1. Key Concepts and Formulas

• Skew Lines: Two lines in 3D space that are neither parallel nor intersecting are called skew lines. The shortest distance between them is the length of the unique line segment that is perpendicular to both lines.
• Vector Form of Lines: A line passing through a point with position vector $\vec{a}$ and parallel to a vector $\vec{v}$ can be represented as $\vec{r} = \vec{a} + \lambda \vec{v}$.
• Shortest Distance Formula: For two skew lines $L_1: \vec{r} = \vec{a_1} + \lambda \vec{v_1}$ and $L_2: \vec{r} = \vec{a_2} + \mu \vec{v_2}$, the shortest distance ($d$) between them is given by: $d = \left| \frac{(\vec{a_2} - \vec{a_1}) \cdot (\vec{v_1} \times \vec{v_2})}{|\vec{v_1} \times \vec{v_2}|} \right|$ This formula is derived from the scalar triple product, representing the volume of a parallelepiped formed by the three vectors, divided by the area of its base.

2. Step-by-Step Solution

Step 1: Extract Points and Direction Vectors from the Given Lines We begin by converting the given Cartesian equations of the lines into their vector components ($\vec{a}$ and $\vec{v}$).

For Line 1 ($L_1$): $\frac{x}{2} = \frac{y}{2} = \frac{z}{1}$ This can be written as $\frac{x - 0}{2} = \frac{y - 0}{2} = \frac{z - 0}{1}$.

• A point on $L_1$ is $\vec{a_1} = (0, 0, 0)$.
• The direction vector of $L_1$ is $\vec{v_1} = (2, 2, 1)$.

For Line 2 ($L_2$): $\frac{x + 2}{-1} = \frac{y - 4}{8} = \frac{z - 5}{4}$ This can be written as $\frac{x - (-2)}{-1} = \frac{y - 4}{8} = \frac{z - 5}{4}$.

• A point on $L_2$ is $\vec{a_2} = (-2, 4, 5)$.
• The direction vector of $L_2$ is $\vec{v_2} = (-1, 8, 4)$.

Step 2: Check for Parallelism Before applying the skew lines formula, it's good practice to check if the lines are parallel. If $\vec{v_1}$ is a scalar multiple of $\vec{v_2}$, the lines are parallel. $\vec{v_1} = (2, 2, 1)$ and $\vec{v_2} = (-1, 8, 4)$. Clearly, $(2, 2, 1) \neq k(-1, 8, 4)$ for any scalar $k$. Thus, the lines are not parallel, confirming they are skew lines (or intersecting).

Step 3: Calculate the Vector Connecting the Points ($\vec{a_2} - \vec{a_1}$) This vector represents the displacement from a point on $L_1$ to a point on $L_2$. $\vec{a_2} - \vec{a_1} = (-2 - 0)\hat{i} + (4 - 0)\hat{j} + (5 - 0)\hat{k}$ $\vec{a_2} - \vec{a_1} = -2\hat{i} + 4\hat{j} + 5\hat{k}$

Step 4: Calculate the Cross Product of the Direction Vectors ($\vec{v_1} \times \vec{v_2}$) The cross product of the direction vectors yields a vector that is perpendicular to both lines. This vector's direction is the direction of the shortest distance segment. $\vec{v_1} \times \vec{v_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 2 & 1 \\ -1 & 8 & 4 \end{vmatrix}$ $= \hat{i}((2)(4) - (1)(8)) - \hat{j}((2)(4) - (1)(-1)) + \hat{k}((2)(8) - (2)(-1))$ $= \hat{i}(8 - 8) - \hat{j}(8 + 1) + \hat{k}(16 + 2)$ $= 0\hat{i} - 9\hat{j} + 18\hat{k}$ So, $\vec{v_1} \times \vec{v_2} = (0, -9, 18)$.

Step 5: Calculate the Magnitude of the Cross Product ($|\vec{v_1} \times \vec{v_2}|$ ) This magnitude forms the denominator of our shortest distance formula. It's crucial for calculating the height of the parallelepiped. $|\vec{v_1} \times \vec{v_2}| = \sqrt{0^2 + (-9)^2 + 18^2}$ $= \sqrt{0 + 81 + 324}$ $= \sqrt{405}$ To simplify the radical, we find perfect square factors of 405. Since $405 = 81 \times 5$: $|\vec{v_1} \times \vec{v_2}| = \sqrt{81 \times 5} = 9\sqrt{5}$

Step 6: Calculate the Scalar Triple Product (Numerator) The numerator is the absolute value of the scalar triple product $(\vec{a_2} - \vec{a_1}) \cdot (\vec{v_1} \times \vec{v_2})$. If this value is zero, the lines intersect, and the shortest distance is 0. $(\vec{a_2} - \vec{a_1}) \cdot (\vec{v_1} \times \vec{v_2}) = (-2, 4, 5) \cdot (0, -9, 18)$ $= (-2)(0) + (4)(-9) + (5)(18)$ $= 0 - 36 + 90$ $= 54$ Since the scalar triple product is not zero, the lines are indeed skew.

Step 7: Calculate the Shortest Distance ($d$) Substitute the calculated values into the shortest distance formula: $d = \left| \frac{(\vec{a_2} - \vec{a_1}) \cdot (\vec{v_1} \times \vec{v_2})}{|\vec{v_1} \times \vec{v_2}|} \right|$ $d = \left| \frac{54}{9\sqrt{5}} \right|$ $d = \frac{6}{\sqrt{5}}$ To rationalize the denominator, multiply the numerator and denominator by $\sqrt{5}$: $d = \frac{6\sqrt{5}}{5}$

Step 8: Approximate the Value and Check Options To determine which interval the distance falls into, we approximate its value: Using $\sqrt{5} \approx 2.236$: $d \approx \frac{6 \times 2.236}{5}$ $d \approx \frac{13.416}{5}$ $d \approx 2.6832$

Now, compare this value with the given options: (A) [0, 1) (B) [1, 2) (C) (2, 3] (D) (3, 4]

The calculated shortest distance $d \approx 2.6832$ falls within the interval (2, 3].

3. Common Mistakes & Tips

• Sign Errors: Pay close attention to negative signs when extracting coordinates (e.g., $x+2$ means $x - (-2)$) and during vector operations (especially in the cross product).
• Order of Operations: Ensure the cross product is calculated first, then the dot product, and finally the magnitude.
• Absolute Value: Remember that distance must be non-negative, so always take the absolute value of the numerator.
• Check for Parallelism/Intersection: A quick check for parallel direction vectors or a zero scalar triple product can save time or indicate a different formula is needed.

4. Summary

This problem required finding the shortest distance between two skew lines in 3D space. We first extracted a point and direction vector for each line. Then, we calculated the vector connecting the points, the cross product of the direction vectors, and their respective magnitudes. Finally, we applied the standard formula for the shortest distance between skew lines, which involves the scalar triple product. The calculated distance is $\frac{6\sqrt{5}}{5}$, which approximates to $2.6832$. This value lies within the interval (2, 3].

The final answer is $\boxed{\frac{6\sqrt{5}}{5}}$, which corresponds to option (C).