Determine vector, parametric and Cartesian equations of a line in three dimensions, convert between these forms, and find the point of intersection of two lines or establish that they are parallel or skew

What this dot point is asking

QCAA wants you to describe a line in three dimensions in three equivalent ways, move between them, and decide how two lines relate. In three dimensions lines need not be parallel or intersecting; they can be skew, missing each other entirely. The assessable skill is setting up the equations correctly and testing for a consistent intersection.

The answer

Vector equation of a line

A line through a point with position vector $\mathbf{a}$ and direction $\mathbf{d}$ has vector equation

As $t$ varies, $\mathbf{r}$ sweeps out every point on the line. The direction $\mathbf{d}$ may be scaled by any nonzero constant without changing the line.

Parametric equations

Writing $\mathbf{a} = (a_1, a_2, a_3)$ and $\mathbf{d} = (d_1, d_2, d_3)$ and splitting into components gives

These three equations carry the same information as the single vector equation.

Cartesian equation

Eliminating the parameter $t$ (when each $d_i \neq 0$) gives the symmetric Cartesian form

Each equal ratio equals $t$. If a direction component is zero, that coordinate is constant and is stated separately, for example $z = a_3$.

Relationship between two lines

Given $\mathbf{r}_1 = \mathbf{a} + t\mathbf{d}$ and $\mathbf{r}_2 = \mathbf{b} + s\mathbf{e}$, classify them as follows. If $\mathbf{d}$ and $\mathbf{e}$ are scalar multiples, the lines are parallel (and identical if a point of one lies on the other). Otherwise solve the component equations for $t$ and $s$. If a single pair $(t, s)$ satisfies all three components, the lines intersect at that point. If two components give a pair $(t,s)$ that fails the third, the lines are skew: not parallel and not intersecting.

Finding the intersection point

Equate the parametric forms component by component to get three equations in two unknowns $t$ and $s$. Solve any two, then substitute into the third to test consistency. A consistent solution gives the intersection point by substituting $t$ back into $\mathbf{r}_1$.

Why skew lines matter

In two dimensions, non-parallel lines always meet. In three dimensions this fails, so you must check the third equation rather than assuming intersection. This third-equation check is exactly where QCAA distinguishes intersecting from skew lines.

A line from two points

A common starting point is two given points $A$ and $B$ rather than a point and a direction. Take the direction to be the displacement $\vec{AB} = \mathbf{b} - \mathbf{a}$ and the base point to be either $A$ or $B$; both choices describe the same line. This is why a line has many equivalent vector equations: any point on the line and any nonzero scalar multiple of the direction will do. When checking whether a given point lies on a line, substitute its coordinates into the parametric equations and confirm the same value of $t$ satisfies all three.

The angle between two lines

The angle between two lines is the angle between their direction vectors, found with the dot product:

The absolute value gives the acute angle. Two lines are perpendicular when $\mathbf{d}_1 \cdot \mathbf{d}_2 = 0$, even if the lines are skew and never meet, because the angle is defined by directions alone. This distinction (perpendicular directions versus actually meeting) is a regularly tested subtlety.

Distance from a point to a line

To find how far a point $P$ is from a line $\mathbf{r} = \mathbf{a} + t\mathbf{d}$, take the displacement $\vec{AP}$ from the base point, then use the cross product: the perpendicular distance is $\dfrac{|\vec{AP} \times \mathbf{d}|}{|\mathbf{d}|}$. This works because $|\vec{AP} \times \mathbf{d}| = |\vec{AP}||\mathbf{d}|\sin\theta$ is the area of the parallelogram, and dividing by the base length $|\mathbf{d}|$ leaves the height, which is exactly the perpendicular distance.