QCE Specialist Mathematics - Unit 3 - Vectors in two and three dimensions

Vector and Cartesian Equations | QCE Specialist Mathematics

Learn Specialist Mathematics vector equations, parametric equations, Cartesian equations, spheres, lines, planes and cross products.

Updated 2026-05-18 - 6 min read

Vector equations describe a set of points by starting at a known point and moving in one or more directions. They are especially useful in 3D, where the usual $y=mx+c$ style of equation does not describe a line cleanly.

Lines

A line through a point with position vector $\mathbf a$ and direction vector $\mathbf d$ can be written as:

$ \mathbf r=\mathbf a+\lambda\mathbf d, \quad \lambda\in\mathbb R. $

If

$ \mathbf a=(1,2,-3),\quad \mathbf d=(4,-1,2), $

then the parametric equations are:

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

When the direction components are non-zero, the Cartesian form is:

$ \frac{x-a_1}{d_1}=\frac{y-a_2}{d_2}=\frac{z-a_3}{d_3}. $

If one direction component is zero, do not divide by it. For example, if $\mathbf d=(2,0,-1)$, then $y$ is constant and the Cartesian information is better written as:

$ y=a_2,\quad \frac{x-a_1}{2}=\frac{z-a_3}{-1}. $

For a line segment from $A$ to $B$, use

$ \mathbf r=\mathbf a+\lambda(\mathbf b-\mathbf a), \quad 0\le\lambda\le1. $

The domain of the parameter matters. Without the restriction, you have the whole line, not just the segment.

Spheres

A sphere with centre $(h,k,l)$ and radius $r$ is:

$ (x-h)^2+(y-k)^2+(z-l)^2=r^2. $

This is the 3D version of a circle equation. Every point on the sphere is exactly $r$ units from the centre.

You can also recognise a sphere by completing the square. For example, terms like $x^2-4x$ become $(x-2)^2-4$. If the final right-hand side is positive, it gives $r^2$. If it is zero, the sphere has radius $0$ and is just one point. If it is negative, there is no real sphere.

In vector form, a sphere can be written as:

$ |\mathbf r-\mathbf c|=a, $

where $\mathbf c$ is the centre position vector and $a$ is the radius. Squaring both sides gives:

$ (\mathbf r-\mathbf c)\cdot(\mathbf r-\mathbf c)=a^2. $

This version is often clearer in 3D because it says exactly what a sphere is: all points whose distance from the centre is fixed.

Planes

A plane needs a point and a normal vector. If the plane passes through a point with position vector $\mathbf a$ and has normal vector $\mathbf n$, then:

$ (\mathbf r-\mathbf a)\cdot\mathbf n=0. $

Equivalently:

$ \mathbf r\cdot\mathbf n=\mathbf a\cdot\mathbf n. $

If $\mathbf n=(A,B,C)$, the Cartesian form is:

$ Ax+By+Cz+D=0. $

The reverse conversion is also important. From:

$ 2x-y+3z=12, $

the normal vector is:

$ \mathbf n=(2,-1,3). $

To find a point on the plane, choose convenient values for two variables. Setting $y=0$ and $z=0$ gives $x=6$, so one point is $(6,0,0)$. Therefore a point-normal form is:

$ (\mathbf r-(6,0,0))\cdot(2,-1,3)=0. $

When drawing a plane, the sheet itself only shows two directions. The normal vector is the extra information that tells you which way the plane is facing in 3D. If you know a Cartesian plane equation, the coefficients of $x$, $y$ and $z$ give the normal vector immediately.

Intersections with planes

A line intersects a plane by satisfying both equations at the same time. If the line is

$ \mathbf r=\mathbf p+\lambda\mathbf d $

and the plane is

$ \mathbf r\cdot\mathbf n=\mathbf a\cdot\mathbf n, $

then substitute the line into the plane:

$ (\mathbf p+\lambda\mathbf d)\cdot\mathbf n=\mathbf a\cdot\mathbf n. $

Solving this gives the parameter value $\lambda$, unless $\mathbf d\cdot\mathbf n=0$. If $\mathbf d\cdot\mathbf n=0$, the line is parallel to the plane. It may have no intersection, or it may lie entirely in the plane.

Cross products

If two non-parallel vectors lie in a plane, their cross product gives a vector normal to that plane:

$ \mathbf a\times\mathbf b= \begin{pmatrix} a_2b_3-a_3b_2\\ a_3b_1-a_1b_3\\ a_1b_2-a_2b_1 \end{pmatrix}. $

The magnitude also has a geometric meaning:

$ |\mathbf a\times\mathbf b|=|\mathbf a||\mathbf b|\sin\theta. $

That is the area of the parallelogram formed by $\mathbf a$ and $\mathbf b$.

The area of the triangle formed by the same two vectors is:

$ \frac12|\mathbf a\times\mathbf b|. $

This is a common application because the cross product packages both the angle and the lengths into one vector.

Vector equations of curves and regions

A vector equation can describe more than a straight line. For a parameter $t$,

$ \mathbf r(t)=x(t)\mathbf i+y(t)\mathbf j $

describes a curve in the plane. To find a corresponding Cartesian equation, eliminate $t$. For example, if $x=2\cos t$ and $y=3\sin t$, then:

$ \frac{x^2}{4}+\frac{y^2}{9}=1, $

so the path is an ellipse.

A unit circle can be written as:

$ \mathbf r(t)=\cos t\,\mathbf i+\sin t\,\mathbf j, \quad 0\le t<2\pi. $

A 3D spiral can be written as:

$ \mathbf r(t)=\cos t\,\mathbf i+\sin t\,\mathbf j+t\,\mathbf k. $

The first two components keep the particle on a cylinder of radius $1$, while the $z$ component rises steadily.

For a parabola, a vector equation such as:

$ \mathbf r(t)=t\,\mathbf i+(t^2+1)\mathbf j $

gives:

$ x=t,\quad y=t^2+1, $

so the Cartesian path is $y=x^2+1$. If the parameter is restricted, for example $0\le t\le2$, then the path is only the corresponding arc from $(0,1)$ to $(2,5)$.

Regions in a plane are often written with restrictions on parameters or inequalities. A line segment uses $0\le\lambda\le1$; a ray might use $\lambda\ge0$; a plane region may require two parameters over a restricted domain. The equation gives the possible points, but the parameter restrictions decide how much of that set is actually included.

Planes through three points

If a plane passes through three non-collinear points $A$, $B$ and $C$, you can build its equation from two direction vectors:

$ \overrightarrow{AB}=\mathbf b-\mathbf a,\quad \overrightarrow{AC}=\mathbf c-\mathbf a. $

These vectors lie in the plane. A normal vector is:

$ \mathbf n=\overrightarrow{AB}\times\overrightarrow{AC}. $

Then use:

$ (\mathbf r-\mathbf a)\cdot\mathbf n=0. $

Any of the three points can be used as the reference point. If your cross product is a scalar multiple of someone else's, both normal vectors are valid and will give equivalent plane equations.

Cross product technique

The cross product formula is easy to miscopy, so many students use a determinant-style layout:

$ \mathbf a\times\mathbf b= \begin{vmatrix} \mathbf i&\mathbf j&\mathbf k\\ a_1&a_2&a_3\\ b_1&b_2&b_3 \end{vmatrix}. $

Expanding gives:

$ (a_2b_3-a_3b_2)\mathbf i-(a_1b_3-a_3b_1)\mathbf j+(a_1b_2-a_2b_1)\mathbf k. $

That middle minus sign is the common trap. The component formula in the syllabus writes the same thing as $a_3b_1-a_1b_3$ for the $\mathbf j$ component.

A quick "criss-cross" check is to write the components twice:

$ a_1,\ a_2,\ a_3,\ a_1,\ a_2 $

and

$ b_1,\ b_2,\ b_3,\ b_1,\ b_2. $

The forward diagonal products minus the backward diagonal products give the three components. This is only a memory device; the determinant or component formula is the safer final method to show in working.

Quick check

Sources

• QCAA Specialist Mathematics subject page
• QCAA Specialist Mathematics 2025 syllabus
• OpenStax Calculus Volume 3: Equations of Lines and Planes in Space
• OpenStax Calculus Volume 3: The Cross Product