Find a vector that is normal to lines L1 and L2 and passes through their common point of intersection where L1 is the line r = (3,1,1) + u(1,-2,-1) and L2 is the line r = (0,-2,3) + v(-5,1,4) where u and v are scalar values.

We will use the cross product of these two vectors to help us find the perpendicular vector. The cross product works by finding a vector which has no effect on the position of a point with respect the the directions of the other 2 vectors. Its like looking down at a ball on top of a table. Moving the ball up and down won't affect its movement in the plane of the table. We use a 3x3 matrix to find the cross product of the two vectors and find the determinant of that matrix. We define it as: (i j k a b c d e f) where abc is the direction vector of L1 and def is the direction vector of L2. The determinant is then i(bf-ce) -j(af-cd) + k(ae-bd) or in the case of the question i(-7) -j(-1) + k(-9) or -7i + j -9k. To solve for their point of intersection we set the r values equal as this will find the common point. We will solve this equation simultaneously. Therefore (3+u) = (-5v), (1-2u) = (-2+v) and (1-u) = (3+4v) solving by adding equations one and two gives us 4 = 3-v so v = -1. Given we are told the lines intersect we can now use this value to find the point of intersection which will be where v = -1 in the second equation. So the point of intersection is (5,-3,-1). Therefore our final answer is (5,-3,-1) + w((-7,1,-9) where w is a scalar value.