Question
The distance between the lines $$\frac{{x - 4}}{2} = \frac{{y + 1}}{{ - 3}} = \frac{z}{6}$$ and $$\frac{x}{{ - 1}} = \frac{{y - 1}}{{\frac{3}{2}}} = \frac{{z + 1}}{{ - 3}}$$ is :
A.
$$\sqrt {\frac{{629}}{7}} $$
B.
$$\sqrt {\frac{{39}}{7}} $$
C.
$$\frac{{\sqrt {629} }}{7}$$
D.
none of these
Answer :
$$\frac{{\sqrt {629} }}{7}$$
Solution :
$$\eqalign{
& {\text{Lines are}}\,\frac{{x - 4}}{2} = \frac{{y + 1}}{{ - 3}} = \frac{z}{6} \cr
& \Rightarrow \frac{{x - 4}}{{ - 1}} = \frac{{y + 1}}{{\frac{3}{2}}} = \frac{z}{{ - 3}}\, \to \left( 1 \right) \cr
& {\text{and }}\frac{x}{{ - 1}} = \frac{{y - 1}}{{\frac{3}{2}}} = \frac{{z + 1}}{{ - 3}}\, \to \left( 2 \right) \cr
& {\text{Clearly both lines are parallel with}} \cr
& \overrightarrow {{a_1}} = 4\hat i - \hat j,\,\overrightarrow {{a_2}} = \hat j - \hat k{\text{ and }}\overrightarrow b = - \hat i - \frac{3}{2}\hat j - 3\hat k \cr
& {\text{Using shortest distance between parallel lines}} \cr} $$
\[\begin{array}{l}
{\rm{distance}} = \frac{{\left| {\overrightarrow b \times \left( {\overrightarrow {{a_2}} - \overrightarrow {{a_1}} } \right)} \right|}}{{\left| {\overrightarrow b } \right|}}\\
= \frac{{\left| \begin{array}{l}
\,\,\,\overrightarrow i \,\,\,\,\,\overrightarrow j \,\,\,\,\,\,\overrightarrow k \\
- 1\,\,\,\,\,\,\frac{3}{2}\,\,\,\, - 3\\
- 4\,\,\,\,\,\,\,2\,\,\,\,\, - 1
\end{array} \right|}}{{\sqrt {\frac{7}{2}} }}\\
= \frac{{\left| {\frac{9}{2}\hat i - 11\hat j + 4\hat k} \right|}}{{\frac{7}{2}}}\\
= \frac{{\sqrt {629} }}{7}
\end{array}\]