Key Concepts and Formulas

• Equation of a Line in 3D Space: A line passing through a point $A_1$ with position vector $\vec{a_1} = x_1\hat{i} + y_1\hat{j} + z_1\hat{k}$ and having a direction vector $\vec{v_1} = a\hat{i} + b\hat{j} + c\hat{k}$ can be represented in vector form as $\vec{r} = \vec{a_1} + \lambda \vec{v_1}$ or in Cartesian (symmetric) form as $\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c}$.
• Direction Vector of a Line Perpendicular to a Plane: If a line is perpendicular to a plane given by $Ax+By+Cz=D$, its direction vector is parallel to the normal vector of the plane, which is $\vec{n} = A\hat{i} + B\hat{j} + C\hat{k}$. Thus, the direction vector of the line can be taken as $\vec{v} = \vec{n}$.
• Shortest Distance Between Two Skew Lines: Two lines are skew if they are not parallel and do not intersect. The shortest distance $d$ between two skew lines $L_1: \vec{r} = \vec{a_1} + \lambda \vec{v_1}$ and $L_2: \vec{r} = \vec{a_2} + \mu \vec{v_2}$ is given by the formula: $d = \left| \frac{(\vec{a_2} - \vec{a_1}) \cdot (\vec{v_1} \times \vec{v_2})}{|\vec{v_1} \times \vec{v_2}|} \right|$ Here, $(\vec{a_2} - \vec{a_1})$ is a vector connecting a point on $L_1$ to a point on $L_2$, and $(\vec{v_1} \times \vec{v_2})$ is a vector perpendicular to both lines, representing the direction of the common perpendicular. The scalar triple product in the numerator represents the volume of a parallelepiped, and dividing by the magnitude of the cross product (area of the base) gives the height, which is the shortest distance.

Step-by-Step Solution

1. Identify Parameters for Line $l_1$ Line $l_1$ passes through the point (2, 6, 2) and is perpendicular to the plane $2x+y-2z=10$.

• Finding the position vector $\vec{a_1}$: The line passes through the point $(2, 6, 2)$. So, $\vec{a_1} = 2\hat{i} + 6\hat{j} + 2\hat{k}$.
• Finding the direction vector $\vec{v_1}$: Since $l_1$ is perpendicular to the plane $2x+y-2z=10$, its direction vector must be parallel to the normal vector of the plane. The normal vector to the plane $Ax+By+Cz=D$ is $\vec{n} = A\hat{i} + B\hat{j} + C\hat{k}$. For the plane $2x+y-2z=10$, the normal vector is $\vec{n} = 2\hat{i} + 1\hat{j} - 2\hat{k}$. Therefore, the direction vector of $l_1$ is $\vec{v_1} = 2\hat{i} + \hat{j} - 2\hat{k}$.

2. Identify Parameters for Line $l_2$ Line $l_2$ is given by the symmetric equation $\frac{x+1}{2}=\frac{y+4}{-3}=\frac{z}{2}$.

• Finding the position vector $\vec{a_2}$: The standard symmetric form of a line is $\frac{x-x_0}{a} = \frac{y-y_0}{b} = \frac{z-z_0}{c}$, where $(x_0, y_0, z_0)$ is a point on the line. Comparing with $\frac{x-(-1)}{2}=\frac{y-(-4)}{-3}=\frac{z-0}{2}$, we find a point on $l_2$ is $(-1, -4, 0)$. So, $\vec{a_2} = -1\hat{i} - 4\hat{j} + 0\hat{k} = -\hat{i} - 4\hat{j}$.
• Finding the direction vector $\vec{v_2}$: From the denominators of the symmetric equation, the direction vector of $l_2$ is $(2, -3, 2)$. So, $\vec{v_2} = 2\hat{i} - 3\hat{j} + 2\hat{k}$.

3. Calculate the Vector $(\vec{a_2} - \vec{a_1})$ This vector connects a point on $l_1$ to a point on $l_2$, forming part of the numerator in our distance formula. Given $\vec{a_1} = (2, 6, 2)$ and $\vec{a_2} = (-1, -4, 0)$: $\vec{a_2} - \vec{a_1} = (-1 - 2)\hat{i} + (-4 - 6)\hat{j} + (0 - 2)\hat{k}$ $\vec{a_2} - \vec{a_1} = -3\hat{i} - 10\hat{j} - 2\hat{k}$

4. Calculate the Cross Product $(\vec{v_1} \times \vec{v_2})$ This vector gives the direction of the common perpendicular between the two lines and is crucial for both the numerator and denominator of the shortest distance formula. Given $\vec{v_1} = (2, 1, -2)$ and $\vec{v_2} = (2, -3, 2)$: $\vec{v_1} \times \vec{v_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & -2 \\ 2 & -3 & 2 \end{vmatrix}$ $= \hat{i}((1)(2) - (-2)(-3)) - \hat{j}((2)(2) - (-2)(2)) + \hat{k}((2)(-3) - (1)(2))$ $= \hat{i}(2 - 6) - \hat{j}(4 - (-4)) + \hat{k}(-6 - 2)$ $= -4\hat{i} - 8\hat{j} - 8\hat{k}$

5. Calculate the Magnitude of the Cross Product $|\vec{v_1} \times \vec{v_2}|$ This is the denominator of the shortest distance formula. $|\vec{v_1} \times \vec{v_2}| = \sqrt{(-4)^2 + (-8)^2 + (-8)^2}$ $= \sqrt{16 + 64 + 64} = \sqrt{144} = 12$

6. Calculate the Scalar Triple Product $(\vec{a_2} - \vec{a_1}) \cdot (\vec{v_1} \times \vec{v_2})$ This forms the numerator of the shortest distance formula. We perform the dot product of $\vec{a_2} - \vec{a_1} = (-3, -10, -2)$ and $\vec{v_1} \times \vec{v_2} = (-4, -8, -8)$: $(\vec{a_2} - \vec{a_1}) \cdot (\vec{v_1} \times \vec{v_2}) = (-3)(-4) + (-10)(-8) + (-2)(-8)$ $= 12 + 80 + 16 = 108$

7. Calculate the Shortest Distance Now, substitute the calculated values into the shortest distance formula: $d = \left| \frac{(\vec{a_2} - \vec{a_1}) \cdot (\vec{v_1} \times \vec{v_2})}{|\vec{v_1} \times \vec{v_2}|} \right|$ $d = \left| \frac{108}{12} \right|$ $d = |9|$ $d = 9$

Common Mistakes & Tips

• Sign Errors: Be extremely careful with signs when extracting points from symmetric equations (e.g., $x+1$ implies $x_0 = -1$) and during vector component calculations (dot products, cross products, and subtraction).
• Cross Product Calculation: A common mistake is miscalculating the determinant for the cross product. Double-check each component.
• Absolute Value: Remember that distance must always be non-negative. The absolute value in the shortest distance formula is essential.
• Check for Parallelism: Before using the skew lines formula, it's good practice to quickly check if $\vec{v_1}$ and $\vec{v_2}$ are parallel (i.e., if one is a scalar multiple of the other, or if $\vec{v_1} \times \vec{v_2} = \vec{0}$). If they were parallel, a different formula for parallel lines would be used. In this case, $\vec{v_1} \times \vec{v_2} = -4\hat{i} - 8\hat{j} - 8\hat{k} \neq \vec{0}$, confirming they are skew.

Summary

To find the shortest distance between the two skew lines, we first determined the position vectors and direction vectors for each line based on the given information. For $l_1$, the direction vector was derived from the normal of the perpendicular plane. We then calculated the vector connecting points on the two lines, the cross product of their direction vectors, and the magnitude of this cross product. Finally, we substituted these values into the shortest distance formula for skew lines, which yielded a distance of 9 units.

The final answer is $\boxed{9}$, which corresponds to option (A).