Equation of Lines in Two-Space and Three-Space
Two-Space
Method 1: Use Paper and Pencil
The equation of this line is \(-x + 2y - 10 = 0\).
- Copy the graph.
- Draw a vector that is parallel to the line. Label the vector \(m\)
- Draw a position vector starting at the origin with its tip on the line. Label this vector \(\vec{r}_0\)
- Reflect In the equation \(y = mx + b\), the slope, \(m\), changes the orientation of the line; the \(y\)-intercept, \(b\), changes the position of the line
- Draw the vector \(a = [-1, 2]\). Compare the vector a with the equation of the line and with the graph of the line
- On a new grid, graph the line \(3x + 4y - 20 = 0\). On the same grid, draw the vector \(b = [3, 4]\). Compare the vector b with the equation of the line and the graph of the line.
- Reflect Describe how your answers for steps 3 and 4 are similar. Give an example of a different line and vector pair that have the same property as those from steps 3 and 4.
Method 2: Use The Geometer's Sketchpad
Go to www.mcgrawhill.ca/links/calculus12 and follow the links to 8.1. Download the file 8.1Vector2D.gsp. Open the sketch.
- The vector m is parallel to the blue line. Move the vectors \(\vec{m}\) and \(\vec{r}_0\) by dragging their tips. Record your observations.
- Reflect In the equation \(y = mx + b\), the slope, \(m\), changes the orientation of the line, and the \(y\)-intercept, \(b\), changes the position of the line
- Draw the vector \(a = [-1, 2]\). Compare the vector a with the equation of the line and with the graph of the line
- On a new grid, graph the line \(3x + 4y - 20 = 0\). On the same grid, draw the vector \(b = [3, 4]\). Compare the vector b with the equation of the line and the graph of the line.
- Reflect Describe how your answers for steps 3 and 4 are similar. Give an example of a different line and vector pair that have the same property as those from steps 3 and 4.
Three-Space
A line in three-space can also be defined by a vector equation or by parametric equations. It cannot, however, be defined by a scalar equation. In three-space, we will see that a scalar equation defines a plane. A plane is a two-dimensional flat surface that extends infinitely far in all directions.
For a line that passes through the points \(A(1, 4)\) and \(B(3,1)\):
- Write a vector equation.
- Determine three more position vectors to points on the line. Graph the line.
- Draw a closed triangle that represents the vector equation using points \(A\) and \(B\)
- Determine if the point \((2, 3)\) is on the line.
First, we can determine the vector from point \(A\) to point \(B\) to find the direction vector:
\(\vec{m}= \vec{OB}-\vec{OA}\)
\(\vec{m}= [3,1]-[1,4]\)
\(\vec{m} = [2,-3]\)
Next, we can choose one of the points to be the position vector, \(\vec{r}_0\). In this instance, we will choose \(B(3, 1)\).
Then, we can write the vector equation as such:
\(\vec{r}=\vec{r}_0+t\vec{m}\)
\(\vec{r}=[3,1]+t[2,-3]\)
Therefore, we can write our vector equation as \(\vec{r}=[3,1]+t[2,-3]\).
ii. First, we can let \(t = 1, 2,\) and \(-2\) and determine their respective position vectors:
\([x, y] = [3, 1] + (1)[2, -3] = [5, -2]\)
\([x, y] = [3, 1] + (2)[2, -3] = [7, -5]\)
\([x, y] = [3, 1] + (-2)[2, -3] = [-1, 7]\)
Next, we can graph the line as such:
iii. We can draw the closed triangle with points \(A\) and \(B\) as such:
iv. If the point \((2, 3)\) is on the line, then the position vector \((2, 3)\) has its tip on the line. So, there exists a single value of \(t\) that makes the equation true:
\([x, y] = [3, 1] + t[2, -3]\) becomes \([2, 3] = [3, 1] + t[2, -3]\)
Since the t-values are not equal, we can determine the point \((2, -3)\) does not lie on the line.
A line passes through points \(A (2, -1, 5)\) and \(B (3, 6, -4)\).
- Write a vector equation of the line.
- Write parametric equations for the line.
- Determine if the point \(C (0, -15, 9)\) lies on the line.
i. First, we can determine the vector from point \(A\) to point \(B\) to find the direction vector:
\(\vec{m}=\vec{OB}-\vec{OA}\)
\(\vec{m}=[3,6,-4]-[2,-1,5]\)
\(\vec{m}=[1,7,-9]\)
Next, we can choose one of the points to be the position vector, \(\vec{r}_0\). In this instance, we will choose \(A(2,-1,5)\).
\(\vec{r} = \vec{r}_0 + t\vec{m}\)
\([x, y, z] = [2, -1, 5] + t[1, 7, -9]\)
Therefore, we can write our vector equation as \(\boldsymbol{[x, y, z] = [2, -1, 5] + t[1, 7, -9]}\).
ii. We can write the corresponding parametric equations as such:
\(x = 2 + t\)
\(y = -1 + 7t\)
\(z = 5 - 9t\)
iii. We can substitute the coordinates of \(C(0,-15, 9)\) into the parametric equations and solve for \(t\).
We can first solve for \(t\) for the first equation:
\(0 = 2 + t\)
\(t=-2\)
We can then solve for \(t\) for the second equation:
\(-15 = -1 + 7t\)
\(t=-2\)
We can finally solve for \(t\) for the third equation:
\(9 = 5 - 9t\)
\(t = -\cfrac{4}{9}\)
The \(t\)-values are not equal, so the point does not lie on the line. This can be seen on the graph:
A line in two-space has an infinite number of normal vectors that are all parallel to one another. A line in three-space also has an infinite number of normal vectors, but they are not necessarily parallel to one another:
