Method 1: Perpendicular, point to 3d line.

The essence of method 1 is to find the intersection of the line and the plane orthogonal to the line where that plane contains the external point. See the figure in the example below. We are given both the line equation and the point, $P.$ We will need to know one point on the line. It can be deduced from the line equation, or we can input any scalar value $t,$ and calculate a point. Call it $A.$

1. Name a point on the line. Call it $A.$ Get its coordinates.
2. Identify the direction vector of the line. Call it $\mathbf{u}.$ If the line was given as being between points $A$ and $B,$ then \[ \mathbf{u}=(B-A)=\left(\begin{array}{c} B_{x}-A_{x}\\ B_{y}-A_{y}\\ B_{z}-A_{z} \end{array}\right). \]
3. Write the line equation. Even if it was previously given. \begin{equation} \left(\begin{array}{c} x\\ y\\ z \end{array}\right)=\left(\begin{array}{c} A_{x}\\ A_{y}\\ A_{z} \end{array}\right)+t\cdot\left(\begin{array}{c} u_{x}\\ u_{y}\\ u_{z} \end{array}\right).\label{eq:eqn for t} \tag{"the line equation"} \end{equation}
4. Write the plane equation that has normal, $\mathbf{u}$ and passes through $P.$ \[ \mathbf{X}\cdot\mathbf{u}=\mathbf{u}\cdot P. \] That is, \[ \left(\begin{array}{c} x\\ y\\ z \end{array}\right)\cdot\left(\begin{array}{c} u_{x}\\ u_{y}\\ u_{z} \end{array}\right)=\left(\begin{array}{c} u_{x}\\ u_{y}\\ u_{z} \end{array}\right)\cdot\left(\begin{array}{c} P_{x}\\ P_{y}\\ P_{z} \end{array}\right). \]
5. Find the intersection of the line and plane. That point will be the place on the line which when connected to point $P$ forms an orthogonal segment. The length of that segment is the distance from $P$ to the line.
1. Substitute the 3 equations from the line equation into the plane equation. Remember that this is a dot product equation. \[ \left(\begin{array}{c} A_{x}+t\cdot u_{x}\\ A_{y}+t\cdot u_{y}\\ A_{z}+t\cdot u_{z} \end{array}\right)\cdot\left(\begin{array}{c} u_{x}\\ u_{y}\\ u_{z} \end{array}\right)=\left(\begin{array}{c} u_{x}\\ u_{y}\\ u_{z} \end{array}\right)\cdot\left(\begin{array}{c} P_{x}\\ P_{y}\\ P_{z} \end{array}\right) \] \[ (A_{x}+t\cdot u_{x})(u_{x})+(A_{y}+t\cdot u_{y})(u_{y})+(A_{z}+t\cdot u_{z})(u_{z})=u_{x}P_{x}+u_{y}P_{y}+u_{z}P_{z}. \]
2. Solve for $t.$
3. Plug the found value for $t$ back into the line equation, eqn. \ref{eq:eqn for t} and calculate the intersection point. Call it $E.$
6. Calculate the distance between $P$ and $E$ using the distance equation. \[ dist=|P-E|=\sqrt{\left(P_{x}-E_{x}\right)^{2}+\left(P_{y}-E_{y}\right)^{2}+\left(P_{z}-E_{z}\right)^{2}} \]

Example: Short Distance to a line in 3d. Given the line between points $A=(-5,2,3)$ and $B=(-1,4,2),$find the perpendicular distance to the line from point $P=(1,2,1).$
Answer: We will show the final figure here and then go through getting the result.

We need to identify the direction vector of the line. Since we already have two points, it is just $\mathbf{u}=(B-A)$. \[ \mathbf{u}=\left(\begin{array}{c} -1\\ 4\\ 2 \end{array}\right)-\left(\begin{array}{c} -5\\ 2\\ 3 \end{array}\right)=\left(\begin{array}{c} 4\\ 2\\ -1 \end{array}\right) \] Now write the equation of the line going through points $A$ and $B.$ We will use point $A$ as the anchor, although either one will work. \begin{equation} \left(\begin{array}{c} x\\ y\\ z \end{array}\right)=\left(\begin{array}{c} -5\\ 2\\ 3 \end{array}\right)+t\cdot\left(\begin{array}{c} 4\\ 2\\ -1 \end{array}\right)\label{eq:theLine}\tag{the Line} \end{equation} Next we can write the equation of the plane with normal $\mathbf{u}$ and which includes point $P.$ \[ \mathbf{X\cdot u}=\mathbf{u}\cdot P \] \[ \left(\begin{array}{c} x\\ y\\ z \end{array}\right)\cdot\left(\begin{array}{c} 4\\ 2\\ -1 \end{array}\right)=\left(\begin{array}{c} 4\\ 2\\ -1 \end{array}\right)\cdot\left(\begin{array}{c} 1\\ 2\\ 1 \end{array}\right)=4+4-1=7 \] Since we need to use this equation, we will multiply it out. \[ 4x+2y-z=7 \label{eq:thePlane}\tag{the Plane} \] From \ref{eq:theLine}, we will insert the algebra expressions for $x,y,$ and $z,$ into \ref{eq:thePlane} equation. \[ 4(4t-5)+2(2t+2)-1(-t+3)=7 \] and solve for $t.$ \[ t=\frac{26}{21} \] Next, substitute the value for $t$ back into the line equation to compute $(x,y,z).$ \[ \left(\begin{array}{c} -5\\ 2\\ 3 \end{array}\right)+\left(\frac{26}{21}\right)\cdot\left(\begin{array}{c} 4\\ 2\\ -1 \end{array}\right)=\left(\begin{array}{c} -1/21\\ 94/21\\ 37/21 \end{array}\right)=E \] We have identified the intersection point on the line which forms a perpendicular with $P.$ Although we could go directly to the distance formula, we will instead first write the direction vector from $P$ to $E.$ \[ (E-P)=\left(\begin{array}{c} -22/21\\ 52/21\\ 16/21 \end{array}\right) \] and the distance is the modulus of that vector. \[ \left|E-P\right|=\frac{2}{21}\sqrt{861}\approx2.79 \]

Method 2: Perpendicular, Point to Line

The essence of method 2 is to write an expression, $\mathbf{v,}$ for the vector from the external point to the line, $\ell$. Then notice that said vector will dot product with the line direction vector to yield zero. Solve that equation for $t,$and then put the $t$ value into equation $\mathbf{v}$ to get the orthogonal direction vector. Alternatively, put the $t$ value into the line, $\ell,$ to get the intersection point.