We have given that L1&L2 are equation of line whose direction cosines are given as (l1,m1,n1)&(l2,m2,n2) respectively obtained from solution of two given equation as follows,
al+bm+cn=0(1)fmn+gnl+hlm=0(2) where abc=0 ,also given that L1⊥L2⟹l1l2+m1m2+n1n2=0(⋆) .
Let's eliminate one of the variable from (1) ,say l ,thus l=−abm+cn and substitute to (2) we get,
fmn+gn(−abm+cn)+hm(−abm+cn)=0⟹hbm2+(hc+gb−af)mn+gcn2=0⟹hby2+(hc+gb−af)y+gc=0(♣)
where
y=nm
Clearly, equation (♣) is quadratic in y ,hence by relation between roots and coefficients(Vieta's Theorem) we get,
y1+y2=−hb(hc+gb−af)(3)y1y2=hbgc(4) Hence,from equation (4) ,
n1m1n2m2=hbgc⟹bgm1m2=chn1n2(5) In the same manner if we eliminate other two variables n,m (say n ) and do the same calculation as above we get,
bgm1m2=afl1l2(6)Now from equation 5 and 6 we finally get
bgm1m2=afl1l2=chn1n2(7)Thus from equation 7 if we consider bgm1m2=afl1l2=chn1n2=p ,clearly p=0 ,hence,
l1l2+m1m2+n1n2=p(bg+af+ch) but from equation (⋆) we get,
l1l2+m1m2+n1n2=p(bg+af+ch)=0⟹p(bg+af+ch)=0⟹(bg+af+ch)=0(∵p=0) Therefore we are done.
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