The intersection point of the two lines is (2,1,0). The respective direction vectors are L1: <1,2,4> L2: <4,1,15> Since the normal vector of the required plane is perpendicular to both direction vectors, the normal vector of the plane is obtained by the cross product of L1 and L2. i j k 1 2 4 4 1 15 =<30-4, 16-15, 1-8> =<26, 1, -7>
We know that the plane must pass through (2,1,0), the equation of the plane is
26(x-2)+1(y-1)-7(z-0)=0 simplifying, 26x+y-7z=52+1+0=53 or 26x+y-7z=53
Check: Put points on L1 in the plane 26(t+2)+(2t+1)-7(4t+0)=53 ok For L2, 26(4t+2)+(t+1)-7(15t+0)=53 ok
