MATH 23, FALL 2011 REVIEW SHEET SOLUTIONS
Instructions:
These problems represent the types of questions you might nd on the rst 4:00
exam in this term's Math 23. They are taken directly from old exams. The best way to use this
review is to choose a number of these problems to make up a (somewhat long) exam for yourself,
then solve it under exam conditions, no notes, no calculator, and no looking at answers. Then,
using your notes and text, grade your trial exam.
(1) If
A
= 3
i

j
+ 3
k
, and
B
= 2
i
+ 4
j

k
, nd:
(5 points per part)
(a)
2
A

B
,
Solution:
2
A

B
= 2(3
i

j
+ 3
k
)

(2
i
+ 4
j

k
)
= 4
i

6
j
+ 7
k
.
(b) The cosine of the angle between the vectors
A
and
B
.
Solution:
Since
A
·
B
=
k
A
kk
B
k
cos
θ
, where
θ
is the angle between the vectors,
cos
θ
=
A
·
B
k
A
kk
B
k
=
3
·
2 + (

1)
·
4 + 3
·
(

1)
√
9 + 1 + 9
√
4 + 16 + 1
=

1
√
19
√
21
.
(c) The number
t
so that
A
+
t
B
is perpendicular to
i
+
j
.
Solution:
A
+
tB
is perpendicular to
i
+
j
when
(
A
+
tB
)
·
(
i
+
j
) = 0
, which is
when:
0 = (
A
+
tB
)
·
(
i
+
j
)
= ((3
i

j
+ 3
k
) +
t
(2
i
+ 4
j

k
))
·
(
i
+
j
)
= (3 + 2
t
) + (

1 + 4
t
)
= 2 + 6
t,
or
t
=

1
/
3
.
(2) Find a vector perpendicular to
3
i

2
j
+
k
.
(5 points)
Solution:
There are lots of right answers to this one.
2
i
+ 3
j
,
i

3
k
,
or
j
+ 2
k
will all
work. All you need to nd is a vector whose dot product with the given one is 0.
(3) If
A
= 2
i
+
j

3
k
, and
B
=

2
i
+ 5
j
+
k
, nd:
(5 points per part)
(a)
A

3
B
,
Solution::
A

3
B
= 8
i

14
j

6
k
.
(b)
A
·
B
.
Solution::
A
·
B
= (2)(

2) + (1)(5) + (

3)(1) =

2
.
(c)
A
×
B
.
Solution::
A
×
B
=
ﬂ
ﬂ
ﬂ
ﬂ
ﬂ
ﬂ
i
j
k
2
1

3

2 5
1
ﬂ
ﬂ
ﬂ
ﬂ
ﬂ
ﬂ
= 16
i
+ 4
j
+ 12
k
.
(4) Find the equation of the line of intersection of the planes
x
+
y
= 2
and
y

z
= 4
.
1
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document2
MATH 23, FALL 2011 REVIEW SHEET SOLUTIONS
Solution:
The standard way to do this is to nd a vector parallel to both planes, which
is the same as being perpendicular to both normals. So, the crossproduct of the two
normals will work. That is, the vector
A
=
N
1
×
N
2
= (
i
+
j
)
×
(
j

k
)
=
k
+
j

i
=

i
+
j
+
k
will be in the direction of the line of intersection of both planes. In order to nd the
equation of this line, you need a point on the line; a point on both planes. But, if
y
= 0
,
then
x
= 2
and
z
=

4
, so
(2
,
0
,

4)
is on the line. Then, the equation for the line is
r
(
t
) = 2
i

4
k
+
t
(

i
+
j
+
k
)
.
An alternate way to solve this problem is to note that you can set
y
=
t
in the equation
for both planes, and solve for
x
in the rst plane, and
z
in the second, giving
x
=
2

t, y
=
t,
and
z
=
t

4
.
(5) Find an equation of the plane
P
through the points
(2
,
1
,
0)
,
(1
,
1
,
1)
, and
(0
,
3
,

1)
.
Solution:
This is the end of the preview.
Sign up
to
access the rest of the document.
 Spring '06
 YUKICH
 Math, 4k, 5J, 2 L

Click to edit the document details