You should already be familiar with the equation of a straight line in Cartesian form in 2 dimensions, y = ax + b. When we move into 3 dimensions, the Cartesian form becomes a little more awkward. Don't worry, vectors are here to help us out! Once you understand the vector equation of a line, it is really useful for solving all sorts of problems with angles and intersections.
Key Concepts
On this page, you should learn about
vector equations of lines in two and three dimensions in the three different forms
vector form
parametric form
cartesian form
Essentials
The following videos will help you understand all the concepts from this page
Vector Equation of a Line
In the following video we are going to look at the equation of a straight line in vector form and gain a strong conceptual understanding of what this formula means
\(\textbf{ r }\ =\ \overrightarrow { OA } +\lambda \textbf{b}\)
You are probably used to Cartesian form of the equation of a straight line (y = mx + c) and perhaps are wondering why you need vectors to describe a straight line. There are number of answers to this
The Cartesian form is a bit messy when used in 3D.
Finding intersections and the angles which lines meet is easier in vector form.
Describing motion is really helpful using velocity vectors.
Let's start by looking at an example in 2D and then we can move into 3D
In the following video we are going to look at the three different forms of the equation of a straight line. In particular, we are going to look at how we can convert from one form to another.
In the following video we are going to look at a lovely application of the equation of the straight line to find points equidistant from another point. We don't necessarily have to do it in this way, but it might help us really understand what the vector equation of a straight line means.
Here is the example
A(3,-1,2) and B(6,-7,-7) lie on a straight line L. C also lies on the straight line L. Find the coordinates of the point C given that \(\left| \overrightarrow { AC } \right| =\left| \overrightarrow { AB } \right| \).
In the following video we are going to look we will try and gain a conceptual understanding of velocity vectors. One of the key ideas of this topic is to decide if objects collide. It is not enough to consider if their paths cross. We need to think about whether they occupy the same position at the same moment in time.
To get you started, you might like to play this game to give you an idea about what is going on. Try to hit the submarine with the torpedo!
Now let's consider the example below:
A submarine is initially positioned at (0, 5) travels with velocity \(\left( \begin{matrix} 4 \\ -3 \end{matrix} \right) \\ \)ms-1 .
One second later a torpedo is fired from (3, 0) with velocity \(\left( \begin{matrix} 5 \\ 1 \end{matrix} \right) \\ \)ms-1 .
The equation of the line below is given by the equation \(\textbf{r}=\overrightarrow { OA } +\lambda \overrightarrow { AB } \).
To describe the position of C, what is a possible value of \(\lambda\)
The equation of the line below is given by the equation \(\textbf{r}=\overrightarrow { OA } +\lambda \overrightarrow { AB } \).
To describe the position of C, what is a possible value of \(\lambda\)
The equation of the line below is given by the equation \(\textbf{r}=\overrightarrow { OA } +\lambda \overrightarrow { AB } \).
To describe the position of C, what is a possible value of \(\lambda\)
Which of the following points lie on the line \(\textbf{r}=\left( \begin{matrix} 1 \\ -2 \end{matrix} \right) +\lambda \left( \begin{matrix} 3 \\ 4 \end{matrix} \right) \)
What is the value of a so that (-7,-8,4) lies on the line \({ r }=\left( \begin{matrix} a \\ -2 \\ 0 \end{matrix} \right) +\lambda \left( \begin{matrix} 4 \\ 3 \\ -2 \end{matrix} \right) \)
What are the values of a and b so that (-2,-7,4) lies on the line \({ r }=\left( \begin{matrix} 1 \\ -1 \\ -2 \end{matrix} \right) +\lambda \left( \begin{matrix} a \\ b \\ 2 \end{matrix} \right) \)
The vector equation of a line L is \(\textbf{ r }=\left( \begin{matrix} -1 \\ -2 \\ 3 \end{matrix} \right) +\mu \left( \begin{matrix} 1 \\ -2 \\ -4 \end{matrix} \right) \)
The vector \(\left( \begin{matrix} a \\ b \\ 8 \end{matrix} \right) \) is parallel to this line.
Find a and b
a =
b =
The line \(\textbf{ r }=\left( \begin{matrix} { x }_{ 0 } \\ { y }_{ 0 } \\ { z }_{ 0 } \end{matrix} \right) +\mu \left( \begin{matrix} l \\ m \\ n \end{matrix} \right) \)is parallel to \(\left( \begin{matrix} l \\ m \\ n \end{matrix} \right) \)
The following quiz tests your understanding of converting between the different forms of a straight line (vector, parametric and Cartesian).
The vector equation of a line is \({ r }=\left( \begin{matrix} -1 \\ 0 \\ 2 \end{matrix} \right) +\mu \left( \begin{matrix} 3 \\ -2 \\ -1 \end{matrix} \right) \). Which is the correct parametric form of the line.
\(\textbf{ r }=\left( \begin{matrix} { x }_{ 0 } \\ { y }_{ 0 } \\ { z }_{ 0 } \end{matrix} \right) +\mu \left( \begin{matrix} l \\ m \\ n \end{matrix} \right) \)corresponds to \(\frac { x-{ x }_{ 0 } }{ l } =\frac { y-{ y }_{ 0 } }{ m } =\frac { z-{ z }_{ 0 } }{ n } \)
The vector equation of a line L is \(\textbf{ r }=\left( \begin{matrix} 1 \\ -3 \\ 0 \end{matrix} \right) +\mu \left( \begin{matrix} 2 \\ -3 \\ 1 \end{matrix} \right) \)
The Cartesian equation of a line L is \(\frac { x-a }{ 2 } =\frac { y-b }{ -3 } =\frac { z-c }{ 1 } \)
Find a, b and c
a =
b =
c =
\(\textbf{ r }=\left( \begin{matrix} { x }_{ 0 } \\ { y }_{ 0 } \\ { z }_{ 0 } \end{matrix} \right) +\mu \left( \begin{matrix} l \\ m \\ n \end{matrix} \right) \) corresponds to \(\frac { x-{ x }_{ 0 } }{ l } =\frac { y-{ y }_{ 0 } }{ m } =\frac { z-{ z }_{ 0 } }{ n } \)
The vector equation of a line L is \(\textbf{ r }=\left( \begin{matrix} 1 \\ -1 \\ 0 \end{matrix} \right) +\mu \left( \begin{matrix} a \\ b \\ c \end{matrix} \right) \)
The Cartesian equation of a line L is \(\frac { 1-x }{ 3 } =\frac { y+1 }{ 2 } =z\)
Find a, b and c
a =
b =
c =
\(\textbf{ r }=\left( \begin{matrix} { x }_{ 0 } \\ { y }_{ 0 } \\ { z }_{ 0 } \end{matrix} \right) +\mu \left( \begin{matrix} l \\ m \\ n \end{matrix} \right) \) corresponds to \(\frac { x-{ x }_{ 0 } }{ l } =\frac { y-{ y }_{ 0 } }{ m } =\frac { z-{ z }_{ 0 } }{ n } \)
The Cartesian form can be written as follows \(\frac { x-1 }{ -3 } =\frac { y-(-1) }{ 2 }=\frac { z-0 }{ 1 }\)
The Cartesian equation of a line L is \(\frac { x-3 }{ 5 } =2y =z+1\)
The vector equation of a line L is \(\textbf{ r }=\left( \begin{matrix} a \\ b \\ c \end{matrix} \right) +\mu \left( \begin{matrix} d \\ e \\ f \end{matrix} \right) \)
Find a , b , c , d , e and f
a =
b =
c =
d =
e =
f =
\(\textbf{ r }=\left( \begin{matrix} { x }_{ 0 } \\ { y }_{ 0 } \\ { z }_{ 0 } \end{matrix} \right) +\mu \left( \begin{matrix} l \\ m \\ n \end{matrix} \right) \)corresponds to \(\frac { x-{ x }_{ 0 } }{ l } =\frac { y-{ y }_{ 0 } }{ m } =\frac { z-{ z }_{ 0 } }{ n } \)
The Cartesian form can be written as follows \(\frac { x-3 }{ 5 } =\frac { y-0 }{ 0.5 }=\frac { z-(-1) }{ 1 }\)
The Cartesian equation of a line L is \(3x=\frac { y+1 }{ 2 } =1+z\)
The line is parallel to the vector \(\left( \begin{matrix} a \\ b \\ -3 \end{matrix} \right) \).
Find a and b
a =
b =
\(\frac { x-{ x }_{ 0 } }{ l } =\frac { y-{ y }_{ 0 } }{ m } =\frac { z-{ z }_{ 0 } }{ n } \) corresponds to \(\textbf{ r }=\left( \begin{matrix} { x }_{ 0 } \\ { y }_{ 0 } \\ { z }_{ 0 } \end{matrix} \right) +\mu \left( \begin{matrix} l \\ m \\ n \end{matrix} \right) \)which is parallel to \(\left( \begin{matrix} l \\ m \\ n \end{matrix} \right) \)
The Cartesian form can be written as follows \(\frac { x-0 }{ \frac { 1 }{ 3 } } =\frac { y-(-1) }{ 2 }=\frac { z-(-1) }{ 1 }\) which is parallel to \(\left( \begin{matrix} \frac{1}{3} \\ 2 \\ 1 \end{matrix} \right) \)
Exam-style Questions
Question 1
A line L passes through the points A(1,-1,3) and B(3,-4,4)
Point C (x,y,1) also lies on the line L. Find x and y.
Hint
If the equation of the straight line is given by \(\textbf{r}= \textbf{a}+\lambda \textbf{b}\) then a certain value of \(\lambda \) will define the position of the position vector \(\overrightarrow { OC } \). Find this value and use it to find x and y.
A line L passes through the points A(0,2,-4) and B(3,-3,2)
Point C also lies on the line L. Find the coordinates of C given that \(\left| \overrightarrow { AC } \right| =\left| \overrightarrow { AB } \right| \)
Hint
Draw a diagram!
Find the equation of the straight line. What is the value of \(\lambda \) that defines the position of B? Think about what this value should be for C.
A line L passes through the points A(0,2,-4) and B(3,-3,2)
Point C also lies on the line L. Find the possible coordinates of C given that \(\left| \overrightarrow { AC } \right| =2\left| \overrightarrow { AB } \right| \)
Hint
Draw a diagram!
Is there just one answer?
Find the equation of the straight line. What is the value of \(\lambda \) that defines the position of B? Think about what this value should be for C.
You should already be familiar with the equation of a straight line in Cartesian form in 2 dimensions, y = ax + b. When we move into 3 dimensions, the Cartesian form becomes a little more awkward. Don't worry, vectors are here to help us out! Once you understand the vector equation of a line, it is really useful for solving all sorts of problems with angles and intersections.
Notes from the video
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