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 1st line:
⇒
⇒
The Direction Ratios for the 1st line is .
For the 2nd line:
⇒
⇒
The Direction Ratios for the 2nd 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 .