Find the angle between the lines whose direction cosines are given by the equations:

2l+2m–n=0 and mn+ln+lm=0

Given relations are:

⇒ mn+ln+lm=0 ……(1)

⇒ 2l+2m–n=0

⇒ n=2l+2m ……(2)

Substituting (2) in (1) we get,

⇒ m(2l+2m)+l(2l+2m)+lm=0

⇒ 2lm+2m²+2l²+2lm+lm=0

⇒ 2m²+5lm+2l²=0

⇒ 2m²+4lm+lm+2l²=0

⇒ 2m(m+2l)+l(m+2l)=0

⇒ (2m+l)(m+2l)=0

⇒ 2m+l=0 or m+2l=0

⇒ 2m=–l or 2l=–m ……(3)

Substituting the values of (3) in (2), we get

For the 1^st line:

⇒ n=2l–l

⇒ n=l

The Direction Ratios for the first line is

For the 2^nd line:

⇒ n=–m+2m

⇒ n=m

The Direction Ratios for the second line is

We know that the angle between the lines with direction ratios proportional to (a₁,b₁,c₁) and (a₂,b₂,c₂) is given by:

⇒

Using the above formula we calculate the angle between the lines.

Let be the angle between the two lines given in the problem.

⇒

⇒

⇒

⇒

∴ the angle between two lines is .