In this explainer, we will learn how to find the equation of a plane in different forms, such as intercept and parametric forms.

Consider a plane that does not pass through the origin, is not parallel to any of the axes, and thus intersects the three axes at three points with coordinates , , and . We say that the -, -, and -intercepts of the plane are , , and .

Having the coordinates of three points in the plane, we can easily find two noncollinear vectors included in the plane, for instance, and . A normal vector of the plane is then given by

As is also a normal vector of the plane, we can write a general equation of the form

Substituting in the coordinates of one of three known points included in the plane, we find that .

### Definition: Intercept Equation of a Plane

The intercept equation of a plane with -, -, and -intercepts , , and , respectively, is given by

Let us look at the equation of planes that are parallel to one axis. Consider, for instance, a plane parallel to the -axis that intersects the -axis at and the -axis at . Two vectors included in the plane are, for instance, and , a vector parallel to the -axis. A normal vector of the plane is then given by

If we take, as a normal vector of the plane, , we find the intercept equation to be

It is equivalent to the equation of a line in the plane, plus the fact that the -coordinate can take up any value in .

Consider now a plane that is parallel to two axes, for example, the - and -axes. It intersects the third axis, here the -axis, at a point with coordinates . This plane contains all the points that have as the -coordinate; hence, its equation is simply , which can be written in intercept form as .

Let us now find the intercept equation of a plane with the next example.

### Example 1: Finding the Intercept Equation of a Plane given Its Intercepts

Find the equation of the plane whose -, -, and -intercepts are , 3, and respectively.

### Answer

The intercept equation of a plane with -, -, and -intercepts , , and , respectively, is given by

Here, , , and . Hence, we find that the equation of the plane is

Finally, let us convert a general equation of a plane into an intercept equation in the next example.

### Example 2: Converting a General Equation of a Plane into Intercept Form

Write, in intercept form, the equation of the plane .

### Answer

An intercept equation of a plane is of the form

Adding 16 to each side of the given equation, we find

As we want to have 1 on the right-hand side, we divide both sides by 16, which gives

To find this equation, we could also first find , , and , that is, the -, -, and -intercepts of the plane. We find the -intercept by substituting 0 for and into our equation. We find

Hence, . Similarly, we find that and . Substituting these values into the intercept equation, we find

The intercept equation of the plane of general equation is .

Let us now look at another form of equation of a plane, namely, the parametric form.

Any point in the coordinate plane is uniquely defined by its two coordinates. In other words, for any point , its position vector is given by where is the origin of the coordinate system and and are the unit vectors along its two axes. We can write a similar equation with any two noncollinear vectors and in the plane: where and are two real numbers. This means that any vector in the plane can be written as a linear combination of two noncollinear vectors. It follows that two noncollinear vectors allow us to define any point in the plane. We use this property to write parametric equations of a plane in space.

Let us consider a plane in space that contains point and two noncollinear vectors and . For any point in the plane, we have where and are two real numbers.

This gives three equations for the three components of the vector :

By rearranging these equations, we get the parametric equations of a plane.

### Definition: Parametric Equations of a Plane

The parametric equations of a plane in space that contains point and two noncollinear vectors and are a set of three equations of the form where and are two varying real numbers, called the parameters.

By varying the parameters and across , the three equations describe the coordinates of all points in the plane.

Let us look at a first example.

### Example 3: Finding the Parametric Equation of a Plane That Passes through a Given Point and Two Given Vectors

Find, in parametric form, the equation of the plane that passes through the point and the two vectors and .

### Answer

The parametric equations of a plane that passes through the point and two vectors and are of the form where and are two real numbers.

Substituting in , , and , we find

Let us now find the parametric equations of a plane that passes through three given points.

### Example 4: Finding the Parametric Equation of a Plane given Three Noncollinear Points

Find the parametric form of the equation of the plane that passes through the points , , and .

- , ,
- , ,
- , ,
- , ,
- , ,

### Answer

We are given three points, , , and , that are in the plane. Let us first check that these three points are noncollinear by using the cross product, since if two vectors are collinear, then their cross product is zero. With and , we find that hence, the points are not collinear, which means that they define a plane indeed.

We know that the parametric equations of a plane in space that contains point and two noncollinear vectors and are a set of three equations of the form where and are two real numbers.

Let us now identify, for each option, the coordinates and the two sets of vector components and . Our findings are given in the table below.

Option | |||
---|---|---|---|

A | |||

B | |||

C | |||

D | |||

E |

With the three points , , and , we can form the vectors , , and - and of course the opposite vectors , , and as well.

Let us now analyze for each option what point and vectors have been used. In the table below, the cell has been shaded when the point coordinates or vector components do not correspond to any of the information given in the question.

We find that option D is correct. The parametric equations correspond to a plane that contains point and the vectors and .

Let us look now at how to write the equation of a plane in general form from its parametric equations. Recall that the standard form of the equation of a plane is where is a normal vector of the plane and is a point on the plane. The standard from can be easily rearranged to the general form

As the normal vector of a plane is given by the cross product of any two noncollinear vectors lying in the plane, it is possible to write the general form of the equation of a plane from its parametric equations.

### How To: Finding the General Equation of a Plane from Its Parametric Equations

The normal vector of the plane is given by the cross product of any two noncollinear vectors of the plane. Therefore, taking and from the parametric equations we find .

The standard equation is then

It can be rearranged to the general form with

Let us use this method in the next example.

### Example 5: Finding the General Equation of a Plane from Its Parametric Equation

Find the general equation of the plane , , .

### Answer

We are given the parametric equations of the plane. We know that the coefficients of on one side and those of on the other side are the components of two vectors in the plane. Let us call these vectors and . We find and . One normal vector of the plane is thus

We have found that is a normal vector of the plane. Normal vectors of a plane are all parallel, which means that they all can be written as , where is one normal vector and is a real number. Here, all components of are multiples of 4, so we can take as a normal vector to write our general equation.

The general equation of a plane is of the form where , , and are the components of a normal vector of the plane. Substituting these components into the general equation of the plane, we find

To find the constant , we need to know the coordinates of a point that belongs to a plane. We can find the coordinates of a point on the plane using the parametric equations: the constants in each equation correspond to the coordinates of a point on the plane. We find that the point with coordinates belongs to the plane. We substitute these into the equation, which gives

The general equation of the plane is, therefore,

It is worth noting that we could have kept our original normal vector to write the general equationβwe would have found the same final general equation after dividing both sides by 4.

### Key Points

- The intercept equation of a plane with -, -, and -intercepts , , and , respectively, is given by
- The parametric equations of a plane in space that contains point and two noncollinear vectors and are a set of three equations of the form where and are two varying real numbers, called the parameters.
- A normal vector of a plane is given by .