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