Find the angle between the lines whose direction cosines are given by the equations l + m + n = 0, l² + m² – n² = 0.

We are given two equations that represents the direction cosines of two lines,

l + m + n = 0 …(i)

l² + m² – n² = 0 …(ii)

We need to find the angle between the lines whose direction cosines are given by the given equations.

Let us find the value of l, m and n.

From equation (i),

l + m + n = 0

⇒ n = -l – m

⇒ n = -(l + m) …(iii)

Substituting the value of n from (i) in (ii),

l² + m² – n² = 0

⇒ l² + m² – (-(l + m))² = 0

⇒ l² + m² – (l + m)² = 0

⇒ l² + m² – (l² + m² + 2lm) = 0

⇒ l² + m² – l² – m² – 2lm = 0

⇒ l² – l² + m² – m² – 2lm = 0

⇒ -2lm = 0

⇒ lm = 0

⇒ l = 0 or m = 0

First, put l = 0 in equation (i),

Equation (i) ⇒ 0 + m + n = 0

⇒ m + n = 0

⇒ m = -n

If m = λ, then

n = -m

⇒ n = -λ

∴, direction ratios (l, m, n) = (0, λ, -λ)

Now, put m = 0 in equation (i),

Equation (i) ⇒ l + 0 + n = 0

⇒ l + n = 0

⇒ l = -n

If n = λ, then

l = -n

⇒ l = -λ

∴, direction ratios (l, m, n) = (-λ, 0, λ)

By theorem, that says

Angles between two lines whose direction ratio are d₁ and d₂ is θ given by,

Substituting values of d₁ and d₂ in θ, we get

Solving numerator,

Solving denominator,

Substituting the values in θ, we get

[∵, ]

Thus, the required angle between the given lines is π/3.