Dear Arkav,
2 l + 2 m – n = 0
2 ( l + m ) = n ------eqn.1
l m + m n + n l = 0
On substituting eqn.1 in above eqn we get;
l m + m { 2 ( l + m ) } + { 2 ( l + m ) } l = 0
l m + 2 l m + 2 m2 + 2 l2 + 2 l m= 0
2 m2 + 5 l m + 2 l2 = 0
on sloving we get,
2m ( m + 2l ) + l ( m + 2l ) = 0
(m+2l)(2m+l)=0
here we get two conditions indicating 2 lines;
1.)m = – 2l
on substituting m = – 2l in eqn.1
we get,
n = – 2l.
(l1,m1,n1) = l : - 2l : - 2l = 1 : – 2 : – 2 = (1, -2, -2)-----eqn.a
2.)l = – 2m
on substituting l = – 2m in eqn.1
we get n= – 2m.
(l2,m2,n2)= – 2m : m : – 2m = – 2: 1 : –2 = (-2,1,-2)------eqn. b
a and b are dr of 2 lines optained on solving the eqns.
For them to be perpendicular;
l1l2 + m1m2 + n1n2 = 0
Hence the condition is satisfied.
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