Question

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

l + m + n = 0 and l² + m² – n² = 0

Answer 1

Given relations are:

⇒ l² + m² – n² = 0 ……(1)

⇒ l + m + n = 0

⇒ l = – m – n ……(2)

Substituting (2) in (1) we get,

⇒ (– m – n)² + m² – n² = 0

⇒ m² + n² + 2mn + m² – n² = 0

⇒ 2m² + 2mn = 0

⇒ 2m(m + n) = 0

⇒ 2m = 0 or m + n = 0

⇒ m = 0 or m = – n ……(3)

Substituting value of m from(3) in (2)

For the 1^st line:

⇒ l = – 0 – n

⇒ l = – n

The Direction Ratios for the first line is (– n, 0, n)

For the 2^nd line:

⇒ l = – (– n) – n

⇒ l = n – n

⇒ l = 0

The Direction Ratios for the second line is (0, –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 given two lines is \(\frac{\pi}{3}\) or \(60^\circ.\)