Question

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

• A. $\frac{372\sqrt{2}}{35}$
• B. $\frac{186}{35}$
• C. $\frac{372}{35\sqrt{2}}$
• D. $\frac{186}{35\sqrt{2}}$

Answer 1

Answer: (C)

Here, $x_1=5$,
$y_1=7$,
$z_1=-3$,
$a_1=4$,
$b_1=-5$,
$c_1=-5$
and $x_2=8$,
$y_2=7$,
$z_2=5$,
$a_2=7$,
$b_2=1$,
$c_2=-3$
$d=\left|\frac{\left|\begin{array}{ccc}3& 0& 8\\ 4& -5& -5\\ 7& 1& -3\end{array}\right|}{\sqrt{{\left(15+5\right)}^2+{\left(-35+12\right)}^2+{\left(4+35\right)}^2}}\right|$

$=\left|\frac{3\left(15+5\right)+8\left(4+35\right)}{\sqrt{400+529+1521}}\right|$

$=\left|\frac{60+312}{\sqrt{2450}}\right|$

$=\frac{372}{35\sqrt{2}}$