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Step-by-Step Solution
Step 1: Express the given lines in parametric form
Line Lā is given in the form
$ \frac{x - a}{l} = \frac{y - 2}{3} = \frac{z - b}{4}.$
We need to find the values of a, b, and l.
Step 2: Use the foot of perpendicular condition
The foot of the perpendicular from M(4, 3, 8) onto Lā is given as N(3, 5, 7). Since N lies on Lā,
$ \frac{3 - a}{l} = \frac{5 - 2}{3} = \frac{7 - b}{4}. $
From the second equality:
$ \frac{5 - 2}{3} = 1 \implies 5 - 2 = 3 \implies 3 = 3, $
which is consistent.
Similarly, using
$ \frac{7 - b}{4} = 1 \implies 7 - b = 4 \implies b = 3.$
And
$ \frac{3 - a}{l} = 1 \implies 3 - a = l. $
Step 3: Enforce the perpendicular condition from M to line Lā
The direction ratios (DRs) of line Lā are $ \langle l, 3, 4 \rangle.$ The vector
$ \overrightarrow{MN} = \langle 3 - 4, 5 - 3, 7 - 8 \rangle = \langle -1, 2, -1 \rangle. $
Since $ \overrightarrow{MN} $ is perpendicular to the direction of Lā, their dot product must be zero:
$ \langle l, 3, 4 \rangle \cdot \langle -1, 2, -1 \rangle = 0.$
That is,
$ l \cdot (-1) + 3 \cdot 2 + 4 \cdot (-1) = 0 \implies -l + 6 - 4 = 0 \implies -l + 2 = 0 \implies l = 2.$
Thus, $ l = 2.$ From $ 3 - a = l, $ we get $ 3 - a = 2 \implies a = 1.$
So $ a = 1, b = 3,$ and $ l = 2.$
Step 4: Write the final equation of line Lā
With these values, line Lā becomes:
$ \frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4}. $
Step 5: Identify direction vectors and points on each line
Line Lā has a point A(1, 2, 3) and direction vector $ \overrightarrow{p} = \langle 2, 3, 4 \rangle.$
Line Lā is given by
$ \frac{x - 2}{3} = \frac{y - 4}{4} = \frac{z - 5}{5}, $
which has a point B(2, 4, 5) and direction vector $ \overrightarrow{q} = \langle 3, 4, 5 \rangle.$
Step 6: Form the vector connecting points on the two lines
$ \overrightarrow{AB} = \langle 2 - 1,\, 4 - 2,\, 5 - 3 \rangle = \langle 1, 2, 2 \rangle. $
Step 7: Calculate the cross product of the direction vectors
Compute
$ \overrightarrow{p} \times \overrightarrow{q} =
\begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
2 & 3 & 4 \\
3 & 4 & 5
\end{vmatrix}. $
This determinant gives:
$ \overrightarrow{p} \times \overrightarrow{q} = \bigl(3 \cdot 5 - 4 \cdot 4\bigr)\mathbf{i} - \bigl(2 \cdot 5 - 4 \cdot 3\bigr)\mathbf{j} + \bigl(2 \cdot 4 - 3 \cdot 3\bigr)\mathbf{k}
= \langle 15 - 16,\, -(10 - 12),\, 8 - 9 \rangle
= \langle -1,\, 2,\, -1 \rangle. $
Step 8: Find the magnitude of $ \overrightarrow{p} \times \overrightarrow{q} $
$ |\overrightarrow{p} \times \overrightarrow{q}|
= \sqrt{(-1)^2 + 2^2 + (-1)^2}
= \sqrt{1 + 4 + 1}
= \sqrt{6}. $
Step 9: Compute the shortest distance between Lā and Lā
The formula for the shortest distance between two skew lines with direction vectors $ \overrightarrow{p}, \overrightarrow{q} $ and a connecting vector $ \overrightarrow{AB} $ is
$ \text{Distance} =
\frac{\bigl|\overrightarrow{AB} \cdot (\overrightarrow{p} \times \overrightarrow{q})\bigr|}{|\overrightarrow{p} \times \overrightarrow{q}|}. $
First, compute the dot product:
$ \overrightarrow{AB} \cdot (\overrightarrow{p} \times \overrightarrow{q})
= \langle 1, 2, 2 \rangle \cdot \langle -1, 2, -1 \rangle
= (1)(-1) + (2)(2) + (2)(-1)
= -1 + 4 - 2
= 1. $
Therefore,
$ \text{Distance} =
\frac{|1|}{\sqrt{6}}
= \frac{1}{\sqrt{6}}. $
Step 10: State the final answer
Thus, the shortest distance between the lines Lā and Lā is
$ \frac{1}{\sqrt{6}}. $
