Math
2940
Solutions, Fall
2011
Section
4
.
4
4
)(a)
Q
=
1 0
0 1
0 0
so
QQ
T
=
1 0
0 1
0 0
±
1 0 0
0 1 0
²
=
1 0 0
0 1 0
0 0 0
. A square matrix
Q
will
never
work.
(b) For this problem one of the two vectors
must
be
~
0. Choose any nonzero vector for the
other.
(c) 1
/
√
3(1
,
1
,
1) is our ﬁrst vector. The other two must lie in the plane
x
+
y
+
z
= 0 (why?)
So use 1
/
√
2(1
,
0
,

1) for the second vector. If the third is (
a, b, c
), then to be perpen
dicular to the ﬁrst, we need
a
+
b
+
c
= 0 and to be perpendicular to the second we need
a

c
= 0 so (1
,

2
,
1) suggest itself. We normalize to get 1
/
√
6(1
,

2
,
1).
5
) We take (1
,

1
,
0) as our ﬁrst, and if (
a, b, c
) is the second, we need
a
+
b
+ 2
c
= 0 and
a
=
b
so (1
,
1
,

1) works. Normalizing we get 1
/
√
2(1
,

1
,
0) and 1
/
√
3(1
,
1
,

1).
10
) (a) Dotting the equation
c
1
~
q
1
+
c
2
~
q
2
+
c
3
~
q
3
=
~
0 with
~
q
i
gives
c
i
= 0. This holds for
i
= 1
,
2
,
3.