Question

Find the shortest distance between the line $x=1+t$, $y=1+6t$, $z=2t$, $t\in R$ and $x=1+2k,y=5+15k,z=-2+6k,k\in R$.

Answer 1

Consider the problem

$x=1+t,y=1+6t,z=2t$

$t=x-1$ ------ $\left(1\right)$

$t=\frac{y-1}{6}$ ----- $\left(2\right)$

$t=\frac{z}{2}$ ----- $\left(3\right)$

from $\left(1\right),\left(2\right)$ and $\left(3\right)$

$x-1=\frac{y-1}{6}=\frac{z}{2}$ ------ $\left(A\right)$

Now,

$x=1+2k,y=5+15k,$

$k=\frac{x-1}{2}$ ----- $\left(4\right)$

$k=\frac{y-5}{15}$ ---- $\left(5\right)$

$z=-2+6k$

$k=\frac{z+2}{6}$ ----- $\left(6\right)$

from $\left(4\right),\left(5\right),\left(6\right)$

$\frac{x-1}{2}=\frac{y-5}{15}=\frac{z+2}{6}$ ----- $\left(B\right)$

Converting equation of lines $\left(A\right)$ and $\left(B\right)$ in the vector form

$\overrightarrow{r}=\overrightarrow{a_1}+\lambda \overrightarrow{b_1}$

$\overrightarrow{r}=\left(\hat{i}+\hat{j}+0\hat{k}\right)+\lambda \left(\hat{i}+6\hat{j}+2\hat{k}\right)$

And

$\overrightarrow{r}=\overrightarrow{a_2}+\mu \overrightarrow{b_2}$

$\overrightarrow{r}=\left(\hat{i}+5\hat{j}-2\hat{k}\right)+\mu \left(2\hat{i}+15\hat{j}+6\hat{k}\right)$

Here,

$\begin{array}{c}\overrightarrow{a_1}=\hat{i}+\hat{j}+0\hat{k},\overrightarrow{b_2}=\hat{i}+6\hat{j}+2\hat{k}\\ \\ \overrightarrow{a_2}=\hat{i}+5\hat{j}-2\hat{k},\overrightarrow{b_2}=2\hat{i}+15\hat{j}+6\hat{k}\end{array}$

$\begin{array}{c}\overrightarrow{b_1}\times \overrightarrow{b_2}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& 6& 2\\ 2& 15& 6\end{array}\right|\\ \\ =6\hat{i}-2\hat{j}+3\hat{k}\\ \\ \left|\overrightarrow{b_1}\times \overrightarrow{b_2}\right|=\sqrt{36+4+9}=\sqrt{49}=7\\ \\ \overrightarrow{a_2}-\overrightarrow{a_1}=0\hat{i}+4\hat{j}-2\hat{k}\\ \\ \left(\overrightarrow{b_1}\times \overrightarrow{b_2}\right).\left(\overrightarrow{a_2}-\overrightarrow{a_1}\right)=(6\times 0)-2(\times 4)+(3\times -2)=-14\end{array}$

Shortest distance between the given lines

$\begin{array}{c}d=\left|\frac{\left(\overrightarrow{b_1}\times \overrightarrow{b_2}\right).\left(\overrightarrow{a_2}-\overrightarrow{a_1}\right)}{\left|\overrightarrow{b_1}\times \overrightarrow{b_2}\right|}\right|\\ \\ =\left|\frac{-14}{7}\right|=2\end{array}$