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Question

Angle between lines whose direction cosine satisfy l+m+n=0,l2+m2n2=0.

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Solution

l+m+n=0(i)l2+m2+n2=1(ii)l2+m2n2=0(iii)(ii)+(iii)2(l2+m2)=12n2=1n=±12l2+m2=12l+m=12l2+(12l)2=12l2+l2+122l=122l22l=0l=0,12m=12,0l2+m2=12l+m=12l2+(12l)2=12l2+12+l2+2l=122l2+2l=0l=0,1212,0
Possible values of (l,m,n) are:
(0,12,12),(12,0,12),(0,12,12),(12,0,12)

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