# Parallel and Perpendicular Planes

## Parallel Planes

Two distinct planes are said to be parallel to each other if they have no common points. Parallel planes do not intersect, meaning they never meet or cross each other, even when extended indefinitely in all directions. In other words, if you were to extend the planes infinitely, they would never touch or intersect.

Formally, two planes are considered parallel if and only if their normal vectors n1 and n2 are parallel or opposite, that is, if their normal vectors are collinear. The normal vector of a plane is a vector that is perpendicular (at a right angle) to the plane. So, the planes are parallel if

$\mathbf{n_1} = \lambda\mathbf{n_2}$

where λ is a real number.

Another way to define parallel planes is through their equations. If the coefficients of the variables in the equations of two planes are proportional, then the planes are parallel.

For example, if the equation of two planes are

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

then the planes are parallel if

$\frac{A_1}{A_2} = \frac{B_1}{B_2} = \frac{C_1}{C_2}$

Two planes are are said to be coincident when their equations are scalar multiples of each other:

$\frac{A_1}{A_2} = \frac{B_1}{B_2} = \frac{C_1}{C_2} = \frac{D_1}{D_2}$

## Perpendicular Planes

Perpendicular planes are planes in three-dimensional space that intersect each other at right angles (90 degrees). In other words, the normal vectors to the planes are orthogonal to each other.

Mathematically, if two planes have normal vectors n1 and n2, then they are perpendicular if and only if the dot product of these normal vectors is zero:

$\mathbf{n_1} \cdot \mathbf{n_2} = 0$

This implies that the angle between the normal vectors, and consequently between the planes, is 90 degrees.

For example, the yz-plane (the plane defined by the equation x = 0) and the xz-plane (the plane defined by the equation y = 0) are perpendicular because their normal vectors i and j are mutually orthogonal.

$\mathbf{i} \perp \mathbf{j} \;\;\text{ or }\;\;\mathbf{i} \cdot \mathbf{j} = 0$

In coordinate form, if n1(A1, B1, C1) and n2(A2, B2, C2) are the normal vectors of the planes, the condition for perpendicularity is

$A_1A_2 + B_1B_2 + C_1C_2 = 0$

## Solved Problems

### Example 1.

Find the distance between two parallel planes $$3x - 4y + 5z - 4 = 0$$ and $$3x - 4y + 5z + 16 = 0.$$

Solution.

Take an arbitrary point $$M_1\left({x_1,y_1,z_1}\right)$$ in the first plane. Let, for example, $$x_1 = 0,$$ $$y_1 = 0.$$ Then

$3 \cdot 0 - 4\cdot 0 + 5\cdot z_1 - 4 = 0, \Rightarrow 5z_1 - 4 = 0,\Rightarrow z_1 = \frac{4}{5}.$

Now letâ€™s find the distance from point $$M_1\left({0,0,\frac{4}{5}}\right)$$ to the second plane using the formula

$d = \frac{\left|{A_2x_1 + B_2y_1 + C_2z_1 + D_2}\right|}{\sqrt{A_2^2 + B_2^2 + C_2^2}}.$

Substitute the coefficients of the equation of the second plane and the coordinates of point M1:

$d = \frac{\left|{3 \cdot 0 - 4 \cdot 0 + 5\cdot \frac{4}{5} + 16}\right|}{\sqrt{3^2 + \left({-4}\right)^2 + 5^2}} = \frac{\left|{4 + 16}\right|}{\sqrt{9 + 16 + 25}} = \frac{20}{\sqrt{50}} = \frac{20}{5\sqrt{2}} = \frac{4}{\sqrt{2}} = 2\sqrt{2}.$

### Example 2.

Find the equation of a plane passing through the point $$M\left({1,2,-3}\right)$$ and parallel to the plane $$2x - y + 3z = 0.$$

Solution.

The coefficients of the given plane are

$A_1 = 2,\; B_1 = -1,\; C_1 = 3,\; D_1 = 0.$

Let the new plane be described by the equation

$A_2x + B_2y + C_2z + D_2 = 0.$

Since both planes are parallel, the following relation holds:

$\frac{A_2}{A_1} = \frac{B_2}{B_1} = \frac{C_2}{C_1} = \lambda.$

Then the previous equation can be written as

$\lambda A_1x + \lambda B_1y + \lambda C_1z + D_2 = 0,$

or

$A_1x + B_1y + C_1z + \frac{D_2}{\lambda} = 0.$

So the equation of the parallel plane has the same coefficients $$A_1,$$ $$B_1,$$ $$C_1$$ and differs only in the free term $$\frac{D_2}{\lambda},$$ which we simply denote by $$D.$$

Let's find the value of $$D$$ knowing that the plane passes through the point $$M\left({1,2,-3}\right):$$

$2\cdot 1 2 + 3\cdot\left({-3}\right) + D = 0, \Rightarrow D = 9.$

So the equation of the plane has the form

$2x - y + 3z + 9 = 0.$

### Example 3.

Write the equation of a plane passing through the origin and perpendicular to the planes $$2x + 3y - z = 0$$ and $$3x - y + z + 4 = 0.$$

Solution.

The plane passing through the origin is described by the general equation

$Ax + By + Cz = 0.$

The normal vectors of two given planes have coordinates

$\mathbf{n_1}\left({2,3,-1}\right),\;\;\mathbf{n_2}\left({3,-1,1}\right).$

Let's find the coordinates of the normal vector n to our third plane. Since

$\mathbf{n} \perp \mathbf{n_1}\;\text{ and }\;\mathbf{n} \perp \mathbf{n_2},$

the vector n can be found using the cross product:

$\mathbf{n} = \mathbf{n_1} \times \mathbf{n_2}.$

In coordinate form:

$\mathbf{n} = \mathbf{n_1} \times \mathbf{n_2} = \left| {\begin{array}{*{20}{c}} \mathbf{i} & \mathbf{j} & \mathbf{k}\\ {2} & {3} & {-1}\\ {3} & {-1} & {1} \end{array}} \right| = \left| {\begin{array}{*{20}{r}} {3} & {-1}\\ {-1} & {1} \end{array}} \right|\mathbf{i} - \left| {\begin{array}{*{20}{r}} {2} & {-1}\\ {3} & {1} \end{array}} \right|\mathbf{j} + \left| {\begin{array}{*{20}{r}} {2} & {3}\\ {3} & {-1} \end{array}} \right|\mathbf{k} = \left({3-1}\right)\mathbf{i} - \left({2+3}\right)\mathbf{j} + \left({-2-9}\right)\mathbf{k} = 2\mathbf{i} - 5\mathbf{j} - 11\mathbf{k}.$

Therefore the normal vector n is equal to

$\mathbf{n}\left({2,-5,-11}\right).$

The plane equation is written as

$2x - 5y - 11z = 0.$

### Example 4.

Write the equations of planes parallel to the plane $$x + 2y - 2z + 5 = 0$$ which are at the distance of $$d = 4$$ away from it.

Solution.

It is obvious that the equations of parallel planes will differ only by the coefficient D. That is, their equations has the form

$x + 2y - 2z + D = 0.$

Take an arbitrary point $$M_0\left({x_0,y_0,z_0}\right)$$ in the original plane. Let for simplicity $$x_0 = 0,$$ $$y_0 = 0.$$ Find the $$z_0-$$coordinate:

$0 + 2\cdot 0 - 2z_0 + 5 = 0,\;\Rightarrow 2z_0 = 5,\;\Rightarrow z_0 = \frac{5}{2}.$

That is, the point M0 has coordinates $$\left({0,0,\frac{5}{2}}\right).$$

The distance from a point to a plane is determined by the formula

$d = \frac{\left|{Ax_0 + By_0 + Cz_0 + D}\right|}{\sqrt{A^2 + B^2 + C^2}}.$

Substituting the known values on the right side we get

$d = \frac{\left|{0 + 2\cdot 0 - 2\cdot\frac{5}{2} + D}\right|}{\sqrt{1^2 + 2^2 + \left({-2}\right)^2}} = \frac{\left|{-2 + D}\right|}{\sqrt{1+4+4}} = \frac{\left|{D - 2}\right|}{3}.$

By condition $$d = 4.$$ We therefore obtain the following equation for determining the coefficient D:

$4 = \frac{\left|{D - 2}\right|}{3},\;\Rightarrow \left|{D - 2}\right| = 12.$

This equation has two solutions, that is, two values of D:

1. $$D - 2 = 12,$$ $$\Rightarrow D_1 = 14.$$
2. $$D - 2 = -12,$$ $$\Rightarrow D_2 = -10.$$

So the equations of parallel planes have the form

$x + 2y - 2z + 14 = 0,\;\;x + 2y - 2z - 10 = 0.$

### Example 5.

Find the equation of a plane passing through the point $$M\left({1,2,-1}\right)$$ and perpendicular to the planes $$x - y + 2z + 3 = 0$$ and $$x + 2y - z - 4 = 0.$$

Solution.

Let n1 and n2 be the normal vectors of two given planes. Their coordinates are equal

$\mathbf{n_1}\left({1,-1,2}\right),\;\;\mathbf{n_2}\left({1,2,-1}\right).$

By condition, the new plane is perpendicular to both given planes. Therefore its normal vector n must be perpendicular to both n1 and n2:

$\mathbf{n} \perp \mathbf{n_1},\;\;\mathbf{n} \perp \mathbf{n_2}.$

Vector n can be found using the cross product:

$\mathbf{n} = \mathbf{n_1} \times \mathbf{n_2}.$

Let's calculate it through the determinant:

$\mathbf{n} = \mathbf{n_1} \times \mathbf{n_2} = \left| {\begin{array}{*{20}{c}} \mathbf{i} & \mathbf{j} & \mathbf{k}\\ {{A_1}} & {{B_1}} & {{C_1}}\\ {{A_2}} & {{B_2}} & {{C_2}} \end{array}} \right| = \left| {\begin{array}{*{20}{c}} \mathbf{i} & \mathbf{j} & \mathbf{k}\\ {1} & {-1} & {2}\\ {1} & {2} & {-1} \end{array}} \right| = \left| {\begin{array}{*{20}{r}} {-1} & {2}\\ {2} & {-1} \end{array}} \right|\mathbf{i} - \left| {\begin{array}{*{20}{r}} {1} & {2}\\ {1} & {-1} \end{array}} \right|\mathbf{j} + \left| {\begin{array}{*{20}{r}} {1} & {-1}\\ {1} & {2} \end{array}} \right|\mathbf{k} = \left({1-4}\right)\mathbf{i} - \left({-1-2}\right)\mathbf{j} + \left({2+1}\right)\mathbf{k} = -3\mathbf{i} + 3\mathbf{j} + 3\mathbf{k}.$

Therefore the normal vector n is given by

$\mathbf{n}\left({-3,3,3}\right)\;\text{ or }\;\mathbf{n}\left({-1,1,1}\right).$

The plane equation looks like

$-x + y + z + D = 0.$

We determine the coefficient D from the condition that the plane passes through the point $$M\left({1,2,-1}\right):$$

$-1 \cdot 1 + 2 - 1 + D = 0,\;\Rightarrow D = 0.$

$x - y - z = 0.$