Point-Line Distance--3-Dimensional
Let a line in three dimensions be specified by two points
and
lying on it, so a vector
along the line is given by
![]() |
(1)
|
The squared distance between a point on the line with parameter
and a point
is therefore
|
(2)
|
To minimize the distance, set
and
solve for
to obtain
|
(3)
|
where
denotes the dot
product. The minimum distance can then be found by plugging
back into (2)
to obtain
![]() |
(4)
|
|
(5)
| |
|
(6)
|
Using the vector quadruple product
|
(7)
|
where
denotes the cross
product then gives
|
(8)
|
and taking the square root results in the beautiful formula
|
(9)
| |||
|
(10)
| |||
|
(11)
|
Here, the numerator is simply twice the area of the triangle formed by points
,
, and
, and the denominator is the length of one of the bases of the
triangle, which follows since, from the usual triangle
area formula,
.
![v=[x_1+(x_2-x_1)t; y_1+(y_2-y_1)t; z_1+(z_2-z_1)t].](/National_Library/20161222123739im_/https://mathworld.wolfram.com/images/equations/Point-LineDistance3-Dimensional/NumberedEquation1.gif)
![d^2=(x_1-x_0)^2+(y_1-y_0)^2+(z_1-z_0)^2+2t[(x_2-x_1)(x_1-x_0)+(y_2-y_1)(y_1-y_0)+(z_2-z_1)(z_1-z_0)]+t^2[(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2]](/National_Library/20161222123739im_/https://mathworld.wolfram.com/images/equations/Point-LineDistance3-Dimensional/Inline9.gif)
point-line distance—3-dimensional

