Hyperplane - Description

   Here we show why an equation in variables , ... , of the form

+ + ... +

describes a hyperplane in . - First, here is the precise statement.

Proposition   Given a non-zero vector  n  in    and a point    with position vector  p, the hyperplane perpendicular to  n  through    is the set of all those  x  in    with

(x p) n 0.

Proof   We use the picture below to guide our argument.

If  x  is the position vector of an arbitrary point in  , let  y x p, as shown. Then   x  belongs to the hyperplane perpendicular to  n  through    if and only if  y is perpendicular to  n; i.e. if and only if

(x p) n 0.

This is what we wanted to show.

   So how are the solutions of  (x – p) n 0  the same as the solutions of the equation in variables

+ + ... + ?

Answer   we rewrite the above equation using coordinates of the vectors x, p, and n:

x (, ... ,),    n (, ... ,),    p (, ... ,).

Then we obtain

(x – p) n    x n – p n
     ( + ... + ) – ( + ... + )

The symbols  , ... , and , ... ,  represent fixed numbers. So we obtain a constant

+ ... + .

With this, the equation  (x – p) n 0  is now seen to be equivalent to

+ + ... + .

This is what we wanted to show.