## Calculus: Early Transcendentals 8th Edition

$\frac{13}{\sqrt {69}}$
Line 1 goes from the origin and $(2,0,-1)$ Line 1 goes from $(1,-1,1)$ and $(4,1,3)$ Direction vector of first line is $v_1=\lt 2,0,-1\gt$ and Direction vector of second line is $v_2=\lt 3,2,2 \gt$ Two skew lines are lying on two parallel planes. $n=v_1 \times v_2=\lt 2,-7,4\gt$ Use the point $(1,-1,1)$ from line 2 and $n$ to form the plane equation. From the plane equation, we have $2x-7y+4z-13=0$ Use point $(0,0,0)$ for line 1. Distance formula between a point and a plane is given as: $D=\frac{|ax_1+by_1+cz_1+d|}{\sqrt {a^2+b^2+c^2}}$ $D=\frac{|2(0)-7(0)+4(0)-13|}{\sqrt {2^2+(-7)^2+4^2}}$ $=\frac{13}{\sqrt {69}}$