## Calculus 8th Edition

Skew, $\frac{1}{\sqrt 6}$
Direction vector of first line is$v_1=\lt 1,1,1\gt$ and Direction vector of second line is$v_2=\lt 1,2,3\gt$ so lines are not parallel because the components are not proportional. For line 1: $x=t,y=t,z=t$ For line 2: $x=-1+s,y=2s,z=3s$ Set the corresponding $x,y,z$ equations equals to each other. $s=-1$ and $t=2(-1)=-2$ and $-2\ne 3(-1) = -3$ Thus, there is no solution so the lines do not intersect.Hence, they are skew. Distance formula between a point and a plane is given as: $D=\frac{|ax_1+by_1+cz_1+d|}{\sqrt {a^2+b^2+c^2}}$ $D=\frac{|1(0)-2(0)+1(0)+1|}{\sqrt {1^2+(-2)^2+1^2}}$ $=\frac{1}{\sqrt 6}$