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

Vectors in Three Dimensions | QCE Specialist Mathematics

Learn QCE Specialist three-dimensional vectors, ordered triples, unit vectors, magnitudes and altitude angles.

Updated 2026-05-18 - 5 min read

Two-dimensional vectors already let you describe horizontal and vertical movement. Three-dimensional vectors add a third direction, so a point or displacement needs three components instead of two.

The point $P(2,-1,4)$ can be represented by the position vector

$ \overrightarrow{OP}= \begin{pmatrix} 2\\ -1\\ 4 \end{pmatrix}. $

The same vector can also be written with the standard unit vectors:

$ 2\mathbf i-\mathbf j+4\mathbf k. $

Components and standard unit vectors

The vectors $\mathbf i$, $\mathbf j$ and $\mathbf k$ each have length $1$ and point in the positive $x$, $y$ and $z$ directions:

$ \mathbf i= \begin{pmatrix} 1\\0\\0 \end{pmatrix}, \quad \mathbf j= \begin{pmatrix} 0\\1\\0 \end{pmatrix}, \quad \mathbf k= \begin{pmatrix} 0\\0\\1 \end{pmatrix}. $

So

$ \begin{pmatrix} a\\b\\c \end{pmatrix}=a\mathbf i+b\mathbf j+c\mathbf k. $

This is called Cartesian or component form. It is handy because addition, subtraction and scalar multiplication can be done component by component. The same position can also be described as an ordered triple $(a,b,c)$, but the vector form makes the direction and displacement interpretation clearer.

Magnitude and unit vectors

The length of a three-dimensional vector is the 3D version of Pythagoras:

$ |\mathbf a|=\sqrt{a_1^2+a_2^2+a_3^2}. $

For example,

$ \left| \begin{pmatrix} 2\\-1\\4 \end{pmatrix} \right| =\sqrt{2^2+(-1)^2+4^2}=\sqrt{21}. $

A unit vector in the direction of $\mathbf a$ is found by dividing by the magnitude:

$ \hat{\mathbf a}=\frac{\mathbf a}{|\mathbf a|}. $

This keeps the direction but changes the length to $1$.

If $\mathbf a=\mathbf 0$, a unit vector in its direction is undefined because the zero vector has no direction. In exam work, this is a small check but an important one before dividing by $|\mathbf a|$.

Altitude angle

In 2D, one angle is enough to describe direction. In 3D, you need more information because a vector can rise above or fall below the $xy$-plane.

QCAA refers to the altitude angle $\varphi$. A useful way to read it is as the angle between the vector and its projection onto the $xy$-plane. With this convention,

$ \sin\varphi=\frac{z}{|\mathbf a|}. $

If $\varphi$ is positive, the vector points upward from the $xy$-plane. If $\varphi$ is negative, it points downward.

You may also see the horizontal projection length:

$ \sqrt{x^2+y^2}. $

That creates a right triangle with vertical side $z$ and hypotenuse $|\mathbf a|$, so:

$ \tan\varphi=\frac{z}{\sqrt{x^2+y^2}}. $

This is often the cleanest formula when the vector is given in components.

The side-view diagram is a useful exam shortcut. Instead of trying to measure the angle from a full 3D sketch, choose a vertical plane that contains the vector and the $z$-axis direction. The horizontal leg is not just $x$ or $y$ unless the vector lies in one of the coordinate planes. In general it is the length of the projection onto the $xy$-plane, so it is $\sqrt{x^2+y^2}$.

Sketching in 3D

A quick sketch usually starts with the $xy$-plane, then moves up or down in the $z$ direction. For $P(2,-1,4)$, move $2$ in the $x$ direction, $-1$ in the $y$ direction, then $4$ vertically. The position vector is the arrow from the origin to that final point.

For vectors that do not start at the origin, such as the vector from $A$ to $B$, subtract coordinates:

$ \overrightarrow{AB}=\mathbf b-\mathbf a. $

That displacement can then be drawn anywhere parallel to itself, because vectors are not fixed to one location unless the question specifically treats them as position vectors.

In textbook diagrams, 3D axes are often drawn in either an isometric style or a more orthogonal "box" style. The drawing is not to scale in the same way a 2D graph is. Use the labels and components, not the apparent lengths on the page, to decide where the point is.

For a rectangular-prism workflow, plot the shadow of the point on the $xy$-plane first. From there, move parallel to the $z$-axis. This makes points such as $(3,4,5)$ easier to visualise: $(3,4,0)$ is the base point, then $5$ units upward reaches the final point.

Two angle descriptions in 3D

One angle is not enough to describe a direction in 3D. You need a horizontal direction and a vertical direction.

The horizontal angle $\theta$ comes from projecting the vector onto the $xy$-plane:

$ \tan\theta=\frac{y}{x}. $

As with complex arguments, the quadrant matters. A calculator inverse tangent gives a reference angle, not always the actual direction.

For the altitude angle $\varphi$, QCE commonly treats it as the angle above or below the $xy$-plane:

$ \sin\varphi=\frac{z}{r}, \quad r=|\mathbf a|. $

Some sources instead measure a polar angle down from the positive $z$-axis:

$ \cos\varphi=\frac{z}{r}. $

Both conventions can describe the same vector, but they are not interchangeable. The question's diagram or wording decides which one is being used.

Using the altitude-from-horizontal convention, a vector with length $r$ can be reconstructed from:

$ x=r\cos\theta\cos\varphi,\quad y=r\sin\theta\cos\varphi,\quad z=r\sin\varphi. $

If the angle $\phi$ is measured down from the positive $z$-axis instead, the reconstruction changes to:

$ x=r\sin\phi\cos\theta,\quad y=r\sin\phi\sin\theta,\quad z=r\cos\phi. $

This is the convention often used in spherical coordinates. Notice the switch: altitude from the $xy$-plane uses $z=r\sin\varphi$, while polar angle from the $z$-axis uses $z=r\cos\phi$.

Quick check

Sources

• QCAA Specialist Mathematics subject page
• QCAA Specialist Mathematics 2025 syllabus
• OpenStax Calculus Volume 3: Vectors in Three Dimensions
• OpenStax Calculus Volume 3: Vectors in the Plane