MA 265 HOMEWORK ASSIGNMENT #6 SOLUTIONS
3
Then we have the value
x
3
=
2
1
6
3
2

2
1
1

4

A

=
(2)
2

2
1

4

(1)
3

2
1

4
+ (6)
3
2
1
1
5
=
4
5
#5. Page 187; Exercise 2.
Determine the head of the vector

2
5
whose tail is (

3
,
2). Make a sketch.
Solution:
Say that the tail is
P
(

3
,
2) and the head is
Q
(
x, y
). Then we have the directed line segment
→
PQ
=
x

(

3)
y

2
=

2
5
.
Since
x
+ 3 =

2 and
y

2 = 5, we see that
x
=

5 and
y
= 7. Hence
the head is (

5
,
7).
A sketch can
be found below.
y
•
(

5
,
7)

7


5


3
•
U
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
(

3
,
2)


1
/


5



3



1

1


3


5
x
O
#6. Page 187; Exercise 5.
For what values of
a
and
b
are the vectors
a

b
2
and
4
a
+
b
equal?
Solution:
Upon equating both components, we find the system of equations
a

b
=
4
a
+
b
=
2
Adding these two equations together gives 2
a
= 6, so that
a
= 3. Similarly, subtracting the two equations
gives

2
b
= 2, so that
b
=

1. Hence the values are
a
= 3 and
b
=

1.