Let a line in three dimensions be specified by two points 
and 
lying on it, so a vector along the line is
given by
![v=[x_1+(x_2-x_1)t; y_1+(y_2-y_1)t; z_1+(z_2-z_1)t].](/images/equations/Point-LineDistance3-Dimensional/NumberedEquation1.svg) |
(1)
|
The squared distance between a point on the line with parameter 
and a point 
is therefore
![d^2=[(x_1-x_0)+(x_2-x_1)t]^2+[(y_1-y_0)+(y_2-y_1)t]^2+[(z_1-z_0)+(z_2-z_1)t]^2.](/images/equations/Point-LineDistance3-Dimensional/NumberedEquation2.svg) |
(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
![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]](/images/equations/Point-LineDistance3-Dimensional/Inline9.svg) |
(4)
|
![=|x_1-x_0|^2-2([(x_1-x_0)·(x_2-x_1)]^2)/(|x_2-x_1|^2)+([(x_1-x_0)·(x_2-x_1)]^2)/(|x_2-x_1|^2)](/images/equations/Point-LineDistance3-Dimensional/Inline10.svg) |
(5)
|
![=(|x_1-x_0|^2|x_2-x_1|^2-[(x_1-x_0)·(x_2-x_1)]^2)/(|x_2-x_1|^2).](/images/equations/Point-LineDistance3-Dimensional/Inline11.svg) |
(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, 
.
