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 toward 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. The same would be true if you switched the $x$ and $y$ axes and used your right hand.) 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.

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 $\vec{v} = \langle 2, -1, 3\rangle.$

First we find the length of $\vec{v}:$

$$ \begin{align} L &= \sqrt{2^2 + (-1)^2 + 3^2} \\[5pt] &= \sqrt{4 + 1 + 9} \\[5pt] &= \sqrt{14} \end{align}$$

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.

Practice problems

Add the two vectors to obtain $\vec{v},$ then normalize the resulting vector to obtain $\hat{v}.$

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:

$$ \begin{align} (-3, 4, 8) &= -3(1, 0, 0) + 4((0, 1, 0) + 8(0, 0, 1) \\[5pt] &= -3 \hat{i} + 4 \hat{j} + 8 \hat{k} \end{align}$$

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 $\vec r_o = \langle x_o, \; y_o, \; z_o\rangle$ 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

$$\vec Q - \vec{P} = \langle x - x_0, \; y - y_0, \; z - z_0 \rangle $$

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 $\vec{v} = \hat{i} + 4 \hat{j} - 2 \hat{k}.$

Solution: Our target equation is

$$(x, y, z) = (x_0, \; y_0, \; z_0) + t \cdot \langle a, \; b, \; c\rangle,$$

where

$$ \begin{align} (x_0, \; y_0, \; z_0) &= (5, \; 1, \; 3) \\[5pt] \langle a, \; b, \;c\rangle &= \langle 1, \; 4, -2\rangle \end{align}$$

So our equation is

$$(x, y, z) = (5, 1, 3) + t\langle 1, 4, -2\rangle $$

We can write this as a list of parametric equations:

$$ \begin{align} x &= 5 + t \\[5pt] y &= 1 + 4t \\[5pt] z &= 3 - 2t \end{align}$$

Example 2

Find the vector equation of the line passing through points $P_1 = (2, -1, -2)$ and $P_2 = (1, -2, 3).$

Solution: Either one of these points will suffice as our anchor point, so we'll just pick the first one, $P_1 = (2, -1, -2).$ Now we can get a direction vector by subtracting $P_1$ from $P_2$:

$$ \begin{align} P_2 - P_1 &= (1, -2, 3) - (2, -1, -2) \\[5pt] &= \langle 1 - 2, -2 - (-1), 3 - (-2)\rangle \\[5pt] &= \langle -1, -1, 5\rangle \end{align}$$

So our direction vector is $\vec{r_0} = \langle -1, -1, 5\rangle.$

The initial equation of our line is then

$$(x, y, z) = (2, -1, -2) + t\langle -1, -1, 5\rangle $$

The parametric equations are:

$$ \begin{align} x &= 2 - t \\[5pt] y &= -1 - t \\[5pt] z &= -2 + 5t \end{align}$$

Practice problems

Find an equation of the line passing through the point and parallel to the given vector

Find an equation of the line passing through the points given.

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.

Example 3

Find the intersection, if there is one, between the lines   $(2, 1, 0) + t(-1, -1, -1)$   and   $(3, 0, 5) + t\langle 2, 0, 6\rangle$.

Solution: The key here is to re-express each line using two different parameters, say $s$ and $t$:

$$ \begin{align} (2, 1, 0) &+ s\langle -1, -1, -1\rangle \\[5pt] &= (2-s, 1-s, -s) \\[9pt] (3, 0, 5) &+ t\langle 2, 0, 6\rangle \\[5pt] &= (3+2t, 0, 5+6t) \end{align}$$

Then set corresponding terms of each vector equal:

$$ \begin{align} 2 - s &= 3 + 2t \\[5pt] 1 - s &= 0 \\[5pt] -s &= 5 + 6t \end{align}$$

Example 4

Determine whether the lines   $(2s - 3, s - 1, s - 4)$   and   $(t - 3, 3t, 5 - 2t)$   intersect.

Solution: 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

$$ \begin{align} 1 + 3t &= 9 - 2t \\[5pt] 5t &= 8 \\[5pt] t &= \frac{8}{5}. \end{align}$$

Example 5

Consider the parametric lines

$$ \begin{align} \vec r_1 &= \langle t, \; t-1, \; 2t+3 \rangle \\[5pt] \vec r_2 &= \langle 2s, \; 2s-1, \; 4s+3 \rangle \end{align}$$

Learn what you can from these two lines: How are they oriented with respect to each other? Do they intersect? If so, where?

Solution: We can reverse-engineer the parametric forms of these lines like this:

$$ \begin{align} \vec r_1 &= (0, -1, \; 3) + t \langle 1, \; 1, \; 2 \rangle \\[5pt] \vec r_2 &= (0, -1, \; 3) + s \langle 2, \; 2, \; 4 \rangle \end{align}$$

These lines have the same starting point and the same direction (notice that $\langle 2, \; 2, \; 4 \rangle$ is just a scalar multiple of $\langle 1, \; 1, \; 2 \rangle$, so both lines have the same direction. You could see this more clearly if you normalized both. So they represent the same line.

Practice problems

1. Show that the parametric equations

   $$ \begin{align} x &= 2s+1, \; y = s+2, \; z = s-2 \\[5pt] x &= 6t+1, \; y = 3t+2, \; z = 3t-2 \end{align}$$

   represent the same line.

   Solution

   We can reconstruct the parametric vector equations of these lines like this:

   $$ \begin{align} \vec r_1 &= (1, \; 2, -2) + s\langle 2, \; 1, \; 1 \rangle \\[5pt] \vec r_2 &= (1, \; 2, -2) + s\langle 6, \; 3, \; 3 \rangle \end{align}$$

   We can see that the starting point of both lines is the point $P = (1, \; 2, -2)$, so both lines go through the same point. The direction vectors are actually multiples of one another: $\langle 6, \; 3, \; 3 \rangle = 3 \langle 2, \; 1, \; 1 \rangle$, so they both also point in the same direction. Thus, they are the same line. You can prove to yourself that the normalized versions of both direction vectors are the same.




2. 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:

   $$ \begin{align} L_1: \; (0, -3, 1) + t\langle 2, 1, -1\rangle \\[5pt] L_2: \; (0, 1, -2) + s\langle 1, 1, 3\rangle \\[5pt] \end{align}$$

   Now the dot product of the direction vectors (the ones multiplied by the parameters) is

   $$ \langle 2, \; 1,-1 \rangle \cdot \langle 1,\; 1, \; 3 \rangle = 2 + 1 - 3 = 0$$

   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.