AOS5 Topic 7: Vector and Cartesian Equations

Vector Equations of Lines

Vector equation of a line given by a point and a direction:

A line in two- or three-dimensional space may be described using two vectors:

  • the position vector \( \vec{a} \) of a point A on the line
  • a vector \( \vec{d} \) parallel to the line.

We can describe the line as:

\[ \vec{r} = \vec{a} + t\vec{d} \] for some \( t \in \mathbb{R} \)

Usually we omit the set notation. We write \( \vec{r}(t) \) for the position vector of a point P on the line, and therefore \( \vec{r}(t) = \vec{a} + t\vec{d} \), \( t \in \mathbb{R} \).

This is a vector equation of the line \( \vec{r} \).

As the value of \( t \) varies over the real numbers, the position vector \( \vec{r}(t) \) varies over all the points on the line \( \vec{r} \).

We sometimes express this idea by saying that \( t \) is a parameter and that \( \vec{r}(t) \) is a parameterization of the line \( \vec{r} \).


Created with GeoGebra, by Cliff-Lee, Link

Note: There is no unique vector equation of a given line. We can choose any point A as the 'starting point' on the line and any vector \( \vec{d} \) parallel to the line.

Vector Equation of a Line Given by Two Points

If the position vectors \( \vec{a} = \overrightarrow{OA} \) and \( \vec{b} = \overrightarrow{OB} \) of two points on a line \( \vec{r} \) are known, then the line may be described by:

\[ \vec{r}(t) = \overrightarrow{OA} + t(\overrightarrow{AB}) = \vec{a} + t(\vec{b} - \vec{a}), \quad t \in \mathbb{R} \]

This is also a vector equation of the line \( \vec{r} \).

This vector equation can be rewritten as:

\[ \vec{r}(t) = (1 - t)\vec{a} + t\vec{b}, \quad t \in \mathbb{R} \]

Cartesian Equation of a Line in Two Dimensions

From a Vector Equation to the Cartesian Equation

For example, start with the vector equation:

\[ \vec{r} = \vec{i} + 5\vec{j} + t(\vec{i} + 2\vec{j}), \quad t \in \mathbb{R} \]

Rearrange this equation as:

\[ \vec{r} = (1 + t)\vec{i} + (5 + 2t)\vec{j} \]

Let \( P(x, y) \) be the point on the line with position vector \( \vec{r} \), so that \( \vec{r} = xi + yj \). Then, by equating coefficients of \( \vec{i} \) and \( \vec{j} \), we have:

  • \( x = 1 + t \)
  • \( y = 5 + 2t \)

These are parametric equations for the line.

Now eliminate \( t \) to find \( y \) in terms of \( x \). We have \( t = x - 1 \), so \( y = 5 + 2(x - 1) = 2x + 3 \).

The Cartesian equation of the line is \( y = 2x + 3 \).

Conversion from Cartesian Equation to Vector Equation

For example, start with the Cartesian equation \( y = 2x + 3 \).

A point on the line is \( (0, 3) \), with position vector \( 3\vec{j} \). The line has gradient 2, so a vector parallel to the line is \( \vec{i} + 2\vec{j} \).

Therefore, a vector equation of the line is:

\[ \vec{r} = 3\vec{j} + t(\vec{i} + 2\vec{j}), \quad t \in \mathbb{R} \]

Note: For a line with equation \( y = mx + c \), you can choose the point \( (0, c) \) on the line and the vector \( \vec{i} + m\vec{j} \) parallel to the line.

Cartesian Form for a Line in Three Dimensions

From a Vector Equation to Cartesian Form

For example, the line through the point \( (5, -2, 4) \) that is parallel to the vector \( 2\vec{i} - \vec{j} + 3\vec{k} \) can be described by the vector equation:

\[ \vec{r} = 5\vec{i} - 2\vec{j} + 4\vec{k} + t(2\vec{i} - \vec{j} + 3\vec{k}), \quad t \in \mathbb{R} \]

Let \( P(x, y, z) \) be the point on the line with position vector \( \vec{r} \). Then we can write the vector equation as:

\[ xi + yj + zk = (5 + 2t)i + (-2 - t)j + (4 + 3t)k \]

The corresponding parametric equations are:

  • \( x = 5 + 2t \)
  • \( y = -2 - t \)
  • \( z = 4 + 3t \)

Solving each of these equations for \( t \), we have:

\( \frac{x - 5}{2} = \frac{y + 2}{-1} = \frac{z - 4}{3} = t \)

This is in Cartesian form. You cannot describe a line in three dimensions using a single linear Cartesian equation.

From Cartesian Form to a Vector Equation

To convert from Cartesian form to a vector equation, we can perform these steps in the reverse order.

Lines in Three Dimensions

A line in three-dimensional space can be described in the following three ways, where \( \vec{a} = a_1\vec{i} + a_2\vec{j} + a_3\vec{k} \) is the position vector of a point \( A \) on the line, and \( \vec{d} = d_1\vec{i} + d_2\vec{j} + d_3\vec{k} \) is a vector parallel to the line.

Vector Equation Parametric Equations Cartesian Form
\( \vec{r} = \vec{a} + t\vec{d} \), \( t \in \mathbb{R} \) \( x = a_1 + d_1t \)
\( y = a_2 + d_2t \)
\( z = a_3 + d_3t \)
\( \frac{x - a_1}{d_1} = \frac{y - a_2}{d_2} = \frac{z - a_3}{d_3} \)

Parallel and Perpendicular Lines

For two lines `1 : \( \vec{r}_1 = \vec{a}_1 + t\vec{d}_1 \), \( t \in \mathbb{R} \), and `2 : \( \vec{r}_2 = \vec{a}_2 + s\vec{d}_2 \), \( s \in \mathbb{R} \):

  • The lines `1 and `2 are parallel if and only if \( \vec{d}_1 \) is parallel to \( \vec{d}_2 \).
  • The lines `1 and `2 are perpendicular if and only if \( \vec{d}_1 \) is perpendicular to \( \vec{d}_2 \).
Example 1

Verification of Point on Line

Verify whether the point \( P(-7, 4, -14) \) lies on the line represented by the vector equation:

\[ \vec{r}(t) = 5\vec{i} - 2\vec{j} + 4\vec{k} + t(2\vec{i} - \vec{j} + 3\vec{k}), \quad t \in \mathbb{R} \]

Solution:

The point \( P(-7, 4, -14) \) has position vector \( -7\vec{i} + 4\vec{j} - 14\vec{k} \).

By equating coefficients of \( \vec{i} \), \( \vec{j} \), and \( \vec{k} \), we can see that the point \( P \) lies on the line if there exists \( t \in \mathbb{R} \) such that:

  • \( 5 + 2t = -7 \)
  • \( -2 - t = 4 \)
  • \( 4 + 3t = -14 \)

A solution for each of these equations is \( t = -6 \). Hence \( P \) lies on the line.

Example 2

Find a Vector Equation for Each of the Following Lines

a) The line through A(1, 2) that is parallel to \( 2\vec{i} + 3\vec{j} \)

b) The line passing through the points A(3, -5, 4) and B(-4, 3, 10)

Solution

a) Point A has position vector \( \vec{i} + 2\vec{j} \). So, a vector equation of the line is:

\[ \vec{r}(t) = \vec{i} + 2\vec{j} + t(2\vec{i} + 3\vec{j}), \quad t \in \mathbb{R} \]

b) The points A and B have position vectors \( \vec{a} = 3\vec{i} - 5\vec{j} + 4\vec{k} \) and \( \vec{b} = -4\vec{i} + 3\vec{j} + 10\vec{k} \) respectively. So, a vector equation of the line is:

\[ \vec{r}(t) = \vec{a} + t(\vec{b} - \vec{a}) \]

\[ = 3\vec{i} - 5\vec{j} + 4\vec{k} + t((-4\vec{i} + 3\vec{j} + 10\vec{k}) - (3\vec{i} - 5\vec{j} + 4\vec{k})) \]

\[ = 3\vec{i} - 5\vec{j} + 4\vec{k} + t(-7\vec{i} + 8\vec{j} + 6\vec{k}), \quad t \in \mathbb{R} \]

Example 3

Vector Equations of Lines

Let ` be the line with vector equation:

\[ \vec{r} = \vec{i} + 2\vec{j} + 3\vec{k} + t(-\vec{i} - 3\vec{j}), \quad t \in \mathbb{R} \]

a) Find a vector equation of the line through \( A(1, 3, 2) \) that is parallel to the line `.

b) Find a vector equation of the line through \( A(1, 3, 2) \) that is perpendicular to the line ` and parallel to the x–y plane.

Solution

a) Line Parallel to ` and Passing Through A(1, 3, 2)

The position vector of \( A \) is \( \vec{i} + 3\vec{j} + 2\vec{k} \), and a vector parallel to ` is \( -\vec{i} - 3\vec{j} \).

Therefore, a vector equation of the line through \( A \) parallel to ` is:

\[ \vec{r} = \vec{i} + 3\vec{j} + 2\vec{k} + s(-\vec{i} - 3\vec{j}), \quad s \in \mathbb{R} \]

b) Line Perpendicular to ` and Parallel to the x–y Plane

If a vector is parallel to the x–y plane, then its \( k \)-component is zero. So we want to find a vector \( \vec{d} = d_1\vec{i} + d_2\vec{j} \) that is perpendicular to \( -\vec{i} - 3\vec{j} \).

Therefore, we require:

\[ (\vec{d}) \cdot (-\vec{i} - 3\vec{j}) = 0 \]

i.e. \( -d_1 - 3d_2 = 0 \)

We see that we can choose \( d_1 = 3 \) and \( d_2 = -1 \). So \( \vec{d} = 3\vec{i} - \vec{j} \).

Hence, a vector equation of the required line is:

\[ \vec{r} = \vec{i} + 3\vec{j} + 2\vec{k} + s(3\vec{i} - \vec{j}), \quad s \in \mathbb{R} \]

Example 4

Question: Vector Equations of Lines

Let \(l\) be the line with vector equation:

\[ \vec{r} = \vec{i} + 2\vec{j} + 3\vec{k} + t(-\vec{i} - 3\vec{j}), \quad t \in \mathbb{R} \]

a) Find a vector equation of the line through \( A(1, 3, 2) \) that is parallel to the line `.

b) Find a vector equation of the line through \( A(1, 3, 2) \) that is perpendicular to the line ` and parallel to the x–y plane.

Solution

a) The position vector of \( A \) is \( \vec{i} + 3\vec{j} + 2\vec{k} \), and a vector parallel to ` is \( -\vec{i} - 3\vec{j} \).

Therefore, a vector equation of the line through \( A \) parallel to ` is:

\[ \vec{r} = \vec{i} + 3\vec{j} + 2\vec{k} + s(-\vec{i} - 3\vec{j}), \quad s \in \mathbb{R} \]

b) If a vector is parallel to the x–y plane, then its \( k \)-component is zero. So we want to find a vector \( \vec{d} = d_1\vec{i} + d_2\vec{j} \) that is perpendicular to \( -\vec{i} - 3\vec{j} \).

Therefore, we require:

\[ (\vec{d}) \cdot (-\vec{i} - 3\vec{j}) = 0 \]

i.e. \( -d_1 - 3d_2 = 0 \)

We see that we can choose \( d_1 = 3 \) and \( d_2 = -1 \). So \( \vec{d} = 3\vec{i} - \vec{j} \).

Hence, a vector equation of the required line is:

\[ \vec{r} = \vec{i} + 3\vec{j} + 2\vec{k} + s(3\vec{i} - \vec{j}), \quad s \in \mathbb{R} \]

Exercise &&1&& (&&1&& Question)

Find a vector equation of the line \( AB \), where the points \( A \) and \( B \) have position vectors \( \overrightarrow{OA} = \vec{i} + \vec{j} - 2\vec{k} \) and \( \overrightarrow{OB} = 2\vec{i} - \vec{j} - \vec{k} \) respectively.

1
Submit

Exercise &&2&& (&&1&& Question)

Given the vector equation:

\[ \vec{r} = \vec{i} + 5\vec{j} + t(\vec{i} + 2\vec{j}), \quad t \in \mathbb{R} \]

What will be the Cartesian equation of the line represented by it?

2
Submit

Exercise &&3&& (&&1&& Question)

In this vector equation \( \vec{r}(t) = \vec{a} + t\vec{d} \), where \( t \in \mathbb{R} \), what does \( \vec{r}(t) \) represent?

3
Submit

Exercise &&4&& (&&1&& Question)

Find the distance to the line \( \vec{r}(t) = (1 - t)\vec{i} + (2 - 3t)\vec{j} + 2\vec{k} \), \( t \in \mathbb{R} \), from the origin.

4
Submit