QCE Specialist Mathematics - Unit 3 - Vectors in two and three dimensions
Algebra of Vectors in Three Dimensions | QCE Specialist Mathematics
Revise QCE Specialist vector algebra, dot products, projections, parallel vectors, perpendicular vectors and geometric applications.
Updated 2026-05-18 - 5 min read
QCAA official coverage - Specialist Mathematics 2025 v1.4
Exact syllabus points covered
- Examine and use addition and subtraction of vectors in Cartesian form.
- Use multiplication by a scalar of a vector in Cartesian form.
- Determine a vector between two points.
- Use a vector representing a section of a line segment, including the midpoint of a line segment.
- Use the scalar dot product: $\mathbf a\cdot\mathbf b=|\mathbf a||\mathbf b|\cos(\theta)$ and $\begin{pmatrix}a_1\\a_2\\a_3\end{pmatrix}\cdot\begin{pmatrix}b_1\\b_2\\b_3\end{pmatrix}=a_1b_1+a_2b_2+a_3b_3$.
- Examine properties of parallel and perpendicular vectors and determine if two vectors are parallel or perpendicular.
- Use scalar and vector projections of vectors: scalar projection of $\mathbf a$ on $\mathbf b$ is $|\mathbf a|\cos(\theta)=\mathbf a\cdot\hat{\mathbf b}$ and vector projection of $\mathbf a$ on $\mathbf b$ is $(\mathbf a\cdot\hat{\mathbf b})\hat{\mathbf b}=\left(\frac{\mathbf a\cdot\mathbf b}{\mathbf b\cdot\mathbf b}\right)\mathbf b$.
- Apply the scalar product to vectors expressed in Cartesian form.
- Model and solve problems that involve displacement, force, velocity and relative velocity using the above concepts.
- Use vectors to prove geometric results in two dimensions, other than those listed in Unit 2 Topic 3, and in three dimensions.
Vector algebra in 3D is mostly the same set of moves you used in 2D: add components, subtract components and multiply every component by a scalar. The extra component does not change the logic. It just gives the vector one more direction to track.
If
$ \mathbf a= \begin{pmatrix} 2\\-1\\5 \end{pmatrix}, \quad \mathbf b= \begin{pmatrix} 4\\3\\-2 \end{pmatrix}, $
then
$ \mathbf a+\mathbf b= \begin{pmatrix} 6\\2\\3 \end{pmatrix}, \quad 3\mathbf a= \begin{pmatrix} 6\\-3\\15 \end{pmatrix}. $
Vectors between points
If $A(a_1,a_2,a_3)$ and $B(b_1,b_2,b_3)$, then
$ \overrightarrow{AB}=\mathbf b-\mathbf a = \begin{pmatrix} b_1-a_1\\ b_2-a_2\\ b_3-a_3 \end{pmatrix}. $
This is one of the most common moves in the whole topic. Points give positions. Subtracting position vectors gives the displacement from one point to the other.
Midpoints and section points are the same idea with a weighted average. The midpoint of $A$ and $B$ is:
$ \frac{\mathbf a+\mathbf b}{2}. $
If a point divides the segment from $A$ to $B$ in the ratio $m:n$, its position vector is:
$ \frac{n\mathbf a+m\mathbf b}{m+n}. $
This formula weights the endpoint more heavily when the point is closer to the other endpoint. A quick diagram is usually the easiest way to avoid reversing $m$ and $n$.
Parallel and perpendicular vectors
Two non-zero vectors are parallel if one is a scalar multiple of the other:
$ \mathbf a=\lambda\mathbf b. $
They are perpendicular if their dot product is zero:
$ \mathbf a\cdot\mathbf b=0. $
The dot product can be calculated component by component:
$ \mathbf a\cdot\mathbf b=a_1b_1+a_2b_2+a_3b_3. $
It also connects to the angle between vectors:
$ \mathbf a\cdot\mathbf b=|\mathbf a||\mathbf b|\cos\theta. $
So if the dot product is positive, the angle is acute. If it is negative, the angle is obtuse. If it is zero, the vectors are perpendicular.
The same formula gives the angle directly:
$ \cos\theta=\frac{\mathbf a\cdot\mathbf b}{|\mathbf a||\mathbf b|}. $
Use this only for non-zero vectors. If either vector is $\mathbf 0$, the angle is undefined.
Projections
The scalar projection of $\mathbf a$ onto $\mathbf b$ is:
$ \frac{\mathbf a\cdot\mathbf b}{|\mathbf b|}. $
The vector projection is:
$ \operatorname{proj}_{\mathbf b}\mathbf a = \left(\frac{\mathbf a\cdot\mathbf b}{|\mathbf b|^2}\right)\mathbf b. $
This is the part of $\mathbf a$ that points along $\mathbf b$. In physical questions, projections often appear when resolving forces, velocities or displacements into a useful direction.
Original Sylligence diagram for specialist vector projection.
A projection can be read as "push $\mathbf a$ straight down onto the line in the direction of $\mathbf b
quot;. The dashed perpendicular in the diagram is the part being removed. Algebraically, this is why$ \mathbf a-\operatorname{proj}_{\mathbf b}\mathbf a $
is perpendicular to $\mathbf b$.
Modelling with vector algebra
Displacement, velocity and force are vector quantities, so the same algebra applies in context. Relative velocity is especially common:
$ \mathbf v_{A/B}=\mathbf v_A-\mathbf v_B. $
This means the velocity of $A$ as observed from $B$. If $\mathbf v_{A/B}=\mathbf 0$, the objects have the same velocity even if they are not at the same position.
For force problems, a resultant force is a vector sum. If two forces act at an angle, the dot product can find the angle between them, while projections can resolve a force into a component along a direction. A proof-style geometry question may use the same tools to show that two sides are perpendicular, parallel or equal in length.
Triangle rule and scalar meaning
Vector addition still has the same geometric meaning as in two dimensions. If $\overrightarrow{AB}=\mathbf u$ and $\overrightarrow{BC}=\mathbf v$, then:
$ \overrightarrow{AC}=\mathbf u+\mathbf v. $
This triangle rule is often the cleanest way to interpret displacement questions. Component addition is the calculation; the triangle rule is the picture behind it.
Scalar multiplication changes length and sometimes direction. If $\lambda>0$, then $\lambda\mathbf a$ points in the same direction as $\mathbf a$. If $\lambda<0$, it points in the opposite direction. If two non-zero vectors are scalar multiples, they are parallel:
$ \mathbf a=\lambda\mathbf b. $
This is also how collinearity is proved. Points $A$, $B$ and $C$ are collinear if $\overrightarrow{AB}$ and $\overrightarrow{AC}$ are parallel.
Projection interpretation
The vector projection:
$ \operatorname{proj}_{\mathbf b}\mathbf a=\left(\frac{\mathbf a\cdot\mathbf b}{\mathbf b\cdot\mathbf b}\right)\mathbf b $
is the part of $\mathbf a$ that lies along $\mathbf b$. The leftover component:
$ \mathbf a-\operatorname{proj}_{\mathbf b}\mathbf a $
is perpendicular to $\mathbf b$. This decomposition is useful in geometry proofs and in physics contexts where a force must be resolved along and perpendicular to a surface.