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EEL3472HA2S - P roblern3.4 GivenA =*2 i3 21 and B = tB y2...

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Problern 3.4 Given A = *2- i3 + 21 and B = tB' +y2 + 2Bz: (a) find B' and B, if A is Parallel to B; (b) find a rel4tion between -B' and B, If Lis perpendicular to B' Solution: (a) If A is parallel to B, then their directions are equal or opposite: it : lin, or From they-comPonent, A/lAl : +B/lBl, ?2-93 +2 , *,Bx+92+28, ___-- --frT - - JT+4TE . l a - ) - L m:@ which can only be solved for the minus sign (which means that A and B must point in opposite directions for fhem to be parallel).Solving fot B! + Bl, / -2 ,---'\ 2 20 ni+ ni: | -\/r4 | -4: ; . \ - J / 9 -8, ^ -2\66 -4 : : I : - - v/i6D) "r 3Jt4 3 From the )c-component, 2 J14 and, from the z-comPonent, , Ar:A' This is consistent with our result for ni + A? These results could also have been obtained by assuming 04 was 0o or l80o and solving lAllBl : fA'B, or by solving A x B = 0' (rii if a is perpendicular to B, then their dot product is zero(see Section 3-1.4). Using Eq. (3.21), 0 : A 'B :2Br- 6+ Bz, or B,--6-28,. There are an infinite number of vectors which could be B and be perpendicular to A, but their x- and z-components must satisff this relation' This result could have also been obtained by assuming 0AB :90" and calculating lAllBl :lAxBl. / a \ r
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Problem 3.6 Given vectors A : i2-9* 23 andB : t3 -tZ,finda vector C whose magnitude is 9 and whose direction is perpendicular to both A and B. Solution: The cross product of two vectors produces a new vector which is perpendicular to both of the original vectors. Two vectors exist which have a magnitude of 9 and are orthogonal to both A and B: one which is 9 units l6ng in the direction of theunit vector parallel to A x B, and one in theopposite direction.
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