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The orthogonal projection of the origin onto the line occurs at $t_{\ast }=-p\cdot u/|u{|}^2=-1/5$, giving closest point $(2,4/5,-2/5)$. Hence the distance from a point to a line in three dimensions is

The common real variable gives the parametric equation of the line:Write $p=(2,-1,4)$ and $u=(1,3,2)$. The point closest to the origin is orthogonal to $u$, so $t_{\ast }=-p\cdot u/|u{|}^2=-7/14=-1/2$. Its position is $(3/2,-5/2,3)$. The distance from a point to a line in three dimensions is therefore

From the line point $r_0=(1,-1,0)$ to $P=(2,2,1)$, take $w=P-r_0=(1,3,1)$ and line direction $d=(2,3,1)$. Orthogonal projection gives the nearest parameter $t=(w\cdot d)/(d\cdot d)=12/14=6/7$. The perpendicular displacement has squared Euclidean normHence the distance from a point to a line in three dimensions isThe nearest point is $(19/7,11/7,6/7)$; subtracting it from $P$ gives a vector orthogonal to $d$, verifying the minimum.

Assume the two line-defining points are distinct. Put $\mathbf{u}={\mathbf{r}}_1-{\mathbf{r}}_0$ and $\mathbf{v}={\mathbf{r}}_2-{\mathbf{r}}_0$. The triangle formed by the three points has area $|\mathbf{u}\times \mathbf{v}|/2$. The same area is half the base length $|{\mathbf{r}}_2-{\mathbf{r}}_1|$ times the perpendicular height $D$ from the third point to the line. Hence the distance from a point to a line in three dimensions isThis remains valid when the nearest point lies outside the segment, since the base line is infinite. If all three points are collinear, the numerator and distance are both zero.