91Ó°ÊÓ

The given lines are$\frac{x-1}{2}=\frac{y+2}{3}=\frac{2z-1}{4}\mathrm{and}\frac{\mathrm{x}+2}{-1}=\frac{2-\mathrm{y}}{2},\mathrm{z}=\frac{1}{2}.$.

Introduction of a unit vector perpendicular to both the lines and then solving for the value of the distance between the given vector and the unit vector gives the distance between the given lines.

Then $\left|PQ.n\right|$where n is a unit vector perpendicular to both lines, is the distance d . Now if A and A are vectors along the two lines, then $\mathrm{A}×\mathrm{B}$is perpendicular to both, and n is just $\mathrm{A}×\mathrm{B}$divided by $\left|\mathrm{A}×\mathrm{B}\right|$. In this problem, consider the two points as following:

The symmetric form role="math" localid="1658815111418" $\frac{x-x_0}{a}=\frac{y-y_0}{b}=\frac{z-z_0}{c}$.

So, deduce that $\mathrm{P}=\left(1,-2\frac{1}{2}\right)$and $Q=\left(-2,2,\frac{1}{2}\right)$which mean

$PQ=\left(1,-2,\frac{1}{2}\right)-\left(-2,2,\frac{1}{2}\right)=\left(-3,4,0,\right)$

Given that a vector V=ai+bj+ck is parallel to the line described by the equation

$\frac{x-x_0}{a}=\frac{y-y_0}{b}=\frac{z-z_0}{c}$

The two vectors along the lines are as follows:

A=2i+3j+2k

B=-i-2j

$$\begin{gathered} \mathrm{A}×\mathrm{B}=\left(\begin{array}{ccc}\mathrm{i}& \mathrm{j}& \mathrm{k}\\ 2& 3& 2\\ -1& -2& 0\end{array}\right) \\ =4\mathrm{i}-2\mathrm{j}-\mathrm{k} \\ \mathrm{n}=\frac{\mathrm{A}×\mathrm{B}}{\left|\mathrm{A}×\mathrm{B}\right|} \\ =\frac{4\mathrm{i}-2\mathrm{j}-\mathrm{k}}{\sqrt{{\left(4\right)}^2+{\left(-2\right)}^2+{\left(-1\right)}^2}} \\ =\frac{4\mathrm{i}-2\mathrm{j}-\mathrm{k}}{\sqrt{21}} \end{gathered}$$

The formula for distance between two lines is given as follows:

$\mathrm{d}=\left|\mathrm{PQ}.\mathrm{n}=\frac{1}{\sqrt{21}}\left|\left[\left(-3\right)\left(4\right)+\left(4\right)\left(-2\right)+\left(0\right)\left(-1\right)\right]\right|=\frac{1}{\sqrt{21}}-20\right|=\frac{20}{\sqrt{21}}$

The distance between the two lines is $\frac{20}{\sqrt{21}}$.