## V6 Vector equation of a line

If you want to uniquely define a line, you need to pin it between two points in 3-dimensional space. You can also define a point with a 3-dimensional vector through it. This process uses three types of equations.

Learn how to find the vector equation, the parametric equation, and the symmetric equation of a line in three-dimensional space.

### Introduction

You have probably been taught that a line in the $$x-y$$ plane can be represented in the form \begin{align*} y & =mx+c \end{align*} where $$m$$ is the gradient ( or slope) of the line and $$c$$ is the $$y-$$intercept. But how do we define a line in three dimensions (3D)? 1 Three dimensional space (3D) is a geometric setting in which three co-ordinates are required to determine the position of a point. This is different to when we are dealing with points in the $$x-y$$ plane where only two co-ordinates are required.

This module deals with the equation of a line in 3D.

### Vector Equation of a Line in 3D

A line in 3D space can be described by a point and a vector along the line. There is only one line that can be drawn through a given point in the direction of a given vector.

Consider the figure below:

Here

• $$\vec{r_{1}}$$ and $$\vec{r}_{2}$$ are two points on the line;

• $$P_{0}\left(x,y,z\right)$$ is a point on the line and $$\vec{P}_{0}$$ is itâ€™s position vector;

• $$\vec{V}$$ is a vector along the line;

• $$t$$ is a constant. That is $$t\in\mathbb{R}$$. To get to the point $$R_{1}$$ you first need to get onto the line. You do this by traveling along $$\vec{P}_{0}$$ to the point $$P_{0}$$ on the line and then traveling a distance along the line in the direction of vector $$\vec{V}$$ . So we could write \begin{align*} \vec{r}_{1} & =\vec{P}_{0}+t\vec{V} \end{align*} where $$t$$ is a positive constant. The quantity $$t\vec{V}$$ represents the distance and the direction we need to go along the line to get to the point $$R_{1}$$ from $$P_{0}$$ .

To get to the point $$R_{2}$$ you first need to get onto the line. You do this by traveling along $$\vec{P}_{0}$$ to the point $$P_{0}$$ on the line and then traveling a distance along the line in the opposite direction of vector $$\vec{V}$$ . So we could write \begin{align*} \vec{r}_{1} & =\vec{P}_{0}+t\vec{V} \end{align*} where $$t$$ is a negative constant (because we are going in the direction opposite to $$\vec{V}$$).

Considering the above, we hope you understand the equation of a line, in vector form, is \begin{align*} \vec{r} & =\vec{P_{0}}+t\vec{V} & \left(1\right) \end{align*} where $$t\in\mathbb{R}$$ is a constant.

It is also possible to define the line in other forms as described below.

### Parametric Equation of a Line in 3D

Suppose we wish to find the line that goes through the point $$P_{0}\left(x_{0},y_{0},z_{0}\right)$$ in the direction of the vector $$\overrightarrow{V}=a\hat{i}+b\hat{j}+c\hat{k}$$ . Consider a general point $$P\left(x,y,z\right)$$ that lies on the line that goes through the point $$P_{0}\left(x_{0},y_{0},z_{0}\right)$$ as shown at right. We can define a vector $$\overrightarrow{P_{0}P}$$ by:

\begin{align*} \overrightarrow{P_{0}P} & =\left(x-x_{0}\right)\hat{i}+\left(y-y_{0}\right)\hat{j}+\left(z-z_{0}\right)\hat{k}. & \left(2\right) \end{align*}

This vector will obviously be in the same direction as $$\overrightarrow{V}$$ but will, in all likelihood have a different magnitude. In other words $$\overrightarrow{P_{0}P}$$ will be a multiple of the direction vector $$\overrightarrow{V}.$$ So $$\overrightarrow{P_{0}P}=t\overrightarrow{V}$$ where $$t\in\mathbb{R}$$ is some number.

Hence:

\begin{alignat*}{2} \overrightarrow{P_{0}P} & =t\overrightarrow{V}\\ & =t\left(a\hat{i}+b\hat{j}+c\hat{k}\right)\\ & =at\hat{i}+bt\hat{j}+ct\hat{k}. & \left(3\right) \end{alignat*}

Equating components of $$\left(2\right)$$ and $$\left(3\right)$$ we have, \begin{align*} \left(x-x_{0}\right) & =at\\ x & =x_{0}+at & \left(4\right) \end{align*}

and

\begin{align*} \left(y-y_{0}\right) & =bt\\ y & =y_{0}+bt & \left(5\right) \end{align*}

and

\begin{align*} \left(z-z_{0}\right) & =ct\\ z & =z_{0}+ct. & \left(6\right) \end{align*}

These equations $$x=x_{0}+at$$, $$y=y_{0}+bt$$ and $$z=z_{0}+ct$$ are called the parametric equations of the line that contains the point $$\left(x_{0},y_{0},z_{0}\right)$$ and has the direction vector $$\overrightarrow{V}=a\hat{i}+b\hat{j}+c\hat{k}$$. The variable $$t\in\mathbb{R}$$ is called a parameter.

### Symmetric Form of the Line

By rearranging $$\left(4\right)$$ to $$\left(6\right)$$ above we find: \begin{align*} t & =\frac{x-x_{0}}{a}\\ t & =\frac{y-y_{0}}{b}\\ t & =\frac{z-z_{0}}{c}. \end{align*}

Since they are all equal, we can say that:

\begin{align*} \frac{x-x_{0}}{a} & =\frac{y-y_{0}}{b}=\frac{z-z_{0}}{c} & \left(7\right) \end{align*}

This is called the symmetric form of the equation of the line.

### Summary

A line in 3D may be represented:

1. In vector form (see $$\left(1\right)$$ above)

2. In parametric form (see $$\left(4-6\right)$$ above)

3. In symmetric form (see $$\left(7\right)$$above).

#### Example $$1$$

Find the parametric and symmetric equations of the line that goes through the point $$(3,2,3)$$ and is in the direction of the vector $$2\hat{i}+\hat{j}-5\hat{k}$$ .

Solution:

Using equations $$\left(4-6\right)$$ the parametric equations will be :

\begin{align*} x & =3+2t\\ y & =2+t\\ z & =3-5t. \end{align*}

Using equation $$\left(7\right)$$ the symmetric equations will be: \begin{alignat*}{1} \frac{x-3}{2} & =\frac{y-2}{1}=\frac{z-3}{-5} \end{alignat*}

or

\begin{alignat*}{1} \frac{x-3}{2} & =y-2=\frac{3-z}{5}. \end{alignat*}

#### Example $$2$$

Find the parametric and symmetric equations of the line that goes through the point $$(1,3,2)$$ in the direction of the vector $$\overrightarrow{V}=\hat{j}-2\hat{k}$$ .

Solution:

The parametric equations will be:

\begin{align*} x & =1+0t\\ & =1\\ y & =3+t\\ z & =2-2t. \end{align*}

The symmetric equation will be: \begin{align*} x=1 & ;\,y-3=\frac{z-2}{-2} \end{align*}

or

\begin{align*} x=1 & ;\,y-3=-\frac{1}{2}z+1. \end{align*}

#### Example $$3$$

Find the direction vector of the line : \begin{align*} \frac{x-6}{2} & =\frac{y-2}{-1}=\frac{z}{3}. \end{align*}

Solution:

The line is in symmetric form and so \begin{align*} \frac{x-6}{2} & =t\\ \frac{y-2}{-1} & =t\\ \frac{z}{3} & =t \end{align*} for any $$t\in\mathbb{R}.$$ Rearranging gives the parametric form:

\begin{align*} x & =6+2t\\ y & =2-t\\ z & =3t. \end{align*} This may be written in vector form as \begin{align*} x\hat{i}+y\hat{j}+z\hat{k} & =\left(6\hat{i}+2\hat{j}+0\hat{k}\right)+t\left(2\hat{i}-\hat{j}+3\hat{k}\right) \end{align*} which is a line through the point $$\left(6,2,0\right)$$ in the direction $$\left(2\hat{i}-\hat{j}+3\hat{k}\right)$$.

Hence the direction vector is $$\overrightarrow{V}=2\hat{i}-\hat{j}+3\hat{k}$$. Note that the direction vector of the line could be any multiple of this vector.

### Example 4

A line passes through the points $$A(1,2,4)$$ and $$B(3,-1,-2)$$. Find the equation of the line in a) vector format b) in parametric format and c) in symmetric format.

Solution:

1. For vector format we need to find a vector along the line. This can be

2 Note that it is also valid to take the vector $$\overrightarrow{BA}$$.\begin{alignat*}{1} \overrightarrow{AB} & =\left(3-1\right)\hat{i}+\left(-1-2\right)\hat{j}+\left(-2-4\right)\hat{k}\\ & =2\hat{i}-3\hat{j}-6\hat{k} \end{alignat*}

The vector $$\overrightarrow{AB}$$ is the direction vector along the line. We can now use either point, $$A$$ or $$B$$ to find the equation of the line in vector format. So the possible vector forms are:

\begin{align*} \vec{r} & =\left(\hat{i}+2\hat{j}+4\hat{k}\right)+t\left(2\hat{i}-3\hat{j}-6\hat{k}\right) & \left(8\right) \end{align*} where $$t\in\mathbb{R}$$ and we have used point $$A$$, or \begin{align*} \vec{r} & =\left(3\hat{i}-\hat{j}-2\hat{k}\right)+t\left(2\hat{i}-3\hat{j}-6\hat{k}\right) & \left(9\right) \end{align*} where we have used point $$B$$. Either is acceptable.

1. For the parametric equations, you can use either $$\left(8\right)$$ or $$\left(9\right)$$. We will use $$\left(8\right)$$.
Equating components, we have

3 Remember \begin{align*} \vec{r} & =x\hat{i}+y\hat{j}+z\hat{k} \end{align*} where $$x,$$ $$y$$ and $$z$$ are the coordinates of any point on the line. \begin{align*} \vec{r} & =\left(1\hat{i}+2\hat{j}+4\hat{k}\right)+t\left(2\hat{i}-3\hat{j}-6\hat{k}\right)\\ x\hat{i}+y\hat{j}+z\hat{k} & =\left(1\hat{i}+2\hat{j}+4\hat{k}\right)+t\left(2\hat{i}-3\hat{j}-6\hat{k}\right) \end{align*} and so the parametric equations are \begin{align*} x & =1+2t & \left(10\right)\\ y & =2-3t & \left(11\right)\\ z & =4-6t & \left(12\right) \end{align*} are the parametric equations.

1. The symmetric equations are obtained by solving $$\left(10-12\right)$$ for $$t$$ to get

\begin{align*} t & =\frac{x-1}{2}\\ t & =\frac{2-y}{3}\\ t & =\frac{4-z}{6}. \end{align*}

So the symmetric equations are: \begin{alignat*}{1} \frac{x-1}{2} & =\frac{2-y}{3}=\frac{4-z}{6}. \end{alignat*}

Note that you will get different parametric and symmetric equations if you use $$\left(9\right)$$ above. However they are acceptable solutions.

### Exercise:

Find the symmetric and parametric equations of the lines that satisfy the given conditions:

1. Contains the point $$\left(1,-1,2\right)$$ with the direction vector $$2\hat{i}-2\hat{j}+3\hat{k}\,$$?
Answer: $$\frac{x-1}{2}=\frac{y+1}{-2}=\frac{z-2}{3}$$ and $$x=1+2t$$ ; $$y=-1-2t$$ ; $$z=2+3t$$

2. Contains the point $$\left(3,4,-1\right)$$ with the direction vector $$\hat{i}+\hat{j}+5\hat{k}\,$$?
Answer: $$\frac{x-3}{1}=\frac{y-4}{1}=\frac{z+1}{5}$$ and $$x=3+t$$; $$y=4+t$$; $$z=5t-1$$

3. Contains the point $$\left(2,3,-1\right)$$ parallel to $$\frac{x-1}{2}=\frac{y-3}{-1}=\frac{x+1}{3}\,$$?
Answer: $$\frac{x-2}{2}=\frac{y-3}{-1}=\frac{z+1}{3}$$ and $$x=2+2t$$ ; $$y=3-t$$ ; $$z=-1+3t$$

4. Contains the point $$\left(2,2,1\right)$$ and $$\left(1,1,3\right)\,$$?
Answer: $$\frac{x-1}{1}=\frac{y-1}{1}=\frac{z-3}{-2}$$ and $$x=1+t$$; $$y=1+t$$; $$z=3-2t$$
or
$$\frac{x-2}{1}=\frac{y-2}{1}=\frac{z-1}{-2}\text{ and x=2+t;y=2+t;z=1-2t }$$

What's next... V7 Intersecting lines in 3D

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