miller (zdm77) – assignment 4 – luecke – (58600)
1
This printout should have 15 questions.
Multiplechoice questions may continue on
the next column or page – fnd all choices
beFore answering.
001
10.0 points
±ind the angle between the vectors
a
=
a
1
,
2
√
3
A
,
b
=
a−
√
3
,
7
A
.
1.
angle =
π
6
correct
2.
angle =
3
π
4
3.
angle =
π
4
4.
angle =
5
π
6
5.
angle =
2
π
3
6.
angle =
π
3
Explanation:
Since the dot product oF vectors
a
and
b
can be written as
a
.
b
=

a

b

cos
θ ,
0
≤
θ
≤
π,
where
θ
is the angle between the vectors, we
see that
cos
θ
=
a
.
b

a
b

,
0
≤
θ
≤
π .
But For the given vectors,
a
·
b
= (1)(
−
√
3) + (2
√
3)(7) = 13
√
3
,
while

a

=
√
13
,

b

=
√
52
.
Consequently,
cos
θ
=
13
√
3
√
13
·
2
√
13
=
√
3
2
where 0
≤
θ
≤
π
. Thus
angle =
π
6
.
keywords: vectors, dot product, angle be
tween vectors
002
10.0 points
A triangle Δ
PQR
in 3space has vertices
P
(1
,
−
6
,
3)
, Q
(2
,
−
3
,
1)
, R
(6
,
−
5
,
2)
.
Use vectors to decide which one oF the Follow
ing properties the triangle has.
1.
rightangled at
R
2.
not rightangled at
P, Q,
or
R
correct
3.
rightangled at
P
4.
rightangled at
Q
Explanation:
Vectors
a
and
b
are perpendicular when
a
·
b
= 0. Thus Δ
will be
(1)
rightangled at
P
when
−−→
QP
·
−→
RP
= 0,
(2)
rightangled at
Q
when
PQ
·
RQ
= 0,
(3)
rightangled at
R
when
PR
·
QR
= 0.
But For the vertices
P
(1
,
−
6
,
3)
, Q
(2
,
−
3
,
1)
, R
(6
,
−
5
,
2)
we see that
=
a
1
,
3
,
−
2
A
,
QR
=
a
4
,
−
2
,
1
A
,
while
RP
=
5
,
−
1
,
1
A
.
Thus
QP
·
RP
= 10
,
·
RQ
= 4
,
and
·
QR
= 17
.
Consequently, Δ
is
not rightangled at
P, Q,
or
R
.
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2
keywords: vectors, dot product, right trian
gle, perpendicular,
003
10.0 points
Find the vector projection of
b
onto
a
when
b
=
a−
2
,
4
A
,
a
=
a
1
,
−
4
A
.
1.
vector projection =
−
1
a
1
,
−
4
A
2.
vector projection =
−
18
√
17
2
,
4
A
3.
vector projection =
−
17
√
17
a
1
,
−
4
A
4.
vector projection =
−
17
√
17
2
,
4
A
5.
vector projection =
−
18
17
2
,
4
A
6.
vector projection =
−
18
17
a
1
,
−
4
A
cor
rect
Explanation:
The vector projection of
b
onto
a
is given
in terms of the dot product by
proj
a
b
=
p
a
·
b

a

2
P
a
.
Now when
b
=
2
,
4
A
,
a
=
a
1
,
−
4
A
,
we see that
a
·
b
=
−
18
,

a

=
r
(1)
2
+ (
−
4)
2
.
Consequently,
proj
a
b
=
−
18
17
a
1
,
−
4
A
.
keywords:
004
10.0 points
Find the vector projection of
b
onto
a
when
b
=
i
+ 3
j
−
k
,
a
= 2
i
+
j
+ 3
k
.
1.
vector projection =
3
14
( 2
i
+
j
+ 3
k
)
2.
vector projection =
2
7
(
i
+ 3
j
−
k
)
3.
vector projection =
3
14
(
i
+ 3
j
−
k
)
4.
vector projection =
1
7
(
i
+ 3
j
−
k
)
5.
vector projection =
1
7
( 2
i
+
j
+ 3
k
)
correct
6.
vector projection =
2
7
( 2
i
+
j
+ 3
k
)
Explanation:
The vector projection of
b
onto
a
is given
in terms of the dot product by
proj
a
b
=
p
a
·
b

a

2
P
a
.
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 Fall '07
 Gilbert
 Dot Product, vector projection

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