since
,
,
are the p
th,Q
th,R
th terms of G.P
=ar
p-1
=arQ-1 =arR-1 [R-1 in power] considering the two given vectors
as A &B A.B=(Q-R)log
3+(R-P)log
3+(P-Q)log
3 =3[(Q-R)l og+(R-P)log+(P-Q)log] log=loga+logr(P-1); log=loga+(Q-1)logr ; log =loga+(R-1)logr A.B=3[(Q-R)(loga+logr{P-1})+(R-P)(loga+(Q-1)logr+(P-Q)(loga+(R-1)logr)]
=3[loga(Q-R+R-P+P-Q)+logr{(Q-R)(p-1)+(R-P)(Q-1)+(P-Q)(R-1)}]
=3[0 +logr(0)]=0
A.B=0 IMPLIES the angle between them is 90 degrees
Moderation Team
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