Question

1. Let a=(ay, ay) and b= (6,62) be vectors in Rº. a. Verify that <a, b >= 20,62 +5 a,b2 satisfies the inner product axioms. b. a=(-1,3), b= (2,5) Find da,b).

Answer 1

a 4, ut f a solution: Let v be field f. Let be arbitrary. Thes < > ; VRV inner product on V are satisfiedi lis. 59,62 = <b, a vector space over a and @,5,CEV function f is said to be if 6/οωίνης φκου, le complex conjugate of < b,9 > di) < 9, Q) so O 18 a = 0 we <a, 9 (iv) sua + vb,c> u<0,0) + veb, (> Now Given that as 19, , G2) and b = (big bz) be vectors in IR² |(a) and < 9, b) = 2 a, b, a 5 az bz (1) will prove that axioms ci-jo) are satisfied for (I) is consider <b,as = qbia, + 5b292 = 2b,a, + 5 b2 92 = 2 b, a, + 5 bz az but since (9., 92), (bi, bz) ie 2, 5 ER :. a, ; 5 = b; az = 9; Tos = 62. 27 2; 5=5 EIR² ا ورط ,4 ,Q 9 ai ii <b, a> 2 b, a, t5 ba az = 2a, b, + 5aabz =<a, b>

since non 2 Thus, we have <9, b) = <b, Q> in consider <a, at 29,. , +592.92 29,² +59² square of number is always negative, therefore 9,² zo and azo 329,2 zo and 59 2 20 29,2 +592 zo <a, a> zo Now <a, as co iff 29,9, + 5Q92 = 0 iff 29,1 + 592 Since Xo iff 21=0 for every number and if ng y zo and aty=0 y=0 599 = 0 1 (۱۱ن 2 every real 2100 9 29,2 29,2 20 and 5 92=0 2 and a, az co itt ift iff ije a, =0 and 92 -0 (0,92) - (0,0) 1 iff a = iff a=0 Thus aa, a> o 42), b=lbi, bi) c=1492) lig let a ca > are in IR² scalars and u, v lere

u <a, 129, bit 5925 consider < U G + vb, c) = <u(9, 19) +11b,,b),(, {2)) <149,, 492) +lub, g Vb2),((,, (2) = < (ua, tub, , 492 tuba), (G, (2) = 2 lua,+ uby) c, + 51uq, fubz) C₂ qua, c, + zreb, c, + 5u92C2 + subala =ulaa, 6, +5 9₂ (2) tulab, c, +5626 C> + L <b,c> Thus cu at ub, c) = uca,c> trueb, c> Thus all four axions of inner product are satisfied by U) (b) now, have G= (-1) 31 12,5) need to find (9,5) know that distance in toms of inner product is given by doolg y) = :: dia,b) = Evca, b> = √2(-1) (2) + 513) (5) 71 ;. dia, b) Jai we b= we we S<n, y> = -4 + 75 1

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