Find the shortest distance between the lines whose vector equations are

Question:

Find the shortest distance between the lines whose vector equations are

$\vec{r}=(1-t) \hat{i}+(t-2) \hat{j}+(3-2 t) \hat{k}$ and

$\vec{r}=(s+1) \hat{i}+(2 s-1) \hat{j}-(2 s+1) \hat{k}$

Solution:

The given lines are

$\vec{r}=(1-t) \hat{i}+(t-2) \hat{j}+(3-2 t) \hat{k}$

$\Rightarrow \vec{r}=(\hat{i}-2 \hat{j}+3 \hat{k})+t(-\hat{i}+\hat{j}-2 \hat{k})$          ...(1)

$\vec{r}=(s+1) \hat{i}+(2 s-1) \hat{j}-(2 s+1) \hat{k}$

$\Rightarrow \vec{r}=(\hat{i}-\hat{j}-\hat{k})+s(\hat{i}+2 \hat{j}-2 \hat{k})$              ...(2)

It is known that the shortest distance between the lines, $\vec{r}=\vec{a}_{1}+\lambda \vec{b}_{1}$ and $\vec{r}=\vec{a}_{2}+\mu \vec{b}_{2}$, is given by,

$d=\left|\frac{\left(\vec{b}_{1} \times \vec{b}_{2}\right) \cdot\left(\vec{a}_{2}-\vec{a}_{2}\right)}{\left|\vec{b}_{1} \times \vec{b}_{2}\right|}\right|$                ...(3)

For the given equations,

$\vec{a}_{1}=\hat{i}-2 \hat{j}+3 \hat{k}$

$\vec{b}_{1}=-\hat{i}+\hat{j}-2 \hat{k}$

$\vec{a}_{2}=\hat{i}-\hat{j}-\hat{k}$

$\vec{b}_{2}=\hat{i}+2 \hat{j}-2 \hat{k}$

$\vec{a}_{2}-\vec{a}_{1}=(\hat{i}-\hat{j}-\hat{k})-(\hat{i}-2 \hat{j}+3 \hat{k})=\hat{j}-4 \hat{k}$

$\vec{b}_{1} \times \vec{b}_{2}=\left|\begin{array}{lll}\hat{i} & \hat{j} & \hat{k} \\ -1 & 1 & -2 \\ 1 & 2 & -2\end{array}\right|=(-2+4) \hat{i}-(2+2) \hat{j}+(-2-1) \hat{k}=2 \hat{i}-4 \hat{j}-3 \hat{k}$

$\Rightarrow\left|\vec{b}_{1} \times \vec{b}_{2}\right|=\sqrt{(2)^{2}+(-4)^{2}+(-3)^{2}}=\sqrt{4+16+9}=\sqrt{29}$

$\therefore\left(\vec{b}_{1} \times \vec{b}_{2}\right) \cdot\left(\vec{a}_{2}-\vec{a}_{1}\right)=(2 \hat{i}-4 \hat{j}-3 \hat{k}) \cdot(\hat{j}-4 \hat{k})=-4+12=8$

Substituting all the values in equation (3), we obtain

$d=\left|\frac{8}{\sqrt{29}}\right|=\frac{8}{\sqrt{29}}$

Therefore, the shortest distance between the lines is $\frac{8}{\sqrt{29}}$ units.