Find the shortest distance between the given lines.

Find the shortest distance between the given lines.

Given equations:

$\overline{\mathrm{r}}=\left(\stackrel{^}{ı}+\stackrel{^}{ȷ}\right)+\lambda \left(2\stackrel{^}{ı}–\stackrel{^}{ȷ}+\stackrel{^}{\mathrm{k}}\right)$
\overline{\mathrm{r}}=(\hat{\imath}+\hat{\jmath})+\lambda(2 \hat{\imath}-\hat{\jmath}+\hat{\mathrm{k}})
$\overline{\mathrm{r}}=\left(2\stackrel{^}{\mathrm{i}}+\stackrel{^}{\mathrm{ȷ}}–\stackrel{^}{\mathrm{k}}\right)+\mu \left(3\stackrel{^}{\mathrm{i}}–5\stackrel{^}{\mathrm{ȷ}}+2\stackrel{^}{\mathrm{k}}\right)$
\overline{\mathrm{r}}=(2 \hat{\mathrm{i}}+\hat{\mathrm{\jmath}}-\hat{\mathrm{k}})+\mu(3 \hat{\mathrm{i}}-5 \hat{\mathrm{\jmath}}+2 \hat{\mathrm{k}})

To Find :
Formula :
1. Cross Product :
If are two vectors

$\begin{array}{l}\overline{\mathrm{a}}={\mathrm{a}}_{1}\stackrel{^}{ı}+{\mathrm{a}}_{2}\stackrel{^}{\mathrm{l}}+{\mathrm{a}}_{3}\stackrel{^}{\mathrm{k}}\\ \overline{\mathrm{b}}={\mathrm{b}}_{1}\stackrel{^}{\mathrm{l}}+{\mathrm{b}}_{2}\stackrel{^}{\mathrm{ȷ}}+{\mathrm{b}}_{3}\stackrel{^}{\mathrm{k}}\end{array}$
\begin{array}{l}
\overline{\mathrm{a}}=\mathrm{a}_{1} \hat{\imath}+\mathrm{a}_{2} \hat{\mathrm{l}}+\mathrm{a}_{3} \hat{\mathrm{k}} \\
\overline{\mathrm{b}}=\mathrm{b}_{1} \hat{\mathrm{l}}+\mathrm{b}_{2} \hat{\mathrm{\jmath}}+\mathrm{b}_{3} \hat{\mathrm{k}}
\end{array}

then,

\overline{\mathrm{a}} \times \overline{\mathrm{b}}=\left|\begin{array}{ccc}
\hat{\mathrm{l}} & \hat{\mathrm{l}} & \hat{\mathrm{k}} \\
\mathrm{a}_{1} & \mathrm{a}_{2} & \mathrm{a}_{3} \\
\mathrm{~b}_{1} & \mathrm{~b}_{2} & \mathrm{~b}_{3}
\end{array}\right|

2. Dot Product :
If are two vectors

$\begin{array}{l}\overline{\mathrm{a}}={\mathrm{a}}_{1}\stackrel{^}{ı}+{\mathrm{a}}_{2}\stackrel{^}{ȷ}+{\mathrm{a}}_{3}\stackrel{^}{\mathrm{k}}\\ \overline{\mathrm{b}}={\mathrm{b}}_{1}\stackrel{^}{ı}+{\mathrm{b}}_{2}\stackrel{^}{ȷ}+{\mathrm{b}}_{3}\stackrel{^}{\mathrm{k}}\end{array}$
\begin{array}{l}
\overline{\mathrm{a}}=\mathrm{a}_{1} \hat{\imath}+\mathrm{a}_{2} \hat{\jmath}+\mathrm{a}_{3} \hat{\mathrm{k}} \\
\overline{\mathrm{b}}=\mathrm{b}_{1} \hat{\imath}+\mathrm{b}_{2} \hat{\jmath}+\mathrm{b}_{3} \hat{\mathrm{k}}
\end{array}

then,

3. Shortest distance between two lines :
The shortest distance between the skew lines and is given by,

$\mathrm{d}=\left|\frac{\left(\overline{{\mathrm{b}}_{1}}×\overline{{\mathrm{b}}_{2}}\right)·\left(\overline{{\mathrm{a}}_{2}}–\overline{{\mathrm{a}}_{1}}\right)}{\left|\overline{{\mathrm{b}}_{1}}×\overline{{\mathrm{b}}_{2}}\right|}\right|$
\mathrm{d}=\left|\frac{\left(\overline{\mathrm{b}_{1}} \times \overline{\mathrm{b}_{2}}\right) \cdot\left(\overline{\mathrm{a}_{2}}-\overline{\mathrm{a}_{1}}\right)}{\left|\overline{\mathrm{b}_{1}} \times \overline{\mathrm{b}_{2}}\right|}\right|

For given lines,

$\begin{array}{l}\overline{\mathrm{r}}=\left(\stackrel{^}{\mathrm{i}}+\stackrel{^}{\mathrm{j}}\right)+\lambda \left(2\stackrel{^}{\mathrm{i}}–\stackrel{^}{\mathrm{j}}+\stackrel{^}{\mathrm{k}}\right)\\ \overline{\mathrm{r}}=\left(2\stackrel{^}{\mathrm{i}}+\stackrel{^}{\mathrm{j}}–\stackrel{^}{\mathrm{k}}\right)+\mu \left(3\stackrel{^}{ı}–5\stackrel{^}{\mathrm{j}}+2\stackrel{^}{\mathrm{k}}\right)\end{array}$
\begin{array}{l}
\overline{\mathrm{r}}=(\hat{\mathrm{i}}+\hat{\mathrm{j}})+\lambda(2 \hat{\mathrm{i}}-\hat{\mathrm{j}}+\hat{\mathrm{k}}) \\
\overline{\mathrm{r}}=(2 \hat{\mathrm{i}}+\hat{\mathrm{j}}-\hat{\mathrm{k}})+\mu(3 \hat{\imath}-5 \hat{\mathrm{j}}+2 \hat{\mathrm{k}})
\end{array}

Here,

$\begin{array}{l}\overline{{\mathrm{a}}_{1}}=\stackrel{^}{\mathrm{l}}+\stackrel{^}{\mathrm{ȷ}}\\ \overline{{\mathrm{b}}_{1}}=2\stackrel{^}{\mathrm{l}}–\stackrel{^}{\mathrm{ȷ}}+\stackrel{^}{\mathrm{k}}\end{array}$
\begin{array}{l}
\overline{\mathrm{a}_{1}}=\hat{\mathrm{l}}+\hat{\mathrm{\jmath}} \\
\overline{\mathrm{b}_{1}}=2 \hat{\mathrm{l}}-\hat{\mathrm{\jmath}}+\hat{\mathrm{k}}
\end{array}
$\begin{array}{l}\overline{{\mathrm{a}}_{2}}=2\stackrel{^}{\mathrm{l}}+\stackrel{^}{\mathrm{ȷ}}–\stackrel{^}{\mathrm{k}}\\ \overline{{\mathrm{b}}_{2}}=3\stackrel{^}{\mathrm{i}}–5\stackrel{^}{\mathrm{j}}+2\stackrel{^}{\mathrm{k}}\end{array}$
\begin{array}{l}
\overline{\mathrm{a}_{2}}=2 \hat{\mathrm{l}}+\hat{\mathrm{\jmath}}-\hat{\mathrm{k}} \\
\overline{\mathrm{b}_{2}}=3 \hat{\mathrm{i}}-5 \hat{\mathrm{j}}+2 \hat{\mathrm{k}}
\end{array}

Therefore,

$\begin{array}{l}\overline{{\mathrm{b}}_{1}}×\overline{{\mathrm{b}}_{2}}=\left|\begin{array}{ccc}\stackrel{^}{ı}& \stackrel{^}{ȷ}& \stackrel{^}{\mathrm{k}}\\ 2& –1& 1\\ 3& –5& 2\end{array}\right|\\ =\stackrel{^}{\mathrm{i}}\left(–2+5\right)–\stackrel{^}{\mathrm{ȷ}}\left(4–3\right)+\stackrel{^}{\mathrm{k}}\left(–10+3\right)\\ \therefore \overline{{\mathrm{b}}_{1}}×\overline{{\mathrm{b}}_{2}}=3\stackrel{^}{\mathrm{i}}–\stackrel{^}{\mathrm{ȷ}}–7\stackrel{^}{\mathrm{k}}\\ \therefore \left|\overline{{\mathrm{b}}_{1}}×\overline{{\mathrm{b}}_{2}}\right|=\sqrt{{3}^{2}+\left(–1{\right)}^{2}+\left(–7{\right)}^{2}}\\ =\sqrt{9+1+49}\\ =\sqrt{59}\\ \overline{{\mathrm{a}}_{2}}–\overline{{\mathrm{a}}_{1}}=\left(2–1\right)\stackrel{^}{\mathrm{i}}+\left(1–1\right)\stackrel{^}{\mathrm{j}}+\left(–1–0\right)\stackrel{^}{\mathrm{k}}\\ \therefore \overline{{\mathrm{a}}_{2}}–\overline{{\mathrm{a}}_{1}}=\stackrel{^}{\mathrm{l}}+0\stackrel{^}{\mathrm{ȷ}}–\stackrel{^}{\mathrm{k}}\end{array}$
\begin{array}{l}
\overline{\mathrm{b}_{1}} \times \overline{\mathrm{b}_{2}}=\left|\begin{array}{ccc}
\hat{\imath} & \hat{\jmath} & \hat{\mathrm{k}} \\
2 & -1 & 1 \\
3 & -5 & 2
\end{array}\right| \\
=\hat{\mathrm{i}}(-2+5)-\hat{\mathrm{\jmath}}(4-3)+\hat{\mathrm{k}}(-10+3) \\
\therefore \overline{\mathrm{b}_{1}} \times \overline{\mathrm{b}_{2}}=3 \hat{\mathrm{i}}-\hat{\mathrm{\jmath}}-7 \hat{\mathrm{k}} \\
\therefore\left|\overline{\mathrm{b}_{1}} \times \overline{\mathrm{b}_{2}}\right|=\sqrt{3^{2}+(-1)^{2}+(-7)^{2}} \\
=\sqrt{9+1+49} \\
=\sqrt{59} \\
\overline{\mathrm{a}_{2}}-\overline{\mathrm{a}_{1}}=(2-1) \hat{\mathrm{i}}+(1-1) \hat{\mathrm{j}}+(-1-0) \hat{\mathrm{k}} \\
\therefore \overline{\mathrm{a}_{2}}-\overline{\mathrm{a}_{1}}=\hat{\mathrm{l}}+0 \hat{\mathrm{\jmath}}-\hat{\mathrm{k}}
\end{array}

Now,

$\begin{array}{l}\left(\overline{{b}_{1}}×\overline{{b}_{2}}\right)·\left(\overline{{a}_{2}}–\overline{{a}_{1}}\right)=\left(3\stackrel{^}{ı}–\stackrel{^}{ȷ}–7\stackrel{^}{k}\right)·\left(\stackrel{^}{ı}+0\stackrel{^}{ȷ}–\stackrel{^}{k}\right)\\ =\left(3×1\right)+\left(\left(–1\right)×0\right)+\left(\left(–7\right)×\left(–1\right)\right)\\ =3+0+7\end{array}$
\begin{array}{l}
\left(\overline{b_{1}} \times \overline{b_{2}}\right) \cdot\left(\overline{a_{2}}-\overline{a_{1}}\right)=(3 \hat{\imath}-\hat{\jmath}-7 \hat{k}) \cdot(\hat{\imath}+0 \hat{\jmath}-\hat{k}) \\
=(3 \times 1)+((-1) \times 0)+((-7) \times(-1)) \\
=3+0+7
\end{array}
$=10$
=10

Therefore, the shortest distance between the given lines is

$\begin{array}{l}\mathrm{d}=\left|\frac{\left(\overline{{\mathrm{b}}_{1}}×\overline{{\mathrm{b}}_{2}}\right)·\left(\overline{{\mathrm{a}}_{2}}–\overline{{\mathrm{a}}_{1}}\right)}{\left|\overline{{\mathrm{b}}_{1}}×\overline{{\mathrm{b}}_{2}}\right|}\right|\\ \therefore \mathrm{d}=\left|\frac{10}{\sqrt{59}}\right|\end{array}$
\begin{array}{l}
\mathrm{d}=\left|\frac{\left(\overline{\mathrm{b}_{1}} \times \overline{\mathrm{b}_{2}}\right) \cdot\left(\overline{\mathrm{a}_{2}}-\overline{\mathrm{a}_{1}}\right)}{\left|\overline{\mathrm{b}_{1}} \times \overline{\mathrm{b}_{2}}\right|}\right| \\
\therefore \mathrm{d}=\left|\frac{10}{\sqrt{59}}\right|
\end{array}