Problems graded: 2.1.7, 2.2.2, 2.3.5 (15 points each); the other twelve prob-
lems were graded based on completeness (to be worth four points each) and
seven points were added to everyone’s score to bring the total to 100.
Section 2.1
1. (a) The dot product of v= (1,2,−1) and w= (−1,0,3) is 1 ∗ −1 + 2 ∗
0 + −1∗3 = −4.
(b) The cross product of vand wis the formal determinant of the matrix
whose first row is (i, j, k), whose second row is v, and whose third row is w; this
yields (2∗3−−1∗0)i+(−1∗−1−1∗3)j+(1∗0−2∗−1)k= 6i−2j+2k= (6,−2,2).
(c) As ||v|| =p12+ 22+ (−1)2=√6, v
||v|| = ( √6
6,√6
3,−√6
6).
As ||w|| =p(−1)2+ 02+ 32=√10, v
||v|| = ( −√10
10 ,0,3√10
10 ).
(d) The cross product of vand w, (6,−2,2), has norm p62+ (−2)2+ 22=
2√11.
(e) Using the formula v•w=||v||||w||cos θ(where θis the angle between v
and w), the cosine of the angle between vand wis −4
√60 =−2√15
15 .
3. To show that e1, e2,and e3constitute a frame it suffices to show that
they are orthonormal: as (1,2,1) has norm √6,||e1|| =6
6= 1 and similarly for
e2(as (−2,0,2) has norm √8) and e3(as (1,−1,1) has norm √3) so it suffices
to show orthogonalilty, for which we can clear the denominators.
The dot product of e1and e2is equal to 1
√48 (−1∗2 + 2 ∗0 + 1 ∗2) = 0, the
dot product of e1and e3is equal to 1
√18 (1 ∗1 + 2 ∗−1 + 1 ∗1) = 0, and the dot
product of e2and e3is equal to 1
√24 (−2∗1 + 0 ∗ −1 + 2 ∗1) = 0 so the three
vectors form an orthonormal set in R3and therefore a frame.
Setting v= (6,1,−1), we note that v= (v•e1)e1+ (v•e2)e2+ (v•e3)e3;
here (v•e1) = 6∗1+1∗2+−1∗1
√6=7√6
6, (v•e2) = 6∗−2+1∗0+−1∗2
√8=−7√2
2, and
(v•e3) = 6∗1+1∗−1+−1∗1
√3=4√3
3so
v=7√6
6e1+−7√2
2e2+4√3
3e3.
This can be checked by noting that 7√6
6e1= (7
6,7
3,7
6), −7√2
2e2= (7
2,0,−7
2),
and 4√3
3e3= (4
3,−4
3,4
3); adding these three indeed gives v.
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