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

Answer 1

Answer: (C)

Shortest distance
$\frac{\begin{vmatrix}x_{2}-x_{1}&y_{2}-y_{1}&z_{2}-z_{1}\\ a_{1}&b_{1}&c_{1}\\ a_{2}&b_{2}&c_{2}\end{vmatrix}}{\begin{Vmatrix}i&j&k\\ a_{1}&b_{1}&c_{1}\\ a_{2}&b_{2}&c_{2}\end{Vmatrix}}$
$=\frac{\begin{vmatrix}-2&4&5\\ 2&2&1\\ -1&3&4\end{vmatrix}}{\begin{vmatrix}i&j&k\\ 2&2&1\\ -1 &8&4\end{vmatrix}}=\frac{254}{20.1}=2.7$