18.02 HOMEWORK #2, DUE SEPTEMBER 23, 2010
BJORN POONEN
1.
Part A
(After Sept. 16) 1E1, 1E2, 1E3, 1E6, 1E7* (15 pts.) (assume that the lines are not
parallel)
Solution.
Suppose that the ﬁrst line consists of the points
r
1
(
t
) :=
a
1
+
t
v
1
for
t
∈
R
, where
a
1
and
v
1
are ﬁxed vectors with
v
1
6
= 0, and that the second line is similarly parameterized
as
r
2
(
u
) :=
a
2
+
u
v
2
for
u
∈
R
. There are now several ways of ﬁnishing the problem.
METHOD 1: The vector
a
2

a
1
goes from a point on the ﬁrst line to a point on the second
line. The distance between the lines is the absolute value of the component of this vector
in a direction perpendicular to both lines, i.e., perpendicular to
v
1
and
v
2
. This direction
can be obtained by taking
v
1
×
v
2
(which is nonzero since
v
1
and
v
2
are not parallel) and
dividing by its length to get a unit vector
u
:=
v
1
×
v
2

v
1
×
v
2

.
Then the answer is

(
a
2

a
1
)
·
u

.
METHOD 2: Geometry shows that the distance

r
1
(
t
)

r
2
(
u
)

is minimized when
t
and
u
are such that
r
1
(
t
)

r
2
(
u
) is perpendicular to both lines, i.e., perpendicular to both
v
1
and
v
2
. (This can also be seen analytically by setting the partial derivatives of the function
(
r
1
(
t
)

r
2
(
u
))
·
(
r
1
(
t
)

r
2
(
u
)) equal to 0.)
This leads to the system
((
a
1
+
t
v
1
)

(
a
2
+
u
v
2
))
·
v
1
= 0
((
a
1
+
t
v
1
)

(
a
2
+
u
v
2
))
·
v
2
= 0
,
which when expanded, is a system of two equations in two unknowns
t
and
u
. It can also be
written in matrix form as
±
v
1
·
v
1

v
1
·
v
2
v
1
·
v
2

v
2
·
v
2
²±
t
u
²
=
±
(
a
1

a
2
)
·
v
1
(
a
1

a
2
)
·
v
2
.
²
If
θ
is the angle between
v
1
and
v
2
, then the determinant of the ﬁrst matrix is

v
1

2

v
2

2
+ (

v
1

v
2

cos
θ
)
2
=

v
1

2

v
2

2
sin
2
θ,
which is nonzero since the lines are not parallel. Thus the system has a unique solution.
Once
t
and
u
are obtained, the minimum distance is the value

r
1
(
t
)

r
2
(
u
)

.
±
(After Sept. 16) p. 650: 34* (15 pts.)
Solution.
The point
C
is moving along a circle of radius
a

b
centered at
O
. After it has
moved
t
radians counterclockwise, its position vector is
C
= ((
a

b
) cos
t,
(
a

b
) sin
t
)
.
1
This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentLet
B
be the point of tangency of the two circles. Let
θ
be the measure of angle
PCB
. The
arc length of the big circle from
A
to
B
is
at
(radius times arc measure), and since there is
no slipping, this should equal the arc length of the smaller circle from
B
to
P
, which is
bθ
,
so
θ
=
at/b
. But
P
is clockwise from
B
, so the direction of
This is the end of the preview.
Sign up
to
access the rest of the document.
 Fall '08
 Auroux
 Multivariable Calculus, Surface normal

Click to edit the document details