Question

The distance between lines $\frac{x+5}{4}=\frac{y+3}{-18}=\frac{z-6}{-4}$ and $\frac{x-3}{2}=\frac{y+10}{-9}=\frac{z-1}{-2}$(approx.) is

• A. $6$
• B. $7.5$
• C. $8$
• D. $9$

Answer 1

The correct option is B $7.5$

Given lines

$\frac{x+5}{4}=\frac{y+3}{-18}=\frac{z-6}{-4}$-----(1)

$\frac{x-3}{2}=\frac{y+10}{-9}=\frac{z-1}{-2}$------(2)

on comparing both lines with general cartesian form

normal vector of line

$\overrightarrow{b}=2\hat{i}-9\hat{j}-2\hat{k}$

position vector of line eq (1)

$\overrightarrow{a}=-5\hat{i}-3\hat{j}+6\hat{k}$

position vector of line (2)

$\overrightarrow{c}=3\hat{i}-10\hat{j}+\hat{k}$

$\overrightarrow{c}-\overrightarrow{a}=3\hat{i}-10\hat{j}+\hat{k}-(-5\hat{i}-3\hat{j}+6\hat{k})$

$\overrightarrow{c}-\overrightarrow{a}=3\hat{i}-10\hat{j}+\hat{k}+5\hat{i}+3\hat{j}-6\hat{k}$

$\overrightarrow{c}-\overrightarrow{a}=8\hat{i}-7\hat{j}-6\hat{k}$

$(\overrightarrow{c}-\overrightarrow{a})\times \overrightarrow{b}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 8& -7& -6\\ 2& -9& -2\end{array}\right|$

$(\overrightarrow{c}-\overrightarrow{a})\times \overrightarrow{b}=\hat{i}(14-54)-\hat{j}(-16+12)+\hat{k}(-72+14)$

$(\overrightarrow{c}-\overrightarrow{a})\times \overrightarrow{b}=-40\hat{i}+4\hat{j}-58\hat{k}$

formula to find distance between parallel lines

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

$SD=\frac{|-40\hat{i}+4\hat{j}-58\hat{k}|}{\sqrt{2^2+(-9{)}^2+(-2{)}^2}}$

$SD=\frac{\sqrt{(-40{)}^2+4^2+(-58{)}^2}}{\sqrt{4+81+4}}$

$SD=\frac{\sqrt{1600+16+3364}}{\sqrt{89}}$

$SD=\frac{\sqrt{4980}}{\sqrt{89}}$

$SD=\sqrt{55.95}$

$SD=\sqrt{55.95}$

$SD=7.5$ approx