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You CAN Ace Calculus

17calculus > vectors > lines in 3-space

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Lines in 3-Space Using Vectors

Up until now, you have worked with lines in two dimensions. It was convenient to define such lines using slope ($$m$$) and y-intercept ($$b$$) and write them in slope-intercept form $$y = mx + b$$. The only kind of lines that can not be written in this form are vertical lines, which we write as $$x = c$$. Another way to write the equation of a line is in general form $$Ax +By + C = 0$$. All lines in the plane can be written in this general form.

Now we are going to work with lines in three dimensions (sometimes called 3-space or just space).

When you studied lines in 2-dimensions, not only could you describe a line using the slope-intercept form, as mentioned above, but you could also describe it using parametric equations. An example might be to describe $$y=mx+b$$ in parametric form, we could use parameter t and write $$x = t$$ and $$y = mt + b$$.

At a minimum, a line in space is uniquely defined by two points on the line. However, there are several ways to specify lines in 3-dimensions. Let's use the graph on the right to get these forms. Here are a few things to notice about this graph.

- We want to specify the line that goes through the two points $$P_1$$ and $$P_2$$. The two points must be distinct to define a line, i.e. $$P_1 \neq P_2$$.
- The points $$P_1$$ and $$P_2$$ are known, i.e. we have the actual $$(x,y,z)$$ values for both of these points.
- Let $$P_1$$ be specified as $$P_1 = (x_1,y_1,z_1)$$.
- Let $$P_2$$ be specified as $$P_2 = (x_2,y_2,z_2)$$.
- The point P represents any point on the line and we write it as $$P = (x,y,z)$$.(1)
- Vector $$\vec{v}_1$$ (labeled v1 in the graph) is the vector from the origin to the point $$P_1$$. So $$\vec{v}_1 = \langle x_1, y_1, z_1 \rangle$$ or $$\vec{v}_1 = x_1\hat{i} + y_1\hat{j} + z_1\hat{k}$$.
- Vector $$\vec{v}_2$$ (labeled v2 in the graph) is the vector from the origin to the point $$P_2$$. So $$\vec{v}_2 = \langle x_2, y_2, z_2 \rangle$$ or $$\vec{v}_2 = x_2\hat{i} + y_2\hat{j} + z_2\hat{k}$$.
- Vector $$\vec{v}$$ is the vector from the origin to the point P. So $$\vec{v} = \langle x, y, z \rangle$$ or $$\vec{v} = x\hat{i} + y\hat{j} + z\hat{k}$$.

Notes
(1) Subscript Notation - It is common in mathematics to use subscripts when we are talking about specific or known values and to leave them off when a general (nonspecific) value is represented. Notice for the point P there are no subscripts in (x,y,z). This means that no specific point is defined.

Here is a very good video clip explaining this idea and paralleling it with the slope-intercept form of a line that you already know.

 PatrickJMT - lines and planes

Parametric Equations

One of the easiest ways to specify 3-dimensional lines is using parametric equations like we did for 2 dimensions. Using the above graph, let's derive one parametric equation. (There are an infinite number of sets of parametric equations.)

Define vector $$\vec{a}$$ as the vector from $$P_1$$ to $$P_2$$. We can write this as $$\vec{a} = \vec{v}_2 - \vec{v}_1$$. Defined this way, vector $$\vec{a}$$ ends up to be a vector parallel to the line.

Using the parameter t, any other point P on the line can be defined as $$\vec{v} = \vec{v}_1 + t \vec{a}$$. In this case, the domain of t is the set of all real numbers (t can be positive, negative or zero).

Specify vector $$\vec{a}$$ as $$\vec{a} = \langle a,b,c \rangle$$.
[Note: Using the letter a as both the vector name and as a scalar in the first component of the vector is not the best use of notation here. However, there should be no confusion since the vector is written with the vector arrow above the name, the scalar is written without it. This is an example of where notation is critically important. Mathematicians do this often, when the context and notation make it clear which a we are talking about.]

Using the equations $$\vec{v} = \vec{v}_1 + t \vec{a}$$ and $$\vec{a} = \langle a,b,c \rangle$$ we can write $$\langle x,y,z \rangle = \langle x_1,y_1,z_1 \rangle + t \langle a,b,c \rangle$$. If we equate the individual componenets we get the parametric equations

 $$x = x_1 + at$$ $$y = y_1 + bt$$ $$z = z_1 + ct$$

1. These equations are not unique since the vector $$\vec{a}$$ is dependent on the choices of $$\vec{v}_1$$ and $$\vec{v}_2$$.
2. We can emphasize that these parametric equations are functions of t by writing them as

 $$x(t) = x_1 + at$$ $$y(t) = y_1 + bt$$ $$z(t) = z_1 + ct$$

3. Another way to write the symmetric equations is as column vectors. This closely parallels the vector form $$\langle x,y,z \rangle = \langle x_1,y_1,z_1 \rangle + t \langle a,b,c \rangle$$.
$$\begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} x_1 \\ y_2 \\ z_3 \end{bmatrix} + t \begin{bmatrix} a \\ b \\ c \end{bmatrix}$$

Extended Parametric Discussion - Describing Lines

Here is a video discussing various ways to describe a line, a line segment and a ray. The question he poses, and answers in the video, is this.
Let A and B be points with respective position vectors $$\vec{a}$$ and $$\vec{b}$$. Determine a parametric vector form for
1. the line segment AB;
2. the ray from B and passing through A;
3. all points P that lie on the line connecting A and B with A between P and B;
4. all points Q that lie on the line through A and B which lie closer to B than A.
This video will help you understand lines and vectors much more deeply.

 Dr Chris Tisdell - Equations of line segments and rays

Vector Function

We can write a vector function using the parametric equations as $$\vec{V}(t) = (x_1+at)\hat{i} + (y_1+bt)\hat{j} + (z_1+ct)\hat{k}$$ or more simply as $$\vec{V}(t) = x(t)\hat{i} + y(t)\hat{j} + z(t)\hat{k}$$. Of course, since this vector function is dependent on non-unique parametric equations, there are many other (in fact, an infinite number) vector functions to describe any given line.

Symmetric Equations

Using the parametric equations above, we solve for t and equate them to get the symmetric equations as follows.

 $$\displaystyle{ x = x_1 + at ~~~ \to ~~~ t = \frac{x-x_1}{a} }$$ $$\displaystyle{ y = y_1 + bt ~~~ \to ~~~ t = \frac{y-y_1}{b} }$$ $$\displaystyle{ z = z_1 + ct ~~~ \to ~~~ t = \frac{z-z_1}{c} }$$

Since t is the same in all three equations, we know that the other side of the equations are equal. This gives us these symmetric equations.

$$\displaystyle{ \frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c} }$$

1. Just as with the parametric equations, these equations are not unique but dependent on the choices of $$\vec{v}_1$$ and $$\vec{v}_2$$.
2. The values $$a,b,c$$ are called the direction numbers of the line. The name comes from the fact that the vector $$\vec{a} = \langle a,b,c \rangle$$ gives the direction of the line.
3. Of course, we are assuming here that $$a$$, $$b$$ and $$c$$ are all nonzero. If one of the values is zero, we can still write the symmetric equations and combine it with the corresponding parametric equation. If, for example, $$c=0$$, from the parametric equation $$z = z_1 + ct$$ we can see that $$z = z_1$$. So we write our symmetric equations as

$$\displaystyle{ \frac{x-x_1}{a} = \frac{y-y_1}{b} }$$;     $$z = z_1$$

Okay, time for some practice problems.

Search 17Calculus

Practice 1

Find a vector equation for the line that goes through $$(1,3,2)$$ and $$(-4,3,0)$$.

solution

Practice 2

Find the symmetric equations of a line through the point $$P(2,3,-4)$$ in the same direction as the vector $$\vec{v} = \langle 1,-1,-2 \rangle$$.

solution

Practice 3

Find the parametric equations of the line that passes through the point $$P(-1,2,3)$$ and is parallel to the vector $$\vec{v}=\langle4,5,6\rangle$$.

solution

Practice 4

Find the scalar equation of the line through the point $$P(0,0,0)$$ in the direction $$\vec{v} = \hat{i}+2\hat{j}+3\hat{k}$$.

solution

Practice 5

Find the scalar equation of the line through the point $$P(4,13,-3)$$ in the direction $$\vec{v} = 2\hat{i}-3\hat{k}$$.

solution

Practice 6

Find the scalar equation of the line through the points $$P(0,0,0)$$ and $$Q(-6,3,5)$$.

solution

Practice 7

Find the scalar equation of the line through the points $$P(3,5,7)$$ and $$P(6,5,4)$$.

solution

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