# Angle Between Two Planes

Let the planes be given by the general equations

$A_1x + B_1y + C_1z + D_1 = 0,\;\;A_2x + B_2y + C_2z + D_2 = 0.$

The angle between two planes is defined as the angle between their respective normal vectors.

Once you have the normal vectors for the two planes, you can compute the dot product of these vectors. The dot product of two vectors n1 and n2 is given by

$\mathbf{n_1} \cdot \mathbf{n_2} = \left|{\mathbf{n_1}}\right| \cdot \left|{\mathbf{n_2}}\right| \cdot \cos\varphi$

where φ is the angle between the vectors and, accordingly, the dihedral angle between the planes.

Using the dot product formula, you can solve for φ and express the cosine of φ in coordinate form:

$\cos\varphi = \frac{\mathbf{n_1} \cdot \mathbf{n_2}}{\left|{\mathbf{n_1}}\right| \cdot \left|{\mathbf{n_2}}\right|} = \frac{A_1A_2 + B_1B_2 + C_1C_2}{\sqrt{A_1^2 + B_1^2 + C_1^2}\sqrt{A_2^2 + B_2^2 + C_2^2}}$

Finally, you can find the angle φ by taking the arccosine (inverse cosine) of the result.

## Solved Problems

### Example 1.

Find the angle between the planes $${x + 4y + z + 3 = 0}$$ and $${x + 2y - 5 = 0}.$$

Solution.

We use the formula

$\cos\varphi = \frac{A_1A_2 + B_1B_2 + C_1C_2}{\sqrt{A_1^2 + B_1^2 + C_1^2}\sqrt{A_2^2 + B_2^2 + C_2^2}}.$

Let's substitute the known coefficients:

$\cos\varphi = \frac{1\cdot 1 + 4\cdot 2 + 1\cdot 0}{\sqrt{1^2 + 4^2 + 1^2}\,\sqrt{1^2 + 2^2}} = \frac{1 + 8}{\sqrt{18}\sqrt{5}} = \frac{9}{3\sqrt{10}} = \frac{3}{\sqrt{10}}.$

Then the angle φ between the planes is equal to

$\varphi = \arccos\frac{3}{\sqrt{10}} \approx \arccos{\left(0.9487\right) = 0.3218\;rad = 18.43^\circ}$

### Example 2.

Find the equation of a plane that passes through the $$z-$$axis and makes an angle of $$\frac{\pi}{3}$$ with the plane $${2x + y - \sqrt{5}z - 7 = 0}.$$

Solution.

If a plane passes through the $$z-$$axis then its equation has the form

$Ax + By = 0.$

Let's write down what is the angle $$\varphi$$ between the two planes:

$\cos\varphi = \frac{2\cdot A + B \cdot 1 + \left({-\sqrt{5}}\right)\cdot 0}{\sqrt{A^2 + B^2}\,\sqrt{2^2 + 1^2 + \left({-\sqrt{5}}\right)^2}} = \frac{2A + B}{\sqrt{A^2 + B^2}\,\sqrt{4 + 1 + 5}} = \frac{2A + B}{\sqrt{A^2 + B^2}\,\sqrt{10}}.$

Since $$\varphi = \frac{\pi}{3},$$ we get the following relation:

$\cos\frac{\pi}{3} = \frac{1}{2} = \frac{2A + B}{\sqrt{A^2 + B^2}\,\sqrt{10}},\;\text{ or }\;\frac{2A + B}{\sqrt{A^2 + B^2}} = \frac{\sqrt{10}}{2}.$

We just need to find the ratio of coefficients $$A$$ and $$B.$$ So we put $$B = 1$$ and get

$\frac{2A + 1}{\sqrt{A^2 + 1}} = \frac{\sqrt{10}}{2}.$

Let's square both sides of the equation and find $$A:$$

$\frac{\left({2A + 1}\right)^2}{A^2+1} = \frac{10}{4} = \frac{5}{2},\;\Rightarrow \left({2A + 1}\right)^2 = \frac{5}{2}\left({A^2+1}\right),\;\Rightarrow 2\left({4A^2 + 4A + 1}\right) = 5\left({A^2+1}\right),$
$\Rightarrow 8A^2 + 8A + 2 = 5A^2 + 5,\;\Rightarrow 3A^2 + 8A - 3 = 0.$

Find the roots of the quadratic equation:

$D = 8^2 -4\cdot 3 \cdot \left({-3}\right) = 64 + 36 = 100, \;\Rightarrow A_{1,2} = \frac{-8 \pm \sqrt{100}}{2\cdot 3} = \frac{-8\pm 10}{6} = -3,\frac{1}{3}.$

So we got two sets of coefficients $$A$$ and $$B$$ that correspond to two possible planes:

1. $$A = -3, B=1:$$
$-3x + y = 0, \;\Rightarrow 3x - y = 0.$
2. $$A = \frac{1}{3}, B=1:$$
$\frac{1}{3}x + y = 0, \;\Rightarrow x + 3y = 0.$

### Example 3.

Find the equation of a plane bisecting the dihedral angles between two planes $${6x - 3y + 2z - 1 = 0}$$ and $${2x + 6y - 3z + 4 = 0}.$$

Solution.

Let point $$M\left({x,y,z}\right)$$ be an arbitrary point of our plane. The distance from point $$M$$ to two given planes must be the same. Therefore we can write the following equation

$\frac{\left|{A_1x + B_1y + C_1z + D_1}\right|}{\sqrt{A_1^2 + B_1^2 + C_1^2}} = \frac{\left|{A_2x + B_2y + C_2z + D_2}\right|}{\sqrt{A_2^2 + B_2^2 + C_2^2}}.$

Substituting the coefficients we get

$\frac{\left|{6x - 3y + 2z - 1}\right|}{\sqrt{6^2 + \left({-3}\right)^2 + 2^2}} = \frac{\left|{2x + 6y - 3z + 4}\right|}{\sqrt{2^2 + 6^2 + \left({-3}\right)^2}}, \;\Rightarrow \frac{\left|{6x - 3y + 2z - 1}\right|}{7} = \frac{\left|{2x + 6y - 3z + 4}\right|}{7},$

or

$\left|{6x - 3y + 2z - 1}\right| = \left|{2x + 6y - 3z + 4}\right|.$

This equation contains two solutions. In the first case we have

$6x - 3y + 2z - 1 = 2x + 6y - 3z + 4, \;\Rightarrow 4x - 9y + 5z - 5 = 0.$

The second solution, that is, another plane is described by the equation

$6x - 3y + 2z - 1 = - \left({2x + 6y - 3z + 4}\right), \;\Rightarrow 6x - 3y + 2z - 1 = -2x - 6y + 3z - 4,\;\Rightarrow 8x + 3y - z + 3 = 0.$

### Example 4.

A triangular pyramid has vertices at points $$A\left({4,0,0}\right),$$ $$B\left({0,3,0}\right),$$ $$C\left({5,4,0}\right),$$ $$D\left({4,0,0}\right).$$ Find the angle between the base of the pyramid $$ABC$$ and the face $$BCD.$$

Solution.

The base $$ABC$$ lies in the $$xy-$$plane, so its equation is $$z = 0.$$

Let's find the equation of the $$BCD$$ face given three points:

$\left| {\begin{array}{*{20}{c}} {{x - x_D}}&{{y - y_D}}&{{z - z_D}}\\ {{x_B - x_D}}&{{y_B - y_D}}&{{z_B - z_D}}\\ {{x_C - x_D}}&{{y_C - y_D}}&{{z_C - z_D}} \end{array}} \right| = 0.$

Substitute the coordinates of points $$B,$$ $$C,$$ $$D:$$

$\left| {\begin{array}{*{20}{l}} {{x - 2}}&{{y - 2}}&{{z - 5}}\\ {{0 - 2}}&{{3 - 2}}&{{0 - 5}}\\ {{5 - 2}}&{{4 - 2}}&{{0 - 5}} \end{array}} \right| = 0,\;\Rightarrow \left| {\begin{array}{*{20}{c}} {{x - 2}}&{{y - 2}}&{{z - 5}}\\ {{- 2}}&{{1}}&{{- 5}}\\ {{3}}&{{2}}&{{- 5}} \end{array}} \right| = 0.$

Expand the determinant along the first row:

$\left({x-2}\right) \left| {\begin{array}{*{20}{c}} {{1}}&{{-5}}\\ {{2}}&{{-5}} \end{array}} \right| - \left({y-2}\right) \left| {\begin{array}{*{20}{c}} {{-2}}&{{-5}}\\ {{3}}&{{-5}} \end{array}} \right| + \left({z-5}\right) \left| {\begin{array}{*{20}{c}} {{-2}}&{{1}}\\ {{3}}&{{2}} \end{array}} \right|= 0,$
$\left({x-2}\right)\left({-5+10}\right) - \left({y-2}\right)\left({10+15}\right) + \left({z-5}\right)\left({-4-3}\right) = 0,$
$5\left({x-2}\right) - 25\left({y-2}\right) - 7\left({z-5}\right) = 0,$
$5x - 10 - 25y + 50 - 7z + 35 = 0,\;\Rightarrow 5x - 25y - 7z + 75 = 0.$

Find the angle between planes $$ABC$$ and $$BCD$$ using the formula

$\cos\varphi = \frac{A_1A_2 + B_1B_2 + C_1C_2}{\sqrt{A_1^2 + B_1^2 + C_1^2}\sqrt{A_2^2 + B_2^2 + C_2^2}}.$

Hence

$\cos\varphi = \frac{0\cdot 5 + 0\cdot \left({-25}\right) + 1\cdot \left({-7}\right)}{\sqrt{0^2 + 0^2 + 1^2}\sqrt{5^2 + \left({-25}\right)^2 + \left({-7}\right)^2}} = \frac{-7}{\sqrt{699}}.$

Since cosine is negative, we have found an obtuse angle $$\varphi_1$$ between the planes. The corresponding acute angle $$\varphi_2$$ between the planes is equal to

$\varphi_2 = 180^\circ - \varphi_1 = \arccos\frac{7}{\sqrt{699}} \approx 74.6^\circ$