Distance calculation of arbitrary point to super plane in space [reprint]

See link in the original: http://http//bubblexc.com/y2011/310/slightly revised.

Straight, flat

Before talking about the super plane, let's talk about the straight line and the surface in Rn space. A point p and a nonnegative vector vin the given Rn space? Meet

　　　　　　　　　　　　　　　　i=TV? +P

The set of point I is called a straight line in Rn space.  T in the above formula is a scalar, vector v? Determines the direction of the line. 1 is shown below:

Fig. 1:line Figure illustration

Relative, a point p in the given Rn space and two linearly independent vector v? ,W? Meet

　　　　　　　　　　i=TV? +SW? +P

The set of point I is called a plane in Rn space. In the upper-middle t,s are scalar. 2 is shown below:

Fig. 2:plane Figure illustration

More generally, a point in the given Rn space p and linearly independent vector v1→,v2→,...,vk→, satisfies

　　　　　　I=t1v1→+T2v2→+...+tkvk→+p

The set of point I is called a K-dimensional affine subspace of Rn space (k-dimensional
Affine subspace). Thus, a straight line is a 1-dimensional affine subspace, and a plane is a 2-dimensional affine subspace.

·
Another representation of a line

Suppose the point set in R2 space i= (x,y) satisfies the equation

Ax+by +D= 0 (1)

Where a,b,D are scalar, and a,b at least one is not 0. Assuming b is not 0, the

y=−abx−db

Set x=t,−∞<t<∞, then the point set I can be expressed as

I= (x,y) = (t, −abt−db) =t(1,−AB ) + (0,−db)

This is actually a straight line L with a point (0,−db) direction (1,−ab).

Further, we set n? = (a,b), the (1) formula can be expressed as

n? ∗i+D= 0 (2)

Set p= (p1,P2) to a point on the line, substituting (2) in the formula I, can get d=−n? ∗P, the (2) type can be expressed as

n? ∗ (i−p) =0 (3)

can be seen, n?  is actually the normal vector of the straight L , and the point set i= (x,y) is those with the p difference vector with n? The orthogonal point.

·
Super plane

Having said so much, now comes the definition of the ultra-plane: A point p in the given Rn space and a nonzero vector n? Meet

n? ∗ (i−p) =0 (4)

The point set i is referred to as the hyper plane passing the point p .  Vector n? is the normal vector for the hyper-plane. According to this definition, a straight line is The super plane of R2 space, a plane is the super Plane of R 3 space,the super plane of Rn space is rn Space
A n-1-dimensional affine subspace. set n? = (a1,a2,...,an),i= (i1,i2,...,in), The (4) formula can be expressed as

a1i1+a2i2+...+anin+D= 0 (5)

Which,d=−n? ∗P

It is important to take advantage of a super plane where we can divide the point of space into two parts (the value of the formula (4) is greater than or equal to 0 or less than 0).  At the same time, the use of formula (4) We can calculate the aspect of the space within a point to the super-plane distance: Set the space of a point Q,Q to the super-plane distance is q−p in vector n? (3), as shown in the projection. The distance between the Q point and the super Plane H is expressed as:

D = | (q−p) ∗u|=| q∗n? −p∗n? | / ||  N| | = | q∗n?  +D| /  ||   N| |  where u?  is n? Vector unit vector, | | N| | As a model of vectors, the L2 paradigm is expanded into algebra.

We can calculate the distance from Q to the plane.

Distance calculation of arbitrary point to super plane in space [reprint]