One normal vector of the plane x-5y+2z=-1 is n1=<1, -5, 2> and one normal vector of the plane x+2z=0 is n2=<1,0,2>. Note that the intersection line of the planes x-5y+2z=-1 and x+2z=0 is perpendicular to both n1 and n2. So we can choose n=n1Xn2, the cross product of n1 and n2, as a direction vector of the intersection line.
| i j k |
Now n=<1,-5,2>X<1,0,2>= | 1 -5 2 | = (-10-0)i - (2-2)j +(0+5)k = <-10, 0 ,5> is a direction vector
| 1 0 2 |
of the line. Furthermore, we need to find one point on the line. To do so, we can set z=0 and get a system of linear equations in x and y: x-5y=-1 and x=0. Therefore, x=0, y=1/5. As a result, the point (0, 1/5, 0) is a point on the intersection line. Hence the vector form of the line is
r(t) = (0, 1/5, 0 ) + t(-10, 0, 5).
