Question

Find the shortest distance between the following pairs of parallel lines whose equations are:  $$\overrightarrow{r}=(\hat{i}+2\hat{j}+3\hat{k})+\lambda (\hat{i}-\hat{j}+\hat{k})\text{ and }\overrightarrow{r}=(2\hat{i}-\hat{j}-\hat{k})+\mu (-\hat{i}+\hat{j}-\hat{k})$$

Answer 1

 The vector equations of the given lines are

$$\overrightarrow{r}=(\hat{i}+2\hat{j}+3\hat{k})+\lambda (\hat{i}-\hat{j}+\hat{k})...(1)$$

$$\overrightarrow{r}=(2\hat{i}-\hat{j}-\hat{k})+\mu (-\hat{i}+\hat{j}-\hat{k})$$

$$=(2\hat{i}-\hat{j}-\hat{k})-\mu (\hat{i}-\hat{j}+\hat{k})...(2)$$

These two lines pass through the points having position vectors $$\overrightarrow{a_1}=\hat{i}+2\hat{j}+3\hat{k}\text{ and }\overrightarrow{a_2}=2\hat{i}-\hat{j}-\hat{k}$$ and are parallel to the vector  $$\overrightarrow{b}=\hat{i}-\hat{j}+\hat{k}$$

Now,

$$\overrightarrow{a_2}-\overrightarrow{a_1}=\hat{i}-3\hat{j}-4\hat{k}$$

and 

$$(\overrightarrow{a_2}-\overrightarrow{a_1})\times \overrightarrow{b}=(\hat{i}-3\hat{j}-4\hat{k})\times (\hat{i}-\hat{j}+\hat{k})$$

$$=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& -3& -4\\ 1& -1& 1\end{array}\right|$$

$$=-7\hat{i}-5\hat{j}+2\hat{k}$$

$$\Rightarrow |(\overrightarrow{a_2}-\overrightarrow{a_1})\times \overrightarrow{b}|=\sqrt{{(-7)}^2+{(-5)}^2+2^2}$$

$$=\sqrt{49+25+4}$$

$$=\sqrt{78}$$

The shortest distance between the two lines is given by 

$$\frac{|(\overrightarrow{a_2}-\overrightarrow{a_1})\times \overrightarrow{b}|}{\left|\overrightarrow{b}\right|}=\frac{\sqrt{78}}{\sqrt{3}}=\sqrt{26}$$