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Step-by-Step Solution
1. Express the Given Lines in Vector Form
Line 1 is given by
\frac{x-5}{4} = \frac{y-4}{1} = \frac{z-5}{3} .
Let the parameter be t . Then in parametric form:
x = 5 + 4t,\quad y = 4 + t,\quad z = 5 + 3t.
Therefore, a general point on Line 1 can be written as:
\vec{r}_1 = (5,\,4,\,5) + t\,(4,\,1,\,3).
So,
\vec{a_1} = (5,\,4,\,5), \quad \vec{d_1} = (4,\,1,\,3).
Line 2 is given by
\frac{x+8}{12} = \frac{y+2}{5} = \frac{z+11}{9} .
Let the parameter be s . Then in parametric form:
x = -8 + 12s,\quad y = -2 + 5s,\quad z = -11 + 9s.
Hence, a general point on Line 2 can be written as:
\vec{r}_2 = (-8,\,-2,\,-11) + s\,(12,\,5,\,9).
So,
\vec{a_2} = (-8,\,-2,\,-11), \quad \vec{d_2} = (12,\,5,\,9).
2. Set Coordinates for Points M and N
Let M lie on Line 1 and N lie on Line 2. Then:
\vec{OM} = \vec{a_1} + t\,\vec{d_1} = (5+4t,\;4+t,\;5+3t),
\vec{ON} = \vec{a_2} + s\,\vec{d_2} = (-8+12s,\;-2+5s,\;-11+9s).
3. Impose the Perpendicularity Conditions
For MN to be the shortest segment between the two skew lines, the vector
\overrightarrow{MN} = \vec{OM} - \vec{ON}
must be perpendicular to both direction vectors
\vec{d_1}
and
\vec{d_2} .
(\vec{OM} - \vec{ON}) \cdot \vec{d_1} = 0.
(\vec{OM} - \vec{ON}) \cdot \vec{d_2} = 0.
3.1. Compute \vec{OM} - \vec{ON}
\overrightarrow{MN}
= \bigl[(5 + 4t) - (-8 + 12s),\; (4 + t) - (-2 + 5s),\; (5 + 3t) - (-11 + 9s)\bigr].
Simplifying each component:
\overrightarrow{MN} = (13 + 4t - 12s,\; 6 + t - 5s,\; 16 + 3t - 9s).
3.2. First Condition: Perpendicular to \vec{d_1} = (4,\,1,\,3)
(13 + 4t - 12s,\; 6 + t - 5s,\; 16 + 3t - 9s)\,\cdot\,(4,\,1,\,3) = 0.
Which gives:
4(13 + 4t - 12s) + 1(6 + t - 5s) + 3(16 + 3t - 9s) = 0.
Combine like terms:
52 + 16t - 48s + 6 + t - 5s + 48 + 9t - 27s = 0,
106 + 26t - 80s = 0. \quad (1)
3.3. Second Condition: Perpendicular to \vec{d_2} = (12,\,5,\,9)
(13 + 4t - 12s,\; 6 + t - 5s,\; 16 + 3t - 9s)\,\cdot\,(12,\,5,\,9) = 0.
Which gives:
12(13 + 4t - 12s) + 5(6 + t - 5s) + 9(16 + 3t - 9s) = 0.
Combine like terms:
156 + 48t - 144s + 30 + 5t - 25s + 144 + 27t - 81s = 0,
330 + 80t - 250s = 0. \quad (2)
4. Solve for t and s
We have the system of equations:
\begin{cases}
106 + 26t - 80s = 0, \\
330 + 80t - 250s = 0.
\end{cases}
4.1. From the First Equation
26t = 80s - 106 \quad\Longrightarrow\quad t = \frac{80s - 106}{26} = \frac{40s - 53}{13}.
4.2. Substitute into the Second Equation
330 + 80\Bigl(\frac{40s - 53}{13}\Bigr) - 250s = 0.
Multiply throughout by 13:
13 \cdot 330 + 80(40s - 53) - 13 \cdot 250s = 0.
4290 + 3200s - 4240 - 3250s = 0.
50 - 50s = 0 \quad\Longrightarrow\quad s = 1.
Then,
t = \frac{40(1) - 53}{13} = \frac{-13}{13} = -1.
5. Find Coordinates of M and N
5.1. Point M (on Line 1)
\vec{OM} = (5,\,4,\,5) + (-1)\,(4,\,1,\,3) = (5-4,\; 4-1,\; 5-3) = (1,\; 3,\; 2).
5.2. Point N (on Line 2)
\vec{ON} = (-8,\, -2,\, -11) + (1)\,(12,\,5,\,9) = (-8+12,\; -2+5,\; -11+9) = (4,\; 3,\;-2).
6. Compute the Dot Product \overrightarrow{OM} \cdot \overrightarrow{ON}
\overrightarrow{OM} = (1,\,3,\,2), \quad \overrightarrow{ON} = (4,\,3,\,-2).
Their dot product is:
(1)(4) + (3)(3) + (2)(-2) = 4 + 9 - 4 = 9.
Final Answer
\overrightarrow{OM} \cdot \overrightarrow{ON} = 9.
