Question

The shortest distance between the lines $\frac{x-2}{2}=\frac{y-3}{2}=\frac{z-0}{1}$ and $\frac{x+4}{-1}=\frac{y-7}{8}=\frac{z-5}{4}$ lies in the interval

• A. $[0,1]$
• B. $[1,2]$
• C. $[2,3]$
• D. $[3,4]$

Answer 1

Answer: (C)

Let, $\overrightarrow{a}=2\hat{i}+3\hat{j}$
$\overrightarrow{b}=-4\hat{i}+7\hat{j}+5\hat{k}$ $\Rightarrow \overrightarrow{b}-\overrightarrow{a}=-6\hat{i}+4\hat{j}+5\hat{k}$
$\overrightarrow{c}=2\hat{i}+2\hat{j}+\hat{k}$ $\overrightarrow{d}=-\hat{i}+8\hat{j}+4\hat{k}$
$\Rightarrow \overrightarrow{c}\times \overrightarrow{d}=-9\hat{j}+18\hat{k}$
Hence, shortest distance $=\left|\frac{\left(\overrightarrow{b}-\overrightarrow{a}\right)\cdot \left(\overrightarrow{c}\times \overrightarrow{d}\right)}{\left|\overrightarrow{c}\times \overrightarrow{d}\right|}\right|$
$=\left|\frac{-36+90}{9\sqrt{5}}\right|=\frac{6}{\sqrt{5}}\in (2,3]$