Lines and vectors

Before delving deep into the relationship of lines and vectors, there are key differences that you should know when comparing the two:

You can describe a line using vectors, and that is what we will explore in some of the equations below.

Vector equation of a line

In a 2D setting (or $\mathbb {R}$), to describe a line in terms of vectors, we need one vector that represents a point on the line and another vector used to describe the direction of the line instead of the slope.

$\vec {r} = \vec {r_0} + t \vec{m} \quad \text {or} \quad (x, \ y) = (x_0, \ y_0) + t(m_1, \ m_2)$

Where:

• $t \in \mathbb {R}$

• $\vec {r} = (x, \ y)$ is a position vector at any point on the line.

• $\vec {r_0} = (x_0, \ y_0)$ is a position vector at some point on the line.

• $\vec {m} = (m_1, \ m_2)$ is a direction vector parallel to the line.

For a line in $\mathbb {R}^3$, the vector equation would be in the form $(x, \ y, \ z) = (x_0, \ y_0, \ z_0) + t(m_1, \ m_2, \ m_3)$

Here is what the vector equation $(x, \ y) = (-1, \ 1) + t(1, \ 1)$ looks like compared to the line $y = x + 2$:

• The vector $\vec {r_0}$ represents a point on $y = x + 2$

• The vector $\vec {m}$ represents the slope $m = \frac {m_2} {m_1} = \frac {1} {1} = 1$

Parametric equations of a line

The parametric equations of a line are equations formed from the individual coordinates of the vectors of the vector equation of a line.

From the vector equation $(x, \ y) = (x_0, \ y_0) + t(m_1, \ m_2)$, we get:

$x = x_0 + t m_1 \quad \text {and} \quad y = y_0 + t m_2$

If the line is in $\mathbb{R}^3$, you also have $z = z_0 + t m_3$.

Cartesian equation of a line

The cartesian (or scalar) equation of a line is what you already know: $y = mx + b$, but with a twist!

You must rearrange the equation and multiply it by the common denominators to get an equation of the form:

$Ax + By + C = 0$

$A$, $B$ and $C$ must be whole numbers, and $\vec {n} = (A, \ B, \ C)$ is the normal vector to the line. The normal vector is a vector that is perpendicular to another.

Here's the graph of the cartesian equation $-3x + 4y - 8 = 0$ looks like compared to its graph $y = \frac {3} {4} x + 2$:

Again, if the line is in $\mathbb {R}^3$, the equation would be in the form $Ax + By + Cz + D = 0$

Example: Vector to cartesian

Convert the vector equation of the line $\vec {r} = (3, \ -5) + t(-1, \ -4)$ to its cartesian form by:

a) Using the vector equation

b) Using the parametric equations

answer

a) We can use the point $\vec {r_0} = (3, \ -5)$, as well as calculate the slope using $\vec {m} = (-1, \ -4)$

$m = \frac {m_2} {m_1} = \frac {-4} {-1} = 4$

Now, substitute these values into the equation of a line:

$\quad \quad \quad \, y - y_0 = m(x - x_0)$

$\quad \quad y - (-5) = (4)(x - 3)$

$\quad \quad \quad \ \, \, y + 5 = 4x - 12$

$-4x + y + 17 = 0$

$\therefore$ The cartesian equation of the line is $-4x + y + 17 = 0$

b) We can use the vector equation $\vec {r} = (3, \ -5) + t(-1, \ -4)$ to form the following parametric equations:

$x = 3 - t$

$y = -5 - 4t$

Now, rearrange for $t$:

$t = 3 - x$

$t = \frac {-5 \ - \ y} {4}$

Finally, plug the equations into each other to get the cartesian equation:

$\quad \quad \quad \ \, \frac {-5 \ - \ y} {4} = 3 - x$

$\quad \quad \ \ \ -5 - y = 12 - 4x$

$\ \ \ 4x - y - 17 = 0$

$-4x + y + 17 = 0$

$\therefore$ The cartesian equation of the line is $-4x + y + 17 = 0$

Example: cartesian to vector form

Convert the cartesian equation of the plane $x + 2y - z + 6 = 0$ to:

a) Its vector form

b) Its parametric equations

answer

a) To convert to vector form, we need one point, and two direction vectors.

We can get a point by selecting a value for two variables, and solving for the other.

Say $x = 0$ and $y = 0$, then:

$\quad \ \, x + 2y - z + 6 = 0$

$(0) + 2(0) - z + 6 = 0$

$\quad \quad \quad \quad \quad \quad \quad \, \, z = 6$

So, the point $(0, \ 0, \ 6)$ is on the plane.

To get the two direction vectors, we need to use the plane's normal vector:

$\vec {n} = (1, \ 2, \ -1)$

We can use the dot product to generate vectors that are perpendicular to the normal. These vectors are the direction vectors of the plane.

$a(1) + b(2) + c(-1) = 0$ where $(a, \ b, \ c)$ is a direction vector for the plane.

If $a = 1$, $b = 0$ and $c = 1$: $(1)(1) + (0)(2) + (1)(-1) = 1 - 1 = 0 \ \Rightarrow \$ So $\vec {d_1} = (1, \ 0, \ 1)$

If $a = 0$, $b = 1$ and $c = 2$: $(0)(1) + (1)(2) + (2)(-1) = 2 - 2 = 0 \ \Rightarrow \$ So $\vec {d_2} = (0, \ 1, \ 2)$

Combining the information we just calculated, we get the vector equation:

$\vec { \pi } = (0, \ 0, \ 6) + s(1, \ 0, \ 1) + t(0, \ 1, \ 2)$

$\therefore$ The plane's vector equation is $\vec { \pi } = (0, \ 0, \ 6) + s(1, \ 0, \ 1) + t(0, \ 1, \ 2)$

b) The parametric equations can be derived from the vector equation we just obtained:

$x = s$

$y = t$

$y = 6 + s + 2t$

Symmetric equation of a line in 3D

Note that the vector equation of a line in $\mathbb {R}^3$ is $(x, \ y, \ z) = (x_0, \ y_0, \ z_0) + t(m_1, \ m_2, \ m_3)$

Well, we can rearrange the parametric equations for $t$:

$x = x_0 + t m_1 \ \Rightarrow \ t = \frac { x \ - \ x_0 } { m_1 }, m_1 \ne 0$

$y = y_0 + t m_2 \ \Rightarrow \ t = \frac { y \ - \ y_0 } { m_2 }, m_2 \ne 0$

$z = z_0 + t m_3 \ \Rightarrow \ t = \frac { z \ - \ z_0 } { m_3 }, m_3 \ne 0$

Now, we just equate them to get the symmetric equation:

$\frac { x \ - \ x_0 } { m_1 } = \frac { y \ - \ y_0 } { m_2 } = \frac { z \ - \ z_0 } { m_3 } \quad \text {where} \quad m_1 \ne 0, m_2 \ne 0, m_3 \ne 0$