Math ↠ Relative position of planes ↠ Point and plane (point on the plane)

Point on the plain?

Similar to checking whether a specific point lies on a line you can also check whether the point lies on the plane.

i

Method

The procedure varies depending on the plane equation:

  1. Insert the position vector of the point (P/V) or its coordinates (C).
  2. Solve equation (V/C) or system of equations (P)
  3. Check if solvable

P - Parametric equation
V - Vector equation of a plane
C - Cartesian equation

!

Remember

The point lies on the plane only if the equation or system of equations can be solved.

Example (parametric form)

$P(2|1|1)$,
$\text{E: } \vec{x} = \begin{pmatrix} 3 \\ 0 \\ 0 \end{pmatrix} + r \cdot \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix}$ $+ s \cdot \begin{pmatrix} 1 \\ 5 \\ 2 \end{pmatrix}$

  1. Insert $P$

    The position vector (vector with the coordinates of the point) of $P$ is used for $\vec{x}$ in $E$.

    $\begin{pmatrix} 2 \\ 1 \\ 1 \end{pmatrix}$ $=\begin{pmatrix} 3 \\ 0 \\ 0 \end{pmatrix} + r \cdot \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix}$ $+ s \cdot \begin{pmatrix} 1 \\ 5 \\ 2 \end{pmatrix}$

  2. Set up system of equations

    Now we set up an equation system and solve it. Every line is an equation.
    1. $2=3+r+s$
    2. $1=r+5s$
    3. $1=2s$

    From III. you get $s=\frac12$, which is used in II..

    $1=r+5\cdot\frac12\quad|-\frac52$
    $r=-\frac32$

  3. Test with I.

    $r$ and $s$ are used for the sample in the unused equation (here: I.).

    $2=3+r+s$
    $2=3-\frac32+\frac12$
    $2=2$

    Since there is no contradiction and it is a true statement, the point lies on the plane.

Example (Vector form)

$P(2|1|-1)$,
$\text{E: } \left(\vec{x} - \begin{pmatrix} 2 \\ 1 \\ 1 \end{pmatrix}\right) \cdot \begin{pmatrix} 2 \\ -2 \\ 4 \end{pmatrix}=0$

  1. Insert $P$

    The position vector (vector with the coordinates of the point) of $P$ is used for $\vec{x}$ in $E$.

    $\left(\begin{pmatrix} 2 \\ 1 \\ -1 \end{pmatrix} - \begin{pmatrix} 2 \\ 1 \\ 1 \end{pmatrix}\right) \cdot \begin{pmatrix} 2 \\ -2 \\ 4 \end{pmatrix}$ $=0$

  2. Solve equation

    The equation can be simplified first.

    $\begin{pmatrix} 2-2 \\ 1-1 \\ -1-1 \end{pmatrix} \cdot \begin{pmatrix} 2 \\ -2 \\ 4 \end{pmatrix}=0$

    $\begin{pmatrix} 0 \\ 0 \\ -2 \end{pmatrix} \cdot \begin{pmatrix} 2 \\ -2 \\ 4 \end{pmatrix}=0$

    Now apply the scalar product on the left side of the equation.

    $0\cdot2+0\cdot(-2)+(-2)\cdot4$ $=0$

    $-8\neq0$

    => Contradiction, point is not on the plane

Example (cartesian form)

$P(2|1|1)$,
$\text{E: } 2x-2y+4z=6$

  1. Insert coordinates of $P$

    The individual coordinates of $P$ are used for x, y and z.

    $2\cdot2-2\cdot1+4\cdot1=6$

  2. Solve equation

    The equation can be solved very easily.

    $2\cdot2-2\cdot1+4\cdot1=6$
    $6=6$

    => true statement, the point lies on the plane