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

l+2m+3n=0 and 3lm–4ln+mn=0

Given relations are:

⇒ 3lm–4ln+mn=0 ……(1)

⇒ l+2m+3n=0

⇒ l=–2m–3n ……(2)

Substituting (2) in (1) we get,

⇒ 3(–2m–3n)m –4(–2m–3n)n +mn =0

⇒ 3(–2m²–3mn) –4(–2mn–3n²) +mn=0

⇒ –6m²–9mn+8mn+12n²+mn=0

⇒ 12n²–6m²=0

⇒ m²–2n²=0

⇒

⇒

⇒ ……(3)

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

For the 1^st line:

⇒

⇒

The Direction Ratios for the 1^st line is .

For the 2^nd line:

⇒

⇒

The Direction Ratios for the 2^nd 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 .