Find the direction cosines of the lines, connected by the relations: l+ m+ n = 0 and 2lm+ 2ln– mn =0.
Given relations are:
⇒ 2lm+ 2ln– mn =0 ……(1)
⇒ l+ m+ n =0
⇒ l = (–m– n) ……(2)
Substituting (2) in (1) we get,
⇒ 2(–m–n)m + 2(–m–n)n – mn =0
⇒ 2(–m2–mn) + 2(–mn–n2) – mn =0
⇒ –2m2 –2mn –2mn –2n2 –mn =0
⇒ –2m2–5mn–2n2 =0
⇒ 2m2+5mn+2n2=0
⇒ 2m2+4mn+mn+2n2=0
⇒ 2m(m+2n)+n(m+2n)=0
⇒ (2m+n)(m+2n)=0
⇒ 2m+n=0 or m+2n=0
⇒ 2m=–n or m=–2n
⇒ 
……(3)
Substituting the values of (3) in eq(2), we get
For 1st line:
⇒ 
⇒ 
⇒ 
The direction ratios for the first line is 
.
Let us assume l1,m1,n1 be the direction cosines of 1st line.
We know that for a line of direction ratios r1, r2, r3 and having direction cosines 
has the following property.
⇒ 
⇒ 
⇒ 
Using the above formulas we get,
⇒ 
⇒ 
⇒ 
⇒ 
⇒ 
⇒ 
⇒ 
⇒ 
⇒ 
⇒ 
⇒ 
⇒ 
The Direction cosines for the 1st line is 
For 2nd line:
⇒ l=–(–2n)–n
⇒ l=2n–n
⇒ l=n
The direction ratios for the second line is 
.
Let us assume l2,m2,n2 be the direction cosines of 1st line.
We know that for a line of direction ratios r1, r2, r3 and having direction cosines 
has the following property.
⇒ 
⇒ 
⇒ 
Using the above formulas we get,
⇒ 
⇒ 
⇒ 
⇒ 
⇒ 
⇒ 
⇒ 
⇒ 
⇒ 
⇒ 
⇒ 
⇒ 
⇒ 
⇒ 
⇒ 
The Direction Cosines for the 2nd line is 
.