# Find the distance between the skew lines r_(1)(t) =(1; 1;0) + (1, 6,2) t and r_(2)(s) =(1, 5, 2) +(2, 15, 6)s.

Find the distance between the skew lines
${r}_{1}\left(t\right)=\left(1;1;0\right)+\left(1,6,2\right)t$ and ${r}_{2}\left(s\right)=\left(1,5,2\right)+\left(2,15,6\right)s$.
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dasse9
We really need to find the distance between parallel planes.
any two skew lines lie on parallel planes, the normal direction of those planes is orthogonal to both lines thus use cross product to find it.
$|\begin{array}{ccc}i& j& k\\ 1& 6& 2\\ 2& 15& 6\end{array}|=<6,-2,3>$
choosing s = 0 from the second line we find the point (1, 5, 2) andusing this point with the normal direction of the plane we find theequation of its plane.
6(x-1)-2(y-5)+3(z-2)=0
6x-2y+3z-2=0
now choosing the point where t = 0 on the first line we have the point (1, 1, 0)
from the equation for the distance from a point to a plane.
$D=\frac{|a{x}_{1}+b{y}_{1}+c{z}_{1}+d|}{\sqrt{{a}^{2}+{b}^{2}+{c}^{2}}}$
where a, b, c, and d are 6, -2, 3, and -2 respectively (from theplane equation) and $\left({x}_{1},{y}_{1},{z}_{1}\right)=\left(1,1,0\right)$ (from the point)
$D=\frac{|\left(6\right)\left(1\right)+\left(-2\right)\left(1\right)+\left(3\right)\left(0\right)+\left(-2\right)|}{\sqrt{{6}^{2}+{2}^{2}+{3}^{2}}}=\frac{2}{\sqrt{49}}=\frac{2}{7}$