Mastering the Equation of a Line in 3D Using Vectors

In two-dimensional geometry, we are accustomed to describing lines using the slope-intercept form $y = mx + c$. However, as we move into three-dimensional space, this approach becomes insufficient. To define a line in 3D, we require a more robust framework: vectors. Understanding how to represent a line using vectors is a cornerstone of Further Maths and essential for success in your exams.

In this article, you will learn how to construct the vector equation of a line, convert it into parametric and Cartesian forms, and apply these concepts to solve geometric problems. Mastering these techniques will allow you to navigate 3D space with confidence.

The Core Concept: Point and Direction

To define a unique line in 3D space, you only need two pieces of information: a fixed point on the line and the direction in which the line travels.

Let $\mathbf{a}$ be the position vector of a known point $A$ on the line. Let $\mathbf{d}$ be a direction vector that is parallel to the line. Any point $P$ on the line, with position vector $\mathbf{r}$, can be reached by starting at $A$ and moving some scalar multiple of the direction vector $\mathbf{d}$. We represent this scalar using a parameter, typically denoted by $\lambda$ or $t$. The vector equation is:

$$\mathbf{r} = \mathbf{a} + \lambda\mathbf{d}$$

Here, $\mathbf{r} = \begin{pmatrix} x \ y \ z \end{pmatrix}$, $\mathbf{a} = \begin{pmatrix} x_0 \ y_0 \ z_0 \end{pmatrix}$, and $\mathbf{d} = \begin{pmatrix} d_1 \ d_2 \ d_3 \end{pmatrix}$.

Parametric Equations

By expanding the vector equation $\mathbf{r} = \mathbf{a} + \lambda\mathbf{d}$, we can separate the components into three individual equations. This is known as the parametric form:

$$x = x_0 + \lambda d_1$$ $$y = y_0 + \lambda d_2$$ $$z = z_0 + \lambda d_3$$

These equations describe the coordinates of any point on the line in terms of the parameter $\lambda$. This is particularly useful when finding the intersection of two lines or determining if a point lies on a specific path.

Cartesian Form

If we rearrange the parametric equations to make $\lambda$ the subject, we obtain the Cartesian (or symmetric) form of the line. Assuming $d_1, d_2, d_3 \neq 0$:

$$\frac{x - x_0}{d_1} = \frac{y - y_0}{d_2} = \frac{z - z_0}{d_3} = \lambda$$

This form is elegant because it eliminates the parameter $\lambda$ entirely, providing a direct relationship between the $x, y,$ and $z$ coordinates.

Worked Example 1: Finding the Equation

Question: Find the vector and Cartesian equations of the line passing through the point $A(2, -1, 3)$ and parallel to the vector $\mathbf{d} = \begin{pmatrix} 1 \ 4 \ -2 \end{pmatrix}$.

Solution:

1. Vector Form: Using $\mathbf{r} = \mathbf{a} + \lambda\mathbf{d}$, we substitute our values: $$\mathbf{r} = \begin{pmatrix} 2 \ -1 \ 3 \end{pmatrix} + \lambda \begin{pmatrix} 1 \ 4 \ -2 \end{pmatrix}$$
2. Cartesian Form: Using the formula $\frac{x - x_0}{d_1} = \frac{y - y_0}{d_2} = \frac{z - z_0}{d_3}$: $$\frac{x - 2}{1} = \frac{y + 1}{4} = \frac{z - 3}{-2}$$

Worked Example 2: Line Through Two Points

Question: Find the vector equation of the line passing through $P(1, 0, 2)$ and $Q(3, 2, 1)$.

Solution:

1. Find the direction vector: The vector $\vec{PQ} = \vec{OQ} - \vec{OP} = \begin{pmatrix} 3-1 \ 2-0 \ 1-2 \end{pmatrix} = \begin{pmatrix} 2 \ 2 \ -1 \end{pmatrix}$.
2. Write the equation: Using point $P$ as our fixed point $\mathbf{a}$: $$\mathbf{r} = \begin{pmatrix} 1 \ 0 \ 2 \end{pmatrix} + \lambda \begin{pmatrix} 2 \ 2 \ -1 \end{pmatrix}$$

Common Mistakes

• Confusing points and vectors: Remember that $\mathbf{a}$ is a position vector (a point), while $\mathbf{d}$ is a direction vector. Never add a point to a direction vector without the scalar parameter $\lambda$.
• Sign errors in Cartesian form: When writing $\frac{x - x_0}{d_1}$, ensure you subtract the coordinates of the point. If the point is $(2, -1, 3)$, the term for $y$ becomes $y - (-1) = y + 1$.
• Non-unique representations: A line can be represented by infinitely many equations. If your answer looks different from a mark scheme, check if your direction vector is a scalar multiple of theirs.

FAQ

Can a line have more than one vector equation? Yes. You can choose any point on the line as your starting point and any vector parallel to the line as your direction vector.

What if one of the direction components is zero? If $d_1 = 0$, you cannot write the Cartesian form as a single fraction. You would write $x = x_0$ and $\frac{y - y_0}{d_2} = \frac{z - z_0}{d_3}$.

How do I check if a point lies on a line? Substitute the point's coordinates into the parametric equations. If you find a single value of $\lambda$ that satisfies all three equations, the point is on the line.

Conclusion

Understanding the vector equation of a line is essential for mastering 3D geometry. By practising the conversion between vector, parametric, and Cartesian forms, you will be well-prepared for your exams. To see these concepts in motion, visit MathInstructor AI to generate a free, narrated animated lesson on this topic.