The direction cosines of a line satisfy the relations λ(l+m)=n and mn+nl+lm=0. The value of λ, for which the two lines are perpendicular to each other, is

Eliminating n, we get λ(l+m)2+lm=0⇒λl2m2+(2λ+1)lm+λ=0∴l1l2m1m2=1 [product of roots l1m1&l2m2]Further eliminating m, we get λl2−ln−n2=0⇒l1l2n1n2=1λ [product of roots l1n1&l2n2]give the lines with DC's (l1,m1,n1) and (l2,m2,n2) are perpendicular we have l1l2+m1m2+n1n2=0⇒1+1−λ=0⇒λ=2