Key Concepts and Formulas

• Vector Equation of a Line: A line passing through a point with position vector $\vec{a}$ and parallel to a direction vector $\vec{b}$ can be represented as $\vec{r} = \vec{a} + \lambda \vec{b}$, where $\lambda$ is a scalar parameter.
• Shortest Distance Between Skew Lines: Skew lines are non-parallel and non-intersecting lines in 3D space. The shortest distance, $d$, between two skew lines $L_1: \vec{r} = \vec{a_1} + \lambda \vec{b_1}$ and $L_2: \vec{r} = \vec{a_2} + \mu \vec{b_2}$ is given by the formula: $d = \left| {{(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2})} \over {|\vec{b_1} \times \vec{b_2}|}} \right|$ This formula represents the projection of the vector connecting points on the two lines ($\vec{a_2} - \vec{a_1}$) onto the common perpendicular direction ($\vec{b_1} \times \vec{b_2}$).
• Cartesian to Vector Form: A line given in Cartesian form ${{x - x_0} \over l} = {{y - y_0} \over m} = {{z - z_0} \over n}$ has a point $(x_0, y_0, z_0)$ and a direction vector $(l, m, n)$.

Step-by-Step Solution

1. Identify Points and Direction Vectors from Cartesian Equations

The given lines are: $L_1: {{x - 1} \over 2} = {{y + 8} \over -7} = {{z - 4} \over 5}$ $L_2: {{x - 1} \over 2} = {{y - 2} \over 1} = {{z - 6} \over { - 3}}$

From $L_1$:

• A point on $L_1$ is $A_1(1, -8, 4)$. So, $\vec{a_1} = \mathbf{i} - 8\mathbf{j} + 4\mathbf{k}$.
  • Note: The term $y+8$ in the denominator means $y-(-8)$, so the y-coordinate of the point is $-8$.
• The direction vector of $L_1$ is $\vec{b_1} = 2\mathbf{i} - 7\mathbf{j} + 5\mathbf{k}$.

From $L_2$:

• A point on $L_2$ is $A_2(1, 2, 6)$. So, $\vec{a_2} = \mathbf{i} + 2\mathbf{j} + 6\mathbf{k}$.
• The direction vector of $L_2$ is $\vec{b_2} = 2\mathbf{i} + \mathbf{j} - 3\mathbf{k}$.

Self-check: We observe that $\vec{b_1}$ and $\vec{b_2}$ are not proportional ($2/2 \neq -7/1 \neq 5/-3$), which confirms that the lines are not parallel. Thus, the formula for skew lines is applicable.

2. Calculate $\vec{a_2} - \vec{a_1}$

This vector represents the displacement from a point on $L_1$ to a point on $L_2$. $\vec{a_2} - \vec{a_1} = (\mathbf{i} + 2\mathbf{j} + 6\mathbf{k}) - (\mathbf{i} - 8\mathbf{j} + 4\mathbf{k})$ $\vec{a_2} - \vec{a_1} = (1-1)\mathbf{i} + (2-(-8))\mathbf{j} + (6-4)\mathbf{k}$ $\vec{a_2} - \vec{a_1} = 0\mathbf{i} + 10\mathbf{j} + 2\mathbf{k}$

3. Calculate the Cross Product $\vec{b_1} \times \vec{b_2}$

The cross product of the direction vectors yields a vector that is perpendicular to both $\vec{b_1}$ and $\vec{b_2}$. This vector defines the direction of the shortest distance between the lines. $\vec{b_1} \times \vec{b_2} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & -7 & 5 \\ 2 & 1 & -3 \end{vmatrix}$ $= \mathbf{i}((-7)(-3) - (5)(1)) - \mathbf{j}((2)(-3) - (5)(2)) + \mathbf{k}((2)(1) - (-7)(2))$ $= \mathbf{i}(21 - 5) - \mathbf{j}(-6 - 10) + \mathbf{k}(2 - (-14))$ $= \mathbf{i}(16) - \mathbf{j}(-16) + \mathbf{k}(16)$ $\vec{b_1} \times \vec{b_2} = 16\mathbf{i} + 16\mathbf{j} + 16\mathbf{k}$

4. Calculate the Magnitude of the Cross Product $|\vec{b_1} \times \vec{b_2}|$

This magnitude serves as the denominator in the shortest distance formula. $|\vec{b_1} \times \vec{b_2}| = \sqrt{16^2 + 16^2 + 16^2}$ $= \sqrt{3 \times 16^2}$ $= 16\sqrt{3}$

5. Calculate the Scalar Triple Product $(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2})$

This is the dot product of the vector connecting the points on the lines with the common perpendicular direction vector. This forms the numerator of our formula. $(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2}) = (0\mathbf{i} + 10\mathbf{j} + 2\mathbf{k}) \cdot (16\mathbf{i} + 16\mathbf{j} + 16\mathbf{k})$ $= (0)(16) + (10)(16) + (2)(16)$ $= 0 + 80 + 16$ $= 96$

6. Substitute Values into the Shortest Distance Formula

Now, we substitute all the calculated values into the formula: $d = \left| {{(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2})} \over {|\vec{b_1} \times \vec{b_2}|}} \right|$ $d = \left| {{96} \over {16\sqrt{3}}} \right|$ Since distance must be positive, we can drop the absolute value sign. $d = {{96} \over {16\sqrt{3}}}$ $d = {{6} \over {\sqrt{3}}}$

7. Rationalize the Denominator

To present the answer in a standard mathematical form, we rationalize the denominator by multiplying the numerator and denominator by $\sqrt{3}$: $d = {{6} \over {\sqrt{3}}} \times {{\sqrt{3}} \over {\sqrt{3}}}$ $d = {{6\sqrt{3}} \over 3}$ $d = 2\sqrt{3}$

The shortest distance between the given lines is $2\sqrt{3}$ units.

Common Mistakes & Tips

• Sign Errors in Point Extraction: Be extremely careful when extracting the coordinates of points from the Cartesian form, especially with terms like $y+8$ (which implies $y-(-8)$).
• Cross Product Calculation: This is a frequent source of errors. Double-check your determinant calculations, especially the signs for the $\mathbf{j}$ component.
• Parallel Lines Check: Always briefly check if the direction vectors are proportional. If they are, the lines are parallel, and a different formula for shortest distance applies: $d = \left| {{(\vec{a_2} - \vec{a_1}) \times \vec{b}} \over {|\vec{b}|}} \right|$, where $\vec{b}$ is the common direction vector. In this problem, they are not parallel.

Summary

The shortest distance between two skew lines is a fundamental concept in 3D geometry. The most efficient way to calculate it is by using the vector formula $d = \left| {{(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2})} \over {|\vec{b_1} \times \vec{b_2}|}} \right|$. The process involves carefully identifying the points and direction vectors from the given line equations, performing vector operations (subtraction, cross product, dot product), and finally substituting these values into the formula. Accuracy in vector algebra is key to arriving at the correct answer.

The final answer is \boxed{\text{2\sqrt3}}, which corresponds to option (A).