Key Concepts and Formulas

• Skew Lines: Two lines in 3D space are called skew lines if they are neither parallel nor intersecting. They lie on different planes.
• Vector Form of a Line: A line passing through a point with position vector $\vec{a}$ and parallel to a direction vector $\vec{b}$ can be represented as $\vec{r} = \vec{a} + \lambda \vec{b}$, where $\lambda$ is a scalar parameter. For a line given in Cartesian form $\frac{x-x_1}{l} = \frac{y-y_1}{m} = \frac{z-z_1}{n}$, the position vector is $\vec{a} = x_1\hat{i} + y_1\hat{j} + z_1\hat{k}$ and the direction vector is $\vec{b} = l\hat{i} + m\hat{j} + n\hat{k}$.
• Shortest Distance Formula: The shortest distance ($d$) between two skew lines $\vec{r} = \vec{a_1} + \lambda \vec{b_1}$ and $\vec{r} = \vec{a_2} + \mu \vec{b_2}$ 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 represents the scalar projection of the vector connecting points on the two lines onto their common perpendicular direction.

Step-by-Step Solution

Step 1: Convert Cartesian Equations to Vector Form To apply the shortest distance formula, we first express the given lines in their vector form $\vec{r} = \vec{a} + \lambda \vec{b}$.

For Line 1: ${{x - 5} \over 1} = {{y - 2} \over 2} = {{z - 4} \over { - 3}}$ A point on this line is $(x_1, y_1, z_1) = (5, 2, 4)$, so its position vector is $\vec{a_1} = 5\hat{i} + 2\hat{j} + 4\hat{k}$. The direction ratios are $(l, m, n) = (1, 2, -3)$, so its direction vector is $\vec{b_1} = \hat{i} + 2\hat{j} - 3\hat{k}$.

For Line 2: ${{x + 3} \over 1} = {{y + 5} \over 4} = {{z - 1} \over { - 5}}$ We rewrite the numerators as $(x - (-3))$, $(y - (-5))$, $(z - 1)$. A point on this line is $(x_2, y_2, z_2) = (-3, -5, 1)$, so its position vector is $\vec{a_2} = -3\hat{i} - 5\hat{j} + \hat{k}$. The direction ratios are $(l, m, n) = (1, 4, -5)$, so its direction vector is $\vec{b_2} = \hat{i} + 4\hat{j} - 5\hat{k}$.

Step 2: Calculate the Vector $(\vec{a_2} - \vec{a_1})$ This vector connects a point on Line 1 to a point on Line 2, and its components are crucial for determining the relative displacement between the lines. $\vec{a_2} - \vec{a_1} = (-3\hat{i} - 5\hat{j} + \hat{k}) - (5\hat{i} + 2\hat{j} + 4\hat{k})$ $\vec{a_2} - \vec{a_1} = (-3 - 5)\hat{i} + (-5 - 2)\hat{j} + (1 - 4)\hat{k}$ $\vec{a_2} - \vec{a_1} = -8\hat{i} - 7\hat{j} - 6\hat{k}$

Step 3: Calculate the Cross Product $(\vec{b_1} \times \vec{b_2})$ The cross product of the direction vectors $\vec{b_1}$ and $\vec{b_2}$ gives a vector that is perpendicular to both lines. This vector defines the direction along which the shortest distance is measured. $\vec{b_1} \times \vec{b_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & -3 \\ 1 & 4 & -5 \end{vmatrix}$ $\vec{b_1} \times \vec{b_2} = \hat{i}((2)(-5) - (-3)(4)) - \hat{j}((1)(-5) - (-3)(1)) + \hat{k}((1)(4) - (2)(1))$ $\vec{b_1} \times \vec{b_2} = \hat{i}(-10 + 12) - \hat{j}(-5 + 3) + \hat{k}(4 - 2)$ $\vec{b_1} \times \vec{b_2} = 2\hat{i} - \hat{j}(-2) + 2\hat{k}$ $\vec{b_1} \times \vec{b_2} = 2\hat{i} + 2\hat{j} + 2\hat{k}$

Step 4: Calculate the Scalar Triple Product $(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2})$ This dot product gives the scalar projection of the vector $(\vec{a_2} - \vec{a_1})$ onto the common perpendicular direction. Its magnitude is the numerator of our shortest distance formula. $(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2}) = (-8\hat{i} - 7\hat{j} - 6\hat{k}) \cdot (2\hat{i} + 2\hat{j} + 2\hat{k})$ $(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2}) = (-8)(2) + (-7)(2) + (-6)(2)$ $(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2}) = -16 - 14 - 12$ $(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2}) = -42$

Step 5: Calculate the Magnitude of the Cross Product $|\vec{b_1} \times \vec{b_2}|$ The magnitude of the cross product serves as the denominator, normalizing the scalar triple product to give the actual perpendicular distance. $|\vec{b_1} \times \vec{b_2}| = |2\hat{i} + 2\hat{j} + 2\hat{k}|$ $|\vec{b_1} \times \vec{b_2}| = \sqrt{(2)^2 + (2)^2 + (2)^2}$ $|\vec{b_1} \times \vec{b_2}| = \sqrt{4 + 4 + 4}$ $|\vec{b_1} \times \vec{b_2}| = \sqrt{12}$ $|\vec{b_1} \times \vec{b_2}| = \sqrt{4 \times 3} = 2\sqrt{3}$

Step 6: Apply the Shortest Distance Formula Substitute the calculated numerator and denominator into the shortest distance formula. $d = \left| \frac{(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2})}{|\vec{b_1} \times \vec{b_2}|} \right|$ $d = \left| \frac{-42}{2\sqrt{3}} \right|$ $d = \left| \frac{-21}{\sqrt{3}} \right|$ To rationalize the denominator, multiply the numerator and denominator by $\sqrt{3}$: $d = \left| \frac{-21\sqrt{3}}{3} \right|$ $d = |-7\sqrt{3}|$ $d = 7\sqrt{3}$

Common Mistakes & Tips

• Sign Errors in Vector Components: Be extremely careful when extracting the components of $\vec{a_1}$ and $\vec{a_2}$ from the Cartesian equations, especially with terms like $(x+a)$, which should be read as $(x - (-a))$.
• Arithmetic Precision: Errors in vector addition/subtraction, cross products, or dot products can significantly affect the final answer. Double-check each calculation.
• Order of Cross Product: While the magnitude $|\vec{b_1} \times \vec{b_2}|$ is the same as $|\vec{b_2} \times \vec{b_1}|$, the vector direction is opposite. However, since the formula involves an absolute value for the distance, the sign of the numerator is ultimately resolved. Consistency in the order is still good practice.
• Check for Parallel Lines: Before applying the skew line formula, quickly check if the lines are parallel. If $\vec{b_1}$ is a scalar multiple of $\vec{b_2}$, the lines are parallel (or coincident), and a different formula for parallel lines is used: $d = \left| \frac{(\vec{a_2} - \vec{a_1}) \times \vec{b}}{|\vec{b}|} \right|$. In this problem, the direction vectors are not proportional, so the lines are skew.

Summary

To find the shortest distance between two skew lines, we first convert their Cartesian equations into vector form, identifying 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 calculate the vector connecting the points $(\vec{a_2} - \vec{a_1})$ and the common perpendicular direction vector $(\vec{b_1} \times \vec{b_2})$. The shortest distance is found by taking the absolute value of the scalar projection of $(\vec{a_2} - \vec{a_1})$ onto $(\vec{b_1} \times \vec{b_2})$, which is given by the formula $d = \left| \frac{(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2})}{|\vec{b_1} \times \vec{b_2}|} \right|$. Careful execution of vector arithmetic leads to the final distance.

The final answer is \boxed{\text{7\sqrt{3}}}, which corresponds to option (A).