Linear geometry in $\mathbb{R}^3$
On this page we'll use our knowledge of vectors to determine the equations of lines and planes in 3-D space.
In all of what follows, we'll be using a right-handed 3-dimensional Cartesian coordinate system like the one shown here. "Right-handed" in this sense means that if we trace the direction of the positive end of the $x$-axis as it rotates into the positive end of the $y$-axis with the fingers of our right hand, our right thumb must point in the positive direction of the $z$-axis. (If you did this with your left hand, $z$ would point downward.) Take a minute to make sure you can visualize the plotting of points on such a coordinate system. It's not always easy, but the more effort you put into it, the better you'll get at it.
One point, $(x, y, z) = (2, 3, 3)$ is shown. To locate the point, we move 2 units toward us (out of the plane of the screen), then 3 units right and 3 units upward.
Vectors – a refresher
A vector is a directed line segment that can be used to represent some physical quantity like velocity or force, or that can be used to point to a specific location in some n-dimensional space, where n can be anything larger than 1. A one-dimensional vector is a scalar, a number without direction.
The only two important properties of a vector are its length and its direction. In the graph below, the 2-D vector $\vec{A}$ is written in several locations on the plane. Each version represents the same vector because the length and direction is preserved; its location is irrelevant.
Any vector can be labeled with a set of coordinates, like the $(4, 2)$ on the graph. It's easy to find the coordinates of a vector when its origin (the non-pointed end) is at the graph origin. No matter where the vector is located, however, we can simply mentally translate the origin of the axis system and count along the axes to find the same coordinates.
The length of a two-dimensional vector can be found using the Pythagorean theorem. The length of vector (2, 4) in the figure is
$$ \begin{align} L &= \sqrt{x^2 + y^2} \\[5pt] &= \sqrt{2^2 + 4^2} = \sqrt{20} \\[5pt] &= 2 \sqrt{5} \end{align}$$
This formula for the length of a vector can be generalized to vectors of any number of dimensions. The length of an n-dimensional vector, $(v_1, v_2, \dots , v_n)$ is
$$L = \sqrt{n_1^2 + n_2^2 + \dots + n_n^2}$$
Normalizing vectors
To normalize a vector means to keep its direction, but reduce its length to one, which we call unit length. This is done by dividing each coordinate of the vector by its length. Here's an example. Let's normalize the vector $\vec{v} = (2, -1, 3).$
First we find the length of $\vec{v}:$
$$ \begin{align} L &= \sqrt{2^2 + (-1)^2 + 3^2} \\[5pt] &= \sqrt{4 + 1 + 9} \\[5pt] &= \sqrt{14} \end{align}$$
Now the normalized version of $\vec{v}$ is
$$\hat{v} = \left( \frac{2}{\sqrt{14}}, \frac{-1}{\sqrt{14}}, \frac{3}{\sqrt{14}} \right)$$
Notice that, once normalized, we wrote the vector using the "hat" symbol, $\hat{v},$ which we reserve for unit vectors. You can check that the length of this vector is one.
This normalization process can be used on vectors of any dimension.
Adding vectors
Geometrically, we add two vectors by joining them in "head-to-tail" fashion. Here's a picture of the addition of two 2-D vectors.
In the graph we add two vectors, $\vec A$ and $\vec B$ graphically. The blue vector is the resultant, the sum of the two. Numerically, we simply add the $x$ and $y$ coordinates to get the coordinates of the sum vector: $(4+2, \; 2+3) = (6, \, 5)$.
The sum of two n-dimensional vectors, $\vec{v_1} = (x_1, y_1, z_1)$ and $\vec{v_2} = (x_2, y_2, z_2)$ can be added by adding their respective terms:
$$\vec{v_1} + \vec{v_2} = (x_1 + x_2, y_1 + y_2, z_1 + z_2)$$
Only vectors of the same dimension can be summed, but vector addition works the same way for any number of dimensions. Subtraction of one vector from another can be viewed as addition of the negative, where $-\vec{v}$ is the same as $\vec{v},$ but pointing in the opposite direction.
Practice problems
Add the two vectors to obtain $\vec{v},$ then normalize the resulting vector to obtain $\hat{v}.$
-
$\vec{v} = (-2, 2, 3) + (2, -4, 5)$
Solution
$$S = (-2 + 2, 2 - 4, 3 + 5) = (0, -2, 8)$$
$$ \begin{align} \text{length} &= \sqrt{0^2 + (-2)^2 + 8^2} \\[5pt] &= \sqrt{4 + 64} = \sqrt{68} \\[5pt] &= 2 \sqrt{17} \end{align}$$
$$\hat{v} = \left( 0, \frac{-1}{\sqrt{17}}, \frac{4}{\sqrt{17}} \right)$$
-
$\vec{v} = (1, 4, -4) + (2, -3, 7)$
Solution
$$S = (1 + 2, 4 - 3, -4 + 7) = (3, 1, 3)$$
$$ \begin{align} \text{length} &= \sqrt{3^2 + 1^2 + 3^2} \\[5pt] &= \sqrt{9 + 1 + 9} = \sqrt{19} \end{align}$$
$$\hat{v} = \left( \frac{3}{\sqrt{19}}, \frac{1}{\sqrt{19}}, \frac{3}{\sqrt{19}} \right)$$
-
$\vec{v} = (-3, 4, -3) - (2, 2, -1)$
Solution
$$S = (-3 + 2, 4 + 2, -3 - 1) = (-1, 6, -4)$$
$$ \begin{align} \text{length} &= \sqrt{(-1)^2 + 6^2 + (-4)^2} \\[5pt] &= \sqrt{1 + 36 + 16} = \sqrt{53} \end{align}$$
$$\hat{v} = \left( \frac{-1}{\sqrt{53}}, \frac{6}{\sqrt{53}}, \frac{-4}{\sqrt{53}} \right)$$
-
$\vec{v} = (1, -2, -4, 3) + (2, -2, 1, -4)$
Solution
$$ \begin{align} S &= (1 + 2, -2 - 2, -4 + 1, 3 - 4) \\[5pt] &= (3, -4, -3, -1) \end{align}$$
$$ \begin{align} \text{length} &= \sqrt{3^2 + (-4)^2 + (-3)^2 + (-1)^2} \\[5pt] &= \sqrt{9 + 16 + 9 + 1} = \sqrt{35} \end{align}$$
$$\hat{v} = \left( \frac{3}{\sqrt{35}}, \frac{-4}{\sqrt{35}}, \frac{-3}{\sqrt{35}}, \frac{-1}{\sqrt{35}} \right)$$
3-D unit vectors: $\hat{i}, \; \hat{j}, \; \hat{k}$
Before we begin talking about multidimensional vectors in earnest, let's talk about the three Cartesian basis vectors, $\hat{i}, \; \hat{j}, \; \hat{k}.$ They are
$$ \begin{align} \hat{i} &= (1, 0, 0) \\[5pt] \hat{j} &= (0, 1, 0) \\[5pt] \hat{k} &= (0, 0, 1) \end{align}$$
Using unit vectors
Let's express the 3-D vector $(-3, 4, 8)$ in terms of the unit vectors:
$$ \begin{align} (-3, 4, 8) &= -3(1, 0, 0) + 4((0, 1, 0) + 8(0, 0, 1) \\[5pt] &= -3 \hat{i} + 4 \hat{j} + 8 \hat{k} \end{align}$$
Using unit vectors
The 3-D unit vectors form a basis for 3-D space. Any vector $(a, b, c)$ can be written as a sum of parameters multiplied by each basis vector:
$$ \begin{align} (a, b, c) &= a(1, 0, 0) + b(0, 1, 0) + c(0, 0, 1) \\[5pt] &= a \hat{i} + b \hat{j} + c \hat{k} \end{align}$$
Pro tip 1: Often unit vectors like $\hat{i}$ are referred to as "i hat" because the symbol $\hat{}$ looks like a hat.
Pro tip 2: In higher dimensional systems, we usually use the letter $e$ to denote the set of unit vectors. In an $n$-dimensional coordinate system, the unit vectors are $e = \{\hat e_1, \hat e_2, \dots , \hat e_n \}$.
Equations of lines
Just as the equation of a line in a plane can be determined using a point and the slope of the line, we can determine the equation of a line in 3-D space. To do so we need a single point and a vector pointing in the same direction as the line.
The setup looks like this:
We have a point, $P = (x_o, y_o, z_o),$ and a direction vector $\vec{v}$ and we will draw the vector $r_o = (x_o, y_o, z_o)$ to point P, as shown in the second graph.
Now because two points define a line, let's imagine a second point $Q = (x, y, z)$ on our line. Then the vector from Q to P, $\overrightarrow{QP},$ is
$$\vec{Q} - \vec{P} = (x - x_0, y - y_0, z - z_0)$$
We've also labeled the coordinates of vector $\vec{v}$ as $\vec{v} = (a, b, c).$
Now if $\overrightarrow{QP}$ is parallel to our direction vector $r_0 ,$ then the two differ by only a constant or parameter, $t.$
$$\overrightarrow{QP} = t \vec{v}.$$
If the coordinates of vector $\vec{r_0}$ are $(a, b, c),$ then the equation of our line is
$$(x, y, z) - (x_0, y_0, z_0) = t \cdot (a, b, c)$$
We can rearrange to
$$ \begin{align} (x, y, z) &= (x_0, y_0, z_0) + t(a, b, c) \\[5pt] \phantom{000} \vec{r} \phantom{000} &= \phantom{000} \vec{r_0} \phantom{000} + \phantom{000} t \vec{v} \end{align}$$
This is analogous to the equation for a line in the x-y plane, $y = mx + b,$ where y is the dependent variable, m is the slope and b is a point on the line a specific one in this case, the y-axis intercept, but nonetheless, just a point. Bu analogy in the equation $\vec{r} = \vec{r_0} + t \vec{v},$ we have the variable vector $\vec{r}$ as our independent variable, the point $(x_0, y_0, z_0)$ as our fixed, known point, and vector $\vec{v} = (a, b, c)$ is our slope.
We can break this equation up into three parametric equations:
$$ \begin{align} x &= x_0 + ta \\[5pt] y &= y_0 + tb \\[5pt] z &= z_0 + tc \end{align}$$
Finally, we can express the equation of the line in parametric form like this:
$$(x, y, z) = (x_0 + ta, y_0 + tb, z_0 + tc)$$
Now let's do a couple of examples to see how this all comes together in practice.
Example 1
Find the vector equation of the line passing through $P = (5, 1, 3)$ and parallel to (or in the direction of) the vector $\vec{v} = \vec{i} + 4 \vec{j} - 2 \vec{k}.$
$$(x, y, z) = (x_0, y_0, z_0) + t \cdot (a, b, c),$$
where
$$ \begin{align} (x_0, y_0, z_0) &= (5, 1, 3) \text{ and} \\[5pt] (a, b, c) &= (1, 4, -2) \end{align}$$
So our equation is
$$(x, y, z) = (5, 1, 3) + t(1, 4, -2)$$
We can write this as a list of parametric equations:
$$ \begin{align} x &= 5 + t \\[5pt] y &= 1 + 4t \\[5pt] z &= 3 - 2t \end{align}$$
or as a parameterized vector:
$$(x, y, z) = (5 + t, 1 + 4t, 3 - 2t)$$
Example 2
Find the vector equation of the line passing through points $P_1 = (2, -1, -2)$ and $P_2 = (1, -2, 3).$
$$ \begin{align} P_2 - P_1 &= (1, -2, 3) - (2, -1, -2) \\[5pt] &= (1 - 2, -2 - (-1), 3 - (-2)) \\[5pt] &= (-1, -1, 5) \end{align}$$
So our direction vector is $\vec{r_0} = (-1, -1, 5).$
The initial equation of our line is then
$$(x, y, z) = (2, -1, -2) + t(-1, -1, 5)$$
The parametric equations are:
$$ \begin{align} x &= 2 - t \\[5pt] y &= -1 - t \\[5pt] z &= -2 + 5t \end{align}$$
and the parameterized vector is
$$(x, y, z) = (2 - t, -1 - t, -2 + 5t)$$
Practice problems
Find an equation of the line passing through the point and parallel to the given vector
-
$(-2, 2, 3)$ parallel to $\hat{i} - \hat{j} + 2 \hat{k}$
Solution
$$ \begin{align} (x, y, z) &= (-2, 2, 3) + t(1, -1, 2) \\[5pt] &= (-2 + t, 2 - t, 3 + 2t) \end{align}$$
-
$(1, 3, 3)$ parallel to $2\hat{i} + \hat{j} - \hat{k}$
Solution
$$ \begin{align} (x, y, z) &= (1, 3, 3) + t(2, 1, -1) \\[5pt] &= (1 + 2t, 3 + t, 3 - t) \end{align}$$
-
$(1, 4, -3)$ parallel to $-2\hat{i} - 3\hat{j} + \hat{k}$
Solution
$$ \begin{align} (x, y, z) &= (1, 4, -3) + t(-2, -3, 1) \\[5pt] &= (1 - 2t, 4 - 3t, -3 + t) \end{align}$$
-
$(1, 1, -4)$ parallel to $-\hat{i} - \hat{j} + 2\hat{k}$
Solution
$$ \begin{align} (x, y, z) &= (1, 1, -4) + t(-1, -1, 2) \\[5pt] &= (1 - t, 1 - t, -4 + 2t) \end{align}$$
Find an equation of the line passing through the points given.
-
$(-2, 2, -1)$ and $(-2, -3, 1)$
Solution
point: $(-2, 2, -1),$ direction:
$$ \begin{align} \vec{v} &= (-2 - (-2), 2 - (-3), -1 - 1) \\[5pt] &= (0, 5, -2) \end{align}$$
$$ \begin{align} (x, y, z) &= (-2, 2, -1) + t(0, 5, -2) \\[5pt] &= (-2, 2 + 5t, -1 -2t) \end{align}$$
-
$(4, 2, 5)$ and $(-1, -1, -1)$
Solution
point: $(4, 2, 5),$ direction:
$$ \begin{align} \vec{v} &= (4 - (-1), 2 - (-1), 5-(-1)) \\[5pt] &= (5, 3, 6) \end{align}$$
$$ \begin{align} (x, y, z) &= (4, 2, 5) + t(5, 3, 6) \\[5pt] &= (4 + 5t, 2 + 3t, 5 + 6t) \end{align}$$
-
$(-3, 4, -3)$ and $(2, 2, -1)$
Solution
point: $(2, 2, -1),$ direction:
$$ \begin{align} \vec{v} &= (2 - (-3), 2 - 4, -1 - (-3)) \\[5pt] &= (5, -2, 2) \end{align}$$
$$ \begin{align} (x, y, z) &= (2, 2, -1) + t(5, -2, 2) \\[5pt] &= (2 + 5t, 2 - 2t, -1 + 2t) \end{align}$$
-
$(3, 4, -1)$ and $(-1, -2, 4)$
Solution
point: $(3, 4, -1),$ direction:
$$ \begin{align} \vec{v} &= (3 - (-1), 4 - (-2), -1 - 4) \\[5pt] &= (4, 6, -5) \end{align}$$
$$ \begin{align} (x, y, z) &= (3, 4, -1) + t(4, 6, -5) \\[5pt] &= (3 + 4t, 4 + 6t, -1 - 5t) \end{align}$$
Relationships between lines in $\mathbb{R}^3$
In $\mathbb{R}^2$ space (the Cartesian plane), we can determine whether lines are parallel, intersect or coincide:
- If two lines have the same slope, they are parallel.
- If they have the same slope and share a point, then they are the same line (they are coincident).
- If two lines have different slopes, they intersect at one and only one point.
Lines in 3-D space can be parallel in that the distance between them remains constant, but they can also not be parallel and never intersect. These are called skew lines.
We can perform some similar tests to determine the relationship(s) between two lines in space. We'll do that through some examples.
Example 3
Find the intersection, if there is one, between the lines $(2, 1, 0) + t(-1, -1, -1)$ and $(3, 0, 5) + t(2, 0, 6).$
The key here is to re-express each line using two different parameters, say $s$ and $t$:
$$ \begin{align} (2, 1, 0) &+ s(-1, -1, -1) \\[5pt] &= (2-s, 1-s, -s) \\[9pt] (3, 0, 5) &+ t(2, 0, 6) \\[5pt] &= (3+2t, 0, 5+6t) \end{align}$$
Then set corresponding terms of each vector equal:
$$ \begin{align} 2 - s &= 3 + 2t \\[5pt] 1 - s &= 0 \\[5pt] -s &= 5 + 6t \end{align}$$
Here we have three equations and two unknowns, so this is a solvable system. The second equation gives $s = 1$. Plugging that into the third gives $t = -1.$ We can confirm both solutions by plugging into the first: $2 - 1 = 3 + 2(-1),$ which is true.
So these lines intersect at $(1, 0, -1).$
By analogy with lines in a plane, if we'd obtained no solution, that would mean that the lines do not cross. If we'd obtained an infinite number of solutions, the lines would be coincident (the same).
Example 4
Determine whether the lines $(2s - 3, s - 1, s - 4)$ and $(t - 3, 3t, 5 - 2t)$ intersect or are the same line.
First equate the $x, y$ and $z$ components of each line:
$$ \begin{align} 2s - 3 &= t - 3 \\[5pt] s - 1 &= 3t \\[5pt] s - 4 &= 5 - 2t \end{align}$$
From the second and third equations, we get
$$ \begin{align} s &= 1 + 3t \\[5pt] s &= 9 - 2t, \end{align}$$
Thus we can eliminate s by the transitive property and solve
$$ \begin{align} 1 + 3t &= 9 - 2t \\[5pt] 5t &= 8 \\[5pt] t &= \frac{8}{5}. \end{align}$$
Subbing into the first equation gives
$$ \require{cancel} \begin{align} 2s - 3 &= \frac{8}{5} - 3 \\[5pt] 2s &= \frac{8}{5} \\[5pt] s &= \frac{4}{5} \end{align}$$
Now we should make sure these work in all three equations. Subbing $s$ and $t$ into the first equation gives:
$$ \begin{align} 2 \left( \frac{8}{5} \right) &= \frac{4}{5} \\[5pt] \frac{16}{5} &\ne \frac{4}{5} \end{align}$$
So that's a contradiction, indicating that these lines have no point in common; they are skew lines.
Practice problems
-
Show that the parametric equations
$$ \begin{align} x &= t, \; y = t-1, \; z = 2t+5 \\[5pt] x &= 2s + 1, \; y = 2s, \; z = 4s + 7 \end{align}$$
represent the same line.
Solution
First equate the parametric versions of each of the $x, y$ and $z$ terms of each line:
$$ \begin{align} t &= 2s + 1 \tag{1} \\[5pt] t - 1 &= 2s \tag{2} \\[5pt] 2t + 5 &= 4s + 7 \tag{3} \end{align}$$
Inserting (1) into (2), gives us
$$ \begin{align} 2s + \cancel{1} \cancel{- 1} &= 2s \\[5pt] s &= 1 \end{align}$$
If we insert $s=1$ into each of (1), (2) and (3) to solve for $t$, we get
$$ \begin{align} t &= 2(1) + 1 \; \rightarrow \; t = 3 \\[5pt] t - 1 &= 2(1) \; \rightarrow \; t = 3 \\[5pt] 2t + 5 &= 4(1) + 7 \; \rightarrow \; t = 3 \end{align}$$
The parametric forms represent the same line (they intersect at every point along their lengths). Another way to think about this is to write the symmetric equations for each line:
$$ \begin{align} x &= y+1 = \frac{z-5}{2} \\[5pt] \frac{x-1}{2} &= \frac{y}{2} = \frac{z-7}{4} \end{align}$$
The first of these can be obtained from the second by multipying each term by 2 and adding 1.
-
Show that the direction vectors of the lines
$$ \begin{align} L_1: \; x &= 2t, \; y = t-3, \; z = 1-t \\[5pt] L_2: \; x &= s, \; y = s+1, \; z = 3s - 2 \end{align}$$
are perpendicular and determine whether the lines intersect.
Solution
We can use the parametric forms to get the vector equations for these lines:
$$ \begin{align} L_1: \; (0, -3, 1) + t(2, 1, -1) \\[5pt] L_2: \; (0, 1, -2) + s(1, 1, 3) \\[5pt] \end{align}$$
Now the dot product of the direction vectors (the ones multiplied by the parameters) is
$$(2,1,-1)\cdot (1, 1, 3) = 2 + 1 - 3 = 0$$
so the direction vectors are perpendicular. Now to see if the lines intersect, we equate the $x, y$ and $z$ components of each:
$$ \begin{align} 2t &= s \tag{1} \\[5pt] t-3 &= s+1 \tag{2} \\[5pt] 1-t &= 3s - 2 \tag{3} \\[5pt] \end{align}$$
Now if we substitute $s = 2t$ for $s$ in (2), we get $t = -4$, and thus $s = -8$. Substituting both of these into (3) gives
$$1-(-4) \ne 3(-8) - 2$$
So the directions are perpendicular, but the lines are on different planes, and are thus skew.
-
Determine whether the lines
$$ \begin{align} L_1: \; x &= 3t+3, \; y = t, \; z = t+3 \\[5pt] L_2: \; x &= s-2, \; y = 2s-5, \; z = s \end{align}$$
intersect
Solution
First equate the parametric versions of each of the $x, y$ and $z$ terms of each line:
$$ \begin{align} 3t + 3 &= s-2 \tag{1} \\[5pt] t &= 2s - 5 \tag{2} \\[5pt] t + 3 &= s \tag{3} \end{align}$$
If we solve (2) and (3) for $t$ and equate them, eliminating $t$, we get
$$ \begin{align} 2s - 5 &= s - 3 \\[5pt] s &= 2 \end{align}$$
Then using (2), we can find $t$:
$$t = 2(2) - 5 = 1$$
Now to see if these work in (1):
$$3(-1) + 3 = 2 - 2 \; \color{green}{\checkmark}$$
If we plug our parameters into their respective vector forms we get the same point, the point of intersection:
$$ \begin{align} (3(-1)+3, -1, -1+3) &= (0, -1, 2) \\[5pt] (2-2, 2(2)-5, 2) &= (0, -1, 2) \end{align}$$
-
Determine whether the lines
$$ \begin{align} L_1: \; x &= 2t-1, \; y = 1-t, \; z = 3t \\[5pt] L_2: \; x &= s+1, \; y = 2s, \; z = 3-2s \end{align}$$
intersect
Solution
First equate the parametric versions of each of the $x, y$ and $z$ terms of each line:
$$ \begin{align} 2t - 1 &= s + 1 \tag{1} \\[5pt] 1 - t &= 2s \tag{2} \\[5pt] 3t &= 3 - 2s \tag{3} \end{align}$$
If we solve (1) for $s$ we get $s = 2t - 2$. Plugging that into (2) gives us
$$ \begin{align} 1 - t &= 2(2t - 2) \\[5pt] -t &= 4t - 4 - 1 \\[5pt] -5t &= -5 \; \rightarrow \; t = 1 \end{align}$$
Plugging $t = 1$ back into (1) gives $s = 0$. Now to see if these work in (3)
$$3(1) = 3 - 0 \; \color{green}{\checkmark}$$
If we plug our parameters into their respective vector forms we get the same point, the point of intersection:
$$ \begin{align} (2(1)-1, 1-1, 3(1)) &= (1, 0, 3) \\[5pt] (0+1, 2(0), 3-2(0)) &= (1, 0, 3) \end{align}$$
Planes: the normal vector, $\vec{n}$
Before we get to discussing planes in 3D space, we need to develop (or review) the concept of the normal vector, $\vec{n}.$ As a reminder, a normal to a line, L, is just a vector that is perpendicular to it. A normal to a curve, C, at some point T on the curve, is a line perpendicular to the line tangent to C at T. Both of these situations are illustrated below.
In the top graph (above), the normal vector is perpendicular to line L. In the bottom graph, we must first draw the (dashed) tangent line to the point of interest on the curve (T), then draw a vector $\vec{n}$ perpendicular to it from that point. $\vec{n}$ is normal to curve C at point T.
A normal vector $(\vec{n})$ to a plane is a vector that is perpendicular to a plane in every direction. As illustrated below, vector $\vec{n}$ is perpendicular to plane P at point Q.
The requirement that the normal vector be perpendicular when viewed from any direction is why we use the new word "normal" instead of perpendicular.
We can also find normal vectors ("normals") to curved surfaces as well. The normal to a curved surface is just the normal to a plane tangent to that point. This is illustrated below: plane P is tangent to the sphere at point T. The normal to P at T is normal to the sphere at that point.
Equations of planes
We saw above that all that is needed to specify a line in space is to know a point on it and direction vector. Likewise, we can specify a plane (that is, we can determine an equation of the plane) if we know one point in the plane and a normal vector to that plane. If the plane is rotated around our fixed point, the normal vector will rotate accordingly.
Let's let $\vec n$ be a normal vector to plane $P$ with coordinates $\vec n = (A, B, C)$. There are an infinite number of such vectors, of course. Now let's take a point $Q$ on plane $P$ with coordinates $Q(x_o, y_o, z_o)$. If we draw a vector from $Q$ to $R = (x, y, z)$, then if $\hat n$ and $\vec{QR}$ are normal.
If $\vec n$ and $\vec{QR}$ are normal, then their dot product must equal zero:
$$ \begin{align} \vec n \cdot \vec{QR} &= (A, B, C) \cdot (x-x_o, y-y_o, z-z_o) \\[5pt] &= A(x-x_o) + B(y-y_o) + C(z-z_o) \\[5pt] &= Ax + By + Cz - (Ax_o + By_o + Cz_o) = 0 \end{align}$$
Now if we let the quantity in parentheses be a constant, $D$, we have a general linear equation for a plane:
$$Ax + By + Cz + D = 0$$
This equation should remind you of the standard for of the equation of a line in $\mathbb{R}^2$, $Ax + By = C$. Notice that the vector $\vec n = (A, B, C)$ is a normal vector of the plane that we can just read off.
Equation of a plane in $\mathbb{R}^3$
The standard-form equation of a plane in $\mathbb{R}^3$ space is
$$Ax + By + Cz = D$$
where $A, B, C$, and $D$ are parameters, and $\vec n = (A, B, C)$ is a normal vector of the plane. The dot product between any vector in the plane and a normal to the plane will be equal to zero.
Example 5
Determine the equation, in the form $Ax + By + Cz = 0$ for the plane that contains the vector $\vec v = (-1, 2, 2)$ and that is normal to $\vec n = (2, 1, 1)$.
$$A(x-x_o) + B(y-y_o) + C(z-z_o) = 0$$
where $\vec v = (x_o, y_o, z_o)$.
An equation of our plane is
$$ \begin{align} 2(x-(-1)) + 1(y-2) + 1(z-2) &= 0 \\[5pt] 2x + 2 + y - 2 + z - 2 &= 0 \\[5pt] 2x + y + z &= 2 \end{align}$$
Example 6
Determine the equation of a plane containing vectors $\vec v_1 = (2, 1, -4)$ and $\vec v_2 = (1, 2, 3)$.
$$ \begin{align} cos(\theta) &= \frac{\vec v_1 \cdot \vec v_2}{|\vec v_1||\vec v_2|} \\[5pt] &= \frac{2(1)+1(2)-4(3)}{\sqrt{2^2 + 1^2 + (-4)^2} \sqrt{1^2 + 2^2 + 3^2}} \\[5pt] &= \frac{-8}{\sqrt{21} \sqrt{15}} \\[5pt] &= \frac{-8}{3 \sqrt{35}} \ne 1 \end{align}$$
So the angle between them is not zero (cos(0) = 1), so they'll define a plane just fine.
Now we'll need a normal vector, which we can get by finding the cross product $\vec v_1 \times \vec v_2$. We'll do that using the determinant mnemonic:
$$\text{det}\left( \begin{array}{rrr|rr} 2 & 1 & -4 & 2 & 1 \\[5pt] 1 & 2 & 3 & 1 & 2\\[5pt] x & y & z & x & y \end{array} \right)$$
$$ \begin{align} \vec v_1 \times \vec v_2 &= 3x - 4y + 4z -z +8x -6y \\[5pt] &= 11x - 10y + 3z \end{align}$$
So a normal vector is $\vec n = (11, -10, -3)$.
An equation of our plane (using $\vec n$ and $\vec v_1$) is
$$ \begin{align} (11, -10, -3) \cdot (2-x, 1-y, -4-z) &= 0 \\[5pt] 11(2-x) - 10(1-y) -3(-4-z) &= 0 \\[5pt] 22 - 11x -10 +10y +12 +3z &= 0 \\[5pt] -11x + 10y + 3z + 24 &= 0 \end{align}$$
Here's a plot of a part of our plane showing $\vec v_1$, $\vec v_2$ and $\vec n$.
Practice problems
Find an equation of the plane using the information given.
-
$Q = (-2, 1, 3)$ lies in plane $P$, which is normal to $\vec n = (3, 3, -1)$.
Solution
The equation of the plane is
$$ \begin{align} (3, 3, -1) \cdot (x+2, y-2, z-3) &= 0 \\[5pt] 3(x+2) + 3(y-1) - 1(z-3) &= 0 \\[5pt] 3x + 6 + 3y - 3 - z + 3 &= 0 \\[5pt] 3x + 3y - z &= -6 \end{align}$$
It's worth playing around with this plane just a bit. Let's determine two vectors that lie in the plane, make sure they are normal to $\vec n$ and plot everything on 3D axes. Let's choose two points on the plane, the first with $x, y = 1, 2$ and the second with $x, y = 0, 3$. These are chosen completely at random. Now we'll plug those into the equation of the plane to find the $z$-coordinates for each:
$$ \begin{align} 3x + 3y - z &= -6 \\[5pt] 3(1) + 3(2) -z &= -6 \\[5pt] 9 - z &= -6 \\[5pt] z &= 15 \; \rightarrow (1, 2, 15) \\[9pt] 3(0) + 3(3) -z &= -6 \\[5pt] 9 - z &= -6 \\[5pt] z &= 15 \; \rightarrow (0, 3, 15) \\[5pt] \end{align}$$
Now from these points, and using our original point $Q = (-2,1,3)$, we can form –by subtraction– two vectors in our plane:
$$ \begin{align} \vec v_1 &= (1-(-2), 2-1, 15-3) = (3, 1, 12) \\[5pt] \vec v_2 &= (0-(-2), 3-1, 15-3) = (2, 2, 12) \end{align}$$
We can check that both of these vectors lie in the plane by showing that their dot products with $\vec n$ are zero:
$$ \begin{align} \vec v_1 \cdot \vec n &= (3, 1, 12)\cdot (3, 3, -1) \\[5pt] &= 9 + 3 -12 = 0 \; \color{green}{\checkmark} \\[5pt] \vec v_2 \cdot \vec n &= (2, 2, 12)\cdot (3, 3, -1) \\[5pt] &= 6 + 6 -12 = 0 \; \color{green}{\checkmark} \end{align}$$
Finally, we can plot our graph, including all of our vectors for reference:
-
$Q = (2, 1, -3)$ lies in plane $P$, which is normal to $\vec n = (1, -3, -2)$
Solution
The equation of the plane is
$$ \begin{align} (1, -3, -2) \cdot (2-x, 1-y, -3-z) &= 0 \\[5pt] 2(2-x) - 3(1-y) -2(-3-z) &= 0 \\[5pt] 4 - 2x -3 + 3y + 6 + 2z &= 0 \\[5pt] -x + 3y + 2z + 5 &= 0 \\[5pt] x - 3y - 2z &= 5 \end{align}$$
We're done, but it's worth playing around with this plane just a bit. Let's determine two vectors that lie in the plane, make sure they are normal to $\vec n$ and plot everything on 3D axes. Let's choose two points on the plane, the first with $x, y = 1, 1$ and the second with $x, y = 1, -2$. These are chosen completely at random. Now we'll plug those into the equation of the plane to find the $z$-coordinates for each:
$$ \begin{align} x - 3y - 2z &= 5 \\[5pt] 1 - 3(1) -2z &= 5 \\[5pt] -2 -2z &= 5 \\[5pt] z &= -\frac{7}{2} \; \rightarrow \left(1, 1, -\frac{7}{2} \right) \\[9pt] 1 - 3(-2) - 2z &= 5 \\[5pt] 7-2z &= 5 \\[5pt] z &= 1 \; \rightarrow (1,-2, 1) \\[5pt] \end{align}$$
Now from these points, and using our original point $Q = (-2,1,3)$, we can form –by subtraction– two vectors in our plane:
$$ \begin{align} \vec v_1 &= (1-2, -2-1, 1+3) = (-1,-3,4) \\[5pt] \vec v_2 &= (1-2, 1-1, -7/2+3) = \left( -1, 0, -\frac{1}{2} \right) \end{align}$$
We can check that both of these vectors lie in the plane by showing that their dot products with $\vec n$ are zero:
$$ \begin{align} \vec v_1 \cdot \vec n &= (-1,-3,4)\cdot (1, -3, -2) \\[5pt] &= -1+9-8 = 0 \; \color{green}{\checkmark} \\[5pt] \vec v_2 \cdot \vec n &= \left( -1, 0, -\frac{1}{2} \right)\cdot (1, -3, -2) \\[5pt] &= -1+0+1 = 0 \; \color{green}{\checkmark} \end{align}$$
-
The plane contains vector $\vec v_1 = (2, -3, -4)$ and points $A = (1, 1, -3)$ and $B = (2, 2, 2)$.
Solution
We can use points $A$ and $B$ to form a new vector in the plane, $\vec v_2 = (2-1, 2-1, 2-(-3)) = (1, 1, 5)$.
Now we can find the normal to these vectors using the cross product (Here I'll use the 3×3 determinant method):
$$ \begin{array}{rrr|rr} 2 & -3 & -4 & 2 & -3 \\[5pt] 1 & 1 & 5 & 1 & 1 \\[5pt] x & y & z & x & y \end{array}$$
$$ \begin{align} \vec v &= 2z -15x - 4y -(-4x) - 10y - (-3z) \\[5pt] &= -11x -14y + 5z \\[5pt] &\rightarrow \vec n = (-11,-14, 5) \end{align}$$
Now the equation of the plane can be found:
$$ \begin{align} (-11,-14,5) \cdot (2-x, -3-y, -4-z) &= 0 \\[5pt] -11(2-x) - 14(-3-y) + 5(-4-z) &= 0 \\[5pt] -22 + 11x + 42 + 14y -20 -5z &= 0 \\[5pt] 11x + 14y - 5z &= 0 \end{align}$$
Notice that this plane contains the origin.
-
The plane contains vectors $\vec v_1 = (-2, 3, -1)$ and $\vec v_2 = (-2, -3, 4)$
Solution
First let's find a vector normal to these using the cross product.
$$\vec v_1 \times \vec v_2 = (-2, 3, -1) \times (-2, -3, 4)$$
$$ \begin{array}{rrr|rr} -2 & 3 & -1 & -2 & 3 \\[5pt] -2 & -3 & 4 & -2 & -3 \\[5pt] x & y & z & x & y \end{array}$$
$$ \begin{align} \vec v_1 \times \vec v_2 &= (-2)(-3)z + 3(4)x +(-1)(-2)y \\[5pt] &- (-1)(-3)x - 4(-2)y - 3(-2)z \\[5pt] &= 6z + 12x +2y -3x +8y +6z \\[5pt] &= 9x + 10y + 12z \\[5pt] &\rightarrow \; \vec n = (9, 10, 12) \end{align}$$
An equation of our plane is
$$ \begin{align} (9, 10, 12) \cdot (-2-x, 3-y, -1-z) &= 0 \\[5pt] 9(-2-x) + 10(3-y) + 12(-1-z) &= 0 \\[5pt] -18 -9x + 30 -10y + 12 -12z &= 0 \\[5pt] -9x - 10y - 12z + 24 &= 0 \\[5pt] 9x + 10y + 12z &= -24 \end{align}$$
Here is a graph of our plane and the vectors.
-
The plane contains vectors $\vec v_1 = (1, 1, -4)$ and $\vec v_2 = (4, 1, 1)$
Solution
First let's find a vector normal to these using the cross product.
$$\vec v_1 \times \vec v_2 = (1, 1, -4) \times (4, 1, 1)$$
$$ \begin{array}{rrr|rr} 1 & 1 & 4 & 1 & 4 \\[5pt] 4 & 1 & 1 & 4 & 1 \\[5pt] x & y & z & x & y \end{array}$$
$$ \begin{align} \vec v_1 \times \vec v_2 &= z + x +16y \\[5pt] &- 4x - y - 16z \\[5pt] &= -3x + 15y - 15z \\[5pt] &\rightarrow \; \vec n = (-3, 15, -15) \end{align}$$
An equation of our plane is
$$ \begin{align} (-3, 15, -15) \cdot (1-x, 1-y, -4-z) &= 0 \\[5pt] -3(1-x) + 15(1-y) - 15(-4-z) &= 0 \\[5pt] -3 + 3x + 15 -15y +60 + 15z &= 0 \\[5pt] 3x -15y + 15z +72 &= 0 \\[5pt] 3x -15y + 15z &= -72 \; \text{ or} \\[5pt] x - 3y + 3z &= -24 \end{align}$$
Here is a graph of our plane and the vectors.
-
The plane defined by these two points: $P = (-2, -3, -2), \; Q = (-2, 1, 4)$ and $R = (-3, -2, 5)$.
Solution
We can define two vectors from these three points by subtraction. The choice is arbitrary, but let's use $\vec v_1 = Q-P$ and $\vec v_2 = R-P$:
$$ \begin{align} Q - P &= (-2-(-2), 1-(-3), 4-(-2)) \\[5pt] \vec v_1 &= (0, 4, 6) \\[8pt] R - P &= (-3-(-2), -2-(-3), 5-(-2)) \\[5pt] \vec v_2 &= (-1, 1, 7) \end{align}$$
Now we take the cross product of these vectors to find a vector normal to our plane:
$$\vec v_1 \times \vec v_2 = (1, 1, -4) \times (4, 1, 1)$$
$$ \begin{array}{rrr|rr} 0 & 4 & 6 & 0 & 4 \\[5pt] -1 & 1 & 7 & -1 & 1 \\[5pt] x & y & z & x & y \end{array}$$
$$ \begin{align} \vec v_1 \times \vec v_2 &= 0z + 28x -6y \\[5pt] &- 6x - 0y + 4z \\[5pt] &= 22x - 6y + 4z \\[5pt] &\rightarrow \; \vec n = (22, -6, 4) \end{align}$$
An equation of our plane is
$$ \begin{align} (22,-6,4) \cdot (0-x, 4-y, 6-z) &= 0 \\[5pt] -22(0-x) - 6(4-y) + 4(6-z) &= 0 \\[5pt] 22x \cancel{- 24} + 6y + \cancel{24} -4z &= 0 \\[5pt] 22x + 6y - 4z &= 0 \\[5pt] 11x + 3y - 2z &= 0 \end{align}$$
Vector product or cross product
Finding a vector orthogonal to two other vectors in a plane.
Quadric surfaces
Learn about a few basic 3D graphs, including cylinders, ellipsoids, paraboloids and hyperboloids.
Parameter
A parameter is an adjustable constant in the definition of a function that is different from the independent variable(s). Parameters are not independent variables. For example, in the quadratic function
f(x) = Ax2 + Bx + C
A, B and C are parameters which change the shape of the graph of the function. x is the independent variable. A, B and C are fixed for any particular version of f(x), but x can range from -&inf; to +&inf;
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