Point-Line Distance--3-D

A line in 3-D is given by the parametric Vector

$${\bf v}=\left[{\matrix{x_0+at\cr y_0+bt\cr z_0+ct\cr}}\right].$$ (1)

The distance between a point on the line with parameter $t$ and the point $(x_1, y_1, z_1)$ is therefore

$$r^2=(x_1-x_0-at)^2+(y_1-y_0-bt)^2+(z_1-z_0-ct)^2.$$ (2)

To minimize the distance, take

$${\partial (r^2)\over \partial t} = -2a(x_1-x_0-at)-2b(y_1-y_0-bt) -2c(z_1-z_0-ct) = 0$$ (3)

$$a(x_1-x_0)+b(y_1-y_0)+c(z_1-z_0)-t(a^2+b^2+c^2)=0$$ (4)

$$t={a(x_1-x_0)+b(y_1-y_0)+c(z_1-z_0)\over a^2+b^2+c^2},$$ (5)

so the minimum distance is found by plugging (5) into (2) and taking the Square Root.

See also Line, Point, Point-Line Distance--2-D