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
Substituting the values of (3) in (2) we get,
For the 1st line:
The Direction Ratios for the 1st line is (-(3 + 2\(\sqrt{2}\))n, \(\sqrt{2}\)n, n).
For the 2nd line:
The Direction Ratios for the 2nd line is ((2\(\sqrt{2}\) - 3)n, -\(\sqrt{2}\)n, n).
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 \(90^\circ.\)