35.

Write the vector equations of the following lines and hence determine the distance between them:

$\frac{ \mathrm{x} -1}{2} = \frac{\mathrm{y} - 2}{3} = \frac{\mathrm{z} + 4}{6}; \frac{\mathrm{x} - 3}{4} = \frac{\mathrm{y} - 3}{6} = \frac{\mathrm{z} + 5}{12}$

$$\begin{gathered} \mathrm{Given} \mathrm{equation} \mathrm{of} \mathrm{line} \mathrm{is} \frac{\mathrm{x} - 1}{2} = \frac{\mathrm{y} - 2}{3} = \frac{\mathrm{z} + 4}{6} \\ \mathrm{This} \mathrm{can} \mathrm{also} \mathrm{be} \mathrm{written} \mathrm{in} \mathrm{the} \mathrm{standard} \mathrm{form} \mathrm{as} \frac{\mathrm{x} - 1}{2} = \frac{\mathrm{y} - 2}{3} = \frac{\mathrm{z} -( - 4 )}{6} \end{gathered}$$

The vector form of the above equation is,

$$\begin{gathered} \overrightarrow{\mathrm{r}} = \left( \hat{\mathrm{i}} + 2 \hat{\mathrm{j}} - 4 k \right) + \mathrm{\lambda } \left( \widehat{2\mathrm{i}} + 3 \hat{\mathrm{j}} + 6k \right) \\ \Rightarrow \overrightarrow{\mathrm{r}} = \overrightarrow{{\mathrm{a}}_1 } + \mathrm{\lambda } \overrightarrow{\mathrm{b}} ......\left(\mathrm{i}\right) \\ \mathrm{where}, \overrightarrow{{\mathrm{a}}_1 } = \hat{\mathrm{i}} + 2 \hat{\mathrm{j}} - 4 k \mathrm{and} \overrightarrow{\mathrm{b}} = \widehat{2\mathrm{i}} + 3 \hat{\mathrm{j}} + 6k \\ \mathrm{The} \mathrm{second} \mathrm{equation} \mathrm{of} \mathrm{line} \mathrm{is} \frac{\mathrm{x} - 3}{4} = \frac{\mathrm{y} - 3}{6} = \frac{\mathrm{z} + 5}{12} \\ \mathrm{The} \mathrm{above} \mathrm{can} \mathrm{also} \mathrm{be} \mathrm{written} \mathrm{as} \frac{\mathrm{x} - 3}{4} = \frac{\mathrm{y} - 3}{6} = \frac{\mathrm{z} - ( - 5 )}{12} \end{gathered}$$

The vector form of this equation is

$$\begin{gathered} \overrightarrow{\mathrm{r}} = \left( 3 \hat{\mathrm{i}} + 3 \hat{\mathrm{j}} - 5 k \right) + \mathrm{\mu } \left( 4 \hat{\mathrm{i}} + 6 \hat{\mathrm{j}} + 12 k \right) \\ \Rightarrow \overrightarrow{\mathrm{r}} = \left( 3 \hat{\mathrm{i}} + 3 \hat{\mathrm{j}} - 5 k \right) + 2 \mathrm{\mu } \left( 2 \hat{\mathrm{i}} + 3 \hat{\mathrm{j}} + 6 k \right) \\ \Rightarrow \overrightarrow{\mathrm{r}} = \overrightarrow{{\mathrm{a}}_{2 }} + 2 \mathrm{\mu } \overrightarrow{\mathrm{b}} .........\left(\mathrm{ii}\right) \\ \mathrm{Where} \overrightarrow{{\mathrm{a}}_{2 }} = 3 \hat{\mathrm{i}} + 3 \hat{\mathrm{j}} - 5 k \mathrm{and} \overrightarrow{\mathrm{b}} = 2 \hat{\mathrm{i}} + 3 \hat{\mathrm{j}} + 6 k \end{gathered}$$

Since  $\overrightarrow{\mathrm{b}}$  is same in equation (i)  and  (ii),  the two lines are parallel.

Distance  d, between the two parallel lines is given by the formula,

$$\begin{gathered} \mathrm{d} = \left| \frac{\overrightarrow{\mathrm{b}} \mathrm{x} \left( \overrightarrow{{\mathrm{a}}_{2 }} - \overrightarrow{{\mathrm{a}}_{1 }} \right) }{\left|\overrightarrow{\mathrm{b}}\right|}\right| \\ \mathrm{Here}, \overrightarrow{\mathrm{b}} = \left( 2 \hat{\mathrm{i}} + 3 \hat{\mathrm{j}} + 6 \mathrm{k} \right) \overrightarrow{{\mathrm{a}}_{2 }} = \left( 3 \hat{\mathrm{i}} + 3 \hat{\mathrm{j}} - 5 \mathrm{k} \right) \mathrm{and} \overrightarrow{{\mathrm{a}}_{1 }} = \left( \hat{\mathrm{i}} + 2 \hat{\mathrm{j}} - 4 \mathrm{k} \right) \end{gathered}$$

On substitution, we get

$$\begin{gathered} \mathrm{d} = \left| \frac{ \left( 2 \hat{\mathrm{i}} + 3 \hat{\mathrm{j}} + 6 \mathrm{k} \right) \mathrm{x} ( 3 \hat{\mathrm{i}} + 3 \hat{\mathrm{j}} - 5 \mathrm{k} - \left( \hat{\mathrm{i}} + 2 \hat{\mathrm{j}} - 4 \mathrm{k} \right) )}{\sqrt{ 4 + 9 + 36}}\right| \\ = \frac{1}{\sqrt{49}} \left| \left( 2 \hat{\mathrm{i}} + 3 \hat{\mathrm{j}} + 6 \mathrm{k} \right) \mathrm{x} \left( 2 \hat{\mathrm{i}} + \hat{\mathrm{j}} - \mathrm{k} \right) \right| \\ = \frac{1}{7} \left| \begin{array}{ccc} \hat{\mathrm{i}}& \hat{\mathrm{j}}& \mathrm{k}\\ 2& 3& 6\\ 2 & 1& - 1\end{array} \right| \\ = \frac{1}{7} \left| \hat{\mathrm{i}} \left( - 3 - 6 \right) - \hat{\mathrm{j}} \left( - 2 - 12 \right) + \mathrm{k} \left( 2 - 6 \right) \right| \\ = \frac{1}{7} \left| - 9 \hat{\mathrm{i}} +14 \hat{\mathrm{j}} - 4 \mathrm{k} \right| \end{gathered}$$

$$\begin{gathered} = \frac{1}{7} \left| \sqrt{ 81 + 196 + 16} \right| \\ = \frac{\sqrt{ 293}}{7} \\ \mathrm{Thus}, \mathrm{the} \mathrm{distance} \mathrm{between} \mathrm{the} \mathrm{two} \mathrm{given} \mathrm{lines} \mathrm{is} \frac{\sqrt{ 293}}{7} \end{gathered}$$