1. Key Concepts and Formulas

• Parametric Form of a Line: A line passing through a point with position vector $\vec{a}$ and having a direction vector $\vec{b}$ can be represented by $\vec{r} = \vec{a} + \lambda \vec{b}$, where $\lambda$ is a scalar parameter.
• Shortest Distance Between Two Skew Lines: The line segment representing the shortest distance between two skew lines is perpendicular to the direction vectors of both lines. If M is a point on the first line and N is a point on the second line such that MN is the shortest distance, then the vector $\overrightarrow{MN}$ is perpendicular to both line direction vectors. This means $\overrightarrow{MN} \cdot \vec{b_1} = 0$ and $\overrightarrow{MN} \cdot \vec{b_2} = 0$.
• Dot Product of Vectors: For two vectors $\vec{u} = (u_x, u_y, u_z)$ and $\vec{v} = (v_x, v_y, v_z)$, their dot product is given by $\vec{u} \cdot \vec{v} = u_x v_x + u_y v_y + u_z v_z$. If O is the origin $(0,0,0)$, then $\overrightarrow{OM}$ is simply the position vector of M, and similarly for N.

2. Step-by-Step Solution

Step 1: Parameterize the given lines and represent points M and N. We are given two lines in symmetric form: Line 1 ($L_1$): $\frac{x-5}{4}=\frac{y-4}{1}=\frac{z-5}{3}$ This line passes through the point $A_1(5, 4, 5)$ and has a direction vector $\vec{b_1} = (4, 1, 3)$. A general point M on $L_1$ can be parameterized as: $M(5+4\lambda, 4+\lambda, 5+3\lambda)$ where $\lambda$ is a scalar parameter.

Line 2 ($L_2$): $\frac{x+8}{12}=\frac{y+2}{5}=\frac{z+11}{9}$ This line passes through the point $A_2(-8, -2, -11)$ and has a direction vector $\vec{b_2} = (12, 5, 9)$. A general point N on $L_2$ can be parameterized as: $N(-8+12\mu, -2+5\mu, -11+9\mu)$ where $\mu$ is a scalar parameter.

Step 2: Form the vector $\overrightarrow{MN}$. The vector connecting point M to point N is $\overrightarrow{MN} = \vec{n} - \vec{m}$. $\overrightarrow{MN} = ((-8+12\mu) - (5+4\lambda), (-2+5\mu) - (4+\lambda), (-11+9\mu) - (5+3\lambda))$ $\overrightarrow{MN} = (12\mu - 4\lambda - 13, 5\mu - \lambda - 6, 9\mu - 3\lambda - 16)$

Step 3: Apply the shortest distance condition to set up a system of equations. Since MN is the shortest distance between the lines, the vector $\overrightarrow{MN}$ must be perpendicular to both direction vectors $\vec{b_1}$ and $\vec{b_2}$. This gives us two dot product equations:

1. $\overrightarrow{MN} \cdot \vec{b_1} = 0$
2. $\overrightarrow{MN} \cdot \vec{b_2} = 0$

Equation 1: $\overrightarrow{MN} \cdot \vec{b_1} = 0$ $(12\mu - 4\lambda - 13)(4) + (5\mu - \lambda - 6)(1) + (9\mu - 3\lambda - 16)(3) = 0$ $(48\mu - 16\lambda - 52) + (5\mu - \lambda - 6) + (27\mu - 9\lambda - 48) = 0$ Combining like terms: $(48+5+27)\mu + (-16-1-9)\lambda + (-52-6-48) = 0$ $80\mu - 26\lambda - 106 = 0$ Dividing by 2: $40\mu - 13\lambda - 53 = 0 \quad \text{(Equation A)}$

Equation 2: $\overrightarrow{MN} \cdot \vec{b_2} = 0$ $(12\mu - 4\lambda - 13)(12) + (5\mu - \lambda - 6)(5) + (9\mu - 3\lambda - 16)(9) = 0$ $(144\mu - 48\lambda - 156) + (25\mu - 5\lambda - 30) + (81\mu - 27\lambda - 144) = 0$ Combining like terms: $(144+25+81)\mu + (-48-5-27)\lambda + (-156-30-144) = 0$ $250\mu - 80\lambda - 330 = 0$ Dividing by 10: $25\mu - 8\lambda - 33 = 0 \quad \text{(Equation B)}$

Step 4: Solve the system of linear equations for $\lambda$ and $\mu$. We have the system:

1. $40\mu - 13\lambda = 53$
2. $25\mu - 8\lambda = 33$

To solve for $\mu$ and $\lambda$, we can use elimination. Multiply Equation A by 8 and Equation B by 13 to eliminate $\lambda$: $8 \times (40\mu - 13\lambda) = 8 \times 53 \implies 320\mu - 104\lambda = 424 \quad \text{(Equation C)}$ $13 \times (25\mu - 8\lambda) = 13 \times 33 \implies 325\mu - 104\lambda = 429 \quad \text{(Equation D)}$

Subtract Equation C from Equation D: $(325\mu - 104\lambda) - (320\mu - 104\lambda) = 429 - 424$ $5\mu = 5$ $\mu = 1$

Substitute $\mu = 1$ into Equation B: $25(1) - 8\lambda - 33 = 0$ $25 - 8\lambda - 33 = 0$ $-8\lambda - 8 = 0$ $-8\lambda = 8$ $\lambda = -1$

Step 5: Find the coordinates of points M and N. Using $\lambda = -1$ for M: $M(5+4(-1), 4+(-1), 5+3(-1)) = M(5-4, 4-1, 5-3) = M(1, 3, 2)$ Using $\mu = 1$ for N: $N(-8+12(1), -2+5(1), -11+9(1)) = N(-8+12, -2+5, -11+9) = N(4, 3, -2)$

Step 6: Calculate $\overrightarrow{OM} \cdot \overrightarrow{ON}$. Since O is the origin $(0,0,0)$: $\overrightarrow{OM} = (1, 3, 2)$ $\overrightarrow{ON} = (4, 3, -2)$

Now, calculate their dot product: $\overrightarrow{OM} \cdot \overrightarrow{ON} = (1)(4) + (3)(3) + (2)(-2)$ $\overrightarrow{OM} \cdot \overrightarrow{ON} = 4 + 9 - 4$ $\overrightarrow{OM} \cdot \overrightarrow{ON} = 9$

3. Common Mistakes & Tips

• Sign Errors: Be extremely careful with signs when writing down the coordinates of points from the symmetric form of the line (e.g., $x+8$ means the line passes through $x=-8$) and during vector subtraction and dot product calculations.
• Algebraic Errors: Solving the system of linear equations for $\lambda$ and $\mu$ is a common place for errors. Double-check your arithmetic.
• Perpendicularity Condition: Remember that the shortest distance vector is perpendicular to both direction vectors, leading to two crucial equations. Neglecting one will lead to an incorrect solution.

4. Summary

We first parameterized the two given lines to express the general coordinates of points M and N in terms of parameters $\lambda$ and $\mu$. Then, we formed the vector $\overrightarrow{MN}$. By applying the shortest distance condition, which states that $\overrightarrow{MN}$ must be perpendicular to the direction vectors of both lines, we set up a system of two linear equations in $\lambda$ and $\mu$. Solving this system yielded $\lambda = -1$ and $\mu = 1$. Substituting these values back into the parameterized equations gave us the exact coordinates of M as $(1, 3, 2)$ and N as $(4, 3, -2)$. Finally, we calculated the dot product $\overrightarrow{OM} \cdot \overrightarrow{ON}$, which is $9$.

5. Final Answer

The final answer is $\boxed{9}$.