# 3.3 Equation of a plane in three-dimensional Euclidean space

We stated in Section 1.7 that the general form of the equation of a plane in ^{3} is

where *a*, *b* and *c* are not all zero. We can prove this now, using the dot product.

How can we specify a plane uniquely in ^{3}? One possibility is to specify three points that lie in the plane. However, this does not enable us to prove equation (3.1) without a significant amount of algebra; so we adopt a different approach.

For a given plane in ^{3}, there is a particular direction that can be used to specify the plane – the direction perpendicular to all the vectors in that plane.

## Definition

A vector that is perpendicular to all the vectors in a given plane is called a **normal vector** to the plane.

*Note:* A normal vector to a plane is also called a *normal* to the plane, and its direction is said to be *normal to* the plane.

If **n** is a normal vector to a given plane, then so is *k***n**, for any non-zero real number *k*. If *k* > 0, then *k***n** is in the same direction as **n**, whereas if *k* < 0, then *k***n** is in the opposite direction to **n**.

A normal vector **n** does not determine a plane uniquely, as there are infinitely many planes that have **n** as a normal; these planes are parallel to one another. However, if we specify both a normal vector and a point that lies in the plane, then the plane is determined uniquely.

## Example 6

Determine the equation of the plane in ^{3} that contains the point *P*(2, 3, 4) and has **n** = (1, 2, −1) as a normal.

### Answer

If the point *Q*(*x*, *y*, *z*) lies in the plane, then the vector must be perpendicular to the normal vector **n**; it follows that must be zero.

Hence

so

(*x* − 2) × 1 + (*y* − 3) × 2 + (*z* − 4) × (−1) = 0.

This equation can be rearranged in the form

*x* + 2*y* − *z* = 4;

this is the equation of the plane.

In general, let **n** = (*a*, *b*, *c*) be normal to a given plane that contains the point *P*(*x*_{1}, *y*_{1}, *z*_{1}). Then an arbitrary point *Q*(*x*, *y*, *z*) lies in the plane if and only if the vectors and **n** are orthogonal; that is, if and only if .

Since , this condition can be written in the form

so

(*x* − *x*_{1}) × *a* + (*y* − *y*_{1}) × *b* + (*z* − *z*_{1}) × *c* = 0.

This equation can be rearranged in the form

*ax* + *by* + *cz* = *ax*_{1} + *by*_{1} + *cz*_{1},

that is,

*ax* + *by* + *cz* = *d*,

where *d* = *ax*_{1} + *by*_{1} + *cz*_{1}.

## Theorem 1

The equation of the plane that contains the point (*x*_{1}, *y*_{1}, *z*_{1})

and has **n** = (*a*, *b*, *c*) as a normal is

*ax* + *by* + *cz* = *d*,

where *d* = *ax*_{1} + *by*_{1} + *cz*_{1}.

Once we know the equation of a plane, we can ‘read off’ the components of a normal vector as they are the coefficients of *x*, *y* and *z* in the equation. For instance, one normal to the plane with equation *x* − 2*y* + 3*z* = 7 is **n** = (1, −2, 3).

We shall not make use of the formula for *d* given in the above theorem. In practice, it is simpler to use the following corollary.

## Corollary to Theorem 1

The equation of the plane that contains the point (*x*_{1}, *y*_{1}, *z*_{1})

and has **n** = (*a*, *b*, *c*) as a normal is

where **x** = (*x*, *y*, *z*) and **p** = (*x*_{1}, *y*_{1}, *z*_{1}).

**Proof**

Since *ax* + *by* + *cz* = *d*, from Theorem 1, it follows that **x · n** = **p · n**.

## Example 7

Determine the equation of the plane in ^{3} that contains the point (1, −1, 4) and has (2, −2, 3) as a normal.

### Answer

The equation of the plane is

where **x** = (*x*, *y*, *z*), **p** = (1, −1, 4) and **n** = (2, −2, 3); in other words, the equation of the plane is

that is,

2*x* − 2*y* + 3*z* = 1 × 2 + (−1) × (−2) + 4 × 3,

or

2*x* − 2*y* + 3*z* = 16.

## Example 45

Determine the equation of each of the following planes:

**(a)**the plane that contains the point (1, 0, 2) and has (2, 3, 1) as a normal;**(b)**the plane that contains the point (−1, 1, 5) and has (4, −2, 1) as a normal.

### Answer

We use the formula

for the equation of a plane, where **x** = (*x*, *y*, *z*), **n** is a normal to the plane and **p** is a point in the plane.

**(a)**Here**n**= (2, 3, 1) and**p**= (1, 0, 2), so the equation of the plane isThis can be expressed in the form

2

*x*+ 3*y*+*z*= 1 × 2 + 0 × 3 + 2 × 1,that is,

2

*x*+ 3*y*+*z*= 4.**(b)**Here**n**= (4, −2, 1) and**p**= (−1, 1, 5), so the equation of the plane isThis can be expressed in the form

4

*x*− 2*y*+*z*= (−1) × 4 + 1 × (−2) + 5 × 1,that is,

4

*x*− 2*y*+*z*= −1.

We can use the general form of the equation of a plane to find the *equation of a plane that contains three given points*, as follows. We assume that the equation of the plane is *ax* + *by* + *cz* = *d*, and then substitute the coordinates of the three points in turn into this equation; this gives three simultaneous equations for *a*, *b* and *c*, which we solve.

## Example 46

Determine the equation of the plane that contains the points (3, 0, 0), (0, 4, 0) and (3, 4, 5).

Write down a vector that is normal to this plane.

### Answer

Let the equation of the plane be

where *a*, *b*, *c* and *d* are real, and *a*, *b* and *c* are not all zero.

Substituting the coordinates of each of the points in turn into equation (S.6), we obtain

From the first two equations we obtain

Substituting these expressions into the third equation, we obtain

so

*d* + *d* + 5*c* = *d*,

hence

Substituting these expressions for *a*, *b* and *c* into equation (S.6), we see that the equation of the plane can be written as

Dividing by *d*, we obtain

It follows that one possible normal vector is .