# The shortest distance between the parallel lines $\large\frac{x-3}{4}=\frac{y-1}{2}=\frac{z-5}{-3}$ and $\large\frac{x-1}{4}=\frac{y-2}{2}=\frac{z-3}{3}$ is

$\begin{array}{1 1}(1)3&(2)2\\(3)1&(4)0\end{array}$

## 1 Answer

The vector equation of the given lines are
$\overrightarrow{r}=(3 \overrightarrow{i}+ \overrightarrow{j} +5 \overrightarrow{k})+t (4 \overrightarrow{i}+2 \overrightarrow{j} - 3 \overrightarrow{k})$ -----(1)
$\overrightarrow{r}=( \overrightarrow{i}+ 2 \overrightarrow{j} +3 \overrightarrow{k})+s(4 \overrightarrow{i}+2 \overrightarrow{j} - 3 \overrightarrow{k})$ -----(2)
$\overrightarrow{a_1}=3\overrightarrow{i}+\overrightarrow{j}+5\overrightarrow{k}$
$\overrightarrow{a_2}=3\overrightarrow{i}+2 \overrightarrow{j}+3\overrightarrow{k}$
$\overrightarrow{a_2} - \overrightarrow{a_1}=-2\overrightarrow{i}+\overrightarrow{j}-2\overrightarrow{k}$
$\overrightarrow{u}=4\overrightarrow{i}+2\overrightarrow{j}-3\overrightarrow{k}$
$\overrightarrow{u} \times (\overrightarrow {a_2} - \overrightarrow{a_1}) = \begin{vmatrix} \overrightarrow{i} & \overrightarrow{j} & \overrightarrow{k} \\ 4 & 2 & -3 \\ -2 & 1 & -2 \end{vmatrix}$
$\qquad=\overrightarrow{i} (-4+3) -\overrightarrow{j}(-8-6) +\overrightarrow{k} (+4+4)$
$\qquad=-\overrightarrow{i} +14 \overrightarrow{j} +8 \overrightarrow{k}$
$|\overrightarrow{u} \times (\overrightarrow {a_2} - \overrightarrow{a_1}) = | -\overrightarrow{i} +14 \overrightarrow{j} + 8 \overrightarrow{k} |$
$\qquad= \sqrt {(-1)^2+14^2+8^2}$
$\qquad= \sqrt {1+196+64}$
$\qquad=\sqrt{261}$
$| \overrightarrow {u} | =\sqrt {4^2+2^2+(-3)^2}$
$\qquad= \sqrt {16+4+9}$
$\qquad={29}$
$d=\large\frac{\sqrt{261}}{\sqrt {29}}$$=\sqrt {9}$
$\qquad=3$
Hence 1 is the correct answer
answered May 13, 2014 by

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