Question

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

• A. $11$
• B. $\frac{\sqrt{38}}{12}$
• C. $13$
• D. $15$

Answer 1

The correct option is B $\frac{\sqrt{38}}{12}$

Given lines

$x-\frac{2}{3}=y-\frac{3}{4}=z-\frac{1}{4}$

$\frac{x-\frac{2}{3}}{1}=\frac{y-\frac{3}{4}}{1}=\frac{z-\frac{1}{4}}{1}$------(1)

$x-\frac{1}{2}=y-\frac{2}{3}=z-\frac{3}{4}$

$\frac{x-\frac{1}{2}}{1}=\frac{y-\frac{2}{3}}{1}=\frac{z-\frac{3}{4}}{1}$------(2)

on comparing both lines with general cartesian form

normal vector

$\overrightarrow{b}=\hat{i}+\hat{j}+\hat{k}$

position vector of line (1)

$\overrightarrow{a}=\frac{2}{3}\hat{i}+\frac{3}{4}\hat{j}+\frac{1}{4}\hat{k}$

position vector of line (2)

$\overrightarrow{c}=\frac{1}{2}\hat{i}+\frac{2}{3}\hat{j}+\frac{3}{4}\hat{k}$

$\overrightarrow{c}-\overrightarrow{a}=\frac{1}{2}\hat{i}+\frac{2}{3}\hat{j}+\frac{3}{4}\hat{k}-(\frac{2}{3}\hat{i}+\frac{3}{4}\hat{j}+\frac{1}{4}\hat{k})$

$\overrightarrow{c}-\overrightarrow{a}=\frac{1}{2}\hat{i}+\frac{2}{3}\hat{j}+\frac{3}{4}\hat{k}-\frac{2}{3}\hat{i}-\frac{3}{4}\hat{j}-\frac{1}{4}\hat{k}$

$\overrightarrow{c}-\overrightarrow{a}=-\frac{1}{6}\hat{i}-\frac{1}{12}\hat{j}+\frac{1}{2}\hat{k}$

$(\overrightarrow{c}-\overrightarrow{a})\times \overrightarrow{b}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ -\frac{1}{6}& -\frac{1}{12}& \frac{1}{2}\\ 1& 1& 1\end{array}\right|$

$(\overrightarrow{c}-\overrightarrow{a})\times \overrightarrow{b}=\hat{i}(-\frac{1}{12}-\frac{1}{2})-\hat{j}(-\frac{1}{6}-\frac{1}{2})+\hat{k}(-\frac{1}{6}+\frac{1}{12})$

$(\overrightarrow{c}-\overrightarrow{a})\times \overrightarrow{b}=-\frac{7}{12}\hat{i}+\frac{2}{3}\hat{j}-\frac{1}{12}\hat{k}$

formula to find distance between parallel lines

$SD=\frac{\left|\right(\overrightarrow{c}-\overrightarrow{a})\times \overrightarrow{b}|}{\left|\overrightarrow{b}\right|}$

$SD=\frac{|-\frac{7}{12}\hat{i}+\frac{2}{3}\hat{j}-\frac{1}{12}\hat{k}|}{\sqrt{1^2+1^2+1^2}}$

$SD=\frac{\sqrt{(-\frac{7}{12}{)}^2+(\frac{2}{3}{)}^2+(-\frac{1}{12}{)}^2}}{\sqrt{3}}$

$SD=\frac{\sqrt{\frac{49}{144}+\frac{4}{9}+\frac{1}{144}}}{\sqrt{3}}$

$SD=\frac{\sqrt{\frac{49}{144}+\frac{4}{9}+\frac{1}{144}}}{\sqrt{3}}$

$SD=\frac{\sqrt{\frac{114}{144}}}{\sqrt{3}}$

$SD=\frac{\sqrt{38}}{12}$