1. Key Concepts and Formulas

• Skew Lines: Two lines in 3D space are called skew if they are neither parallel nor intersecting. The shortest distance between them is the length of the unique line segment that is perpendicular to both lines.
• Vector Form of a Line: A line passing through a point with position vector $\vec{a}$ and parallel to a direction vector $\vec{d}$ can be represented as $\vec{r} = \vec{a} + \lambda \vec{d}$.
• Shortest Distance Formula: For two skew lines $L_1: \vec{r}_1 = \vec{a}_1 + \lambda \vec{p}$ and $L_2: \vec{r}_2 = \vec{a}_2 + \mu \vec{q}$, the shortest distance (SD) between them is given by: $SD = \left| \frac{(\vec{a}_2 - \vec{a}_1) \cdot (\vec{p} \times \vec{q})}{|\vec{p} \times \vec{q}|} \right|$ This formula calculates the projection of the vector connecting a point on $L_1$ to a point on $L_2$ onto the common normal vector ($\vec{p} \times \vec{q}$) to both lines.

2. Step-by-Step Solution

Step 1: Extract Point and Direction Vectors for each Line First, we need to convert the given Cartesian equations into the standard vector form $\vec{r} = \vec{a} + \lambda \vec{d}$. The standard Cartesian form is ${{x - x_1} \over a} = {{y - y_1} \over b} = {{z - z_1} \over c}$, where $(x_1, y_1, z_1)$ is a point on the line and $(a, b, c)$ are the direction ratios.

• For Line 1 ($L_1$): The equation is ${{x - 2} \over 3} = {{y + 1} \over 2} = {{z - 6} \over 2}$ By comparing with the standard form, we identify:

  • A point on $L_1$: $(x_1, y_1, z_1) = (2, -1, 6)$. So, $\vec{a}_1 = 2\hat{i} - \hat{j} + 6\hat{k}$.
  • The direction vector parallel to $L_1$: $(a, b, c) = (3, 2, 2)$. So, $\vec{p} = 3\hat{i} + 2\hat{j} + 2\hat{k}$.

• For Line 2 ($L_2$): The equation is ${{x - 6} \over 3} = {{1 - y} \over 2} = {{z + 8} \over 0}$ We must rewrite the $y$-term to match the standard form $(y - y_1)$. ${{1 - y} \over 2} = {{-(y - 1)} \over 2} = {{y - 1} \over -2}$ So, the equation for $L_2$ becomes: ${{x - 6} \over 3} = {{y - 1} \over -2} = {{z - (-8)} \over 0}$ Now, comparing with the standard form, we identify:

  • A point on $L_2$: $(x_1, y_1, z_1) = (6, 1, -8)$. So, $\vec{a}_2 = 6\hat{i} + \hat{j} - 8\hat{k}$.
  • The direction vector parallel to $L_2$: $(a, b, c) = (3, -2, 0)$. So, $\vec{q} = 3\hat{i} - 2\hat{j} + 0\hat{k}$.

Step 2: Calculate the Vector Connecting Points on the Lines: $\vec{a}_2 - \vec{a}_1$ This vector connects a specific point on the first line to a specific point on the second line. $\vec{a}_2 - \vec{a}_1 = (6\hat{i} + \hat{j} - 8\hat{k}) - (2\hat{i} - \hat{j} + 6\hat{k})$ $= (6 - 2)\hat{i} + (1 - (-1))\hat{j} + (-8 - 6)\hat{k}$ $= 4\hat{i} + 2\hat{j} - 14\hat{k}$

Step 3: Calculate the Cross Product of Direction Vectors: $\vec{p} \times \vec{q}$ The cross product $\vec{p} \times \vec{q}$ yields a vector that is perpendicular to both $\vec{p}$ and $\vec{q}$. This vector gives the direction of the shortest distance between the lines. $\vec{p} \times \vec{q} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 2 & 2 \\ 3 & -2 & 0 \end{vmatrix}$ $= \hat{i}((2)(0) - (2)(-2)) - \hat{j}((3)(0) - (2)(3)) + \hat{k}((3)(-2) - (2)(3))$ $= \hat{i}(0 - (-4)) - \hat{j}(0 - 6) + \hat{k}(-6 - 6)$ $= 4\hat{i} + 6\hat{j} - 12\hat{k}$

Step 4: Calculate the Magnitude of the Cross Product: $|\vec{p} \times \vec{q}|$ This is the magnitude of the common normal vector, which will be the denominator in our shortest distance formula. $|\vec{p} \times \vec{q}| = |4\hat{i} + 6\hat{j} - 12\hat{k}|$ $= \sqrt{4^2 + 6^2 + (-12)^2}$ $= \sqrt{16 + 36 + 144}$ $= \sqrt{196}$ $= 14$

Step 5: Calculate the Scalar Triple Product (Numerator): $(\vec{a}_2 - \vec{a}_1) \cdot (\vec{p} \times \vec{q})$ This is the dot product of the vector connecting the two points and the common normal vector. $(\vec{a}_2 - \vec{a}_1) \cdot (\vec{p} \times \vec{q}) = (4\hat{i} + 2\hat{j} - 14\hat{k}) \cdot (4\hat{i} + 6\hat{j} - 12\hat{k})$ $= (4)(4) + (2)(6) + (-14)(-12)$ $= 16 + 12 + 168$ $= 196$

Step 6: Calculate the Shortest Distance Substitute the calculated values into the shortest distance formula: $SD = \left| \frac{(\vec{a}_2 - \vec{a}_1) \cdot (\vec{p} \times \vec{q})}{|\vec{p} \times \vec{q}|} \right|$ $SD = \left| \frac{196}{14} \right|$ $SD = |14|$ $SD = 14$

3. Common Mistakes & Tips

• Standard Form Conversion: Always ensure the line equations are in the standard form ${{x - x_1} \over a} = {{y - y_1} \over b} = {{z - z_1} \over c}$. Pay close attention to signs, especially terms like $1-y$, which must be rewritten as $-(y-1)$, leading to a sign change in the denominator.
• Calculation Errors: Vector cross products, dot products, and magnitude calculations are prone to arithmetic errors. Double-check each step carefully.
• Parallel Lines Check: Briefly check if the direction vectors are parallel (proportional). If they are, a different formula for parallel lines would be used. In this case, $(3,2,2)$ and $(3,-2,0)$ are clearly not parallel, confirming they are skew lines.

4. Summary

To find the shortest distance between two skew lines, we first extract the position vector of a point on each line ($\vec{a}_1, \vec{a}_2$) and their respective direction vectors ($\vec{p}, \vec{q}$). We then calculate the vector connecting the points ($\vec{a}_2 - \vec{a}_1$) and the common normal vector ($\vec{p} \times \vec{q}$). Finally, we compute the magnitude of the common normal vector and use these values in the shortest distance formula $SD = \left| \frac{(\vec{a}_2 - \vec{a}_1) \cdot (\vec{p} \times \vec{q})}{|\vec{p} \times \vec{q}|} \right|$. Following these steps systematically, the shortest distance between the given lines is found to be 14 units.

5. Final Answer

The final answer is $\boxed{14}$.