Question 1049882
 2x-2y+3z-12 = 0 = 2x+2y+z represent the line that is the intersection of the two planes 2x-2y+3z-12 = 0 and  2x+2y+z = 0.  Similarly,  2x-z = 0 = 5x-2y+9 represent the intersection of the planes 2x-z = 0 and 5x-2y+9 = 0.
 

In the first pair of planes, solving for x and y in terms of z gives x + z = 3 and y - z/2 = -3.  The symmetric equation of the line of intersection is thus {{{(x-3)/(-1) = (y+3)/(1/2) = z}}}.  This line {{{L[1]}}} has direction vector < -1,1/2,1 >, or < -2,1,2 > with the fraction cleared, and passes through the point (3,-3,0).


The second pair of planes would give a line of intersection with symmetric equation {{{x/(1/2) = (y-9/2)/(5/4) = z}}}, with direction vector < 1/2, 5/4,1 >, or < 2,5,4 > with the fractions cleared.  This line {{{L[2]}}} line passes through the point (0,9/2,0).


The vector {{{v[o]}}} starting at the point (0,9/2,0) of {{{L[2]}}} and terminating at the point (3,-3,0) of  {{{L[1]}}} is given by < 3,-15/2,0 >, or < 6,-15,0 >, with the fraction cleared.


Now a vector N perpendicular to both the direction vectors of the two intersection lines is < 1,-2,2 >.  (This can be obtained either by getting the cross product of the two direction vectors, or by solving simultaneously the system < a,b,c >*< -2,1,2 > = 0 and < a,b,c >*< 2,5,4 > = 0 for a,b, and c.)


The distance between the two lines is then just the scalar projection of the vector  {{{v[o]}}} onto the vector N:


{{{(v[o]*N)/abs(N) = (6+30+0)/sqrt(1+4+4) = 36/3 = highlight(12)}}}.