Key Concepts and Formulas

• Vector Form 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. From the Cartesian form $\frac{x-x_1}{l}=\frac{y-y_1}{m}=\frac{z-z_1}{n}$, we have $\vec{a} = x_1\hat{i} + y_1\hat{j} + z_1\hat{k}$ and $\vec{b} = l\hat{i} + m\hat{j} + n\hat{k}$.
• Skew Lines: Two lines in three-dimensional space are called skew lines 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.
• Shortest Distance Formula: For two skew lines given by $\vec{r}_1 = \vec{a}_1 + \lambda \vec{b}_1$ and $\vec{r}_2 = \vec{a}_2 + \mu \vec{b}_2$, the shortest distance $d$ between them is given by: $d = \left| \frac{(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2)}{|\vec{b}_1 \times \vec{b}_2|} \right|$ This formula represents the projection of the vector connecting any point on the first line to any point on the second line, onto the common perpendicular direction vector $(\vec{b}_1 \times \vec{b}_2)$. Geometrically, the numerator is the scalar triple product, representing the volume of a parallelepiped, and the denominator is the area of its base. Their ratio yields the height, which is the shortest distance.

Step-by-Step Solution

Step 1: Extract Position and Direction Vectors from Cartesian Form The first step is to convert the given Cartesian equations of the lines into their vector forms, as the shortest distance formula requires position vectors ($\vec{a}$) and direction vectors ($\vec{b}$).

Given Line 1 ($L_1$): $\frac{x-4}{4}=\frac{y+2}{5}=\frac{z+3}{3}$ Comparing with $\frac{x-x_1}{l} = \frac{y-y_1}{m} = \frac{z-z_1}{n}$:

• The position vector of a point on $L_1$ is $\vec{a}_1 = 4\hat{i} - 2\hat{j} - 3\hat{k}$ (note that $y+2 = y-(-2)$ and $z+3 = z-(-3)$).
• The direction vector parallel to $L_1$ is $\vec{b}_1 = 4\hat{i} + 5\hat{j} + 3\hat{k}$.

Given Line 2 ($L_2$): $\frac{x-1}{3}=\frac{y-3}{4}=\frac{z-4}{2}$ Comparing with $\frac{x-x_2}{l'} = \frac{y-y_2}{m'} = \frac{z-z_2}{n'}$:

• The position vector of a point on $L_2$ is $\vec{a}_2 = 1\hat{i} + 3\hat{j} + 4\hat{k}$.
• The direction vector parallel to $L_2$ is $\vec{b}_2 = 3\hat{i} + 4\hat{j} + 2\hat{k}$.

Step 2: Calculate the Connecting Vector $(\vec{a}_2 - \vec{a}_1)$ This vector connects a point on the first line to a point on the second line, forming a crucial component of the scalar triple product in the numerator of the distance formula. $\vec{a}_2 - \vec{a}_1 = (1\hat{i} + 3\hat{j} + 4\hat{k}) - (4\hat{i} - 2\hat{j} - 3\hat{k})$ $= (1-4)\hat{i} + (3-(-2))\hat{j} + (4-(-3))\hat{k}$ $= -3\hat{i} + 5\hat{j} + 7\hat{k}$

Step 3: Find the Common Perpendicular Direction Vector $(\vec{b}_1 \times \vec{b}_2)$ The cross product of the two direction vectors, $\vec{b}_1 \times \vec{b}_2$, gives a vector that is perpendicular to both lines. This vector defines the direction of the shortest distance segment. $\vec{b}_1 \times \vec{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 4 & 5 & 3 \\ 3 & 4 & 2 \end{vmatrix}$ Expanding the determinant: $= \hat{i}(5 \times 2 - 3 \times 4) - \hat{j}(4 \times 2 - 3 \times 3) + \hat{k}(4 \times 4 - 5 \times 3)$ $= \hat{i}(10 - 12) - \hat{j}(8 - 9) + \hat{k}(16 - 15)$ $= -2\hat{i} - (-1)\hat{j} + 1\hat{k}$ $= -2\hat{i} + \hat{j} + \hat{k}$

Step 4: Calculate the Magnitude of the Common Perpendicular Direction Vector $|\vec{b}_1 \times \vec{b}_2|$ This magnitude forms the denominator of the shortest distance formula. $|\vec{b}_1 \times \vec{b}_2| = \sqrt{(-2)^2 + (1)^2 + (1)^2}$ $= \sqrt{4 + 1 + 1}$ $= \sqrt{6}$

Step 5: Calculate the Scalar Triple Product $(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2)$ This dot product forms the numerator of the shortest distance formula. It effectively projects the connecting vector onto the common perpendicular direction. $(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2) = (-3\hat{i} + 5\hat{j} + 7\hat{k}) \cdot (-2\hat{i} + \hat{j} + \hat{k})$ $= (-3)(-2) + (5)(1) + (7)(1)$ $= 6 + 5 + 7$ $= 18$

Step 6: Apply the Shortest Distance Formula Substitute the calculated values into the formula to find the shortest distance. $d = \left| \frac{(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2)}{|\vec{b}_1 \times \vec{b}_2|} \right|$ $d = \left| \frac{18}{\sqrt{6}} \right|$ To rationalize the denominator and simplify: $d = \frac{18}{\sqrt{6}} \times \frac{\sqrt{6}}{\sqrt{6}}$ $d = \frac{18\sqrt{6}}{6}$ $d = 3\sqrt{6}$

Common Mistakes & Tips

• Sign Errors in Vector Extraction: Be very careful when extracting $x_1, y_1, z_1$ from the Cartesian form, especially with terms like $(y+2)$, which means $y-(-2)$, so $y_1 = -2$.
• Cross Product Calculation: Mistakes in calculating the cross product $(\vec{b}_1 \times \vec{b}_2)$ are common. Pay special attention to the alternating signs for the determinant expansion, particularly the negative sign for the $\hat{j}$ component.
• Arithmetic Errors: Double-check all additions, subtractions, and multiplications, especially in the dot product and magnitude calculations.
• Rationalizing the Denominator: Always rationalize the denominator for a cleaner final answer, as shown in Step 6.

Summary

To find the shortest distance between the given skew lines, we first extracted the position vectors ($\vec{a}_1, \vec{a}_2$) and direction vectors ($\vec{b}_1, \vec{b}_2$) from their Cartesian equations. We then calculated the connecting vector $(\vec{a}_2 - \vec{a}_1)$ and the common perpendicular direction vector $(\vec{b}_1 \times \vec{b}_2)$. Finally, we computed the scalar triple product in the numerator and the magnitude of the cross product in the denominator, substituting these into the shortest distance formula and simplifying to arrive at the final answer.

The final answer is $\boxed{3 \sqrt{6}}$, which corresponds to option (A).