Find the shortest distance between the line L: x=1+t, y=1+2t, z=1-t and the point A: (2,3,4)

First, write the line in vector form r=(1 1 1) + t(1 2 -1). Consider a point P on the line such that the line connecting P and A is perpendicular to L. The vector P->A is (2 3 4) - (1 1 1) - t(1 2 -1) = (1 2 3) - t(1 2 -1).

To make P->A perpendicular to L, the dot product of P->A and the direction vector of L must be zero. This means (1 2 -1) . (1-t 2-2t 3+t)=0 so 1-t+4-4t-3-t=0 or 2-6t=0 so t=1/3. This means that P is (4/3 7/3 -2/3) and P->A is (2/3 4/3 10/3). Therefore the shortest distance between A and L is sqrt(4/9+16/9+100/9)=2sqrt(10/3).