Question

Find the shortest distance between the lines $\frac{x-1}{1}=\frac{y-2}{-1}=\frac{z-1}{1}$ and $\frac{x-2}{2}=\frac{y+1}{1}=\frac{z+1}{2}$.

Answer 1

$SD=\frac{\left|\begin{array}{ccc}{x}_2-x_1& y_2-y_1& z_2-z_1\\ l_1& m_1& n_1\\ l_2& m_2& n_2\end{array}\right|}{\sqrt{(m_1-n_2-m_2-n_1{)}^2+(n_1-l_2-n_2-l_1{)}^2+(l_1{m}_2-l_2{m}_1{)}^2}}$

$=\frac{\left|\begin{array}{ccc}2-1& -1-2& -1-1\\ 1& -1& 1\\ 2& 1& 2\end{array}\right|}{\sqrt{\left(\right(-1\times 2)-(1\times 1){)}^2+(1\times 2-2\times 1{)}^2+\left(\right(1\times 1)-2(1){)}^2}}$

$\frac{\left|\begin{array}{ccc}1& -3& -2\\ 1& -1& 1\\ 2& 1& 2\end{array}\right|}{\sqrt{9+0+1}}$

$=\frac{1}{\sqrt{10}}\left\{1\right(-2-1)+3(2-2)-2(1+2\left)\right\}$

$=\left|\frac{1}{\sqrt{10}}\right\{-3+0.6\left)\right\}|$

$=\frac{9}{\sqrt{10}}$