Distance of a Point from a Hyperplane

   Here we state and prove a formula which allows us to compute the distance of a point from a hyperplane.

Proposition   Let    be the hyperplane of all  x  in    with

(x p) n 0,     p, n  in  ,   n   0.

Then the distance of a point  , with position vector  q, from    is

dist(,)    |(q – p) n|
|n|

The point    on    which is closest to  , has position vector

r   q –  (q – p) n  · n
n n

Proof   While we illustrate the argument in  , the argument itself is valid in all positive dimensions.

   We are given the plane    through    and perpendicular to n. In order to compute the distance of from , the key observation is that

dist(,) | · n|,

where is chosen so that  r + · n .

   To find  |n|, we write  q p + z + n, with  z n. Consequently,

n q p z
|n| (q p z) n (q p) n
|||n| |(q p) n|
dist(,) |n| (1/|n|)|(q p) n|

   It remains to verify that    is indeed the point in    which is closest to  , and to determine the position vector of  . Now we know that the point    belongs to  , and we know its position vector is

r q n q ((q p) n) (n/n n).

Inspection of the picture suggests that    is be the point on    which is closest to . But how can we be sure that what is suggested by a specific 2-dimensional sketch holds true for all choices of    and  and applies in all dimensions? - We must to prove that this is indeed the case!

Let  x  be the position vector of an arbitrary point    on  . Then we can break the vector  q x  joining  x  to  q up into

q x    u + tn,

where  u n. Then we find that the distance of    from    satisfies

dist(,) (q x) (q x)
  (u + n) (u + n)
  |u| + |n|
  |n| dist(,)

Moreover, we see that the distance of    from    is equal to the distance of    from    exactly when the vector  u  is the 0-vector. But this means  , and that’s exactly what we wanted to show.