Question

Find the shortest distance between the lines $\frac{x-1}{1}=\frac{y+1}{3}=z$ and $\frac{x-1}{3}=\frac{y-2}{1},z=2$

Answer 1

Line (1)$\to \frac{x-1}{1}=\frac{y+1}{3}=z$ $l_1=1,m_1=3n_1=1$

Line(2)$\to \frac{x-1}{3}=\frac{y-2}{1}$ $z=2l_2=3m_2=1n_2=0$

$S.D=\left[\begin{array}{ccc}{x}_1-y_2& y_1-y_2& z_1-z_2\\ l_1& m_1& n_1\\ l_2& m_2& n_2\end{array}\right]=\left[\begin{array}{ccc}1-1& -1-2& 0-2\\ 1& 3& 1\\ 3& 1& 0\end{array}\right]$

$S.D=\left[\begin{array}{ccc}0& -3& -2\\ 1& 3& 1\\ 3& 1& 0\end{array}\right]=3\left[\begin{array}{cc}1& 1\\ 3& 0\end{array}\right]-2\left[\begin{array}{cc}1& 3\\ 3& 1\end{array}\right]$

$S.D=-9-2(1-9)$

$S.D=-9+16$

$S.D=7writs$.