# Plane

## General Equation of a Plane

The general or standard equation of a plane in the Cartesian coordinate system is represented by the linear equation

$Ax + By + Cz + D = 0$

where (x, y, z) are coordinates of the points belonging to the plane, and A, B, C are real numbers.

## Normal Vector to a Plane

The coordinates of the normal vector n (A, B, C) to a plane are the coefficients in the general equation of the plane

$Ax + By + Cz + D = 0.$

## Special Cases of the Equation of a Plane

$Ax + By + Cz + D = 0$
• If $$A = 0,$$ the plane is parallel to the $$x-$$axis;
• If $$B = 0,$$ the plane is parallel to the $$y-$$axis;
• If $$C = 0,$$ the plane is parallel to the $$z-$$axis;
• If $$D = 0,$$ the plane passes through the origin;
• If $$A = B = 0,$$ the plane is parallel to the $$xy-$$plane;
• If $$B = C = 0,$$ the plane is parallel to the $$yz-$$plane;
• If $$A = C = 0,$$ the plane is parallel to the $$xz-$$plane.

## Point Direction Form

$A\left({x - x_0}\right) + B\left({y - y_0}\right) + C\left({z - z_0}\right) = 0$

where the point P (x0, y0, z0) lies in the plane and the vector $$\mathbf{n}\left( {A,B,C} \right)$$ is normal to the plane.

## Intercept Form

$\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$

where the point a, b, c are the intercepts on the x, y, and z axes, respectively.

## Three Points Form

$\left| {\begin{array}{*{20}{l}} {{x - x_3}}&{{y - y_3}}&{{z - z_3}}\\ {{x_1 - x_3}}&{{y_1 - y_3}}&{{z_1 - z_3}}\\ {{x_2 - x_3}}&{{y_2 - y_3}}&{{z_2 - z_3}} \end{array}} \right| = 0$

or

$\left| {\begin{array}{*{20}{l}} {{x}}&{{y}}&{{z}}&{{1}}\\ {{x_1}}&{{y_1}}&{{z_1}}&{{1}}\\ {{x_2}}&{{y_2}}&{{z_2}}&{{1}}\\ {{x_3}}&{{y_3}}&{{z_3}}&{{1}} \end{array}} \right| = 0$

where the points A (x1, y1, z1), B (x2, y2, z2), C (x3, y3, z3) lie in the given plane.

## Normal Form

$x\cos\alpha + y\cos\beta + z\cos\gamma - p = 0$

Here p is the distance from the origin to the plane, and cos α, cos β, cos γ are the direction cosines of any straight line normal to the plane.

## Parametric Form

$\left\{ \begin{array}{l} x = x_1 + a_1s + a_2t\\ y = y_1 + b_1s + b_2t\\ z = z_1 + c_1s + c_2t \end{array} \right.$

where (x, y, z) are the coordinates of any point of the plane, s and t are parameters, the point P (x1, y1, z1) lies in this plane, and the vectors u (a1, b1, c1), v (a2, b2, c2) are parallel to the plane (see Figure 6 below).

## Equation of a Plane Given a Point and Two Vectors

The plane passing through the point P (x1, y1, z1) and parallel to two non-collinear vectors u (a1, b1, c1) and v (a2, b2, c2) is determined by the equation

$\left| {\begin{array}{*{20}{c}} {{x - x_1}}&{{y- y_1}}&{{z - z_1}}\\ {{a_1}}&{{b_1}}&{{c_1}}\\ {{a_2}}&{{b_2}}&{{c_2}} \end{array}} \right| = 0$

## Equation of a Plane Given Two Points and a Vector

The plane passing through the points P1(x1, y1, z1) and P2(x2, y2, z2) and parallel to the vector u (a, b, c) is described by the equation

$\left| {\begin{array}{*{20}{c}} {{x - x_1}}&{{y - y_1}}&{{z - z_1}}\\ {{x_2 - x_1}}&{{y_2 - y_1}}&{{z_2 - z_1}}\\ {{a}}&{{b}}&{{c}} \end{array}} \right| = 0$

## Solved Problems

### Example 1.

Find the equation of a plane that cuts off intercepts $$a = 2,$$ $$b = -1,$$ $$c = 4$$ on the coordinate axes.

Solution.

The equation of the plane in intercept form is written as

$\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1.$

Substituting the values of a, b and c, we get

$\frac{x}{2} + \frac{y}{-1} + \frac{z}{4} = 1.$

Multiply both sides by 4 and move all terms to the left side.

$2x - 4y + z = 4, \;\Rightarrow 2z - 4y + z - 4 = 0.$

This is the equation of the given plane in general form.

### Example 2.

Find the equation of a plane that passes through the point $$M\left({1,-3,5}\right)$$ and cuts off equal intercepts on coordinate axes.

Solution.

Let the plane cut off intercepts of length a on the coordinate axes. Then its equation can be written as

$\frac{x}{a} + \frac{y}{a} + \frac{z}{a} = 1.$

Substitute the coordinates of point M and define the segment a:

$\frac{1}{a} - \frac{3}{a} + \frac{5}{a} = 1, \;\Rightarrow 1 - 3 + 5 = a, \;\Rightarrow a = 3.$

Then the equation of the plane looks like this:

$\frac{x}{3} + \frac{y}{3} + \frac{z}{3} = 1,$

or in general form:

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

### Example 3.

Find the equation of a plane that passes through the point $$P\left({1,2,2}\right)$$ and cuts off intercepts $$a = 3,$$ $$b = 4$$ on the $$x-$$axis and $$y-$$axis, respectively.

Solution.

Let's write down the equation of a plane in intercept form:

$\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1.$

We know the values of a and b, so we have

$\frac{x}{3} + \frac{y}{4} + \frac{z}{c} = 1.$

Let's determine the intercept $$c,$$ knowing that the plane passes through the point $$P\left({1,2,2}\right):$$

$\frac{1}{3} + \frac{2}{4} + \frac{2}{c} = 1, \;\Rightarrow \frac{5}{6} + \frac{2}{c} = 1, \;\Rightarrow \frac{2}{c} = \frac{1}{6},\;\Rightarrow c = 12.$

Then the equation of our plane is written as

$\frac{x}{3} + \frac{y}{4} + \frac{z}{12} = 1.$

It can be easily converted into general form:

$4x + 3y + z - 12 = 0.$

### Example 4.

Write the equation of a plane passing through points $$A\left({1,0,-1}\right),$$ $$B\left({2,3,-1}\right),$$ and $$C\left({1,3,2}\right).$$

Solution.

The equation of a plane passing through three points $$A\left({x_1,y_1,z_1}\right),$$ $$B\left({x_2,y_2,z_2}\right),$$ $$C\left({x_3,y_3,z_3}\right)$$ is written using a determinant and has the form

$\left| {\begin{array}{*{20}{c}} {{x - x_1}}&{{y - y_1}}&{{z - z_1}}\\ {{x_2 - x_1}}&{{y_2 - y_1}}&{{z_2 - z_1}}\\ {{x_3 - x_1}}&{{y_3 - y_1}}&{{z_3 - z_1}} \end{array}} \right| = 0.$

Let's substitute the known coordinates:

$\left| {\begin{array}{*{20}{c}} {{x - 1}}&{{y - 0}}&{{z + 1}}\\ {{2 - 1}}&{{3 - 0}}&{{-1 + 1}}\\ {{1 - 1}}&{{3 - 0}}&{{2 + 1}} \end{array}} \right| = 0.$

Hence

$\left| {\begin{array}{*{20}{c}} {{x - 1}}&{{y}}&{{z + 1}}\\ {{1}}&{{3}}&{{0}}\\ {{0}}&{{3}}&{{3}} \end{array}} \right| = 0.$

To calculate this determinant, let's expand it, for example, along the first row:

$\left({x-1}\right)\left| {\begin{array}{*{20}{c}} {{3}}&{{0}}\\ {{3}}&{{3}} \end{array}} \right| - y\left| {\begin{array}{*{20}{c}} {{1}}&{{0}}\\ {{0}}&{{3}} \end{array}} \right| + \left({z+1}\right)\left| {\begin{array}{*{20}{c}} {{1}}&{{3}}\\ {{0}}&{{3}} \end{array}} \right| = 0.$

Simplify this equation:

$\left({x-1}\right)\left({9-3}\right) - y\left({3-0}\right) + \left({z+1}\right)\left({3-0}\right) = 0,$
$6\left({x-1}\right) - 3y + 3\left({z+1}\right) = 0,$
$6x - 6 - 3y + 3z + 3 = 0,$
$6x - 3y + 3z - 3 = 0.$

Dividing both parts by 3 we get the final answer:

$2x - y + z - 1 = 0.$

### Example 5.

Find the equation of a plane in parametric form and in general form if the plane passes through the point $$M\left({1,2,-4}\right)$$ and is parallel to the vectors $$\mathbf{u}\left({-2,3,3}\right)$$ and $$\mathbf{v}\left({1,-2,0}\right).$$

Solution.

The plane equation in parametric form is written as follows:

$\left\{ \begin{array}{l} x = x_1 + a_1s + a_2t\\ y = y_1 + b_1s + b_2t\\ z = z_1 + c_1s + c_2t \end{array} \right.,$

where point M has coordinates (x1, y1, z1), the coordinates of vectors u and v are (a1, b1, c1) and (a2, b2, c2), and s, t are parameters.

Substitute the coordinates of point M and vectors u, v:

$\left\{ \begin{array}{l} x = 1 - 2s + t\\ y = 2 + 3s - 2t\\ z = -4 + 3s \end{array} \right..$

This system represents the equation of the plane in parametric form. Let us now derive the equation of the same plane in general form. Recall that the equation of a plane given a point and two vectors has the following form:

$\left| {\begin{array}{*{20}{c}} {{x - x_1}}&{{y- y_1}}&{{z - z_1}}\\ {{a_1}}&{{b_1}}&{{c_1}}\\ {{a_2}}&{{b_2}}&{{c_2}} \end{array}} \right| = 0.$

Substituting the coordinates of the point and two vectors we get

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

Expand the determinant along the third column:

$\left({z+4}\right)\left| {\begin{array}{*{20}{c}} {{-2}}&{{3}}\\ {{1}}&{{-2}} \end{array}} \right| - 3\left| {\begin{array}{*{20}{c}} {{x-1}}&{{y-2}}\\ {{1}}&{{-2}} \end{array}} \right| = 0,$

and simplify:

$\left({z+4}\right)\left({4-3}\right) - 3\left({-2x+2-y+2}\right) = 0,$
$z + 4 + 6x + 3y - 12 = 0,$

or

$6x + 3y + z - 8 = 0.$