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Unformatted text preview: SECTION 13.3 THE DOT PRODUCT ET SECTION 12.3 From the diagram b2j and 112 j point in opposite directions, so I); = “a2. AB = BC], so lbgl = sin¢ IBCI : sind) AB = la3, and lb1= coed) IBCl = cosd) lAB=a1. b3 k and (13 k have the same direction, as do b1 i and (11 i, so b = (a1, —a2, a3). When the ray hits the other mirrors, similar
arguments show that these reﬂections will reverse the signs of the
other two coordinates, so the ﬁnal reﬂected ray will be (—a1, —a2, 4113) = —a, which is parallel to a. 13.3 The Dot Product ET 12.3 1. (a) a  b is a scalar, and the dot product is deﬁned only for vectors, so (a  b) ~ 0 has no meaning. (b) (a  b) c is a scalar multiple of a vector, so it does have meaning. (c) Both [a] and b  c are scalars, so lal (b  c) is an ordinary product of real numbers, and has meaning. (d) Both a and b + c are vectors, so the dot product a ~ (b + c) has meaning. 7 (e) a  b is a scalar, but c is a vector, and so the two quantities cannot be added and a  b + c has no meaning. (f) a is a scalar, and the dot product is deﬁned only for vectors, so lal . (b + c) has no meaning. 2. Let the vectors be a and b. Then by Theorem 3, a . b = lal b cos0 = (6)(§) cos g 2 E675 = ﬂ. 3. a ~ b = (“2, g} . (—5, 12) : (—2)(—5) + (§)(12) = 10 + 4 = 14 4. ab = (—2,3>  (0.11.2) : (2)(0.7) +(3)(1.2) = 22
Glab :<4,1,4l1>(6,#3,—8):(4)(6)+(1)(3)+(i)(e8): 19 s. a . b = (s, 25, 33>  <t, t, 5t) : (5)05) + (25)(—t) + (3s)(5t) = st — 2st + 153t : 14st
@ab : (ie2j+3k)  (5i+9k) : (1)(5)+(—2)(0)+(3)(9) :32 8. ab : (4j —3k)  (2i+4j+6k) : (0)(2)+(4)(4)+(#3)(6) : —2 9. a . b = a ]b cos6 = (6)(5)c0s 2?" : 30 (—g) = #15 10. UseTheorem3: ab: ablcos0 = (3)(\/6)Cos45° : 3&1?) = g 2\/§:3\/§,~o 5.20 11. 117 v, and w are all unit vectors, so the triangle is an equilateral triangle. Thus the angle between u and v is 60° and u  v : u v cos 60° : (1)(1) = Ifw is moved so it has the same initial point as u, we can see that the angle between them is 120° and we have u  w : u w cos 120O : (1)(1)(*%) : —%. tutti; i . lt‘. mm.ww..,._m , RY OF SPACE ET CHAPTER 12 i C/ [3 232 CHAPTER 13 VECTORS AND THE GEOMET Iv) can be determined by examining the right triangle formed by u and v. 12. u is a unit vector, so w is also a unit vector, and Since the angle between u and v is 45°, we have (V) = (11] cos 45° 2 Then 11  v = [11) v cos 45° = (1) l? : Since u and w are orthogonal, u  w : 0. 13. (a) i j z (1, 0,0)  (0, 1, 0) : (1)(0) + (0)(1) + (0)(0) = 0. Similarly,j  k z (0)(0)+(1)(0)+(0)(1) = 0 and k ~ i = (0X1) + (0X0) + (1X0) = 0 perpendicular, the cosine factor in each dot product (see Theorem 3) Another method: Because i, j, and k are mutually . 1r_
ISCOS2 —— 2 = 1 since i is a unit vector. Similarly, j  j : Lil2 : 1 and (b) By Property 1 of the dot product, i  i z m2 = 1 kkzlkl221. 14. The dot product A  P is
(a,b,c)  (2,1.5, 1) : (1(2) + b(1.5) + C(1)
= (number of hamburgers sold) (price per hamburger)
+ (number of hot dogs sold) (price per hot dog)
+ (number of soft drinks sold) (price per soft drink) so it is equal to the vendor’s total revenue for that day. is. la] = 1/(—8)2 + 62 = 10,1bl = (ﬁf + 32 = 4, anda . b = Ham/7) + (6)(3) = 18 — 8 ﬁ. From Corollary 6,
M) m 95°. ~b — — 4 7 .
we have c086 _ f; lb = 1810.8? = So the angle between a and b IS 0 : cos_1( 20 16. la] 2 + 12 : 2, lb) = «/0 + 2 = 5, anda  b = + (1)(5) = 5. Using Corollary 6, we have and the angle between a and b is 005'1 = 60°. a
8
sh
ll
1”:
E
H
N)
01
ml >l =5.Then @lal : (/32 + (—1)2 + 52 = J33, b : t/(—2)2 + 42 + 32 2 J25, anda  b : (3)(*2) + (—1)(4) + (5)(3)
a b 5 5
= and the an le between a and b is 0 = cos“1 4—) % 81°.
«35  J29 «1015 g (ﬂo'fs' (i? la] = «42 + 02 + 22 = m, 1b) = (/22 + (—1)2 +02 = ﬂ}, andab = (4)(2)+(0)(—1)+(2)(0) = 8. ab 8 zgandﬂzcos’wé) z370. . (/5 5 “’59 2 1a) 1101 ,.]
a“
(D
5
O
8
Q)
H
E
:5:
H
S
O 19. 1a) = «02 + 12 + 12 = ﬂ, M = (/12 + 22 + (—3)2 : $11, and a . b : (0)(1)+(1)(2)+(1)(—3) = —1. ab ——:1——— — ‘1 and0:cosT1 m101°. 2W SECTION 13.3 THE DOT PRODUCT ET SECTION 12.3 C 235 incelZi+3j—6kl=v4+9+3 = 4 :7,Equation58and9givecoscx=%,cosﬁ=%,andcos'y=%§,while 31.5
o  N o __ —1 N o
“5005*1(%)21 73 ,B = cos ~65 ,andv — cos (*2) ~ 149.
32 Sincepij+2k1: «4+ 1 + = \/§ = 3, Equations 8 and9 give cosa : %, cosﬁ = :3—1,andcosry : —3,while
(2:7 : cos1g) m 48° andﬁ : cosT1(—%) % 109°.
1
i— : —— and ,_ W z ‘ = : :
33' “0,6,6”  c +c +0 x/gc [srncec> 0], so cosa c055 (3081 £0 3 azﬂzryzcos1(71§)z55°. 204100523 2 1 #cos2(§) —cos2(§) : 1* — : i. N) 34. Since 0082 a + C082 [3 + cos2 y z 1, cos2 ’7 : 1  cos 27r Thuscos'yzd:% and'y: g or'y ?. . j . b . 4 .
3? [a1 = 1/32 + (—4)2 = 5. The scalar projection ofb onto a 15 compab : ala 3 5 +é ) 0 : 3 and the vector projection ofb onto a is proja b = lal
. . . b 1 ~4 ~1 2
36. lal 2 V12 + 22 = x/g, so the scalar prejection of b onto a is compa b = 3—— : ( ) + 2 : —— and the vector
la J5 V5
. . A . _ a  b a _ 2 1 _ 2 4
prOJection ofb onto ais pr0jab — (Ta‘ >T2—l—i — —E > W (1,2) 4 3, —3
. . . b
37. a 2 V9 + 36 + 4 = 7 so the scalar pr0ject10ri ofb onto a is compab = 8T; : § (3 + 12 — 6) = g. The vector
projection ofb onto a is projab : gr; 2 % %<3,6, #2) = 19?; (3,67 —2) 2 (ﬁg, 3%, :3
a  b 1 37 .
b : —— : 7(—10 — 3 — 24): —7,while the la] 2 «4 + 9 + 36 = 7 so the scalar projection ofb onto a is compa 37 a
__ : #ﬂ . 1 (4,3,4) = 43—3 (2,3r6> = Zara—’23 vector o'ect'on‘s 0' b:
Pr] 1 1Prla 71a] av 39. lal = x/4 + 1 + 1 = v21 so the scalar projection ofb onto a is compa b : T;— : 7.1——
2 . . . , 1 a 1 2i#'+4k . . . .
pTOJCCUOnOfbontoaISpI‘OJab=Tim=ﬁ'———j—§"=i<21~J+4k):ﬁl—%J+%k. 41. (orthab)a:(b—projab)a=ba«(projab)a:b~a#  ~ ‘ 12
a a So they are orthogonal by (7). ...
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This note was uploaded on 05/02/2011 for the course MATH 408D taught by Professor Chu during the Spring '09 term at University of Texas.
 Spring '09
 Chu

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