Key Concepts and Formulas

• Skew Lines: In three-dimensional space, two lines are called skew lines if they are neither parallel nor intersecting. The shortest distance between them is along a line segment that is perpendicular to both lines.
• 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 by the vector equation $\vec{r} = \vec{a} + \lambda \vec{b}$, where $\lambda$ is a scalar parameter.
• Shortest Distance Formula: For two skew lines $L_1: \vec{r} = \vec{a}_1 + \lambda \vec{b}_1$ and $L_2: \vec{r} = \vec{a}_2 + \mu \vec{b}_2$, the shortest distance $D$ between them is given by the formula: $D = \left| \frac{(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2)}{|\vec{b}_1 \times \vec{b}_2|} \right|$ This formula essentially projects the vector connecting a point on $L_1$ to a point on $L_2$ onto the direction vector of the common perpendicular (which is $\vec{b}_1 \times \vec{b}_2$).

---

Step-by-Step Solution

Step 1: Identify the Position Vectors and Direction Vectors for Each Line We begin by extracting the necessary vector information for lines $L_1$ and $L_2$ from the problem statement. This is crucial as these vectors are the inputs for our shortest distance formula.

For line $L_1$:

• It passes through the point $(7, 6, 2)$. The position vector of this point is $\vec{a}_1 = 7 \hat{i} + 6 \hat{j} + 2 \hat{k}$.
• It is parallel to the vector $\overrightarrow{a} = -3 \hat{i} + 2 \hat{j} + 4 \hat{k}$. This is the direction vector for $L_1$, so $\vec{b}_1 = -3 \hat{i} + 2 \hat{j} + 4 \hat{k}$.

For line $L_2$:

• It passes through the point $(5, 3, 4)$. The position vector of this point is $\vec{a}_2 = 5 \hat{i} + 3 \hat{j} + 4 \hat{k}$.
• It is parallel to the vector $\overrightarrow{b} = 2 \hat{i} + \hat{j} + 3 \hat{k}$. This is the direction vector for $L_2$, so $\vec{b}_2 = 2 \hat{i} + \hat{j} + 3 \hat{k}$.

We can observe that $\vec{b}_1$ and $\vec{b}_2$ are not parallel (they are not scalar multiples of each other), confirming that the lines are skew (or intersect). Since the problem asks for the shortest distance, we assume they are skew lines.

Step 2: Calculate the Vector Connecting a Point on $L_1$ to a Point on $L_2$ The vector $(\vec{a}_2 - \vec{a}_1)$ connects a specific point on $L_1$ to a specific point on $L_2$. This vector is a component of the numerator in our shortest distance formula and helps define the "separation" between the lines.

$\vec{a}_2 - \vec{a}_1 = (5 \hat{i} + 3 \hat{j} + 4 \hat{k}) - (7 \hat{i} + 6 \hat{j} + 2 \hat{k})$ $\vec{a}_2 - \vec{a}_1 = (5-7) \hat{i} + (3-6) \hat{j} + (4-2) \hat{k}$ $\vec{a}_2 - \vec{a}_1 = -2 \hat{i} - 3 \hat{j} + 2 \hat{k}$

Step 3: Determine the Direction Vector of the Common Perpendicular The cross product $\vec{b}_1 \times \vec{b}_2$ gives a vector that is simultaneously perpendicular to both $\vec{b}_1$ and $\vec{b}_2$. This vector defines the direction of the shortest distance between the two skew lines.

\hat{i} & \hat{j} & \hat{k} \\ -3 & 2 & 4 \\ 2 & 1 & 3 \end{vmatrix} $$ Expanding the determinant: $$ \vec{b}_1 \times \vec{b}_2 = \hat{i}((2)(3) - (4)(1)) - \hat{j}((-3)(3) - (4)(2)) + \hat{k}((-3)(1) - (2)(2)) $$ $$ \vec{b}_1 \times \vec{b}_2 = \hat{i}(6 - 4) - \hat{j}(-9 - 8) + \hat{k}(-3 - 4) $$ $$ \vec{b}_1 \times \vec{b}_2 = 2 \hat{i} - (-17) \hat{j} - 7 \hat{k} $$ $$ \vec{b}_1 \times \vec{b}_2 = 2 \hat{i} + 17 \hat{j} - 7 \hat{k} $$ **Step 4: Find the Magnitude of the Common Perpendicular Direction Vector** The magnitude of the cross product, $|\vec{b}_1 \times \vec{b}_2|$, forms the denominator of our shortest distance formula. It represents the area of the parallelogram formed by the direction vectors $\vec{b}_1$ and $\vec{b}_2$. $$ |\vec{b}_1 \times \vec{b}_2| = |2 \hat{i} + 17 \hat{j} - 7 \hat{k}| $$ $$ |\vec{b}_1 \times \vec{b}_2| = \sqrt{(2)^2 + (17)^2 + (-7)^2} $$ $$ |\vec{b}_1 \times \vec{b}_2| = \sqrt{4 + 289 + 49} $$ $$ |\vec{b}_1 \times \vec{b}_2| = \sqrt{342} $$ To simplify the radical, we look for perfect square factors: $342 = 9 \times 38$. $$ |\vec{b}_1 \times \vec{b}_2| = \sqrt{9 \times 38} = 3\sqrt{38} $$ **Step 5: Calculate the Scalar Triple Product (Numerator)** The numerator of the formula is the scalar triple product $(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2)$. Geometrically, this represents the volume of the parallelepiped formed by the three vectors $\vec{a}_2 - \vec{a}_1$, $\vec{b}_1$, and $\vec{b}_2$. $$ (\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2) = (-2 \hat{i} - 3 \hat{j} + 2 \hat{k}) \cdot (2 \hat{i} + 17 \hat{j} - 7 \hat{k}) $$ $$ = (-2)(2) + (-3)(17) + (2)(-7) $$ $$ = -4 - 51 - 14 $$ $$ = -69 $$ **Step 6: Compute the Shortest Distance** Finally, we substitute all the calculated components into the shortest distance formula. Remember to take the absolute value of the numerator as distance must always be non-negative. $$ 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{-69}{\sqrt{342}} \right| $$ $$ D = \frac{69}{\sqrt{342}} $$ Using the simplified form of the denominator from Step 4: $$ D = \frac{69}{3\sqrt{38}} $$ $$ D = \frac{23}{\sqrt{38}} $$ --- **Common Mistakes & Tips** * **Sign Errors**: Be vigilant with signs during vector subtraction, cross products, and dot products. A single sign mistake will propagate and lead to an incorrect final answer. * **Absolute Value**: Always remember to take the absolute value of the numerator in the final step, as distance is inherently a non-negative quantity. * **Simplifying Radicals**: Simplify any square roots in the denominator or numerator if possible. This often helps in matching the answer with the given options. * **Checking for Parallel Lines**: Before applying this formula, it's good practice to quickly check if the direction vectors $\vec{b}_1$ and $\vec{b}_2$ are parallel. If they are, the lines are either parallel or coincident, and a different (simpler) formula for the distance between parallel lines should be used. In this case, $\vec{b}_1$ is not a scalar multiple of $\vec{b}_2$, so they are not parallel. --- **Summary** To find the shortest distance between two skew lines, we first identified the position vectors of points on each line ($\vec{a}_1, \vec{a}_2$) and their respective direction vectors ($\vec{b}_1, \vec{b}_2$). We then calculated the vector connecting the points $(\vec{a}_2 - \vec{a}_1)$ and the common perpendicular direction vector $(\vec{b}_1 \times \vec{b}_2)$. Finally, we used the formula $D = \left| \frac{(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2)}{|\vec{b}_1 \times \vec{b}_2|} \right|$ to arrive at the shortest distance. The calculated distance is $\frac{23}{\sqrt{38}}$. The final answer is \boxed{\frac{23}{\sqrt{38}}} which corresponds to option (A).