Find the angle between the lines whose direction cosines are given by…

The correct answer is Option (2) → $\frac{\pi}{3}$ ##

Given, $l + m + n = 0 \dots (i)$

and $l^2 + m^2 - n^2 = 0 \dots (ii)$

On eliminating $n$ from both the equations, we get

$l^2 + m^2 - (-l - m)^2 = 0$

$\Rightarrow l^2 + m^2 - l^2 - m^2 - 2lm = 0 \Rightarrow 2lm = 0 \quad [∵(a - b)^2 = a^2 + b^2 - 2ab]$

$\Rightarrow lm = 0 \Rightarrow (-m - n)m = 0 \quad [∵l = -m - n]$

$\Rightarrow (m + n)m = 0$

$\Rightarrow m = -n \text{ and } m = 0$

$\Rightarrow l = 0, l = -n \quad [\text{from Eq. (i)}]$

Thus, Dr's (Direction Ratios) of two lines are proportional to $0, -n, n$ and $-n, 0, n$ i.e., $0, -1, 1$ and $-1, 0, 1$.

So, the vector parallel to these given lines are $\mathbf{a} = -\hat{j} + \hat{k}$ and $\mathbf{b} = -\hat{i} + \hat{k}$.

Now, $\cos \theta = \frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{a}| |\mathbf{b}|} = \frac{1}{\sqrt{2} \cdot \sqrt{2}}$

$\Rightarrow \cos \theta = \frac{1}{2} = \cos \frac{\pi}{3} \quad \left[ ∵\cos \frac{\pi}{3} = \frac{1}{2} \right]$

$∴\theta = \frac{\pi}{3}$