thanks. is this correct?
v=(-2,1,0) is a vector parallel to the line so it a normal vector to plane containing the point(1,4,5).so using the components of the normal vector and the point(1,4,5) i got equation of the plane as -2x+y-2=0. The parametric equations of the line i got as x=3-2t, y=t, z=1. substituting these values of x,y and z into the plane equation and solving for t=7/5. plugging this value of t back into the parametric equations gives the point.
I have a primitive (non-vector, non-calculus) solution . . .
Let be the line that passes through the points and .
Find the point on the line that is closest to the point
The direction vector of is: .
The parametric equations of line are: .
Let be any point on line
We will minimize:
This is an up-opening parabola; its minimum occurs at its vertex.
Vertex formula: .
We have: .
Hence, the vertex is at: .