# Two lines are given by the equation x = 10 - 8t, y=1+8t, z=15 +2t and x=6-8t, y=2+2t, z=3+6t What is the shortest distance between these two lines?

Two lines are given by the equation x = 10 - 8t, y=1+8t, z=15 +2t and x=6-8t, y=2+2t, z=3+6t What is the shortest distance between these two lines?
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esbalatzaj
any two skew lines lie on parallel planes, and the normal direction of these planes if found using the cross product of the two line directions.
$|\begin{array}{ccc}i& j& k\\ -8& 8& 2\\ -8& 2& 6\end{array}|=<44,32,48>$
choosing t = 0 on the first line yields the point (10,1,15) and using the normal vector <44,32,48> the plane equation is.
44(x - 10) + 32(y - 1) + 48(z - 15) = 0
44x + 32y + 48z - 1192 = 0 divide by 4.
11x + 8y + 12z - 298 = 0
since the other line is parallel to this plane then the orthogonal distance from any point on that line to this plane is the shortest distance between the two lines, t = 0 yields the point (6,2,3)
Plug into the distance equation from a point to a plane.
$D=\frac{|\left(11\right)\left(6\right)+\left(8\right)\left(2\right)+\left(12\right)\left(3\right)+\left(-298\right)|}{\sqrt{{11}^{2}+{8}^{2}+{12}^{2}}}=\frac{|-180|}{\sqrt{329}}=\frac{180}{\sqrt{329}}\cong 9.92$