The closest point in the plane

[bezgin]

Let x + y + 3 z = 7 represent a plane. (it does)

We find the closest point to origin in this plane by [d/[n]^2] * n. In this case n = (1,1,3); d = 7; [n]^2 = 1^2 + 1^2 + 3^2 = 11; then the vector that gives us the closest point is: (7/11, 7/11, 21/11)

I don't understand WHY this operation gives us the closest point and Strang's book doesn't really explain. I'd appreciate if you help.

 

[HallsofIvy]

The line from the origin to the "closest point" in the plane is perpendicular to the plane (to see that, remember that the hypotenuse of a right triangle is always longer than the two legs).

A vector perpendicular to the given plane is <1, 1, 3>. Any line perpendicular to the plane must be of the form x= t+ a, y= t+ b, z= 3t+ c. Take t= 0 to be the value when the line passes through the origin: x= 0+ a= 0 , y= 0+ b= 0, z= 0+ c= 0 so the line is x= t, y= t, z= 3t.

Where does that line intersect the given plane? Replacing x, y, z, by t, t, 3t, respectively in the equation of the plane, t+ t+ 3(3t)= 11t= 7 so t= 7/11. That gives
x= 7/5, y= 7/5, z= 21/5 or (7/5, 7/5, 21/5) as the nearest point just as the formula does.

More generally, suppose the equation of the plane is Ax+ By+ Cz= d. The line perpendicular to that plane AND passing through the origin is x= At, y= Bt, z= Ct. Putting those into the equation of the plane, we had A²t+ B²t+ C²t= d so t= d/(A²+ B²+ C²) or, using your notation,t= d/[n]² for the value of t where that normal line intersects the plane. Putting that into the equations of the line gives x= [d/[n]²]A, y= [d/[n]²]B, z= [d/[n]²]C, once again, exactly the formula's [d/[n]²]*n.

 

[M98Ranger]

The operation of [d/[n]^2] * n allows us to find the point on the plane that is closest to the origin by using the normal vector (n) of the plane. This is because the normal vector is perpendicular to the plane and therefore, the shortest distance from the origin to the plane will be along this vector.

The value of d in this equation represents the signed distance from the origin to the plane. By dividing it by the squared length of the normal vector, we are essentially scaling the normal vector to a unit vector (with length 1). This allows us to easily move along the normal vector in the direction of the plane.

Multiplying this scaled normal vector by the normal vector itself gives us the coordinates of the point on the plane that is closest to the origin. In this case, the coordinates are (7/11, 7/11, 21/11).

I understand that this may seem like a mathematical formula without much explanation, but it is a common method used in linear algebra to find the closest point on a plane. If you are interested in learning more, I suggest looking into the concept of projection in vector spaces. This will give you a better understanding of why this operation works.