Write parametric vector equations of lines and convert between vector and Cartesian forms

What this dot point is asking

NESA wants you to describe a line not by a Cartesian equation but by a starting point plus a direction: $\mathbf{r} = \mathbf{a} + t\mathbf{d}$. You should be able to write this form from a point and a direction (or from two points), convert it to and from Cartesian form, and use it to find where two lines meet, including the difference between two paths crossing and two particles colliding. The parametric form is the natural language for motion, and it handles vertical lines and intersections more cleanly than $y = mx + c$.

The answer

A line is a point plus multiples of a direction

Start at a known point $A$ on the line, with position vector $\mathbf{a}$. Pick any non-zero direction vector $\mathbf{d}$ pointing along the line. Every other point is reached by walking some number of steps $t$ along $\mathbf{d}$. So a general point has position vector

Here $\mathbf{r}$ is the position vector of a variable point on the line and $t$ is the parameter: each value of $t$ picks out one point, and as $t$ runs over all real numbers, $\mathbf{r}$ sweeps out the whole line. (Different textbooks use $\lambda$, $\mu$ or $t$ for the parameter; they all mean the same thing.) Let us build this up.

Stage 1, the position vector of a known point. Fix the origin $O$. The position vector $\mathbf{a} = \overrightarrow{OA}$ pins down one point $A$ that the line passes through.

Stage 2, the direction vector. From $A$, the direction vector $\mathbf{d}$ points along the line. Adding one copy of $\mathbf{d}$ to $\mathbf{a}$ lands on a second point of the line.

Stage 3, different values of $t$ give different points. Walking $t$ steps along $\mathbf{d}$ from $A$ reaches the point $\mathbf{a} + t\mathbf{d}$. At $t = 0$ we are at $A$; $t = 1$ is one step on; $t = 2$ is two steps; $t = -1$ is one step backwards. The parameter is a coordinate along the line.

Stage 4, the whole line. Letting $t$ range over every real number sweeps out the entire line. The position vector $\mathbf{r}$ of a general point $P$ is $\mathbf{a}$ (to get onto the line at $A$) plus $t$ copies of $\mathbf{d}$ (to slide along to $P$).

In components this reads

Line through two points

If a line passes through $A$ and $B$, a direction along it is $\mathbf{d} = \overrightarrow{AB} = \mathbf{b} - \mathbf{a}$. So

The second form is worth noticing: at $t = 0$ you are at $A$, at $t = 1$ you are at $B$, and for $0 < t < 1$ you are on the segment between them (this is exactly the section-formula idea from geometric proofs).

Converting to Cartesian form: eliminate the parameter

To get a Cartesian equation, write the two component equations and eliminate $t$. If $d_1 \ne 0$, solve $x = a_1 + t d_1$ for $t = \dfrac{x - a_1}{d_1}$, then substitute into $y$:

That is the point-slope form, so the gradient is $\dfrac{d_2}{d_1}$, the "rise over run" of the direction vector. Going the other way, a Cartesian line $y = mx + c$ has direction vector $(1, m)$ and passes through $(0, c)$, giving $\mathbf{r} = (0, c) + t(1, m)$.

Vertical and horizontal lines

The parametric form copes with cases that break $y = mx + c$:

• If $d_1 = 0$ (direction straight up), the line is vertical: $x = a_1$ for all $t$, with no defined gradient.
• If $d_2 = 0$ (direction horizontal), the line is horizontal: $y = a_2$.

This robustness, no special-casing vertical lines, is one practical reason the vector form is preferred for motion and intersection problems.

Intersection of two lines

Two lines meet where their position vectors coincide. Use different parameters for the two lines and set them equal:

That is one equation per coordinate, so two equations in the two unknowns $\lambda$ and $\mu$. Solve them, then substitute back to get the meeting point.

Three outcomes are possible. A unique solution means the lines cross at one point. No solution means they are parallel and distinct (directions are scalar multiples, but no shared point). Infinitely many solutions means they are the same line (coincident).

Two readings of the parameter

• Geometric: $t$ just indexes points along the line; its scale depends on how big $\mathbf{d}$ is.
• Time: if a particle starts at $\mathbf{a}$ at time $0$ and moves with velocity $\mathbf{d}$, then $\mathbf{r}(t) = \mathbf{a} + t\mathbf{d}$ is its position at time $t$, and $\mathbf{d}$ is genuinely its velocity vector.

The time reading is what makes collision problems different from intersection problems, covered next.

Collision versus path-crossing

If two particles trace lines that cross, their paths share a point, but the particles only collide if they are at that point at the same time. So for a collision you must find a single value of $t$ that makes both position vectors equal, not two different parameter values. Equate the position vectors using the same $t$:

Solve one coordinate for $t$, then check the other coordinate holds at that same $t$. If it does, they collide (and you have the time and place); if it does not, the paths merely cross.

How exam questions ask about parametric lines

• "Write the vector equation of the line through $A$ with direction $\mathbf{d}$": state $\mathbf{r} = \mathbf{a} + t\mathbf{d}$.
• "... through $A$ and $B$": take $\mathbf{d} = \mathbf{b} - \mathbf{a}$ first.
• "Find the Cartesian equation" or "... in the form $y = mx + c$": eliminate the parameter.
• "Find the point of intersection of $L_1$ and $L_2$": set the forms equal (different parameters), solve, substitute back.
• "Are the lines parallel? Do they meet?": compare directions for parallelism; if parallel, test whether a point of one lies on the other to tell coincident from distinct.
• "Do the two particles collide?" or "At what time do they meet?": equate positions with the same parameter $t$ and require both coordinates to match.