It is known that the solution of a differential equation can be displayed graphically as a family of integral curves in the plane, which is usually called the phase plane. The properties of planes are a subject of study in Calculus III. In this post, we’ll investigate equations of planes, and explain how they can be employed.

Lines and Line Segments in Space

Throughout this post, we will represent an arbitrary point in space as P_0 = (x_0,y_0,z_0) , where one can think of P_0 as the name of the point and (x_0,y_0,z_0) as its position in space. We will use \overline{i}, \overline{j}, \overline{k} to represent directions in the plane (i.e. \overline{i} is a unit vector along the x-axis, \overline{j} is a unit vector along the y-axis, and \overline{k} is a unit vector along the z-axis). Given any two points in space, say P = (x_0,y_0,z_0) and Q = (x_1,y_1,z_1) , we write PQ to represent the line from P to Q. We will begin with a review of lines and line segments in space, which we will then build into the equation of a plane in space. It will be advantageous for the reader to have a strong grasp of the cross product. If necessary, now is a good time to review the cross product, as well as the dot product and their properties.

We defined a line on a plane as a point and a number that gives the slope of the line. In three-dimensional space, we define a line as a point and a vector that gives the direction of the line. Let L be a line in space passing through a point P_0 = (x_0,y_0,z_0) parallel to the vector \overline{v} = v_1 \overline{i} + v_2 \overline{j} + v_3 \overline{k} . Then, L consists of all points P = (x,y,z) for which P_0 P is parallel to \overline{v} . Then, P_0 P = t \overline{v} for some parameter t, which depends on the location of P on the line. We can expand the equation P_0 P = t \overline{v} to obtain

(x-x_0)\overline{i} + (y-y_0)\overline{j} + (z-z_0)\overline{k} = t(v_1 \overline{i} + v_2 \overline{j} + v_3 \overline{k})

which we can rewrite as

x\overline{i} + y\overline{j} + z\overline{k} = x_0\overline{i} + y_0\overline{j} + z_0\overline{k} + t(v_1 \overline{i} + v_2 \overline{j} + v_3 \overline{k})

Let \overline{r}(t) be the position vector of the point P = (x,y,z) (i.e. \overline{r}(t)=x\overline{i}+y\overline{j}+z\overline{k} ), and \overline{r}_0 be the position vector of the point P_0 = (x_0,y_0,z_0) (i.e \overline{r}_0=x_0 \overline{i}+y_0 \overline{j}+z_0 \overline{k} ). Then, we can write the equation for a line in space as

\overline{r}(t) = \overline{r}_0 + t \overline{v},, \quad -\infty < t < \infty

So, we obtain the following parametric equations for line L:

x = x_0 + t v_1 ,,\quad y = y_0 + t v_2 ,,\quad z = z_0 + t v_3 ,,\quad -\infty < t < \infty

Imagine a bee, starting at point P_0 = (x_0,y_0,z_0) and flying along a line \overline{r}(t) in space. It can be very useful to think of \overline{r}(t) as the bee’s path from its starting point in the direction of \overline{v} . Through a little algebraic manipulation, this becomes even more apparent:

\overline{r}(t) = \overline{r}_0 + t \overline{v} = \overline{r}_0 + t |\overline{v}| \dfrac{\overline{v}}{|\overline{v}|} \qquad\qquad (1)

Thus, the bee’s position at time t is its initial position plus the distance it has moved (speed times time) in the direction of \dfrac{\overline{v}}{|\overline{v}|} .

Example 1

A hummingbird, hovering at the origin, sees a flower in the direction (1,1,1) and wants to move in that direction at its maximum flight speed of 60 miles per hour. What is the position of the hummingbird after 5 seconds?

Solution

The bird begins at the origin, so

\overline{r}_0 = \overline{0}

This is not simply the number 0, but a compact way to write 0\overline{i}+0\overline{k}+0\overline{j} . Since the bird is traveling in the direction of (1,1,1) , the bird is flying parallel to the vector

\overline{v} = \overline{i} + \overline{j} + \overline{k}

This vector has magnitude

|\overline{v}| = \sqrt{1^2+1^2+1^2} = \sqrt{3}

Thus, the direction of the bird’s flight, \dfrac{\overline{v}}{|\overline{v}|} , is given by

\dfrac{\overline{v}}{|\overline{v}|} = \dfrac{1}{\sqrt{3}},\overline{i} + \dfrac{1}{\sqrt{3}},\overline{j} + \dfrac{1}{\sqrt{3}},\overline{k}

From equation (1), we see that the position of the bird at any time t is given by

\overline{r}(t) = \overline{r}_0 + t (speed) \dfrac{\overline{v}}{|\overline{v}|} = \overline{0} + t (60) \left( \dfrac{1}{\sqrt{3}},\overline{i} + \dfrac{1}{\sqrt{3}},\overline{j} + \dfrac{1}{\sqrt{3}},\overline{k} \right) = 20\sqrt{3},t ( \overline{i} + \overline{j} + \overline{k} )

So, when t=5, the bird is at the position

\overline{r}(5) = 100\sqrt{3}( \overline{i} + \overline{j} + \overline{k} )

Equations of Planes in Space

Recall that a vector is normal to a plane if it is perpendicular to every point on the plane. Let P_0 = (x_0,y_0,z_0) be a point on a plane, P = (x,y,z) some other arbitrary point on the plane, and \overline{n} = A\overline{i}+B\overline{j}+C\overline{k} a vector normal to the plane. Then, the plane can be defined by the following equations:

Vector equation of a plane: \quad \overline{n}\cdot \overline{P_0 P} = 0

Component equation of a plane: \quad A(x-x_0) + B(y-y_0) + C(z-z_0) = 0

Simplified component equation: \quad A x + B y + C z = D , where D = A x_0 + B y_0 + C z_0

There are several ways to determine the equation of a plane, depending on the information presented to you. Suppose you are given three points that lie in the plane. Then, all we need to do is find a vector that is normal to the plane passing through those points. Alternatively, if we know the equation of a vector normal to the plane, say \overline{n} = A\overline{i}+B\overline{j}+C\overline{k} , and a point on the plane, P_0 = (x_0, y_0, z_0 ) , it becomes rather easy to find the equation of the plane. We use the coordinates of the points and the normal vector and simply input them into the component equation as defined above. However, if you are not given a vector normal to the plane, you must do a little more work to find one.

Example 2

In this example, we investigate how to find the equation of a plane, given three points on the plane.

Find the equation of the plane passing through the points Q = (2,0,0) , R = (0,3,0) , and S = (0,0,4) , which is illustrated in the image below:

Solution

Recall that for any two vectors \overline{v}_1= \alpha\overline{i} + \beta\overline{j} + \gamma\overline{k} and \overline{v}_2= a\overline{i} + b\overline{j} + c\overline{k} , we compute their cross product \overline{v}_1 \times \overline{v}_2 by taking the determinant of the matrix:

\overline{v}_1 \times \overline{v}_2 = \left| \begin{array}{ccc}\overline{i} & \overline{j} & \overline{k} \ \alpha & \beta & \gamma \ a & b & \end{array} \right|

which is given by the following equation:

\left| \begin{array}{ccc} \overline{i} & \overline{j} & \overline{k} \ \alpha & \beta & \gamma \ a & b & c \end{array} \right| = \left| \begin{array}{cc} \beta & \gamma \ b & c \end{array} \right|\overline{i} ,-, \left| \begin{array}{cc} \alpha & \gamma \ a & c \end{array} \right|\overline{j} ,+, \left| \begin{array}{cc} \alpha & \beta \ a & b \end{array} \right|\overline{k} = (\beta c -\gamma b)\overline{i} - (\alpha c - \gamma a)\overline{j} + (\alpha b - \beta a)\overline{k}

From the definition of the equation for a plane, we know that our equation must satisfy

A(x-x_0) + B(y-y_0) + C(z-z_0) = 0

Now, we already have three points to choose from for (x_0,y_0,z_0) , so all we need to do is find a vector \overline{n} = A\overline{i}+B\overline{j}+C\overline{k} that is perpendicular to the plane. Recall that the cross product of two vectors is perpendicular (you might also hear this referred to as orthogonal) to each vector. We don’t have any vectors to work with yet, but we can create two vectors by taking the line segments QR ad SR. It follows that since all three points are in the plane, both of those vectors must be in the plane, and so their cross product must be perpendicular to the plane.

We calculate the vectors \overline{QR} and \overline{SR} as follows:

\overline{QR} = 2 \overline{i} - 3 \overline{j}

\overline{SR} = - 3 \overline{j} + 4 \overline{k}

Finally, we take their cross product to find a vector perpendicular to the plane:

\overline{QR} \times \overline{SR} = \left| \begin{array}{ccc} \overline{i} & \overline{j} & \overline{k} \ 2 & -3 & 0 \ 0 & -3 & 4 \end{array} \right| = \left| \begin{array}{cc} -3 & 0 \ -3 & 4 \end{array} \right|\overline{i} ,-, \left| \begin{array}{cc} 2 & 0 \ 0 & 4 \end{array} \right|\overline{j} ,+, \left| \begin{array}{cc} 2 & -3 \ 0 & -3 \end{array} \right|\overline{k} = -12\overline{i} - 8\overline{j} - 6\overline{k}

So the vector \overline{n} normal to the plane is found to be

\overline{n} = A\overline{i}+B\overline{j}+C\overline{k} = -12\overline{i} - 8\overline{j} - 6\overline{k}

Now, we see that

A = -12,,\quad B = -8,,\quad C = - 6

Finally, inputting these into the equation for the plane and choosing Q = (2,0,0) as (x_0,y_0,z_0), the equation of the plane is given by

-12 (x-2) - 8 (y-0) - 6 (z-0) = 0

-12 x - 8 y - 6 z = - 24

6 x + 4 y + 3 z = 12

Properties of the Cross Product

Before proceeding any further, we must outline a few facts that will be incredibly useful in our study of planes. As opposed to the algebraic approach of taking the determinant of a 3 x 3 matrix, there is a more geometric way to think about the norm of the cross product of two vectors \overline{a} and \overline{b} given by the equation

|\overline{a} \times \overline{b} | = |\overline{a} | | \overline{b} | sin\theta

where \theta is the angle between the vectors.

If \overline{a} and \overline{b} are parallel to one another, then the angle between them is either 0 degrees or 180 degrees.

Since sin\theta = 0 for \theta = 0 and \theta = 180 , this implies that

|\overline{a} \times \overline{b} | = 0

This in turn implies that

\overline{a} \times \overline{b} = 0

This enables us to state some nice facts concerning the cross product and parallel vectors:

\text{Fact 1:} \quad \text{If} \quad \overline{a} \times \overline{b} = 0,, \quad \text{then} \quad \overline{a} \quad \text{and} \quad \overline{b} \quad \text{are parallel vectors}

\text{Fact 2:} \quad \text{Provided} \quad \overline{a} \times \overline{b} \neq 0,, \quad \text{then} \quad \overline{a} \times \overline{b} \quad \text{is orthogonal to both} \quad \overline{a} \quad \text{and} \quad \overline{b}

There are also a few geometric applications to the cross product. Suppose we have three vectors, \overline{a} , \overline{b} , and \overline{c} . Then, we can form the three-dimensional object below, called a parallelepiped:

The area of the parallelogram created by \overline{a} and \overline{c} (that is, the face of the parallelepiped “closest” to the front of the page) is given by

\text{Area} = |\overline{a} \times \overline{c} |

and the volume of the parallelepiped is given by

\text{Volume} = |\overline{a} \cdot (\overline{b} \times \overline{c}) |

With this in mind, it becomes easy to see if three vectors lie in the same plane. If they do, then the volume of the parallelepiped will be 0. If they do not, then the volume will be some positive value. We will see a direct application of this in the following example, in which we check if three vectors lie in the same plane.

Example 3

Determine whether the three vectors \overline{a} = \overline{i}+4\overline{j}-7\overline{k} , \overline{b} = 2\overline{i}-\overline{j}+ 4\overline{k} , and \overline{c} = -9\overline{j}+18\overline{k} lie in the same plane.

Solution

As we saw above, all we need to do in order to prove that the three vectors lie in the same plane is to show that the volume of the parallelepiped formed by the three vectors is 0. If we calculate the volume and it is not zero, then the three vectors do not lie in the same plane. To calculate this volume, we use the formula for the volume of the parallelepiped along with the method used in example 2 to calculate the cross product:

\overline{b} \times \overline{c} = \left| \begin{array}{ccc} \overline{i} & \overline{j} & \overline{k} \ 2 & -1 & 4 \ 0 & -9 & 18 \end{array} \right| = (-18+36) \overline{i} - (36-0)\overline{j} + (-18-0)\overline{k} = 18 \overline{i} - 36 \overline{j} - 18 \overline{k}

|\overline{a} \cdot (\overline{b} \times \overline{c}) | = 1\cdot 18 + 4\cdot(-36) + (-7)\cdot(-18) = 18 - 144 + 126 = 0

The volume of the parallelepiped is 0, so the three vectors lie in the same plane!

Lines of Intersection

We’ve already seen that two lines are parallel if and only if they have the same direction. Similarly, two planes are parallel if their normals are parallel. That is, if \overline{n}_1 is normal to a plane P and \overline{n}_2 is normal to a plane Q, then P is parallel to Q if there exists some scalar k such that \overline{n}_1 = k \overline{n}_2 . If no such scalar exists, then P and Q are not parallel to each other.

Example 4

Find a vector parallel to the line of intersection of the planes 3x-6y-2z=15 and 2x+y-2z=5 .

Solution

The line of intersection of two planes is perpendicular to the normal of each plane. Observe that by the simplified component equation, the normal of the first plane is \overline{n}_1 = 3\overline{i}-6\overline{j}-2\overline{k} , and the normal of the second plane is \overline{n}_2 = 2\overline{i}+\overline{j}-2\overline{k} . Since the line of intersection is perpendicular to both normals, it follows that it must be parallel to their cross product, \overline{n}_1 \times \overline{n}_2 . Thus, \overline{n}_1 \times \overline{n}_2 is a vector parallel to the planes’ line of intersection, or

\overline{n}_1 \times \overline{n}_2 = \left| \begin{array}{ccc} \overline{i} & \overline{j} & \overline{k} \ 3 & -6 & -2 \ 2 & 1 & -2 \end{array} \right| = 14 \overline{i} + 2 \overline{j} + 15 \overline{k}

Distance from a Point to a Plane

Let Q be a point on a plane whose normal is \overline{n} . Then, the distance from any point R in space to the plane is given by

d = \left| \overline{QR} \cdot \dfrac{\overline{n}}{|\overline{n}|} \right|

Example 5

Find the distance from the point S = (1,1,3) to the plane 3x+2y+6z = 6 .

Solution

The main idea is to find a point Q on the plane and then to calculate the length of the vector projection of QS onto a vector \overline{n} normal to the plane. We begin by observing that the vector normal to the plane is given by \overline{n} = 3\overline{i}+2\overline{j}+6\overline{k} and the magnitude of the normal vector is

|\overline{n}| = \sqrt{3^2 + 2^2 + 6^2} = \sqrt{49} = 7

To find a point on the plane, we simply need to choose values x, y, and z that satisfy the equation of the plane. We will choose the point Q = (0,3,0) . Then, the vector \overline{QS} is given by

\overline{QS} = (1-0) \overline{i} + (1-3) \overline{j} + (3-0) \overline{k} = \overline{i} - 2\overline{j} + 3\overline{k}

Now, all that remains is to input this into the equation for the distance from a point to a plane. Finally, we have that the distance from the point S = (1,1,3) to the plane 3x+2y+6z = 6 is

d = \left| \overline{QS} \cdot \dfrac{\overline{n}}{|\overline{n}|} \right| = \dfrac{1}{7},\left| (\overline{i} - 2\overline{j} + 3\overline{k})(3\overline{i}+2\overline{j}+6\overline{k}) \right| = \dfrac{1}{7},(3-4+18) = \dfrac{17}{7}

Wrapping Up Equations of Planes

We will conclude our discussion of equations of planes in space. As you may have noticed, the cross product is a concept that becomes important when studying vectors and planes. The interested student should reread this article and try to prove the two facts and the properties of the cross product to get a better understanding of the mathematics behind them.

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