1. Key Concepts and Formulas

• Direction Cosines: For any line in 3D space, its direction cosines $(l, m, n)$ are the cosines of the angles it makes with the positive x, y, and z axes, respectively. They satisfy the fundamental identity: $l^2 + m^2 + n^2 = 1$
• Angle Between Two Lines: If two lines have direction cosines $(l_1, m_1, n_1)$ and $(l_2, m_2, n_2)$, the cosine of the angle $\alpha$ between them is given by: $\cos \alpha = l_1 l_2 + m_1 m_2 + n_1 n_2$ We usually consider the acute angle, so $\cos \alpha \ge 0$.
• Trigonometric Identity: The expression $\sin^4 \alpha + \cos^4 \alpha$ can be simplified using the identity $\sin^2 \alpha + \cos^2 \alpha = 1$: $\sin^4 \alpha + \cos^4 \alpha = (\sin^2 \alpha + \cos^2 \alpha)^2 - 2 \sin^2 \alpha \cos^2 \alpha = 1 - 2 \sin^2 \alpha \cos^2 \alpha$

2. Step-by-Step Solution

Step 1: Write down the given equations and the fundamental identity. We are given two conditions that the direction cosines $(l, m, n)$ of the lines must satisfy:

1. $l + m - n = 0 \quad \implies \quad n = l+m \quad \text{(Equation A)}$
2. $l^2 + m^2 - n^2 = 0 \quad \implies \quad l^2 + m^2 = n^2 \quad \text{(Equation B)}$ And the fundamental identity for direction cosines:
3. $l^2 + m^2 + n^2 = 1 \quad \text{(Equation C)}$

Step 2: Determine the value of $n$. Our goal is to find the specific values of $l, m, n$ that satisfy these equations. We can start by combining Equation B and Equation C. Substitute $l^2 + m^2 = n^2$ (from Equation B) into Equation C: $n^2 + n^2 = 1$ $2n^2 = 1$ $n^2 = \frac{1}{2}$ $n = \pm \frac{1}{\sqrt{2}}$ This gives us two possible values for the z-component of the direction cosines.

Step 3: Find the relationship between $l$ and $m$. Now, let's use Equation A and Equation B to find a relationship between $l$ and $m$. Substitute $n = l+m$ (from Equation A) into $l^2 + m^2 = n^2$ (from Equation B): $l^2 + m^2 = (l+m)^2$ $l^2 + m^2 = l^2 + 2lm + m^2$ $0 = 2lm$ $lm = 0$ This crucial result implies that either $l=0$ or $m=0$ (or both, but that won't be the case here as it would lead to $n=0$ which contradicts $n^2=1/2$). This means one of the direction cosines (l or m) must be zero for each line.

Step 4: Find the direction cosines for the two lines. We will consider the positive value for $n$ for simplicity, $n = \frac{1}{\sqrt{2}}$. The angle between the lines will be the same regardless of the sign choice for $n$. Using $n = \frac{1}{\sqrt{2}}$ and $l+m=n$: $l+m = \frac{1}{\sqrt{2}}$ Now, apply the condition $lm=0$:

• Case 1: If $l=0$ Substitute $l=0$ into $l+m = \frac{1}{\sqrt{2}}$: $0 + m = \frac{1}{\sqrt{2}} \implies m = \frac{1}{\sqrt{2}}$ So, the direction cosines for the first line are $(l_1, m_1, n_1) = \left(0, \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right)$.

• Case 2: If $m=0$ Substitute $m=0$ into $l+m = \frac{1}{\sqrt{2}}$: $l + 0 = \frac{1}{\sqrt{2}} \implies l = \frac{1}{\sqrt{2}}$ So, the direction cosines for the second line are $(l_2, m_2, n_2) = \left(\frac{1}{\sqrt{2}}, 0, \frac{1}{\sqrt{2}}\right)$.

Step 5: Calculate the cosine of the angle $\alpha$ between the lines. Using the direction cosines found in Step 4: Line 1: $(l_1, m_1, n_1) = \left(0, \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right)$ Line 2: $(l_2, m_2, n_2) = \left(\frac{1}{\sqrt{2}}, 0, \frac{1}{\sqrt{2}}\right)$

Now, use the formula for $\cos \alpha$: $\cos \alpha = l_1 l_2 + m_1 m_2 + n_1 n_2$ $\cos \alpha = (0)\left(\frac{1}{\sqrt{2}}\right) + \left(\frac{1}{\sqrt{2}}\right)(0) + \left(\frac{1}{\sqrt{2}}\right)\left(\frac{1}{\sqrt{2}}\right)$ $\cos \alpha = 0 + 0 + \frac{1}{2}$ $\cos \alpha = \frac{1}{2}$ This implies $\alpha = 60^\circ$ or $\frac{\pi}{3}$ radians.

Step 6: Evaluate the expression $\sin^4 \alpha + \cos^4 \alpha$. We have $\cos \alpha = \frac{1}{2}$. First, find $\sin \alpha$ using $\sin^2 \alpha + \cos^2 \alpha = 1$: $\sin^2 \alpha = 1 - \cos^2 \alpha = 1 - \left(\frac{1}{2}\right)^2 = 1 - \frac{1}{4} = \frac{3}{4}$ $\sin \alpha = \frac{\sqrt{3}}{2} \quad (\text{since } \alpha \text{ is an angle of a triangle, } \sin \alpha > 0)$

Now, substitute $\sin^2 \alpha = \frac{3}{4}$ and $\cos^2 \alpha = \frac{1}{4}$ into the simplified expression for $\sin^4 \alpha + \cos^4 \alpha$: $\sin^4 \alpha + \cos^4 \alpha = 1 - 2 \sin^2 \alpha \cos^2 \alpha$ $\sin^4 \alpha + \cos^4 \alpha = 1 - 2 \left(\frac{3}{4}\right)\left(\frac{1}{4}\right)$ $\sin^4 \alpha + \cos^4 \alpha = 1 - 2 \left(\frac{3}{16}\right)$ $\sin^4 \alpha + \cos^4 \alpha = 1 - \frac{6}{16}$ $\sin^4 \alpha + \cos^4 \alpha = 1 - \frac{3}{8}$ $\sin^4 \alpha + \cos^4 \alpha = \frac{8}{8} - \frac{3}{8}$ $\sin^4 \alpha + \cos^4 \alpha = \frac{5}{8}$

3. Common Mistakes & Tips

• Don't forget the fundamental identity: The relation $l^2+m^2+n^2=1$ is crucial for solving problems involving direction cosines.
• Consider all possibilities: When $lm=0$, remember it means either $l=0$ or $m=0$, leading to two distinct sets of direction cosines.
• Simplify trigonometric expressions: Using identities like $\sin^4 \alpha + \cos^4 \alpha = 1 - 2 \sin^2 \alpha \cos^2 \alpha$ can significantly simplify calculations and reduce the chance of errors.

4. Summary

The problem required us to find the angle between two lines whose direction cosines satisfy two given equations. We combined these equations with the fundamental property of direction cosines ($l^2+m^2+n^2=1$) to determine the specific direction cosines for each line. This led to two sets of direction cosines: $\left(0, \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right)$ and $\left(\frac{1}{\sqrt{2}}, 0, \frac{1}{\sqrt{2}}\right)$. Using the formula for the angle between two lines, we found $\cos \alpha = \frac{1}{2}$. Finally, we evaluated the expression $\sin^4 \alpha + \cos^4 \alpha$ using trigonometric identities, which yielded $\frac{5}{8}$.

5. Final Answer

The final answer is $\boxed{\frac{5}{8}}$, which corresponds to option (D).