This writeup explains how to find the shortest distance
from a point
to a plane
, when both the point and the plane are define
d by vector
s, by an example
Suppose that there is a plane with scalar
This writeup explains how to find the shortest distance
from a point
to a plane
, when both the point and the plane are define
d by vector
s, by an example
Suppose that there is a plane with scalar equation:
r.(2i - j - 3k) = 13
and we want to find the shortest distance from a point P with position vector (3i + j + 16k) to this plane.
First, find the perpendicular distance of the plane from the origin. To do this, consider the position vector r of an arbitrary point on the plane. Now imagine a vector that is perpendicular to the plane and passes through the origin. (I would recommend drawing a diagram.) It is now evident that the vector r, that is the vector from the origin to the arbitrary point on the plane, is the hypotenuse of a triangle with its base on the plane, and the remaining side formed by the perpendicular vector. This perpendicular vector is called n.
Looking at the triangle, we can see that the perpendicular distance of the plane from the origin, call it OA, is |r|cos θ.
Now, using the dot product:
r.n = |r||n| cos θ
which we can rearrange to say that:
|r|cos θ = (r.n)/|n|
So: OA = (r.n)/|n|
The general form of the scalar equation of a plane is r.n = a.n = ρ
OA = ρ/|n|
From the equation of the plane given at the beginning, we know that ρ = 13. We also know that n = 2i - j - 3k. Therefore |n| = sqrt(4 + 1 + 9) = sqrt(14).
Hence, OA = 13/sqrt(14).
Now we need to form a plane containing the point P that is parallel to our original plane. As the planes are parallel, the vector n must be the same for both