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 $x_2= 8$,
$y_2 = 7$,
$z_2 = 5$,
$a_2 = 7$,
$b_2 = 1$,
$c_2 = -3$
$d=\left|\frac{\begin{vmatrix}3&0&8\\ 4&-5&-5\\ 7&1&-3\end{vmatrix}}{\sqrt{\left(15+5\right)^{2}+\left(-35+12\right)^{2}+\left(4+35\right)^{2}}}\right|$

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

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

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