# Find the shortest distance between the lines

Question:

Find the shortest distance between the lines $\frac{x+1}{7}=\frac{y+1}{-6}=\frac{z+1}{1}$ and $\frac{x-3}{1}=\frac{y-5}{-2}=\frac{z-7}{1}$

Solution:

The given lines are $\frac{x+1}{7}=\frac{y+1}{-6}=\frac{z+1}{1}$ and $\frac{x-3}{1}=\frac{y-5}{-2}=\frac{z-7}{1}$

It is known that the shortest distance between the two lines, $\frac{x-x_{1}}{a_{1}}=\frac{y-y_{1}}{b_{1}}=\frac{z-z_{1}}{c_{1}}$ and $\frac{x-x_{2}}{a_{2}}=\frac{y-y_{2}}{b_{2}}=\frac{z-z_{2}}{c_{2}}$, is given by,

$d=\frac{\left|\begin{array}{ccc}x_{2}-x_{1} & y_{2}-y_{1} & z_{2}-z_{1} \\ a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2}\end{array}\right|}{\sqrt{\left(b_{1} c_{2}-b_{2} c_{1}\right)^{2}+\left(c_{1} a_{2}-c_{2} a_{1}\right)^{2}+\left(a_{1} b_{2}-a_{2} b_{1}\right)^{2}}}$          ...(1)

Comparing the given equations, we obtain

$x_{1}=-1, y_{1}=-1, z_{1}=-1$

$a_{1}=7, \quad b_{1}=-6, \quad c_{1}=1$

$x_{2}=3, \quad y_{2}=5, \quad z_{2}=7$

$a_{2}=1, \quad b_{2}=-2, c_{2}=1$

Then, $\left|\begin{array}{ccc}x_{2}-x_{1} & y_{2}-y_{1} & z_{2}-z_{1} \\ a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2}\end{array}\right|=\left|\begin{array}{lll}4 & 6 & 8 \\ 7 & -6 & 1 \\ 1 & -2 & 1\end{array}\right|$

$=4(-6+2)-6(7-1)+8(-14+6)$

$=-16-36-64$

$=-116$

$\Rightarrow \sqrt{\left(b_{1} c_{2}-b_{2} c_{1}\right)^{2}+\left(c_{1} a_{2}-c_{2} a_{1}\right)^{2}+\left(a_{1} b_{2}-a_{2} b_{1}\right)^{2}}=\sqrt{(-6+2)^{2}+(1+7)^{2}+(-14+6)^{2}}$

$=\sqrt{16+36+64}$

$=\sqrt{116}$

$=2 \sqrt{29}$

Substituting all the values in equation (1), we obtain

$d=\frac{-116}{2 \sqrt{29}}=\frac{-58}{\sqrt{29}}=\frac{-2 \times 29}{\sqrt{29}}=-2 \sqrt{29}$

Since distance is always non-negative, the distance between the given lines is $2 \sqrt{29}$ units.