QUIZ 1 (CHAPTER 14) - SOLUTIONS
MATH 252 – FALL 2007 – KUNIYUKI
SCORED OUT OF 125 POINTS
⇒
MULTIPLIED BY 0.84
⇒
105% POSSIBLE
Clearly mark vectors, as we have done in class. I will use boldface, but you don’t!
When describing vectors, you may use either
or “
i
−
j
−
k
” notation.
Assume we are in our usual 2- and 3-dimensional Cartesian coordinate systems.
Give exact answers, unless otherwise specified.
1)
Assume that
a
1
,
a
2
,
p
, and
q
are real numbers.
Prove that, if
a
=
a
1
,
a
2
, then
p
+
q
( )
a
=
p
a
+
q
a
. Show all steps! (10 points)
Let
a
=
a
1
,
a
2
.
p
+
q
( )
a
=
p
+
q
( )
a
1
,
a
2
=
p
+
q
( )
a
1
,
p
+
q
( )
a
2
=
pa
1
+
qa
1
,
pa
2
+
qa
2
=
pa
1
,
pa
2
+
qa
1
,
qa
2
=
p a
1
,
a
2
+
q a
1
,
a
2
=
p
a
+
q
a
Q.E.D.
Note: Many people who scored 4 points did the following:
p
+
q
( )
a
=
p
+
q
( )
a
1
,
a
2
=
p a
1
,
a
2
+
q a
1
,
a
2
←
You're using the property you're trying to prove!
That's circular reasoning!
=
p
a
+
q
a
2)
Write an inequality in
x
,
y
, and/or
z
whose graph in our usual three-dimensional
xyz
-coordinate system consists of the sphere of radius 4 centered at the origin
and all points inside that sphere. (4 points)
The equation of the sphere of radius 4 centered at the origin:
x
2
+
y
2
+
z
2
=
16
.
We also want all points inside that sphere, so our inequality is:
x
2
+
y
2
+
z
2
≤
16
.
Another approach: We want all points whose distance from the origin is at most 4.