Vector Equations of Lines and Intersections

Problem Statement

The familiar Cartesian equation of a line, $y = mx + c$, relies on the gradient $m$. The gradient is defined as "rise over run" ($\Delta y / \Delta x$). But what happens when we move to 3D space? We now have an $x$, $y$, and $z$ axis. A single number $m$ can no longer describe a direction.

To solve this, we use the vector equation of a line: $\mathbf{r} = \mathbf{a} + \lambda\mathbf{b}$. Here, $\mathbf{a}$ is a known position vector on the line (analogous to the $y$-intercept $c$), and $\mathbf{b}$ is the direction vector of the line (analogous to the gradient $m$). The scalar parameter $\lambda$ stretches the direction vector to reach every point on the line.

A major challenge in 3D geometry is determining if two lines intersect. Because we have three dimensions but only one parameter per line, two lines in 3D will usually miss each other completely (these are called skew lines). To prove they intersect, we must solve simultaneous equations.

Consider two lines in 3D space, $L_1$ and $L_2$, defined by the following vector equations:

• $L_1$: $\mathbf{r_1} = (2\mathbf{i} + 3\mathbf{j} - \mathbf{k}) + \lambda(\mathbf{i} - 2\mathbf{j} + 2\mathbf{k})$
• $L_2$: $\mathbf{r_2} = (-1\mathbf{i} + 7\mathbf{j} + 4\mathbf{k}) + \mu(2\mathbf{i} + \mathbf{j} - 3\mathbf{k})$

(a) Write down the parametric equations for the $x, y$, and $z$ coordinates for both $L_1$ and $L_2$.

(b) Determine whether the lines $L_1$ and $L_2$ intersect. If they do, find the position vector of the point of intersection.

Hints

• Part (a): Group the $\mathbf{i}$, $\mathbf{j}$, and $\mathbf{k}$ components together for each line. For example, the $x$-coordinate for $L_1$ is $x_1 = 2 + 1\lambda$.
• Part (b): If the lines intersect, there must exist a specific value of $\lambda$ and a specific value of $\mu$ that yield the exact same $x, y$, and $z$ coordinates. Set $x_1 = x_2$ and $y_1 = y_2$ to form a system of two simultaneous equations with two unknowns ($\lambda, \mu$). Solve for $\lambda$ and $\mu$. Finally, substitute these values into the $z$-coordinate equation ($z_1 = z_2$) to see if it holds true. If it does, they intersect; if not, they are skew.

Solutions

Part (a): Parametric Equations

1. For Line $L_1$: $x_1 = 2 + \lambda$ $y_1 = 3 - 2\lambda$ $z_1 = -1 + 2\lambda$
2. For Line $L_2$: $x_2 = -1 + 2\mu$ $y_2 = 7 + \mu$ $z_2 = 4 - 3\mu$

Part (b): Testing for Intersection

1. Equate the $x$ components: $2 + \lambda = -1 + 2\mu$ $\lambda - 2\mu = -3 \quad \text{--- (Equation 1)}$
2. Equate the $y$ components: $3 - 2\lambda = 7 + \mu$ $-2\lambda - \mu = 4 \quad \text{--- (Equation 2)}$
3. Solve the simultaneous equations. Multiply Equation 1 by 2: $2\lambda - 4\mu = -6$
4. Add this to Equation 2: $(2\lambda - 4\mu) + (-2\lambda - \mu) = -6 + 4$ $-5\mu = -2$ $\mu = \frac{2}{5} = 0.4$
5. Substitute $\mu = 0.4$ back into Equation 1 to find $\lambda$: $\lambda - 2(0.4) = -3$ $\lambda - 0.8 = -3$ $\lambda = -2.2$
6. The Crucial Test: We must now check if these values satisfy the $z$-coordinate equation. If they don't, the lines do not intersect.
   • Test $z_1$ with $\lambda = -2.2$: $z_1 = -1 + 2(-2.2) = -1 - 4.4 = -5.4$
   • Test $z_2$ with $\mu = 0.4$: $z_2 = 4 - 3(0.4) = 4 - 1.2 = 2.8$
7. Since $z_1 \neq z_2$ ($-5.4 \neq 2.8$), there is no point in 3D space where both lines exist simultaneously.
8. Conclusion: The lines $L_1$ and $L_2$ do not intersect. They are skew lines.

Takeaways

• The Power of Parameters: The vector form $\mathbf{r} = \mathbf{a} + \lambda\mathbf{b}$ cleanly separates a line into a fixed starting position ($\mathbf{a}$) and a direction vector ($\mathbf{b}$), overcoming the dimensional limitations of Cartesian gradients.
• 3D Intersections are Rare: In 2D, non-parallel lines must intersect. In 3D, non-parallel lines will almost always miss each other (skew). Proving intersection requires verifying that all three coordinate planes align perfectly for a specific pair of parameters.
• Simultaneous Verification: Always use two coordinate planes (e.g., $x$ and $y$) to solve for the parameters, and reserve the third plane ($z$) purely as a verification check.

Further Readings

• HSC Vectors: https://vumaths.com/booklets/hsc-vectors/
• HSC Collections: https://vumaths.com/booklets/hsc-collections/
• HSC Algebra: https://vumaths.com/booklets/hsc-polynomials/

Connect with me

If you found this breakdown of 3D intersections helpful, there is plenty more where that came from! Head over to Vu's Maths Hub for complete booklets on Extension 2 Vectors. You can also catch my video walkthroughs on YouTube, or read my deeper dives into syllabus changes over on my Substack. Don't forget to follow me on Instagram!