
,
0
,
0) is on the
second line. The direction vectors of these two lines are
±
1
,
1
,
1
²
and
±
1
,
2
,
3
²
respectively.
The vector
=
±
1
,
1
,
1
² × ±
1
,
2
,
3
²
=
±
1
,

4
,
3
²
is orthogonal to both lines. The distance
d
=
±
±
±
comp
±
PQ
±
±
±
=

±
·



=
±
1
,
0
,
0
² · ±
1
,

4
,
3
²
√
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
=
²
4
e
i
5
π/
6
³
5
=4
5
e
i
25
π
6
5
e
iπ/
6
(b) Cube roots of 1i: write

1

i
=
√
2
e
i
5
4
, then the three cube roots are
6
√
2
e
i
5
12
,
6
√
2
e
i
(5
12+2
3)
=
6
√
2
e
i
13
12
, and
6
√
2
e
i
(5
12+4
3)
=
6
√
2
e
i
21
12
1