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 1st line:
⇒ n = 2l – l
⇒ n = l
The Direction Ratios for the first line is \((1, -\frac{1}{2}, 1)\)
For the 2nd line:
⇒ n = –m + 2m
⇒ n = m
The Direction Ratios for the second line is \(\left(\frac{-m}{2}, m, m \right)\)
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 α be the angle between the two lines given in the problem.
∴ the angle between two lines is \(\frac{\pi}{2}\) or \(90^\circ.\)