MATH 550, VECTOR ANALYSIS, EXTRA
HOMEWORK 1 AND SOLUTIONS
Suppose (e1,e2,e3) is an orthonormal basis of the abstract vector space V, and define
A=
−√6
4−2√2+√6
8−√2+2√6
8
−√6
4
2√2−√6
8−√2+2√6
8
−1
2
3
4
√3
4
.
Let (e0
1,e0
2,e0
3)=(e
1
,e
2
,e
3
)Abe another orthonormal basis. Define v=e0
1−2e0
2+3e
0
3
and w=−2e0
1−e0
2−4e0
3. Answer the following:
Problem 1. Find the components of vand wrelative to the basis (e1,e2,e3).
Solution: We know that v=(e
0
1
,e
0
2
,e
0
3
)h1,−2,3iand w=(e
0
1
,e
0
2
,e
0
3
)h−2,−1,−4i.
(Recall that ha, b, ciis an abbreviation for
a
b
c
, which takes up too much room.) Using
the equation (e0
1,e0
2,e0
3)=(e
1
,e
2
,e
3
)Awe have v=(e
1
,e
2
,e
3
)Ah1,−2,3iand w=
(e1,e2,e3)Ah−2,−1,−4i. Computing we get
Ah1,−2,3i=
−√6
4−2√2+√6
8−√2+2√6
8
−√6
4
2√2−√6
8−√2+2√6
8
−1
2
3
4
√3
4
1
−2
3
=
√2+6√6
8
−7√2+6√6
8
−8+3√3
4
,
Ah−2,−1,−4i=
−√6
4−2√2+√6
8−√2+2√6
8
−√6
4
2√2−√6
8−√2+2√6
8
−1
2
3
4
√3
4
−2
−1
−4
=
6√2−3√6
8
2√2+13√6
8
1−4√3
4
.
Thus v=(e
1
,e
2
,e
3
)h
√
2+6√6
8,−7√2+6√6
8,−8+3√3
4iand
w=(e
1
,e
2
,e
3
)h
6
√
2−3
√
6
8,2
√
2+13√6
8,1−4√3
4i.
Problem 2. Compute v·wusing the basis (e1,e2,e3), and also compute v·wusing the
basis (e0
1,e0
2,e0
3). Do you get the same answer?
Solution: Using the basis (e1,e2,e3) we have v·w=(
√
2+6√6
8)(6√2−3√6
8)
+(−7
√
2+6√6
8)(2√2+13√6
8)+(−8+3√3
4)(1−4√3
4)=−12. Using the basis (e0
1,e0
2,e0
3)wehave
v·w= (1)(−2) + (−2)(−1) + (3)(−4) = −12. Clearly we get the same answer.
Problem 3. Compute v×wusing the basis (e1,e2,e3), and also compute v×0wusing
the basis (e0
1,e0
2,e0
3). Do you get the same answer?
Solution: Using the basis (e1,e2,e3) we have v×w=(e
1
,e
2
,e
3
)h(−
7
√
2+6√6
8)(1−4√3
4)
−(−8+3√3
4)(2√2+13√6
8),(−8+3√3
4)(6√2−3√6
8)−(√2+6√6
8)(1−4√3
4),(√2+6√6
8)(2√2+13√6
8)
−(−7√2+6√6
8)(6√2−3√6
8)i=(e
1
,e
2
,e
3
)h
−9
√
2+30√6
8,−√2+10√6
8,28+5√3
4i.
