1. Key Concepts and Formulas

• Vector Form of a Line: A line passing through a point with position vector $\vec{a}$ and having a direction vector $\vec{d}$ can be represented as $\vec{r} = \vec{a} + \lambda \vec{d}$, where $\lambda$ is a scalar parameter.
• Shortest Distance Between Two Skew Lines: For two skew lines, $L_1: \vec{r} = \vec{a_1} + \lambda \vec{d_1}$ and $L_2: \vec{r} = \vec{a_2} + \mu \vec{d_2}$, the shortest distance (S.D.) between them is given by the formula: $\text{S.D.} = \frac{|(\vec{a_2} - \vec{a_1}) \cdot (\vec{d_1} \times \vec{d_2})|}{|\vec{d_1} \times \vec{d_2}|}$ This formula calculates the projection of the vector connecting any point on $L_1$ to any point on $L_2$ onto their common perpendicular direction.
• Perpendicular Lines: Two lines are perpendicular if the dot product of their direction vectors is zero. If $\vec{d_A}$ and $\vec{d_B}$ are the direction vectors of two perpendicular lines, then $\vec{d_A} \cdot \vec{d_B} = 0$.

2. Step-by-Step Solution

Step 1: Express Line $L_1$ in Vector Form

The equation of line $L_1$ is given in Cartesian form: $L_1: \frac{x-1}{2}=\frac{y+1}{-3}=\frac{z+4}{2}$ We compare this with the standard symmetric form $\frac{x-x_1}{d_x}=\frac{y-y_1}{d_y}=\frac{z-z_1}{d_z}$. From this, we can identify a point on $L_1$ and its direction vector:

• The position vector of a point on $L_1$ is $\vec{a_1} = 1\hat{i} - 1\hat{j} - 4\hat{k} = \hat{i} - \hat{j} - 4\hat{k}$.
• The direction vector of $L_1$ is $\vec{d_1} = 2\hat{i} - 3\hat{j} + 2\hat{k}$.

Step 2: Determine Point and Direction Vector for Line $L_2$

Line $L_2$ passes through points A$(-4,4,3)$ and B$(-1,6,3)$, and is perpendicular to line $L_3: \frac{x-3}{-2}=\frac{y}{3}=\frac{z-1}{1}$.

• Finding a point on $L_2$ ($\vec{a_2}$): Since $L_2$ passes through point A$(-4,4,3)$, we can choose this as our reference point for $L_2$: $\vec{a_2} = -4\hat{i} + 4\hat{j} + 3\hat{k}$.

• Finding the direction vector of $L_2$ ($\vec{d_2}$):

  1. The line $L_2$ passes through points A and B. Therefore, its direction vector must be parallel to the vector $\vec{AB}$. Let's calculate $\vec{AB}$: $\vec{AB} = \text{Position Vector of B} - \text{Position Vector of A}$ $\vec{AB} = (-1 - (-4))\hat{i} + (6 - 4)\hat{j} + (3 - 3)\hat{k}$ $\vec{AB} = 3\hat{i} + 2\hat{j} + 0\hat{k}$.

  2. We are also given that $L_2$ is perpendicular to line $L_3: \frac{x-3}{-2}=\frac{y}{3}=\frac{z-1}{1}$. The direction vector of $L_3$ is $\vec{d_3} = -2\hat{i} + 3\hat{j} + 1\hat{k}$. For $L_2$ to be perpendicular to $L_3$, their direction vectors must have a dot product of zero. Let's check if $\vec{AB}$ satisfies this condition: $\vec{AB} \cdot \vec{d_3} = (3\hat{i} + 2\hat{j} + 0\hat{k}) \cdot (-2\hat{i} + 3\hat{j} + \hat{k})$ $= (3)(-2) + (2)(3) + (0)(1)$ $= -6 + 6 + 0 = 0$. Since the dot product is zero, $\vec{AB}$ is indeed perpendicular to $\vec{d_3}$. Thus, we can confidently use $\vec{d_2} = \vec{AB}$ as the direction vector for line $L_2$. So, $\vec{d_2} = 3\hat{i} + 2\hat{j} + 0\hat{k}$.

Step 3: Calculate the Vector Connecting Points on $L_1$ and $L_2$

We need to calculate the vector $\vec{a_2} - \vec{a_1}$: $\vec{a_2} - \vec{a_1} = (-4\hat{i} + 4\hat{j} + 3\hat{k}) - (\hat{i} - \hat{j} - 4\hat{k})$ $= (-4-1)\hat{i} + (4-(-1))\hat{j} + (3-(-4))\hat{k}$ $= -5\hat{i} + 5\hat{j} + 7\hat{k}$

Step 4: Calculate the Common Perpendicular Direction Vector

This is given by the cross product of the direction vectors of $L_1$ and $L_2$, i.e., $\vec{d_1} \times \vec{d_2}$: $\vec{d_1} \times \vec{d_2} = (2\hat{i} - 3\hat{j} + 2\hat{k}) \times (3\hat{i} + 2\hat{j} + 0\hat{k})$ We compute this using the determinant form: $\vec{d_1} \times \vec{d_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -3 & 2 \\ 3 & 2 & 0 \end{vmatrix}$ $= \hat{i}((-3)(0) - (2)(2)) - \hat{j}((2)(0) - (2)(3)) + \hat{k}((2)(2) - (-3)(3))$ $= \hat{i}(0 - 4) - \hat{j}(0 - 6) + \hat{k}(4 - (-9))$ $= -4\hat{i} + 6\hat{j} + 13\hat{k}$

Step 5: Calculate the Scalar Triple Product (Numerator)

This is the dot product of the vector from Step 3 and the vector from Step 4: $(\vec{a_2} - \vec{a_1}) \cdot (\vec{d_1} \times \vec{d_2}) = (-5\hat{i} + 5\hat{j} + 7\hat{k}) \cdot (-4\hat{i} + 6\hat{j} + 13\hat{k})$ $= (-5)(-4) + (5)(6) + (7)(13)$ $= 20 + 30 + 91 = 141$

Step 6: Calculate the Magnitude of the Common Perpendicular Direction Vector (Denominator)

This is the magnitude of the cross product calculated in Step 4: $|\vec{d_1} \times \vec{d_2}| = |-4\hat{i} + 6\hat{j} + 13\hat{k}|$ $= \sqrt{(-4)^2 + (6)^2 + (13)^2}$ $= \sqrt{16 + 36 + 169}$ $= \sqrt{221}$

Step 7: Apply the Shortest Distance Formula

Now, we substitute the calculated values into the shortest distance formula: $\text{S.D.} = \frac{|(\vec{a_2} - \vec{a_1}) \cdot (\vec{d_1} \times \vec{d_2})|}{|\vec{d_1} \times \vec{d_2}|}$ $\text{S.D.} = \frac{|141|}{\sqrt{221}} = \frac{141}{\sqrt{221}}$

3. Common Mistakes & Tips

• Sign Errors: Be extremely careful when extracting coordinates from the Cartesian form of a line (e.g., $y+1$ means $y_1 = -1$) and during vector arithmetic (subtraction, dot product, cross product).
• Vector Interpretation: When a line is defined by multiple conditions (e.g., passing through two points AND perpendicular to another line), ensure that the chosen direction vector satisfies ALL conditions. In this case, $\vec{AB}$ fortunately satisfied the perpendicularity condition.
• Cross Product Calculation: Mistakes in calculating the determinant for the cross product are common. Double-check your signs and arithmetic for each component.
• Absolute Value: Remember that distance must always be a non-negative value, so always take the absolute value of the numerator in the shortest distance formula.

4. Summary

To find the shortest distance between the given lines, we first converted $L_1$ to its vector form by identifying a point $\vec{a_1}$ and direction vector $\vec{d_1}$. For $L_2$, we used point A as $\vec{a_2}$ and derived its direction vector $\vec{d_2}$ by finding the vector $\vec{AB}$ and verifying its perpendicularity to $L_3$. Subsequently, we calculated the vector $(\vec{a_2} - \vec{a_1})$, the cross product $(\vec{d_1} \times \vec{d_2})$, and their dot product. Finally, dividing the absolute value of this dot product by the magnitude of the cross product yielded the shortest distance. The calculated shortest distance is $\frac{141}{\sqrt{221}}$.

5. Final Answer

The shortest distance between lines $L_1$ and $L_2$ is $\frac{141}{\sqrt{221}}$, which corresponds to option (A).

The final answer is $\boxed{\text{(A)}}$.