Q.

Find the angle between the lines whose direction cosines satisfy the equations
 i) l+m+n=0,l2+m2n2=0
 ii) 3l+m+5n=0 and 6mn2nl+5lm=0

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Detailed Solution

i) Given equations are l + m + n = 0 ..... (1)
l2+m2n2=0 ...(2)
From eq (1), l = – m – n
Substituting in equation (2)
(mn)2+m2n2=0
m2+n2+2mn+m2n2=02m2+2mn=02m(m+n)=02m=0 or m+n=0m=0 or m=n
Case (i): put m = 0 in eq (1)
l=nl:m:n=n:0:n=1:0:1
dr’s of first line  are (– 1, 0, 1)
dc’s  of first line are (l1, m1, n1) =
1(1)2+02+12,0(1)2+02+12,1(1)2+02+12)=12,0,12
Case (ii): put m = –n in (1)
l=0 l:m:n=0:n:n=0:1:1
dr’s  of second line are= (a, b, c) = (0, –1, 1)
dc’s  of second line are (l2, m2, n2) =
002+(1)2+12,102+(1)2+12, 102+(-1)2+12 )=0,12,12
Letθ be angle between the lines then
cosθ=l1l2+m1m2+n1n2=12(0)+012+1212=12θ=60
ii) Given equations are
3l + m + 5n = 0 ..... (1)
6mn - 2nl + 5lm = 0 ...... (2)
From eq (1), m = –3l – 5n
Substituting in equation (2)
6(3l5n)n2nl+5l(3l5n)=018ln30n22nl15l225nl=030n245nl15l2=0152n2+3nl+l2=02n2+3nl+l2=02n2+2nl+nl+l2=0
2n(n+l)+l(n+l)=0(n+l)(2n+l)=0n+l=0 (or) 2n+l=0
Case (i): l+n=0l=n
From eq (1), 3n+m+5n=0m=2n
l:m:n=n:2n:n=1:2:1
dr’s of first line are= (a, b, c) = (–1, –2, 1)
dc’s of second line are =
aa2+b2+c2,ba2+b2+c2,ca2+b2+c2
=1(1)2+(2)2+12    ,2(1)2+(2)2+12,1(1)2+(2)2+12 
=16,26,16=l1,m1,n1
Case (ii): If 2n+l=0l=2n
From eq (i), 6n+m+5n=0m=n
l:m:n=2n:n:n=2:1:1
dr’s = (a, b, c) = (–2, 1, 1)
dc’s = (l2, m2, n2) =
2(2)2+12+12,1(2)2+12+12,1(2)2+12+12=26,16,16
 Let θ is the angle between the lines 
cosθ=l1l2+m1m2+n1n2=1626+2616+1616=2626+16=16 θ=cos116

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