Chapter 10
Orthogonality
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Example 10.1 (Dot product)
Find the dot product of
x
= (2
,
2
,
1)
and
y
= (2
,
5
,

3)
.
Solution
The dot product (or inner product or scalar product) of
x
and
y
is given by
x
·
y
= (2
,
2
,
1)
·
(2
,
5
,

3) = 2
·
2 + 2
·
5 + 1
·
(

3) = 11.
That is,
x
·
y
is obtained by multiplying corresponding components and adding the resulting products. The
vectors
x
and
y
are said to be
orthogonal
(or perpendicular) if their dot product is zero, that is, if
x
·
y
= 0.
Therefore, for this example, the two given vectors
x
and
y
are not orthogonal.
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Example 10.2 (Norm)
For
x
= (1
,

2
,
3
,

4
,
5)
, find the norm
k
x
k
.
Solution
The norm of
x
is given by
k
x
k
=
√
x
·
x
=
p
1
2
+ (

2)
2
+ 3
2
+ (

4)
2
+ 5
2
=
√
55.
The norm (or length) of a vector
x
in
R
n
, denoted by
k
x
k
, is defined to be the nonnegative square root of
x
·
x
. In particular, if
x
= (
x
1
, x
2
,
· · ·
, x
n
), then
k
x
k
=
p
x
2
1
+
x
2
2
+
· · ·
+
x
2
n
. That is,
k
x
k
is the square root
of the sum of the squares of the components of
x
. Thus,
k
x
k
>
0, and
k
x
k
= 0 if and only if
x
=
0
.
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Example 10.3 (Normalize a vector)
Find the rescaled vector
x
k
x
k
, where
x
= (2
,
2
,
1)
.
Solution
By finding the norm
k
x
k
=
√
2
2
+ 2
2
+ 1
2
= 3, we could normalize
x
to the following unit vector
x
k
x
k
=
1
3
(2
,
2
,
1) = (
2
3
,
2
3
,
1
3
).
Verify that the norm of the above rescaled vector is
p
(2
/
3)
2
+ (2
/
3)
2
+ (1
/
3)
2
= 1. In general, a vector
x
is called a unit vector if
k
x
k
= 1 or, equivalently, if
x
·
x
= 1. For any nonzero vector
x
in
R
n
, the vector
ˆ
x
= (1
/
k
x
k
)
x
=
x
/
k
x
k
is the unique unit vector in the same direction of
x
. The process of finding ˆ
x
from
x
is called
normalizing
x
.
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