Key Concepts and Formulas

• Skew Lines: Two lines in three-dimensional space are called skew lines if they are neither parallel nor intersecting. They lie in different planes.
• Vector Form of a Line: A line passing through a point with position vector $\vec{a}$ and parallel to 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: If two skew lines are given by their vector equations: Line 1 ($L_1$): $\vec{r} = \vec{a_1} + \lambda \vec{p}$ Line 2 ($L_2$): $\vec{r} = \vec{a_2} + \mu \vec{q}$ where $\vec{a_1}$ and $\vec{a_2}$ are the position vectors of points on $L_1$ and $L_2$ respectively, and $\vec{p}$ and $\vec{q}$ are their respective direction vectors, then the shortest distance (SD) between them is given by the formula: $SD = \left| \frac{(\vec{a_2} - \vec{a_1}) \cdot (\vec{p} \times \vec{q})}{|\vec{p} \times \vec{q}|} \right|$ This formula calculates the scalar projection of the vector connecting any point on $L_1$ to any point on $L_2$ (i.e., $\vec{a_2} - \vec{a_1}$) onto the direction of the common perpendicular to both lines (i.e., $\vec{p} \times \vec{q}$).

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Step-by-Step Solution

Step 1: Extract Position and Direction Vectors from the Given Cartesian Equations The given lines are in Cartesian form, which we need to convert to vector form $\vec{r} = \vec{a} + \lambda \vec{d}$ to use the shortest distance formula. The general Cartesian form is $\frac{x - x_1}{d_x} = \frac{y - y_1}{d_y} = \frac{z - z_1}{d_z}$, where $(x_1, y_1, z_1)$ is a point on the line and $(d_x, d_y, d_z)$ are its direction ratios.

For Line 1 ($L_1$): ${{x - 3} \over 2} = {{y - 2} \over 3} = {{z - 1} \over { - 1}}$

• The line passes through the point $(3, 2, 1)$. So, its position vector $\vec{a_1}$ is: $\vec{a_1} = 3\hat{i} + 2\hat{j} + 1\hat{k}$
• The direction ratios of $L_1$ are $(2, 3, -1)$. So, its direction vector $\vec{p}$ is: $\vec{p} = 2\hat{i} + 3\hat{j} - 1\hat{k}$

For Line 2 ($L_2$): ${{x + 3} \over 2} = {{y - 6} \over 1} = {{z - 5} \over 3}$

• The line passes through the point $(-3, 6, 5)$ (note that $x+3$ implies $x - (-3)$). So, its position vector $\vec{a_2}$ is: $\vec{a_2} = -3\hat{i} + 6\hat{j} + 5\hat{k}$
• The direction ratios of $L_2$ are $(2, 1, 3)$. So, its direction vector $\vec{q}$ is: $\vec{q} = 2\hat{i} + 1\hat{j} + 3\hat{k}$

Step 2: Calculate the Vector Connecting Points on the Lines, $\vec{a_2} - \vec{a_1}$ This vector connects a specific point on $L_1$ to a specific point on $L_2$. $\vec{a_2} - \vec{a_1} = (-3\hat{i} + 6\hat{j} + 5\hat{k}) - (3\hat{i} + 2\hat{j} + 1\hat{k})$ $\vec{a_2} - \vec{a_1} = (-3 - 3)\hat{i} + (6 - 2)\hat{j} + (5 - 1)\hat{k}$ $\vec{a_2} - \vec{a_1} = -6\hat{i} + 4\hat{j} + 4\hat{k}$

Step 3: Calculate the Cross Product of the Direction Vectors, $\vec{p} \times \vec{q}$ The cross product $\vec{p} \times \vec{q}$ yields a vector that is perpendicular to both $\vec{p}$ and $\vec{q}$. This vector represents the direction of the common perpendicular between the two skew lines. $\vec{p} \times \vec{q} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & -1 \\ 2 & 1 & 3 \end{vmatrix}$ Expanding the determinant: $= \hat{i}((3)(3) - (-1)(1)) - \hat{j}((2)(3) - (-1)(2)) + \hat{k}((2)(1) - (3)(2))$ $= \hat{i}(9 + 1) - \hat{j}(6 + 2) + \hat{k}(2 - 6)$ $= 10\hat{i} - 8\hat{j} - 4\hat{k}$

Step 4: Calculate the Magnitude of the Cross Product, $|\vec{p} \times \vec{q}|$ The magnitude of this vector is needed for the denominator of the shortest distance formula. $|\vec{p} \times \vec{q}| = \sqrt{(10)^2 + (-8)^2 + (-4)^2}$ $= \sqrt{100 + 64 + 16}$ $= \sqrt{180}$ To simplify the radical, we find perfect square factors of 180: $180 = 36 \times 5$. $= \sqrt{36 \times 5} = 6\sqrt{5}$

Step 5: Calculate the Scalar Triple Product, $(\vec{a_2} - \vec{a_1}) \cdot (\vec{p} \times \vec{q})$ This is the dot product of the vector connecting the points on the lines (from Step 2) and the common perpendicular direction vector (from Step 3). This value forms the numerator of our shortest distance formula. $(\vec{a_2} - \vec{a_1}) \cdot (\vec{p} \times \vec{q}) = (-6\hat{i} + 4\hat{j} + 4\hat{k}) \cdot (10\hat{i} - 8\hat{j} - 4\hat{k})$ $= (-6)(10) + (4)(-8) + (4)(-4)$ $= -60 - 32 - 16$ $= -108$

Step 6: Apply the Shortest Distance Formula Now, substitute the calculated values from Steps 2, 4, and 5 into the shortest distance formula: $SD = \left| \frac{(\vec{a_2} - \vec{a_1}) \cdot (\vec{p} \times \vec{q})}{|\vec{p} \times \vec{q}|} \right|$ $SD = \left| \frac{-108}{6\sqrt{5}} \right|$ Since distance must be a non-negative value, we take the absolute value: $SD = \frac{108}{6\sqrt{5}}$ Simplify the fraction by dividing 108 by 6: $SD = \frac{18}{\sqrt{5}}$

This result matches one of the options directly.

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Common Mistakes & Tips

• Sign Errors: Pay close attention to signs when extracting coordinates from the Cartesian form. For example, $x+3$ means the x-coordinate of the point is $-3$. Similarly, be careful with signs during vector operations.
• Calculation Accuracy: Vector cross products and dot products involve several arithmetic operations. Double-check all calculations to avoid errors.
• Absolute Value: Always remember that distance is a non-negative quantity. The absolute value in the shortest distance formula is crucial.
• Parallel Lines Check: Before applying the skew lines formula, it's good practice to quickly check if the direction vectors ($\vec{p}$ and $\vec{q}$) are proportional. If they are, the lines are parallel, and a different formula for the distance between parallel lines should be used. In this case, $(2, 3, -1)$ and $(2, 1, 3)$ are not proportional, confirming they are skew (or intersecting, but the formula handles both).

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Summary

To find the shortest distance between two skew lines, we first convert their Cartesian equations into vector form to identify the position vectors of points on each line ($\vec{a_1}, \vec{a_2}$) and their direction vectors ($\vec{p}, \vec{q}$). We then calculate the vector connecting these points ($\vec{a_2} - \vec{a_1}$), determine the common perpendicular direction by taking the cross product of the direction vectors ($\vec{p} \times \vec{q}$), and find its magnitude. Finally, we apply the standard formula for the shortest distance between skew lines, taking the absolute value of the scalar triple product divided by the magnitude of the cross product. This systematic approach leads to the correct distance.

The shortest distance is $\frac{18}{\sqrt{5}}$.

The final answer is $\boxed{{{18} \over {\sqrt 5 }}}$, which corresponds to option (A).