Math Relative position of planes Angle of intersection between two planes

# Angle of intersection between two planes

The angle of intersection between two planes is calculated like that of two lines (or like angles between two vectors).

There are two planes (ideally in normal- or cartesian form for reading the normal vector):

$E_1: (\vec{x} - \vec{a}) \cdot \vec{n_1}=0$

$E_2: (\vec{x} - \vec{b}) \cdot \vec{n_2}=0$

The two normal vectors of the planes are now used to calculate the angle:

$\cos(\gamma) = \frac{|\vec{n_1}\cdot\vec{n_2}|}{|\vec{n_1}|\cdot|\vec{n_2}|}$

$\gamma = \cos^{-1}\left(\frac{|\vec{n_1}\cdot\vec{n_2}|}{|\vec{n_1}|\cdot|\vec{n_2}|}\right)$
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### Method

2. Calculate the scalar product of the normal vectors
3. Calculate normal vector absolute value
4. Insert results in the formula

### Example

$\text{E: } 9x + 8y + 7z = 0$

$\text{F: } x+y+2z=0$

1. #### Read off normal vectors

With the cartesian form, the normal vector can always be read off by the factors before x, y and z.

$\vec{n_1}=\begin{pmatrix}9 \\ 8 \\ 7 \end{pmatrix}$

$\vec{n_2}=\begin{pmatrix}1 \\ 1 \\ 2\end{pmatrix}$
2. #### Calculate scalar product

$\vec{n_1}\cdot\vec{n_2}$ $=\begin{pmatrix}9 \\ 8 \\ 7 \end{pmatrix}\cdot\begin{pmatrix}1 \\ 1 \\ 2\end{pmatrix}$ $=9\cdot1+8\cdot1+7\cdot2$ $=31$
3. #### Calculate normal vector absolute value

$|\vec{n_1}|=\sqrt{9^2+8^2+7^2}$ $=\sqrt{194}$

$|\vec{n_2}|=\sqrt{1^2+1^2+2^2}$ $=\sqrt{6}$
4. #### Insert results in the formula

$\cos(\gamma) = \frac{|\vec{n_1}\cdot\vec{n_2}|}{|\vec{n_1}|\cdot|\vec{n_2}|}$

$\cos(\gamma) = \frac{31}{\sqrt{194}\cdot\sqrt{6}}$

$\gamma = \cos^{-1}\left(\frac{31}{\sqrt{194}\cdot\sqrt{6}}\right)$ $\approx24.68°$