Motivating Questions
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How is a plane defined in terms of measurements of points and vectors?
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What different ways are there to determine a plane through geometric information?
Section 9.6 Planes in Space
Subsection 9.6.1 Introduction
In Section 9.5, we saw how to describe a line in \(\R^n\) by setting an initial point and allowing unrestricted movement in a direction given by a direction vector. In this section’s Preview Activity, we will be looking at what happens when we allow only movement that is perpendicular to a given direction. Since we will specify the given direction by a vector, we will find the perpendicular directions using orthogonal vectors.
Preview Activity 9.6.1.
We will consider what happens when we allow movement in \(\R^3\) with the restriction that the movement must be orthogonal to \(\vv=\langle 1,2,3 \rangle \text{.}\)
(a)
Find values for \(a_0\) and \(b_0\) such that \(\langle 0,a_0,b_0 \rangle\) is orthogonal to \(\langle 1,2,3\rangle\text{.}\)
(b)
Find values for \(c_0\) and \(d_0\) such that \(\langle c_0,d_0,0 \rangle\) is orthogonal to \(\langle 1,2,3\rangle\text{.}\)
(c)
Find values for \(c_1\) and \(d_1\) such that \(\langle c_1,d_1,1 \rangle\) is orthogonal to \(\langle 1,2,3\rangle\text{.}\)
(d)
Find two other values for each of \(c\) and \(d\) such that \(\langle c,d,1 \rangle\) is orthogonal to \(\langle 1,2,3\rangle\text{.}\)
(e)
Verify that each of the following vectors is also orthogonal to \(\langle 1,2,3 \rangle\text{.}\)
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\(\displaystyle \langle -2,-2,2 \rangle\)
(f)
Put all of the vectors you have computed into the interactive below to visually verify that each of them is orthogonal to \(\langle 1,2,3 \rangle\text{.}\) You should put the component values of each vector into this array with each vector corresponding to a row. If you do not have eight distinct vectors from the previous parts, multiply one of your repeated vectors by \(-1\) and enter that instead of entering a vector multiple times.
(g)
Describe what you think the plot of the set of vectors that are orthogonal to \(\langle 1,2,3 \rangle\) will look like.
Subsection 9.6.2 Planes in Space
Now that we have a way of describing lines, we would like to develop a means of describing planes in three dimensions. In Section 9.1, we studied the coordinate planes and planes parallel to them. In particular, \(x=1\text{,}\) \(y=-2\text{,}\) and \(z=\sqrt{3}\) are examples of fundamental planes. In general, the equations of fundamental planes have the form \(coordinate = constant\text{.}\)
As shown in Figure 9.6.2, any vector in a plane with \(x= \text{constant}\) will be orthogonal to the vector \(\langle 1,0,0 \rangle\text{,}\) any vector in a plane with \(y= \text{constant}\) will be orthogonal to the vector \(\langle 0,1,0 \rangle\text{,}\) and any vector in a plane with \(z= \text{constant}\) will be orthogonal to the vector \(\langle 0,0,1 \rangle\text{.}\) We will use this idea to define a plane in general.
Definition 9.6.3.
A plane \(p\) in space is the set of all terminal points of vectors emanating from a given point \(P_0\) perpendicular to a fixed vector \(\vn\text{,}\) as shown in Figure 9.6.4. We call \(\vn\) the normal vector of the plane.
Like the definition of a line, the definition of a plane given above uses a starting point and a vector as the critical pieces of information. For a line, you begin at the starting point and move as much as you want parallel to the given vector (the direction vector). For a plane, you begin at the starting point and move as much as you want orthogonal to the given vector (the normal vector). For a line, you move only in the direction of the given vector whereas on a plane you cannot move at all in the direction of the given vector.
This description of a plane allows us to find the equation of a plane. Assume that \(\vn=\langle a,b,c\rangle\text{,}\) \(P = (x_0, y_0, z_0)\text{,}\) and that \(Q=(x,y,z)\) is an arbitrary point on the plane. Since the vector \(\overrightarrow{PQ}\) lies in the plane, it must be perpendicular to \(\vn\text{.}\) This means that
\begin{align*}
0 =\mathstrut \amp \vn\cdot\overrightarrow{PQ}\\
=\mathstrut \amp \vn\cdot \big(\langle x,y,z \rangle - \langle x_0, y_0,
z_0\rangle\big)\\
=\mathstrut \amp \vn \cdot \langle x-x_0, y-y_0, z-z_0 \rangle\\
=\mathstrut \amp a(x-x_0) + b(y-y_0) + c(z-z_0).
\end{align*}
We may now summarize our new equation for a plane.
Equations of a plane.
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The scalar equation of the plane with normal vector \(\vn =\langle a,b,c \rangle\) containing the point \(P = (x_0, y_0,z_0)\) is\begin{equation} a(x-x_0) + b(y-y_0) + c(z-z_0) = 0\text{.}\tag{9.6.1} \end{equation}
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The vector equation of the plane with normal vector \(\vn =\langle a,b,c \rangle\) containing the points \(P = (x_0, y_0,z_0)\) and \(Q = (x,y,z)\) is\begin{equation} \vn \cdot \overrightarrow{PQ} = 0\text{.}\tag{9.6.2} \end{equation}
We may take the scalar equation of a plane a little further and note that since
\begin{equation*}
a(x-x_0) + b(y-y_0) + c(z-z_0) = 0\text{,}
\end{equation*}
it equivalently follows that
\begin{equation*}
ax + by + cz = ax_0+by_0+cz_0\text{.}
\end{equation*}
That is, we may write an equation of a plane as \(ax+by+cz = d\) where \(d = \vn\cdot\langle x_0,y_0,z_0\rangle\text{.}\)
Before we look at some examples, we will mention a few ideas that relate planes (as a collection of points), equations for these planes, and normal vectors for these planes. As we saw with lines and will see for most other types of graphs, there are multiple different equations that can describe any given plane. This highlights the difference between an equation that describes a graph and the set of points that make up the graph or make the equation true. For instance, \(x+y+z=1\) and \(2x+2y+2z=2\) are different equations but describe the same graph. That is, the equations have the same set of \((x,y,z)\) points that make the equation true. You may be tempted to say “Those are the same equations. They just differ by a scalar multiple!”, but this difference is the same as the difference between parallel vectors.
The good news is that the orientation of a plane is encoded into the normal vector, thus we can use all the powerful tools we developed for vectors. We say that two planes are parallel if their normal vectors are parallel (as vectors). In fact, we can describe the angle between two planes using normal vectors. The angle between two planes is the acute angle between their respective normal vectors. While the angle between two vectors (as defined in Subsection 9.3.3) can be obtuse, the angle between planes will not be greater than \(\frac{\pi}{2}\text{.}\) When using normal vectors to describe the orientation of a plane, we need to consider all the vectors parallel to our choice of the normal vector because any vector parallel to our choice would give an equivalent equations for the plane. We will look at specifics of this in our next example.
Example 9.6.5.
If we would like to describe the plane passing through the point \(P=(4,-2,1)\) and perpendicular to the vector \(\vn = \langle 1, 2, 1 \rangle\text{,}\) we have
\begin{equation*}
\langle 1,2,1 \rangle\cdot \langle x,y,z\rangle = \langle 1,2,1 \rangle\cdot \langle 4,-2,1\rangle
\end{equation*}
or
\begin{equation*}
x + 2y + z = 1\text{.}
\end{equation*}
Notice that the coefficients of \(x\text{,}\) \(y\text{,}\) and \(z\) in this description give a vector perpendicular to the plane.
For instance, if we are presented with the plane
\begin{equation*}
-2x + y - 3z = 4,
\end{equation*}
we know that \(\vn = \langle -2, 1, -3\rangle\) is a vector perpendicular to the plane.
Using our definition of parallel planes, we can see that our plane given by \(x+2y+z=1\) will be parallel to planes given by \(x+2y+z=-2\text{,}\) \(2x+4y+2z=1\text{,}\) and \(-\frac{3}{2}x-3y-\frac{3}{2}z=12,345\) because the respective normal vectors (\(\langle1,2,1\rangle\text{,}\) \(\langle1,2,1\rangle\text{,}\) \(\langle2,4,2\rangle\text{,}\) and \(\langle -\frac{3}{2},-3,-\frac{3}{2}\rangle\)) are all scalar multiples of each other.
If we wanted to find the angle between \(x+2y+z=1\) and \(-2x+y-3z=4\text{,}\) we would need to consider the angle between the normal vectors \(\langle1,2,1\rangle\) and \(\langle -2,1,-3\rangle\text{.}\) The dot product of these normal vectors is \(\langle1,2,1\rangle \cdot \langle -2,1,-3\rangle=-3\text{.}\) This tells us that the angle between these vectors is obtuse. However, the (smallest) angle between the planes is not obtuse. The equation \(x-\frac{1}{2}y+\frac{3}{2}z=2\) will describe the same set of points as \(-2x+y-3z=4\) but the normal vector for \(x-\frac{1}{2}y+\frac{3}{2}z=2\) will be \(\langle 1,-\frac{1}{2},\frac{3}{2} \rangle\text{.}\) This vector makes an acute angle with \(\langle1,2,1\rangle \text{.}\) This allows us to conclude that the angle between \(x+2y+z=1\) and \(-2x+y-3z=4\) is
\begin{equation*}
\arccos\left(\frac{\langle1,2,1\rangle \cdot \langle1,\frac{1}{2},\frac{3}{2} \rangle}{\sqrt{7} \sqrt{\frac{7}{2}}}\right)\approx 1.2373\text{.}
\end{equation*}
Our next activity gives you an opportunity to practice with equations of planes and also to think about interactions between planes and lines.
Activity 9.6.2.
(a)
Write a scalar equation of the plane \(p_1\) passing through the point \((0, 2, 4)\) and perpendicular to the vector \(\vn=\langle 2, -1, 1\rangle\text{.}\)
(b)
(c)
Write a scalar equation of the plane \(p_2\) that is parallel to \(p_1\) and passes through the point \((3, 0, 4)\text{.}\)
(d)
Give parametric equations for the line \(\mathcal{L}\) passing through the point \((2,0,2)\) and perpendicular to the plane \(p_3\) described by the equation \(x+2y-2z = 7\text{.}\)
(e)
Just as two distinct points in space determine a line, three non-collinear points in space determine a plane. Our next example illustrates a method for finding an equation of the plane containing three non-collinear points in \(\R^3\text{.}\)
Example 9.6.9.
Consider three points \(P_0\text{,}\) \(P_1\text{,}\) and \(P_2\) in space, not all lying on the same line, as shown in Figure 9.6.10.
Notice that the vectors \(\overrightarrow{P_0P_1}\) and \(\overrightarrow{P_0P_2}\) both lie in the plane \(p\text{.}\) If we form their cross-product
\begin{equation*}
\vn = \overrightarrow{P_0P_1} \times \overrightarrow{P_0P_2}\text{,}
\end{equation*}
we obtain a normal vector to the plane \(p\text{.}\) Therefore, if \(Q\) is any other point on \(p\text{,}\) it then follows that \(\overrightarrow{P_0Q}\) will be perpendicular to \(\vn\text{,}\) and we have the equation:
\begin{equation}
\vn \cdot \overrightarrow{P_0Q} = 0\tag{9.6.3}
\end{equation}
The final activity of this section allows you to practice with finding an equation of a plane passing through three points.
Activity 9.6.3.
Let \(P_0 = (1,2,-1)\text{,}\) \(P_1 = (1, 0 ,-1)\text{,}\) and \(P_2 = (0,1,3)\) and let \(p\) be the plane containing \(P_0\text{,}\) \(P_1\text{,}\) and \(P_2\text{.}\)
(a)
Determine the components of the vectors \(\overrightarrow{P_0P_1}\) and \(\overrightarrow{P_0P_2}\text{.}\)
(b)
(c)
Find a scalar equation of \(p\text{.}\)
(d)
Consider a second plane \(q\) with scalar equation \(-3(x-1) + 4(y+3) + 2(z-5)=0\text{.}\) Find two different points on \(q\) as well as a vector \(\vm\) that is normal to \(q\text{.}\)
(e)
Summary
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A plane in space is the set of all terminal points of vectors emanating from a given point perpendicular to a fixed vector.
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If \(P_1\text{,}\) \(P_2\text{,}\) and \(P_3\) are non-collinear points in space, the vectors \(\overrightarrow{P_1P_2}\) and and \(\overrightarrow{P_1P_3}\) are vectors in the plane and the vector \(\vn = \overrightarrow{P_1P_2} \times \overrightarrow{P_1P_3}\) is a normal vector to the plane. So any point \(P\) in the plane satisfies the equation \(\overrightarrow{PP_1} \cdot \vn = 0\text{.}\) If we let \(P = (x,y,z)\text{,}\) \(\vn = \langle a,b,c \rangle\) be the normal vector, and \(P_1 = (x_0,y_0,z_0)\text{,}\) we can also represent the plane with the equation\begin{equation*} a(x-x_0) + b(y-y_0) + c(z-z_0) = 0. \end{equation*}
Exercises 9.6.3 Exercises
1.
2.
3.
4.
The table below gives the number of calories burned per minute for someone roller-blading, as a function of the person’s weight in pounds and speed in miles per hour [from the August 28,1994, issue of Parade Magazine].
calories burned per minute
| weight\(\backslash\)speed | 8 | 9 | 10 | 11 |
| 120 | 4.2 | 5.8 | 7.4 | 8.9 |
| 140 | 5.1 | 6.7 | 8.3 | 9.9 |
| 160 | 6.1 | 7.7 | 9.2 | 10.8 |
| 180 | 7 | 8.6 | 10.2 | 11.7 |
| 200 | 7.9 | 9.5 | 11.1 | 12.6 |
(a) Suppose that a 160 lb person and a 200 person both go 10 miles, the first at 9 mph and the second at 8 mph.
How many calories does the 160 lb person burn?
How many calories does the 200 lb person burn?
(b) We might also be interested in the number of calories each person burns per pound of their weight.
How many calories per pound does the 160 lb person burn?
How many calories per pound does the 200 lb person burn?
5.
This exercise explores key relationships between a pair of planes. Consider the following two planes: one with scalar equation \(4x - 5y + z = -2\text{,}\) and the other which passes through the points \((1,1,1)\text{,}\) \((0,1,-1)\text{,}\) and \((4, 2, -1)\text{.}\)
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Find a vector normal to the first plane.
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Find a scalar equation for the second plane.
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Find the angle between the planes, where the angle between them is defined by the angle between their respective normal vectors.
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Find a point that lies on both planes.
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Since these two planes do not have parallel normal vectors, the planes must intersect, and thus must intersect in a line. Observe that the line of intersection lies in both planes, and thus the direction vector of the line must be perpendicular to each of the respective normal vectors of the two planes. Find a direction vector for the line of intersection for the two planes.
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Determine parametric equations for the line of intersection of the two planes.
6.
In this problem, we explore how we can use what we know about vectors and projections to find the distance from a point to a plane.
Let \(p\) be the plane with equation \(z=-4x+3y+4\text{,}\) and let \(Q = (4,-1,8)\text{.}\)
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Find the components of \(\overrightarrow{PQ}\text{.}\) Draw a picture to illustrate the objects found so far.
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Explain why \(|\comp_{\vn} \overrightarrow{PQ}|\) gives the distance from the point \(Q\) to the plane \(p\text{.}\) Find this distance.
