Question

The shortest distance between the lines $\frac{x-5}{4}=\frac{y-7}{-5}=\frac{z+3}{-5}$ and $\frac{x-8}{4}=\frac{y-7}{-5}=\frac{z-5}{-5}$ is

• A. $45$
• B. $46$
• C. $47$
• D. $48$

Answer 1

The correct option is B $46$

Given eq of lines

$\frac{x-5}{4}=\frac{y-7}{-5}=\frac{z+3}{-5}$-----(1)

$\frac{x-8}{4}=\frac{y-7}{-5}=\frac{z-5}{-5}$-----(2)

position vector of line (1)

$\overrightarrow{a}=5\hat{i}+7\hat{j}-3\hat{k}$

position vector of line (2)

$\overrightarrow{c}=8\hat{i}+7\hat{j}+5\hat{k}$

normal vector

$\overrightarrow{n}=4\hat{i}-5\hat{j}-5\hat{k}$

shortest between two parallel line

$SD=\frac{|\overrightarrow{AC}\times (\overrightarrow{n}\left)\right|}{\left|\overrightarrow{n}\right|}$

$AC=\overrightarrow{c}-\overrightarrow{a}$

$AC=8\hat{i}+7\hat{j}+5\hat{k}-(5\hat{i}+7\hat{j}-3\hat{k})$

$AC=3\hat{i}+8\hat{k}$

$\left|\overrightarrow{n}\right|=\sqrt{4^2+(-5{)}^2+(-5{)}^2}$

$\left|\overrightarrow{n}\right|=\sqrt{66}$

putting AC $\overrightarrow{n},\left|\overrightarrow{n}\right|$ in formula

$SD=\frac{|\overrightarrow{AC}\times (\overrightarrow{n}\left)\right|}{\left|\overrightarrow{n}\right|}$

$SD=\frac{\left|\right(3\hat{i}+8\hat{k})\times (4\hat{i}-5\hat{j}-5\hat{k}\left)\right|}{\sqrt{66}}$

$SD=\frac{\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 3& 0& 8\\ 4& -5& -5\end{array}\right|}{\sqrt{66}}$

$SD=\frac{\left|\hat{i}\right(0+40)-\hat{j}(-15-32)+\hat{k}(-15-0\left)\right|}{\sqrt{66}}$

$SD=\frac{|40\hat{i}+47\hat{j}-15\hat{k}|}{\sqrt{66}}$

$SD=\frac{\sqrt{{40}^2+{47}^2+(-15{)}^2}}{\sqrt{66}}$

$SD=\frac{\sqrt{1600+2209+225}}{\sqrt{66}}$

$SD=\frac{\sqrt{3834}}{\sqrt{66}}$

ans is aprroximately 46

hence option B is correct