Finding the Closest Point on a Plane to a Given Point

[Char. Limit]

Homework Statement

Find the point on the plane $x-y+z=4$ closest to the point (1,2,3).

$$d=\frac{2}{\sqrt{3}}$$

Homework Equations

Hmm...

The Attempt at a Solution

As you can see, I've already solved for the shortest distance. But, knowing the plane I'm on and the distance, how do I find the point that lies on that plane, at that distance from (1,2,3)? Help...

 

[Inferior89]

Embed the point into a plane with the same normal vector as the given plane.

The shortest distance from the origin to any plane $$ax +by +cz = d$$ is
$$\frac{|\mathbf{n}|}{d}$$ where $$\mathbf{n}$$ is the normal vector to the plane.

Do this for both the given plane and your new plane from and calculate the difference.