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Unformatted text preview: or “wedge” between v and w. For
example, if v D .1; 0/ and w D .0; 1/, then the cone is the whole quadrant x 0,
y 0. Question: What if w D v? The cone opens to a halfspace. 13 Sum D zero vector. Sum D
p
14
15 16
17 18 19 Solutions to Exercises 3 1
C 1 v C 1 w is the center of the triangle between u; v and w; 1 u C 2 w lies
3
3
2
between u and w
(b) To ﬁll the triangle keep c 0, d 0, e 0, and c C d C e D 1. 20 (a) 1
u
3 21 The sum is .v u / C .w
are in the same plane! v / C .u w/ D zero vector. Those three sides of a triangle 1
22 The vector 2 .u C v C w/ is outside the pyramid because c C d C e D 1
2 C 1 C 1 > 1.
2
2 23 All vectors are combinations of u; v; w as drawn (not in the same plane). Start by seeing that c u C d v ﬁlls a plane, then adding e w ﬁlls all of R3 . 24 The combinations of u and v ﬁll one plane. The combinations of v and w ﬁll another plane. Those planes meet in a line: only the vectors c v are in both planes.
25 (a) For a line, choose u D v D w D any nonzero vector (b) For a plane, choose
u and v in different directions. A combination like w D u C v is in the same plane. 26 Two equations come from the two components: c C 3d D 14 and 2c C d D 8. The solution is c D 2 and d D 4. Then 2.1; 2/ C 4.3; 1/ D .14; 8/. 27 The combinations of i D .1; 0; 0/ and i C j D .1; 1; 0/ ﬁll the xy plane in xyz space. 28 There are 6 unknown numbers v1 ; v2 ; v3 ; w1 ; w2 ; w3 . The six equations come from the components of v C w D .4; 5; 6/ and v
so v D .3; 5; 7/ and w D .1; 0; 1/. w D .2; 5; 8/. Add to ﬁnd 2v D .6; 10; 14/ 29 Two combinations out of inﬁnitely many that produce b D .0; 1/ are 2u C v and
No, three vectors u; v; w in the x y plane could fail to produce b if all
three lie on a line that does not contain b. Yes, if one combination produces b then
two (and inﬁnitely many) combinations will produce b. This is true even if u D 0; the
combinations can have different c u.
1
w
2 1
v.
2 30 The combinations of v and w ﬁll the plane unless v and w lie on the same line through .0; 0/. Four vectors whose combinations ﬁll 4dimensional space: one example is the
“standard basis” .1; 0; 0; 0/; .0; 1; 0; 0/; .0; 0; 1; 0/, and .0; 0; 0; 1/.
31 The equations c u C d v C e w D b are 2c
d
D1
c C2d
eD0
d C2e D 0 So d D 2e
then c D 3e
then 4e D 1 c D 3=4
d D 2=4
e D 1=4 Problem Set 1.2, page 19
1 uv D 1:8 C 3:2 D 1:4, u w D 4 :8 C 4:8 D 0, v w D 24 C 24 D 48 D w v. 2 kuk D 1 and kvk D 5 and kwk D 10. Then 1:4 < .1/.5/ and 48 < .5/.10/, conﬁrming the Schwarz inequality. 3 Unit vectors v=kvk D . 3 ; 4 / D .:6; :8/ and w=kwk D . 4 ; 3 / D .:8; :6/. The cosine
55
55 of is kvk kwk D 24 . The vectors w; u; w make 0ı ; 90ı ; 180ı angles with w.
v
w
25
4 (a) v . v/ D 1
(b) .v C w/ .v w/ D v v C w v v w w w D
1 C . / . / 1 D 0 so D 90ı (notice v w D w v)
(c) .v 2w/ .v C 2w/ D
v v 4w w D 1 4 D 3. Solutions to Exercises 4
p p 5 u1 D v=kvk D .3; 1/= 10 and u2 D p=kwk D .2; 1; 2/=3....
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This note was uploaded on 09/25/2012 for the course PHY 103 taught by Professor Minki during the Spring '12 term at Korea Advanced Institute of Science and Technology.
 Spring '12
 Minki
 Mass

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