Exercise 11.2

Find the shortest distance between the lines whose vector equations are $\begin{array}{l} \vec{r}=(1-t) \hat{i}+(t-2) \hat{j}+(3-2 t) \hat{k} \text { and } \\ \vec{r}=(s+1) \hat{i}+(2 s-1) \hat{j}-(2 s+1) \hat{k} \end{array}$

Solution: Consider the given equations $\begin{array}{l} \Rightarrow \vec{r}=(1-t) \hat{i}+(t-2) \hat{j}+(3-2 t) \hat{k} \\ \vec{r}=\hat{i}-t \hat{i}+t \hat{j}-2 \hat{j}+3 \hat{k}-2 t \hat{k} \\...

read more