110.201 Linear Algebra
4th Quiz
April 8, 2005
Problem 1
Find an orthogonal basis for the plane
x

y
+
z
= 0
,
viewed as a subspace of
R
3
.
Solution
So it is easy to see that the basis for the plane is
v
1
=
1
1
0
, v
2
=
1
0

1
Now let’s use GramSchmidt algorithm to get the orthogonal basis of it.
v
1
=
1
1
0
,
then

v
1

=
√
v
1
·
v
1
=
√
2
So we have
u
1
=
v
1

v
1

=
1
√
2
1
1
0
And
v
⊥
2
=
v
2

Proj
u
1
v
2
=
v
2

(
v
2
·
u
1
)
u
1
=
v
2

1
2
v
1
=
1
2

1
2

1
,
and

v
⊥
2

=
r
3
2
Therefore,
u
2
=
v
⊥
2

v
⊥
2

=
r
2
3
1
2

1
2

1
So the orthogonal basis for the plan is
1
√
2
1
1
0
,
r
2
3
1
2

1
2

1
1
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View Full DocumentProblem 2
Let
~e
1
,~e
2
,~e
3
be the standard basis of
R
3
.
Consider the plane
V
spanned by
~e
1
and
~e
2
.
a. For a given vector
~w
= (
a, b, c
)
∈
R
3
, calculate the vector
~u
∈
V
that
minimizes the distance between
V
and
~w
, i.e. ﬁnd
~u
∈
V
such that
k
~u

~w
k ≤ k
~v

~w
k
∀
~v
∈
V.
Solution
~u
=
Proj
V
w
=
a
b
0
b. In the inequality above, is such a
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 Spring '05
 CONSANI
 Linear Algebra, Algebra, orthogonal basis

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