Math ↠ Distance calculations ↠ Point and line

Distance from point to a line

The distance calculation from a point $P$ to a straight line is somewhat more difficult.

Here we need an auxiliary plane $H$ that is orthogonal (perpendicular) to the line $g$ and contains the point $P$.

The intersection point of the line with the auxiliary plane is called the perpendicular point $F$.

The distance from perpendicular point $F$ to point $P$ corresponds to the distance from the line to the point and can be easily calculated.

i

Method

  1. Set up the normal equation of the auxiliary plane H (with P and direction vector of g)

  2. Calculate perpendicular point F (intersection of line g and plane H)

  3. $d$ corresponds to the distance from P to F, ie $|\vec{PF}|$ (length of vector)

Example

$P(-1|0|3)$

$\text{g: } \vec{x} = \begin{pmatrix} 1 \\ 2 \\ 1 \end{pmatrix} + r \cdot \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix}$

  1. Set up the auxiliary plane

    The auxiliary level should contain point $P$. So that's our support point. We take the direction vector of the straight line as the normal vector $\vec{n}$, since the line and plane should be orthogonal to each other.

    $\text{H: } (\vec{x} - \vec{a}) \cdot \vec{n}=0$

    $\text{H: } (\vec{x} - \begin{pmatrix} -1 \\ 0 \\ 3 \end{pmatrix}) \cdot \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix}=0$

  2. Calculate perpendicular point

    The perpendicular point is the intersection of the line $g$ and the auxiliary plane $H$. To calculate the intersection, the equation of a line for $\vec{x}$ is used in the plane.

    $\left(\color{red}{\begin{pmatrix} 1 \\ 2 \\ 1 \end{pmatrix} + r \cdot \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix}} - \begin{pmatrix} -1 \\ 0 \\ 3 \end{pmatrix}\right)$ $\cdot \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix}=0$

    $\begin{pmatrix} 2+r \\ 2+r \\ -2 \end{pmatrix}$ $\cdot \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix}=0$

    Calculate scalar product

    $(2+r)\cdot1+(2+r)\cdot1$ $+(-2)\cdot0=0$
    $4+2r=0\quad|-4$
    $2r=-4\quad|:2$
    $r=-2$

    Insert $r$ in $g$ to get perpendicular point $F$

    $\vec{OF} = \begin{pmatrix} 1 \\ 2 \\ 1 \end{pmatrix} \color{red}{-2} \cdot \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix}$ $= \begin{pmatrix} 1-2 \\ 2-2 \\ 1-0 \end{pmatrix}$ $= \begin{pmatrix} -1 \\ 0 \\ 1 \end{pmatrix}$


    $F(-1|0|1)$

  3. Calculate the distance between the two points

    The distance from the perpendicular point to point $P$ is also the distance from the line to this point.

    The distance between two points can be easily calculated using vectors.

    $d=|\vec{PF}|$ $=\left| \begin{pmatrix} -1-(-1) \\ 0-0 \\ 1-3 \end{pmatrix}\right|$ $=\left| \begin{pmatrix} 0 \\ 0 \\ -2 \end{pmatrix}\right|$ $=\sqrt{(-2)^2}$ $=2$

    The distance from the line $g$ to the point $P$ is 2 LU.