Find the angle between the lines whose direction cosines are given by the equations:
2l–m+2n=0 and mn+nl+lm=0
Given relations are:
⇒ mn+nl+lm=0 ……(1)
⇒ 2l–m+2n=0
⇒ m=2l+2n ……(2)
Substituting (2) in (1) we get,
⇒ (2l+2n)n+nl+l(2l+2n)=0
⇒ 2ln+2n2+nl+2l2+2ln=0
⇒ 2n2+5ln+2l2=0
⇒ 2n2+4ln+ln+2l2=0
⇒ 2n(n+2l)+l(n+2l)=0
⇒ (2n+l)(n+2l)=0
⇒ 2n+l=0 or n+2l=0
⇒ l=–2n or 2l=–n ……(3)
Substituting the values of(3) in (2) we get,
For the 1st line:
⇒ m = 2(–2n)+2n
⇒ m=–4n+2n
⇒ m=–2n
The direction ratios for the 1st line is (–2n,–2n,n)
For the 2nd line:
⇒ m=–n+2n
⇒ m=n
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 .