HW4Solution - Problem 2.132 By evaluating the cross product...

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Unformatted text preview: Problem 2.132 By evaluating the cross product U x V, prove the identity stow; * 93) = sine}. cos 6; — ms 6: sin 62. Solution: Assume that both U and V lie in the x—_\' plane. The neg)! is to use the definition of the cross product (Eq. 2.28) and the liq. (134), and equate the two. From Eq. (2.28) U x V : lUHVl sin“?! —9_, be. Since the positive z-zixis is out of the paper, and 0 points into - k paper, then e z 7k. Take the dot product of both sides with e, and “flunk-k: LThus (UxV)-k) '"fi ’92) : ’( IUIiVi II: \ectors are: t: 1Ui{icost91 +jsin63), and V = |V|(ioos :92 +jsin03). H cross product is i j k IUi 00561 |U| sin 6] 0 |V| c0562 |V| sin 9;; 0 Uzi-V: : i(0) —j(0) + k(lI]HV|){COSI913iD 92 — cos :9; Sin9|} mimic into the definition to obtain: sinwl 7 63) : sin91c0592 e -6; sinflg. QED. Problem 2.138 The rope AB exerts a SO-N force T on the collar at A. Let rm be the position vector from point C to point A. Determine the cross product rm x T. Solution: We define the appropriate vectors. l'co = (“Mi 7 0.3j + 0.251;) m PC!) 1'01 : (02 m) r I (‘0 :{~0.091ifi 0.137] + 0.] 14k) in 1’03 : (0.5j + 0.1510111 roc : [0.4i Jr 0.3j) m I'AB : ['03 — (For +l‘(',t) = (0.6“ 7 1.22j — 0.30510 "1 rAB T : (50 N) : (7333i + 36.7j + 3.93k) N IPABI Now take the cross product i j rm x T : 70.09] —0.]37 0.] I4 = (—4.72i * 3.48j + —'.". #33] 36.7 3.93 ' 96k) N—m ' Problem 1141* Determine the minimum distance ion] point P to the plane defined by the three points i A. B, and C. . Sohtion: The strategy is to find the unit vector perpendicular to A it plane. The projection of this unit vector on the vector OF: POP A e is it distance from the origin to P along the perpendicular to the plane. the projection on e of any vector into the plane (rm ~e, r03 - e, or ‘ [at he] is the disumce from the origin to the plane along this same perpendicular. Thus the distance of P from the plane is :J:r0p-e—r0A-e. the position vectors are: rm = 31, 1‘03 : Sj, roc = 4k and top = I—6j+5k. The unit vector perpendicular to the plane is found in] the cross product of any two vectors lying in the plane. Noting: I I1- = r05 —r03 : 75.] +4k, and ray; : rgA —I‘()3 : 3i 7 Sj. The units product: i jk ' new“: 0 —5 4 :20i+12j+15k_ 3 —5 0 1E magnitude is II‘gC x r3,” : 27.73, thus the unit vector is I: : Q—Cllifir 0.4327j + 0.5409kr The distance of point P from the plane 65' = rep - e 7 1-0;; -e : t 1.792 — 2.l64 : 9.63 m. The second term "a the distance of the plane from the origin; the vectors r03, or rec ‘ mid have been used instead of two 3 (O, 5, (J) in Z/ C[0.0.4] P 33 (9, 6, 5) In P£9,6,5] O o___.’_ A[3,0,0 J ...
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This note was uploaded on 04/18/2011 for the course ENGR 200 taught by Professor Mullen during the Spring '08 term at Case Western.

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HW4Solution - Problem 2.132 By evaluating the cross product...

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