Find the shortest distance between the given lines.

    \[\begin{array}{l} \square \vec{r}=(\hat{i}+\hat{j})+\lambda(2 \hat{i}-\hat{j}+\hat{k}) \\ \vec{r}=(2 \hat{i}+\hat{j}-\hat{k})+\mu(3 \hat{i}-5 \hat{j}+2 \hat{k}) \end{array}\]

Find the shortest distance between the given lines.

    \[\begin{array}{l} \square \vec{r}=(\hat{i}+\hat{j})+\lambda(2 \hat{i}-\hat{j}+\hat{k}) \\ \vec{r}=(2 \hat{i}+\hat{j}-\hat{k})+\mu(3 \hat{i}-5 \hat{j}+2 \hat{k}) \end{array}\]

Answer
Given equations:

r¯=(ı^+ȷ^)+λ(2ı^ȷ^+k^)
\overline{\mathrm{r}}=(\hat{\imath}+\hat{\jmath})+\lambda(2 \hat{\imath}-\hat{\jmath}+\hat{\mathrm{k}})
r¯=(2i^+ȷ^k^)+μ(3i^5ȷ^+2k^)
\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 : d
Formula :
1. Cross Product :
If \overline{\mathrm{a}} \& \overline{\mathrm{b}} are two vectors

a¯=a1ı^+a2l^+a3k^b¯=b1l^+b2ȷ^+b3k^
\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,

a¯×b¯=l^l^k^a1a2a3 b1 b2 b3
\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 \overline{\mathrm{a}} \& \overline{\mathrm{b}} are two vectors

a¯=a1ı^+a2ȷ^+a3k^b¯=b1ı^+b2ȷ^+b3k^
\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,
\overline{\mathrm{a}} \cdot \overline{\mathrm{b}}=\left(\mathrm{a}_{1} \times \mathrm{b}_{1}\right)+\left(\mathrm{a}_{2} \times \mathrm{b}_{2}\right)+\left(\mathrm{a}_{3} \times \mathrm{b}_{3}\right)
3. Shortest distance between two lines :
The shortest distance between the skew lines \overline{\mathrm{r}}=\overline{\mathrm{a}_{1}}+\lambda \overline{\mathrm{b}_{1}} and \overline{\mathrm{r}}=\overline{\mathrm{a}_{2}}+\lambda \overline{\mathrm{b}_{2}} is given by,

d=b1¯×b2¯·a2¯a1¯b1¯×b2¯
\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|

Answer:
For given lines,

r¯=(i^+j^)+λ(2i^j^+k^)r¯=(2i^+j^k^)+μ(3ı^5j^+2k^)
\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,

a1¯=l^+ȷ^b1¯=2l^ȷ^+k^
\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}
a2¯=2l^+ȷ^k^b2¯=3i^5j^+2k^
\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,

b1¯×b2¯=ı^ȷ^k^211352=i^(2+5)ȷ^(43)+k^(10+3)b1¯×b2¯=3i^ȷ^7k^b1¯×b2¯=32+(1)2+(7)2=9+1+49=59a2¯a1¯=(21)i^+(11)j^+(10)k^a2¯a1¯=l^+0ȷ^k^
\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,

b1¯×b2¯·a2¯a1¯=(3ı^ȷ^7k^)·(ı^+0ȷ^k^)=(3×1)+((1)×0)+((7)×(1))=3+0+7
\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

d=b1¯×b2¯·a2¯a1¯b1¯×b2¯d=1059
\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}