Key Concepts and Formulas

The problem involves finding the shortest distance between two skew lines in 3D space. Skew lines are lines that are neither parallel nor intersecting.

1. Standard Form of a Line: A straight line passing through a point $(x_1, y_1, z_1)$ and having direction ratios $(a, b, c)$ can be represented in Cartesian form as $\frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c}$. In vector form, this is $\vec{r} = \vec{a} + t\vec{b}$, where $\vec{a} = x_1\hat{i} + y_1\hat{j} + z_1\hat{k}$ is the position vector of a point on the line, and $\vec{b} = a\hat{i} + b\hat{j} + c\hat{k}$ is the direction vector of the line.

2. Shortest Distance Between Two Skew Lines: For two skew lines, $L_1: \vec{r} = \vec{a_1} + t\vec{b_1}$ and $L_2: \vec{r} = \vec{a_2} + s\vec{b_2}$, the shortest distance $D$ between them is given by the formula: $D = \frac{|(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2})|}{|\vec{b_1} \times \vec{b_2}|}$ Here:

   • $\vec{a_1}$ and $\vec{a_2}$ are position vectors of points on lines $L_1$ and $L_2$ respectively.
   • $\vec{b_1}$ and $\vec{b_2}$ are the direction vectors of lines $L_1$ and $L_2$ respectively.
   • $(\vec{a_2} - \vec{a_1})$ is a vector connecting a point on $L_1$ to a point on $L_2$.
   • $(\vec{b_1} \times \vec{b_2})$ is a vector perpendicular to both lines, representing the common normal direction. Its magnitude, $|\vec{b_1} \times \vec{b_2}|$, is the length of this common normal vector.
   • The numerator, $|(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2})|$, is the magnitude of the scalar triple product, which geometrically represents the volume of the parallelepiped formed by these three vectors. More importantly, it is the absolute value of the projection of the vector $(\vec{a_2} - \vec{a_1})$ onto the common normal vector $(\vec{b_1} \times \vec{b_2})$.

Step-by-Step Solution

Step 1: Convert the Given Lines into Standard Vector Form

The first crucial step is to rewrite the given Cartesian equations of the lines into their standard symmetric form, $\frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c}$. This allows us to directly identify a point on the line ($\vec{a}$) and its direction vector ($\vec{b}$). Remember that the coefficients of $x, y, z$ in the numerator must be 1.

For Line $L_1$: The given equation is $3(x - 1) = 6(y - 2) = 2(z - 1)$. To achieve the standard form, we divide each part of the equation by the Least Common Multiple (LCM) of the coefficients of the terms in the numerators (3, 6, and 2), which is 6. This normalizes the denominators to represent the direction ratios. $\frac{3(x - 1)}{6} = \frac{6(y - 2)}{6} = \frac{2(z - 1)}{6}$ $\implies \frac{x - 1}{2} = \frac{y - 2}{1} = \frac{z - 1}{3}$ From this standard form, we can identify:

• A point on $L_1$: $\vec{a_1} = 1\hat{i} + 2\hat{j} + 1\hat{k}$.
• The direction vector of $L_1$: $\vec{b_1} = 2\hat{i} + 1\hat{j} + 3\hat{k}$.

For Line $L_2$: The given equation is $4(x - 2) = 2(y - \lambda ) = (z - 3)$. Similarly, we divide by the LCM of the coefficients (4, 2, and 1), which is 4. $\frac{4(x - 2)}{4} = \frac{2(y - \lambda)}{4} = \frac{(z - 3)}{4}$ $\implies \frac{x - 2}{1} = \frac{y - \lambda}{2} = \frac{z - 3}{4}$ From this standard form, we can identify:

• A point on $L_2$: $\vec{a_2} = 2\hat{i} + \lambda\hat{j} + 3\hat{k}$.
• The direction vector of $L_2$: $\vec{b_2} = 1\hat{i} + 2\hat{j} + 4\hat{k}$.

Step 2: Calculate the Necessary Vector Components for the Shortest Distance Formula

Now we compute the specific vector quantities required for the shortest distance formula.

2a. Calculate the vector $(\vec{a_2} - \vec{a_1})$: This vector connects a known point on $L_1$ to a known point on $L_2$. $\vec{a_2} - \vec{a_1} = (2\hat{i} + \lambda\hat{j} + 3\hat{k}) - (1\hat{i} + 2\hat{j} + 1\hat{k})$ $\vec{a_2} - \vec{a_1} = (2 - 1)\hat{i} + (\lambda - 2)\hat{j} + (3 - 1)\hat{k}$ $\vec{a_2} - \vec{a_1} = \hat{i} + (\lambda - 2)\hat{j} + 2\hat{k}$

2b. Calculate the cross product $(\vec{b_1} \times \vec{b_2})$: This vector represents the common normal to both lines, which is essential for determining the direction of the shortest distance. $\vec{b_1} \times \vec{b_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & 3 \\ 1 & 2 & 4 \end{vmatrix}$ Expanding the determinant: $= \hat{i}(1 \cdot 4 - 3 \cdot 2) - \hat{j}(2 \cdot 4 - 3 \cdot 1) + \hat{k}(2 \cdot 2 - 1 \cdot 1)$ $= \hat{i}(4 - 6) - \hat{j}(8 - 3) + \hat{k}(4 - 1)$ $= -2\hat{i} - 5\hat{j} + 3\hat{k}$

2c. Calculate the magnitude of $|\vec{b_1} \times \vec{b_2}|$: This magnitude forms the denominator of the shortest distance formula. $|\vec{b_1} \times \vec{b_2}| = \sqrt{(-2)^2 + (-5)^2 + (3)^2}$ $= \sqrt{4 + 25 + 9} = \sqrt{38}$

2d. Calculate the scalar triple product $(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2})$: This scalar value forms the numerator (before taking the absolute value) and can be computed as the dot product of the vector from Step 2a and the vector from Step 2b. Alternatively, it can be calculated directly as a $3 \times 3$ determinant with the components of the three vectors. $(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2}) = \begin{vmatrix} 1 & \lambda - 2 & 2 \\ 2 & 1 & 3 \\ 1 & 2 & 4 \end{vmatrix}$ Expanding the determinant along the first row: $= 1 \cdot (1 \cdot 4 - 3 \cdot 2) - (\lambda - 2) \cdot (2 \cdot 4 - 3 \cdot 1) + 2 \cdot (2 \cdot 2 - 1 \cdot 1)$ $= 1 \cdot (4 - 6) - (\lambda - 2) \cdot (8 - 3) + 2 \cdot (4 - 1)$ $= 1 \cdot (-2) - (\lambda - 2) \cdot (5) + 2 \cdot (3)$ $= -2 - 5\lambda + 10 + 6$ $= 14 - 5\lambda$

Step 3: Apply the Shortest Distance Formula and Solve for $\lambda$

We are given that the shortest distance $D = \frac{1}{\sqrt{38}}$. Now, we substitute all the calculated components into the shortest distance formula: $D = \frac{|(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2})|}{|\vec{b_1} \times \vec{b_2}|}$ $\frac{1}{\sqrt{38}} = \frac{|14 - 5\lambda|}{\sqrt{38}}$ Since the denominators are equal and non-zero, we can equate the numerators: $|14 - 5\lambda| = 1$ To solve this absolute value equation, we consider two possible cases: Case 1: $14 - 5\lambda = 1$ $5\lambda = 14 - 1$ $5\lambda = 13$ $\lambda = \frac{13}{5}$ Case 2: $14 - 5\lambda = -1$ $5\lambda = 14 + 1$ $5\lambda = 15$ $\lambda = 3$ The problem specifically asks for the integral value of $\lambda$. Comparing the two solutions, $\frac{13}{5}$ (which is 2.6) and $3$, the integral value is $3$.

Common Mistakes & Tips

• Standard Form Conversion: Always ensure the coefficients of $x, y, z$ in the numerator are 1. If you have $Ax - x_1$, rewrite it as $A(x - x_1/A)$ and divide the corresponding denominator by $A$.
• Cross Product and Determinant Calculation: Be meticulous with signs and arithmetic when calculating the cross product and the scalar triple product (determinant). A single error can propagate through the entire solution.
• Absolute Value: Do not forget the absolute value in the shortest distance formula. Ignoring it might lead to only one solution or a negative distance, which is physically impossible.
• Answering the Specific Question: Pay attention to what the question asks for (e.g., "integral value"). Always review your final answers against the question's requirements.

Summary

To find the integral value of $\lambda$ given the shortest distance between two skew lines, we first converted the Cartesian equations of the lines into their standard vector forms to extract the point vectors ($\vec{a_1}, \vec{a_2}$) and direction vectors ($\vec{b_1}, \vec{b_2}$). We then calculated the necessary components: the vector connecting points on the lines ($\vec{a_2} - \vec{a_1}$), the common normal vector ($\vec{b_1} \times \vec{b_2}$), its magnitude, and the scalar triple product. Substituting these into the shortest distance formula and solving the resulting absolute value equation yielded two possible values for $\lambda$. Finally, selecting the integral value among these solutions gave us the answer.

The integral value of $\lambda$ is 3, which corresponds to option (A).

The final answer is $\boxed{3}$.