Vectors- How to find point on a plane closest to a point

Question

I'm really stuck with an Applied Maths problem related to vectors. The problem is as follows:

In a three-dimensional cartesian coordinate system, four points are given by their coordinates as follows: $$P = (−3, 2, 1), Q = (4, 0, −1), R = (−2, 4, 6), S = (1, −3, 1)$$ Find the point on the plane through Q and spanned by QS and QR which is closest to P.

Does anyone have any idea how to solve this problem? I would really appreciate some help! Thanks in advance

Answer 1

HINT

We are looking for the projection $P'$ of $P$ onto the plane $QSR$ and we can proceed as follow

• find the plane equation $QSR$
• find the line equation through $P$ perpendicular to the plane $QSR$
• find the intersection $P'$ between the line and the plane

Answer 2

A point on the plane is given by $Q+\lambda\,QS+\mu\,QR$, and the squared distance to $P$ is given by

$$d^2=(PQ+\lambda\,QS+\mu\,QR)^2.$$

You minimize this expression by cancelling the gradient, which gives

$$QS\cdot(PQ+\lambda\,QS+\mu\,QR)=QR\cdot(PQ+\lambda\,QS+\mu\,QR)=0.$$

This is a linear system in $\lambda,\mu$, the resolution of which is straightforward. The position of the requested point follows.

Answer 3

The nearest point to $P$ is its orthogonal projection onto the plane. One way to compute this point is to find the projection of $P-Q$ onto a normal to the plane and subtract this from $P$. Since you’re working in $\mathbb R^3$, you can find a normal to the plane via a cross product.