Question

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

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

Answer 1

The correct option is D $(2,3]$

Writing the equation of Line in Vector Form

$r_1=\lambda (2i+2j+k)$
$r_2=(-2i+4j+5k)+\lambda (-i+8j+4k)$
$a_2-a_1=(-2i+4j+5k)$
$b_1\times b_2=(2i+2j+k)\times (-i+8j+4k)$
$b_1\times b_2=-9j+18k$

Now,

$(a_2-a_1).(b_1\times b_2)=(-2i+4j+5k).(-9j+18k)=54$
$|b_1\times b_2|=9\sqrt{5}$
Shortest Distance $d=|\frac{(a_2-a_1).(b_1\times b_2)}{|(b_1\times b_2)|}|$

Hence, $d=\frac{54}{9\sqrt{5}}=\frac{6}{\sqrt{5}}$ which lies between $(2,3]$