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Step-by-Step Solution
Step 1: Express the lines Lā and Lā in parametric form
Line Lā is given by
$ \displaystyle \frac{x-1}{2} \;=\; \frac{y-3}{1} \;=\; \frac{z-2}{2}\,. $
Let that common ratio be $ \lambda $. Then the coordinates of any point $P$ on Lā can be written as
$
x = 2\lambda + 1,\quad
y = \lambda + 3,\quad
z = 2\lambda + 2.
$
Line Lā is given by
$ \displaystyle \frac{x - 2}{1} \;=\; \frac{y - 2}{2} \;=\; \frac{z - 3}{3}\,. $
Let that common ratio be $ \mu $. Then the coordinates of any point $Q$ on Lā can be written as
$
x = \mu + 2,\quad
y = 2\mu + 2,\quad
z = 3\mu + 3.
$
Step 2: Use the direction ratios of Lā
The line Lā has direction ratios $ \langle 1,\,-1,\,-2\rangle $. Suppose Lā passes through a point $P$ on Lā and through a point $Q$ on Lā. Since $P$ and $Q$ lie on Lā, the vector $ \overrightarrow{PQ} $ must be parallel to $ \langle 1,\,-1,\,-2\rangle $.
From the coordinates,
$ P(\,2\lambda + 1,\;\lambda + 3,\;2\lambda + 2\,)
\quad\text{and}\quad
Q(\,\mu + 2,\;2\mu + 2,\;3\mu + 3)\,.
$
Step 3: Form the vector PQ and set it proportional to the direction ratios
The vector
$ \overrightarrow{PQ}
= \langle (\mu + 2) - (2\lambda + 1),\; (2\mu + 2) - (\lambda + 3),\; (3\mu + 3) - (2\lambda + 2)\rangle. $
Simplify each component:
$ x $-component: $ \mu + 2 - (2\lambda + 1) = \mu - 2\lambda + 1, $
$ y $-component: $ 2\mu + 2 - (\lambda + 3) = 2\mu - \lambda - 1, $
$ z $-component: $ 3\mu + 3 - (2\lambda + 2) = 3\mu - 2\lambda + 1. $
Because $ \overrightarrow{PQ} $ is parallel to $ \langle 1,\,-1,\,-2\rangle $, we write
$ \frac{\mu - 2\lambda + 1}{1}
\;=\;
\frac{2\mu - \lambda - 1}{-1}
\;=\;
\frac{3\mu - 2\lambda + 1}{-2}. $
Step 4: Solve for Ī» and Ī¼
From
$ \frac{\mu - 2\lambda + 1}{1}
=
\frac{2\mu - \lambda - 1}{-1}, $
we get
$ \mu - 2\lambda + 1
=
-(\,2\mu - \lambda - 1)\,.
$
Simplify:
$ \mu - 2\lambda + 1
=
-2\mu + \lambda + 1
\;\Longrightarrow\;
\mu - 2\lambda
=
-2\mu + \lambda
\;\Longrightarrow\;
3\mu
=
3\lambda
\;\Longrightarrow\;
\mu
=
\lambda.
$
Next, use
$ \frac{\mu - 2\lambda + 1}{1}
=
\frac{3\mu - 2\lambda + 1}{-2}
\,.$
Substitute $ \mu = \lambda $ into that:
Left side $ = \lambda - 2\lambda + 1 = -\lambda + 1,\quad$
Right side $ = \frac{-(\,3\lambda - 2\lambda + 1)}{2}
= \frac{-(\lambda + 1)}{2}
= -\frac{\lambda + 1}{2}\,. $
Equating them:
$ -\lambda + 1
=
-\frac{\lambda + 1}{2}. $
Multiply both sides by 2:
$ -2\lambda + 2
=
-(\lambda + 1). $
Simplify:
$ -2\lambda + 2
=
-\lambda - 1
\;\Longrightarrow\;
-\lambda
=
-3
\;\Longrightarrow\;
\lambda
=
3.
$
Since $ \mu = \lambda = 3. $
Step 5: Find the coordinates of P and Q
Substitute $ \lambda = 3 $ into the parametric form of $ L_1 $:
$ P = (2\cdot 3 + 1,\; 3 + 3,\; 2\cdot 3 + 2) = (7,\;6,\;8). $
Substitute $ \mu = 3 $ into the parametric form of $ L_2 $:
$ Q = (3 + 2,\; 2\cdot 3 + 2,\; 3\cdot 3 + 3) = (5,\;8,\;12). $
Step 6: Compute the distance PQ
The distance between points $P(7,\,6,\,8)$ and $Q(5,\,8,\,12)$ is given by
$
PQ
=
\sqrt{(5 - 7)^2 + (8 - 6)^2 + (12 - 8)^2}
=
\sqrt{(-2)^2 + (2)^2 + (4)^2}
=
\sqrt{4 + 4 + 16}
=
\sqrt{24}
=
2\sqrt{6}.
$
Final Answer
Therefore, the length of the line segment PQ is
$ 2\sqrt{6}. $
