S. F. Ellermeyer
MATH 2203 — Exam 1
January 26, 2004
Name
Instructions. This exam contains seven problems, but only six of them will be graded. You
may choose any six to do. Please write DON’T GRADE on the one that you don’t want me
to grade. In writing your solution to each problem, include sufficient detail and use correct
notation. (For instance, don’t forget to write “=” when you mean to say that two things are
equal.) Your method of solving the problem should be clear to the reader (me). If I have to
struggle to understand what you have written, then you might not get full credit for your
solution even if you get a correct answer.
1. One of the diameters of a sphere has endpoints (−2, 6, 4) and (5, 2, 9).
(a) Find the radius of this sphere.
The diameter of the sphere is
q
√
(5 − (−2))2 + (2 − 6)2 + (9 − 4)2 = 90
so the radius is
√
90/2.
(b) Find the center of this sphere.
The center of the sphere is at the point
¶ µ
¶
µ
3
13
−2 + 5 6 + 2 4 + 9
,
,
=
, 4,
.
2
2
2
2
2
(c) Write an equation for this sphere.
An equation for this sphere is
µ
¶2
¶2
µ
3
13
90
2
x−
+ (y − 4) + z −
= .
2
2
4
(a) Let a and b be the vectors
a= h2, 1i
b= h8, 12i .
Draw pictures of the standard representatives of a and b.
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(b) Find the vector u =proja b and the vector v = b−u.
u = proja b
b·a
a
=
|a|2
28
h2, 1i
=
¿5
À
56 28
=
,
5 5
and
v = b − u = h8, 12i −
¿
56 28
,
5 5
À
¿
À
16 32
= − ,
.
5 5
(c) Show that u + v = b and that u is orthogonal to v. Illustrate this result including
the vectors u and v in the drawing that you made in part a.
u+v =
and
¿
56 28
,
5 5
µ
56
u·v =
5
À
¿
À
16 32
+ − ,
= h8, 12i = b
5 5
¶µ
¶ µ ¶µ ¶
16
28
32
+
=0
−
5
5
5
so u and v are orthogonal (because their dot product is zero).
(a) A child pulls a sled through the snow with a force of 50 Newtons exerted at an
angle of 30◦ above the horizontal. Find the horizontal and vertical components
of this force.
The force vector can be written as
F = 50 cos (30◦ ) i + 50 cos (60◦ ) j
2
so the horizontal component of the force is
√
30 cos (30◦ ) = 25 3 Newtons
and the vertical component of the force is
30 cos (60◦ ) = 25 Newtons.
(b) Assuming that the child pulls the sled in this way a distance of 300 meters, how
much work is done (in Joules)?
The displacement vector is D = 300i so the amount of work done by the force is
³ √ ´
D · F =300 25 3 ≈ 12, 990 Joules.
(c) If a and b are the vectors pictured below, does the vector a × b point into the
page or out of the page? How do you know?
By the right—hand rule, the vector a × b points into the page.
(d) If we were to make vector a be longer (but not change the length of vector b and
not change the angle between a and b), what effect would that have on the vector
a × b. In particular, would a × b become shorter or longer? Would a × b point
into the page or out of the page? How do you know?
The vector a × b would still point into the page by the right—hand rule. Also,
since
|a × b| = |a| |b| sin (θ) ,
the length of the vector a × b would increase.
(e) If we were to increase the angle between a and b (but still keep a and b the same
length and still keep the angle acute), what effect would that have on the vector
a × b? In particular, would a × b become shorter or longer? Would a × b point
into the page or out of the page? How do you know?
By the right—hand rule, the vector a × b would still point into the page. However,
since sin (θ) would increase, the length of the vector a × b would increase.
(f) If we were to increase the angle between a and b to be just slightly less than
180◦ (but still keep a and b the same length), what effect would that have on the
vector a × b? In particular, would a × b become shorter or longer? Would a × b
point into the page or out of the page? How do you know?
3
The vector a × b would still point into the page by the right—hand rule. The
length of a × b would decrease, however, since sin (θ) would be very small.
Find the point at which the line
x=1−t
y=t
z = 1 + 3t
intersects the plane
4x − 2y − z = 6.
The line intersects the plane at the parameter value (t) such that
4 (1 − t) − 2 (t) − (1 + 3t) = 6.
The value of t for which the above equation
¡
¢is true is t = −1/3. Thus the line
and the plane intersect at the point 43 , − 13 , 0 .
2. Find parametric and/or symmetric equations for the line of intersection of the two
planes
5x − 6y − 2z = 12
3x + 8z = 0.
The equations of the planes can be rewritten as
−15x + 18y + 6z = −36
15x + 40z = 0.
Thus, any point lying in both planes must satisfy both equations
9y + 23z = 0
3x + 8z = 0.
Two such points are (0, −2, 0) and (8/3, 5/9, −1). Thus, a vector parallel to the line
of intersection is
¿
À
8 23
v=
, , −1 .
3 9
Parametric equations for this line of intersection are
8
x= t
3
y = −2 +
23
t
9
z = −t.
Symmetric equations for this line are
x
8
3
=
y+2
23
9
4
=
z
.
−1
3. Here are six equations labelled 1—6 and six surfaces labelled A—F. Match each equation
with the surface that it describes.
5
1) x2 + y 2 − z 2 = 1
2) x2 − y 2 − z = 0
3) x2 + y 2 − z = 0
4) x2 + y 2 − z 2 = −1
5) x2 + y 2 − z 2 = 0
6) x2 + y 2 + z 2 = 1
Equation 1 matches surface B.
Equation 2 matches surface C.
Equation 3 matches surface F.
Equation 4 matches surface E.
Equation 5 matches surface D.
Equation 6 matches surface A.
6