Lesson Explainer: Equations of Parallel and Perpendicular Planes Mathematics

In this explainer, we will learn how to find the equation of a plane that is parallel or perpendicular to another plane given its equation or some properties.

Before starting to look at parallel and perpendicular planes, you should already be familiar with finding the equation of a plane. Let us recap the various forms of equations of a plane:

  • The vector form is ⃑𝑛⋅⃑𝑟=⃑𝑛⋅⃑𝐴, where ⃑𝑛=𝑛,𝑛,𝑛 is a nonzero normal vector of a plane, ⃑𝑟=(𝑥,𝑦,𝑧) is the position vector of any point in the plane, and ⃑𝐴=(𝑥,𝑦,𝑧) is the position vector of the point 𝐴 with coordinates (𝑥,𝑦,𝑧) that belongs to the plane.
  • The general form is 𝑛𝑥+𝑛𝑦+𝑛𝑧+𝑑=0, where 𝑛, 𝑛, and 𝑛 are the components of a normal vector of the plane and 𝑑 is a constant.
  • The parametric form is the set of three equations: 𝑥=𝑥+𝑡𝑢+𝑡𝑣,𝑦=𝑦+𝑡𝑢+𝑡𝑣,𝑧=𝑧+𝑡𝑢+𝑡𝑣, where the point 𝐴 of coordinates (𝑥,𝑦,𝑧) belongs to the plane, ⃑𝑢=𝑢,𝑢,𝑢 and ⃑𝑣=𝑣,𝑣,𝑣 are two nonzero and noncollinear vectors in the plane, and 𝑡 and 𝑡 are scalars. A normal vector of the plane is then given as ⃑𝑛=⃑𝑢×⃑𝑣.
  • The intercept form is 𝑥𝑎+𝑦𝑏+𝑧𝑐=1, where 𝑎, 𝑏, and 𝑐 are the 𝑥-, 𝑦-, and 𝑧-intercepts of the plane, provided they all exist, and 1𝑎, 1𝑏, and 1𝑐 are the components of a normal vector of the plane.

Except the parametric equations that define the plane with one point and two nonzero and noncollinear vectors, the equations are based on the fact that any vector in the plane (e.g., 𝐴𝑀, with 𝐴 being a known point in the plane and 𝑀 any point in the plane with coordinates (𝑥,𝑦,𝑧)) is perpendicular to a normal vector of the plane. Therefore, we have ⃑𝑛⋅𝐴𝑀=0.

When defining a plane, we see the importance of its normal vector (or rather, of one normal vector of the plane as any nonzero vector parallel to the normal vector is also a normal vector). The normal vector is also central in identifying whether two planes are parallel or perpendicular.

Definition: Parallel and Perpendicular Planes

Two distinct planes are parallel if they have parallel nonzero normal vectors, which means that they have no points of intersection. Two planes are perpendicular if their normal vectors are perpendicular.

It is worth noting that two coincident planes have parallel nonzero normal vectors as well; the equation of one of the planes is then a multiple of the equation of the other.

This property is illustrated in the figure below, showing (a) the two parallel (left) and perpendicular (right) planes in three dimensions and (b) a top view of both planes 𝑃 and 𝑄, showing their normal vectors.

Let us use this property to complete the equations of two planes so that they are parallel.

Example 1: Finding the Condition for Two Planes to Be Parallel

Given that the plane 𝐾𝑧+2𝑥+3𝑦=−4 is parallel to the plane 𝐿𝑦−2𝑥−2𝑧=3, find the values of 𝐾 and 𝐿.

Answer

Let us first call 𝑃 the plane of equation 𝐾𝑧+2𝑥+3𝑦=−4 and 𝑃 that of equation 𝐿𝑦−2𝑥−2𝑧=3. If planes 𝑃 and 𝑃 are parallel, then their normal vectors must be parallel. Let us call the normal vector of 𝑃⃑𝑛 and that of 𝑃⃑𝑛. The components of these normal vectors can be extracted from the coefficients of the corresponding variables in the general equations of both planes. Beware, in both equations the terms are not given in the usual order.

We find ⃑𝑛=(2,3,𝐾) and ⃑𝑛=(−2,𝐿,−2).

The vectors ⃑𝑛 and ⃑𝑛 are parallel if there is a nonzero scalar 𝑚 such that ⃑𝑛=𝑚⃑𝑛. This vector equation yields three scalar equations when equating the three components of both vectors:

2=−2𝑚,3=𝑚𝐿,𝐾=−2𝑚.(1)(2)(3)

Equation (1) gives 𝑚=−1, from which it follows that 𝐿=−3 and 𝐾=2.

Answer

We find that 𝐾=2 and 𝐿=−3.

We will now find the equation of a plane passing through a given point and parallel to another plane.

Example 2: Finding the General Equation of a Plane That Is Parallel to Another Plane and Passes through a Given Point

Find the equation of the plane passing through the point (𝑎,𝑏,𝑐) and parallel to the plane 𝑥+𝑦+𝑧=0.

  1. 𝑎𝑥+𝑏𝑦+𝑐𝑧=1
  2. 𝑥+𝑦+𝑧=𝑎+𝑏+𝑐
  3. 𝑥+𝑦+𝑧+𝑎+𝑏+𝑐=0
  4. 𝑎𝑥+𝑏𝑦+𝑐𝑧=𝑎+𝑏+𝑐
  5. 𝑥𝑎=𝑦𝑏=𝑧𝑐

Answer

We are given a point on the plane and the equation of another plane that is parallel to the plane we are trying to find the equation of. We can write the general equation of a plane if we know the coordinates of one of its points and its normal vector. We can find the latter by using the fact that parallel planes have parallel normal vectors.

The plane with the general equation 𝑥+𝑦+𝑧=0 has a normal vector of components (1,1,1). Any nonzero vector parallel to this vector is a normal vector to the plane we want to write an equation of. The simplest parallel vector we can find is this very same vector, which gives for the equation of the plane 𝑥+𝑦+𝑧+𝑑=0, where 𝑑 is a constant to be found. For this, we use the coordinates (𝑎,𝑏,𝑐) of the point that is in the plane. Its coordinates must satisfy the equation of the plane, and therefore we have 𝑎+𝑏+𝑐+𝑑=0𝑑=−(𝑎+𝑏+𝑐). This gives, as the final equation, 𝑥+𝑦+𝑧−(𝑎+𝑏+𝑐)=0, or, rearranging, 𝑥+𝑦+𝑧=𝑎+𝑏+𝑐.

Answer

The general equation of the plane is 𝑥+𝑦+𝑧−(𝑎+𝑏+𝑐)=0, which can be written as 𝑥+𝑦+𝑧=𝑎+𝑏+𝑐.

Let us now turn our attention to perpendicular planes.

Example 3: Finding the Condition for Two Planes to Be Perpendicular

Given that the plane 3𝑥−3𝑦−3𝑧=1 is perpendicular to the plane 𝑎𝑥−2𝑦−𝑧=4, find the value of 𝑎.

Answer

If the two planes are perpendicular, then their normal vectors must be perpendicular. It follows that the dot product of both normal vectors is zero. Let us find, from both general equations, the components of both normal vectors. For the first, we find (3,−3,−3) and, for the second, we find (𝑎,−2,−1).

The dot product of these two vectors is zero, which gives (3,−3,−3)⋅(𝑎,−2,−1)=03𝑎+(−3)×(−2)+(−3)×(−1)=03𝑎+9=0𝑎=−3.

Answer

The value of 𝑎 is −3.

While a plane is fully defined if we know that it is parallel to another plane and passes through a given point, there is an infinite number of planes that are perpendicular to another plane, as illustrated in the figure below, where all red planes are perpendicular to 𝑃, and each of them represents a whole family of parallel planes.

Property: Family of Planes Perpendicular to a Given Plane

The normal vectors of all the planes perpendicular to a plane 𝑃 are parallel to 𝑃, and the normal vector of 𝑃 is parallel to all the planes perpendicular to 𝑃.

It follows that to unambiguously define a plane with perpendicularity, the plane must be either of the following:

  • perpendicular to two nonparallel planes and passing through a given point,
  • perpendicular to a plane 𝑃 and passing through two distinct points 𝐴 and 𝐵 such that 𝐴𝐵 and the normal vector of 𝑃 are noncollinear.

Let us look at these two cases with the following two examples.

Example 4: Finding the Equation of a Plane That Is Perpendicular to Two Planes and Passes through a Given Point

Find the general equation of the plane that passes through the point (2,8,1) and is perpendicular to the two planes −6𝑥−4𝑦+6𝑧=−5 and 5𝑥+3𝑦−6𝑧=3.

Answer

The plane we are considering is perpendicular to the two planes of equations −6𝑥−4𝑦+6𝑧=−5 and 5𝑥+3𝑦−6𝑧=3. Their respective normal vectors are (−6,−4,6) and (5,3,−6). Since we are looking for a plane perpendicular to both of these nonparallel planes, its normal vector, ⃑𝑛, must be perpendicular to both of their nonparallel normal vectors. There are two ways to find a vector perpendicular to two nonparallel vectors.

  1. We say that the dot product between ⃑𝑛 and either of these two vectors must be zero. With ⃑𝑛=𝑛,𝑛,𝑛, the normal vector of the plane, it gives −6𝑛−4𝑛+6𝑛=05𝑛+3𝑛−6𝑛=0. To solve this system of two equations, we need to cancel one term. Adding the two equations can cancel the term of 𝑛 and yields −𝑛−𝑛=0,𝑛=−𝑛.i.e. We have found 𝑛 in terms of 𝑛. Substituting 𝑛 with −𝑛 into the second equation and rearranging gives us 𝑛 in terms of 𝑛: 𝑛=𝑛3. The components of the normal vector can now be expressed in terms of 𝑛: ⃑𝑛=𝑛,−𝑛,𝑛3. As 𝑛 can take any nonzero real value, the direction ratios of the components of the normal vector are 1,−1,13. Multiplying the ratios by 3 gives ⃑𝑛=(3,−3,1).
  2. We find the cross product of these vectors, which gives a vector perpendicular to both: ⃑𝑛=(−6,−4,6)×(5,3,−6)=||||⃑𝑖⃑𝑗⃑𝑘−6−4653−6||||=((−4)(−6)−3×6)⃑𝑖−((−6)(−6)−5×6)⃑𝑗+((−6)×3−5×(−4))⃑𝑘=6⃑𝑖−6⃑𝑗+2⃑𝑘=(6,−6,2). This normal vector can be simplified by dividing its components by 2, giving (3,−3,1) as we found before.

Using ⃑𝑛=(3,−3,1) to write a general equation of the plane 𝑛𝑥+𝑛𝑦+𝑛𝑧+𝑑=0 gives 3𝑥−3𝑦+𝑧+𝑑=0.

We know that the plane passes through the point (2,8,1). Therefore, the coordinates (2,8,1) must satisfy the equation of the plane. Substituting these in gives 3×2−3×8+1+𝑑=0𝑑=17.

Answer

The equation of the plane that passes through the point (2,8,1) and is perpendicular to the two planes −6𝑥−4𝑦+6𝑧=−5 and 5𝑥+3𝑦−6𝑧=3 is 3𝑥−3𝑦+𝑧+17=0.

Let us recap the method used in the previous example.

How To: Finding the Equation of a Plane Perpendicular to Two Other Planes

If a plane is perpendicular to two other nonparallel planes with normal vectors ⃑𝑛, and ⃑𝑛, then the components of its normal vector, ⃑𝑛, can be found by two different methods:

  1. solving (in terms of one of its components) the system of equations derived from ⃑𝑛⋅⃑𝑛=0, and ⃑𝑛⋅⃑𝑛=0, then, dividing the components of ⃑𝑛 in terms of one of its components by this component (which was 𝑛 in the previous example), which gives the direction ratio of ⃑𝑛,
  2. finding the cross product of ⃑𝑛 and ⃑𝑛: ⃑𝑛=⃑𝑛×⃑𝑛.

Finally, the equation of the plane is completed using the coordinates of one point of the plane.

Let us see how we can find the equation of a plane that passes through two points and that is perpendicular to another plane in the next example.

Example 5: Finding the Equation of a Plane That Is Perpendicular to Another Plane and Passes through Two Given Points

Find the general form of the equation of the plane that passes through the two points 𝐴(2,5,4) and 𝐵(3,−3,5) and that is perpendicular to the plane 2𝑥−𝑦+2𝑧−2=0.

Answer

To find the general equation of a plane, we need a vector normal to the plane and a point on the plane. However, we are not given a normal vector to the plane we are interested in (let us call this plane 𝑃), but instead we are given the equation of a plane perpendicular to 𝑃; let us call it 𝑄. 𝑄∶2𝑥−𝑦+2𝑧=0.

The normal vector of 𝑄, ⃑𝑛, is parallel to 𝑃. As we will see below, there are then two slightly different ways to find the general equation of plane 𝑃.

Let us first determine the normal vector of the plane 𝑄∶2𝑥−𝑦+2𝑧−2=0; it is the vector ⃑𝑛 with components (2,−1,2). As plane 𝑃 is perpendicular to plane 𝑄, ⃑𝑛 is parallel to plane 𝑃. According to the properties of 3D vectors, it can be considered that plane 𝑃 includes a vector of the same components as those of ⃑𝑛.

Now, we need two nonparallel vectors included in plane 𝑃 to find the normal vector ⃑𝑛, given by their cross product.

We further know that the plane passes through the two points 𝐴(2,5,4) and 𝐵(3,−3,5). The vector 𝐴𝐵=(1,−8,1) is therefore included in the plane.

The first way to find the equation of 𝑃 consists in saying that as 𝐴𝐵 and ⃑𝑛 are noncollinear and both included in plane 𝑃, the normal vector of 𝑃, ⃑𝑛, is given by the cross product of 𝐴𝐵 and ⃑𝑛: ⃑𝑛=𝐴𝐵×⃑𝑛=||||⃑𝑖⃑𝑗⃑𝑘1−812−12||||=(−15,0,15).

To write the general equation, we can take 115⃑𝑛=(−1,0,1) as the normal vector, which gives −𝑥+𝑧+𝑑=0, for some constant 𝑑.

As the points 𝐴(2,5,4) and 𝐵(3,−3,5) are in the plane, their coordinates satisfy the equation. Substituting the coordinates of one of the points, for example, those of point 𝐵, into the equation allows us to find the value of 𝑑 (we would find the same value using the coordinates of 𝐴, of course): −3+5+𝑑=0𝑑=−2.

The equation of the plane is therefore −𝑥+𝑧−2=0.

Another way consists in saying that, for any point 𝑀(𝑥,𝑦,𝑧) in the plane, the dot product of the vector 𝐴𝑀 (or 𝐵𝑀) with the normal vector of the plane (which is given by 𝐴𝐵×⃑𝑛—see above) is zero. If you are familiar with the triple scalar product, this method is equivalent to saying that the vectors 𝐴𝑀 (or 𝐵𝑀), 𝐴𝐵, and ⃑𝑛 are coplanar, which means that their scalar triple product is zero. We have 𝐴𝑀⋅𝐴𝐵×⃑𝑛=0||||𝑥−2𝑦−5𝑧−41−812−12||||=0(𝑥−2)(−16+1)−(𝑦−5)(2−2)+(𝑧−4)(−1+16)=0−15𝑥+15𝑧−30=0.

Dividing both sides of this equation by 15 gives −𝑥+𝑧−2=0.

Answer

The general equation of the plane that passes through the two points 𝐴(2,5,4) and 𝐵(3,−3,5) and that is perpendicular to the plane 2𝑥−𝑦+2𝑧−2=0 is −𝑥+𝑧−2=0.

Let us summarize the two methods used in the last example.

How To: Finding the General Equation of a Plane That Is Perpendicular to Another Plane and Passes through Two Points

If we are told that a plane is perpendicular to another plane of known normal vector and passes through two distinct points 𝐴 and 𝐵 such that 𝐴𝐵 and ⃑𝑛 are noncollinear, then there are two methods to find the equation of the plane.

The 1st method: The normal vector of a plane perpendicular to another plane of normal vector ⃑𝑛 and that passes through two points 𝐴 and 𝐵 is given by ⃑𝑛=𝐴𝐵×⃑𝑛.

Vector ⃑𝑛 is a nonzero vector since 𝐴𝐵 and ⃑𝑛 are noncollinear.

The equation can then be found using the components of ⃑𝑛 and the coordinates of either 𝐴 or 𝐵.

The 2nd method: For any point 𝑀(𝑥,𝑦,𝑧) in a plane perpendicular to another plane of normal vector ⃑𝑛 and that passes through two points 𝐴 and 𝐵 such that 𝐴𝐵 and ⃑𝑛 are noncollinear, the three vectors 𝐴𝑀 (or 𝐵𝑀), 𝐴𝐵, and ⃑𝑛 are coplanar. It means that 𝐴𝑀⋅𝐴𝐵×⃑𝑛=0𝐵𝑀⋅𝐴𝐵×⃑𝑛=0.and

Writing out one of these equations involving a triple scalar product directly gives a general equation of the plane.

Note that the parametric equations of a plane perpendicular to another plane of normal vector ⃑𝑛 and that passes through two points 𝐴 and 𝐵 such that 𝐴𝐵 and ⃑𝑛 are noncollinear can be easily found using the two vectors in the plane, 𝐴𝐵 and ⃑𝑛.

Key Points

  • Two parallel planes 𝑃 and 𝑄 are characterized by parallel normal vectors, which means that ⃑𝑛=𝑘⃑𝑛, where ⃑𝑛 and ⃑𝑛 are the normal vectors of 𝑃 and 𝑄, respectively, and 𝑘 is a nonzero real number. 𝑃 and 𝑄 are parallel if they are distinct, that is, if they do not have any intersection points. Two intersecting planes with parallel normal vectors are coincident.
  • Any two perpendicular planes 𝑃 and 𝑄 have perpendicular normal vectors, which means that the dot product of their normal vectors, ⃑𝑛 and ⃑𝑛, respectively, is zero: ⃑𝑛⋅⃑𝑛=0.
  • The normal vectors of all the planes perpendicular to a plane 𝑃 are parallel to 𝑃.
  • The normal vector of a plane 𝑃 is parallel to all the planes perpendicular to 𝑃.

Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy.