1. Key Concepts and Formulas

• Direction Cosines: The direction cosines $(l, m, n)$ of a line are the cosines of the angles that the line makes with the positive $x, y,$ and $z$ axes. They describe the orientation of the line in 3D space.
• Fundamental Identity of Direction Cosines: For any set of direction cosines $(l, m, n)$, the sum of their squares is always equal to 1: $l^2 + m^2 + n^2 = 1$ This identity is crucial for finding specific values of $l, m, n$.
• Angle Between Two Lines: If two lines have direction cosines $(l_1, m_1, n_1)$ and $(l_2, m_2, n_2)$, the angle $\theta$ between them is given by the dot product formula: $\cos \theta = |l_1 l_2 + m_1 m_2 + n_1 n_2|$ The absolute value ensures that we find the acute angle between the lines ($0 \le \theta \le \pi/2$).

2. Step-by-Step Solution

Step 1: Set up the system of equations for the direction cosines. We are given two conditions that the direction cosines $(l, m, n)$ of the lines must satisfy:

1. $l+m+n=0$
2. $l^2 = m^2 + n^2$

We also know the fundamental identity for direction cosines: 3. $l^2 + m^2 + n^2 = 1$

Our goal is to find two distinct sets of $(l, m, n)$ that satisfy all three equations. Each set will represent one of the lines.

Step 2: Simplify the given conditions to find relationships between $l, m,$ and $n$. From equation (1), we can express $l$ in terms of $m$ and $n$: $l = -(m+n) \quad (*)$ Now, substitute this expression for $l$ into equation (2): $(-(m+n))^2 = m^2 + n^2$ Squaring the left side gives: $(m+n)^2 = m^2 + n^2$ Expand the left side: $m^2 + 2mn + n^2 = m^2 + n^2$ Subtract $m^2 + n^2$ from both sides to simplify: $2mn = 0$ This significant result implies that for any line satisfying the given conditions, either $m=0$ or $n=0$ (or both). This means the lines must lie in specific coordinate planes (the $xz$-plane or the $xy$-plane, respectively).

Step 3: Determine the possible values for $l$. We can use the result from equation (2), $l^2 = m^2+n^2$, and substitute it into the fundamental identity (3), $l^2+m^2+n^2=1$. $l^2 + (l^2) = 1$ $2l^2 = 1$ $l^2 = \frac{1}{2}$ Taking the square root, we find the possible values for $l$: $l = \pm \frac{1}{\sqrt{2}}$

Step 4: Find the two distinct sets of direction cosines. We combine the results from Step 2 ($2mn=0$) and Step 3 ($l = \pm \frac{1}{\sqrt{2}}$) with the relation $l=-(m+n)$.

Case 1: $m=0$ If $m=0$, the condition $2mn=0$ is satisfied. From $l = -(m+n)$, substituting $m=0$ gives: $l = -(0+n) \implies l = -n$ Since we know $l = \pm \frac{1}{\sqrt{2}}$, we have two possibilities for $n$:

• If $l = \frac{1}{\sqrt{2}}$, then $n = -\frac{1}{\sqrt{2}}$. This gives the direction cosines for the first line: $(l_1, m_1, n_1) = \left(\frac{1}{\sqrt{2}}, 0, -\frac{1}{\sqrt{2}}\right)$.
• If $l = -\frac{1}{\sqrt{2}}$, then $n = \frac{1}{\sqrt{2}}$. This gives $\left(-\frac{1}{\sqrt{2}}, 0, \frac{1}{\sqrt{2}}\right)$. This set of direction cosines represents the same line as the first set, just in the opposite direction. We only need one set for each line.

Case 2: $n=0$ If $n=0$, the condition $2mn=0$ is satisfied. From $l = -(m+n)$, substituting $n=0$ gives: $l = -(m+0) \implies l = -m$ Since we know $l = \pm \frac{1}{\sqrt{2}}$, we have two possibilities for $m$:

• If $l = \frac{1}{\sqrt{2}}$, then $m = -\frac{1}{\sqrt{2}}$. This gives the direction cosines for the second line: $(l_2, m_2, n_2) = \left(\frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}}, 0\right)$.
• If $l = -\frac{1}{\sqrt{2}}$, then $m = \frac{1}{\sqrt{2}}$. This gives $\left(-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}, 0\right)$, which represents the same line as the second set, just in the opposite direction.

So, we have found the direction cosines for two distinct lines: Line 1: $(l_1, m_1, n_1) = \left(\frac{1}{\sqrt{2}}, 0, -\frac{1}{\sqrt{2}}\right)$ Line 2: $(l_2, m_2, n_2) = \left(\frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}}, 0\right)$

Step 5: Calculate the angle between the lines. Now we use the formula for the angle $\theta$ between two lines using their direction cosines: $\cos \theta = |l_1 l_2 + m_1 m_2 + n_1 n_2|$ Substitute the values of the direction cosines we found in Step 4: $\cos \theta = \left| \left(\frac{1}{\sqrt{2}}\right) \left(\frac{1}{\sqrt{2}}\right) + (0)\left(-\frac{1}{\sqrt{2}}\right) + \left(-\frac{1}{\sqrt{2}}\right)(0) \right|$ $\cos \theta = \left| \frac{1}{2} + 0 + 0 \right|$ $\cos \theta = \frac{1}{2}$ To find the angle $\theta$, we take the inverse cosine: $\theta = \arccos\left(\frac{1}{2}\right)$ $\theta = \frac{\pi}{3}$

3. Common Mistakes & Tips

• Forgetting $l^2+m^2+n^2=1$: This is the most common pitfall. Without this fundamental identity, you can only find direction ratios, not specific direction cosines.
• Incorrect Algebraic Manipulation: Be careful when squaring expressions or simplifying equations. A small error in $2mn=0$ could lead to vastly different sets of direction cosines.
• Not Identifying Distinct Lines: Remember that $(l, m, n)$ and $(-l, -m, -n)$ represent the same line, just different directions. Ensure you select two truly distinct lines for calculating the angle.
• Acute Angle: The absolute value in the formula for $\cos \theta$ ensures that you find the acute angle, which is typically what is asked for unless specified otherwise.

4. Summary

This problem required us to first determine the direction cosines of the lines by solving a system of three equations: two given conditions and the fundamental identity ($l^2+m^2+n^2=1$). The key simplification came from deriving $2mn=0$, which implies that either $m=0$ or $n=0$. This led to two distinct sets of direction cosines. Finally, the angle between these two lines was calculated using the dot product formula for direction cosines, yielding $\cos \theta = 1/2$.

The final answer is $\boxed{\text{\(\frac{\pi}{3}\)}}$, which corresponds to option (C).