Distance from a point to a plane explained
In Euclidean space, the distance from a point to a plane is the distance between a given point and its orthogonal projection on the plane, the perpendicular distance to the nearest point on the plane.
that is closest to the
origin. The resulting point has
Cartesian coordinates
:
}, \quad \quad \displaystyle y = \frac, \quad \quad \displaystyle z = \frac .The distance between the origin and the point
is
.
Converting general problem to distance-from-origin problem
Suppose we wish to find the nearest point on a plane to the point (
), where the plane is given by
. We define
,
,
, and
, to obtain
as the plane expressed in terms of the transformed variables. Now the problem has become one of finding the nearest point on this plane to the origin, and its distance from the origin. The point on the plane in terms of the original coordinates can be found from this point using the above relationships between
and
, between
and
, and between
and
; the distance in terms of the original coordinates is the same as the distance in terms of the revised coordinates.
Restatement using linear algebra
The formula for the closest point to the origin may be expressed more succinctly using notation from linear algebra. The expression
in the definition of a plane is a
dot product
, and the expression
appearing in the solution is the squared
norm
. Thus, if
is a given vector, the plane may be described as the set of vectors
for which
and the closest point on this plane to the origin is the vector
.
[1] [2] The Euclidean distance from the origin to the plane is the norm of this point,
}.
Why this is the closest point
In either the coordinate or vector formulations, one may verify that the given point lies on the given plane by plugging the point into the equation of the plane.
To see that it is the closest point to the origin on the plane, observe that
is a scalar multiple of the vector
defining the plane, and is therefore orthogonal to the plane.Thus, if
is any point on the plane other than
itself, then the line segments from the origin to
and from
to
form a
right triangle, and by the
Pythagorean theorem the distance from the origin to
is
.Since
must be a positive number, this distance is greater than
, the distance from the origin to
.
[2] Alternatively, it is possible to rewrite the equation of the plane using dot products with
in place of the original dot product with
(because these two vectors are scalar multiples of each other) after which the fact that
is the closest point becomes an immediate consequence of the
Cauchy–Schwarz inequality.
[1] Closest point and distance for a hyperplane and arbitrary point
The vector equation for a hyperplane in
-dimensional
Euclidean space
through a point
with normal vector
is
or
where
.
[3] The corresponding Cartesian form is
where
d=p ⋅ a=a1p1+a2p2+ … anpn
.
[3] The closest point on this hyperplane to an arbitrary point
is
x=y-\left[\dfrac{(y-p) ⋅ a
}\right]\mathbf=\mathbf-\left[\dfrac{\mathbf{y}\cdot\mathbf{a}-d}{\mathbf{a}\cdot\mathbf{a}}\right]\mathbfand the distance from
to the hyperplane is
\left\|x-y\right\|=\left\|\left[\dfrac{(y-p) ⋅ a
}\right]\mathbf\right\|=\dfrac=\dfrac.
[3] Written in Cartesian form, the closest point is given by
for
where
}=\dfrac,and the distance from
to the hyperplane is
\dfrac{\left|a1y1+a2y2+ … anyn-d\right|}{\sqrt{a
.
Thus in
the point on a plane
closest to an arbitrary point
is
given by
\left.\begin{array}{l}x=x1-ka\\y=y1-kb\\z=z1-kc\end{array}\right\}
where
k=\dfrac{ax1+by1+cz
2+c2}
,and the distance from the point to the plane is
\dfrac{\left|ax1+by1+cz
2+c2}}
.
See also
Notes and References
- .
- .
- Book: Cheney. Ward. Kincaid. David. Linear Algebra: Theory and Applications. 2010. Jones & Bartlett Publishers. 9781449613525. 450, 451.