Coordinates of a projected point to a plane ?

Question

Given that the equation of the plane is ax + by + c + d = 0,
and given a point p:(p1, p2, p3) which does not lie on this plane, what are the coordinates of the projected point p' on the plane.

To clarify: Imagine that you dropped a perpendicular from p to the plane. I would like to know the coordinates of the foot of the perpendicular ...

Answer 1

I found a website with good information about this problem (see reference). Ultimately, you can get the point in question by calculating the distance from the point to the plane, then multiplying that distance by the unit normal of the plane and subtracting the resulting vector from the point.

The distance, D, from the point to the plane, as defined in your question, will be D = (a*p1 + b*p2 + c*p3 + d) / sqrt(a^2 + b^2 + c^2). The normal to the plane is (a, b, c), so the unit normal is (a / sqrt(a^2 + b^2 + c^2), b / sqrt(a^2 + b^2 + c^2), c / sqrt(a^2 + b^2 + c^2)). These expressions are a bit ungainly, but they would have simple numerical values if you knew a, b, and c. But if the unit normal is referred to as u (which would be written in bold if we had rich text here on Answers) and we use the original point p and distance D calculated previously, the projected point is p - Du.

It's worth noting that the formula for D gives you a value that may be positive or negative, which is relative to the normal of the plane. So the formula automatically accounts for the point being "above" or "below" the plane (again relative to the normal), and you don't need to worry about that while calculating your answer.

Answer 2

Come on, sweating over school assignments is good for you.