How Do You Convert a Vector Equation to Parametric Form?

Write one equation for each coordinate by equating the corresponding components of the vector equation. The position-vector components become constants, while the direction-vector components become the coefficients of the same parameter.

This is HL-only content from IB Mathematics: Applications and Interpretation, AHL 3.11: Vector equation of a line in 2D and 3D. Assume an HL Paper 1 short-response setting, although the method can also appear in Papers 2 and 3.

A vector equation of a line has the form

$$\mathbf{r}=\mathbf{a}+\lambda\mathbf{b},$$

where $\mathbf{a}$ is the position vector of a point on the line, $\mathbf{b}$ is its direction vector, and $\lambda\in\mathbb{R}$ is a parameter. Expand the vector equation component by component to obtain the parametric equations.

For example, consider

$$\mathbf{r}=\begin{pmatrix}2\\-1\\4\end{pmatrix}+\lambda\begin{pmatrix}3\\5\\-2\end{pmatrix}.$$

Equating components gives

$$x=2+3\lambda,\qquad y=-1+5\lambda,\qquad z=4-2\lambda.$$

The same value of $\lambda$ must be used in all three equations because each value identifies one point on the line.

To interpret the result, choose values of the parameter. When $\lambda=0$, the point is $(2,-1,4)$, matching the position vector. When $\lambda=1$, the point is $(5,4,2)$. The direction vector shows that increasing by changes by , by , and by .

To verify the point $P(8,9,0)$, solve each component: $8=2+3\lambda$, $9=-1+5\lambda$, and . Each equation gives , so lies on the line. If even one equation gave a different value, the point would not lie on the line.

A common misconception is that converting to parametric form requires solving for $\lambda$. It does not: you are simply separating one vector equation into coordinate equations.

Exam technique: Write every coordinate equation and preserve the same parameter throughout. An examiner may also ask you to verify whether a point lies on the line; substitute its coordinates and check that all component equations produce one consistent value of $\lambda$.