Vectors in three dimensions
Master vectors in three dimensions for Year 12 Specialist Mathematics in Queensland (QCAA). Space is described by three perpendicular axes and the unit vectors i, j and k, so every point in space has a position vector written in component form.
You will learn to plot points in space, write vectors in component and column form, find a vector's magnitude and unit vector, and work out the altitude angle — the foundation for the algebra of 3D vectors, lines and planes later in the course.
Theory
Vectors in three dimensions extend plane vectors into space using the three perpendicular unit vectors i, j and k, in Year 12 Specialist Mathematics (QCAA, Queensland). A point \(P(x,y,z)\) has position vector \(\overrightarrow{OP}=x\mathbf{i}+y\mathbf{j}+z\mathbf{k}\); from it you find the magnitude, the unit vector and the altitude angle. This page shows each calculation with full worked examples.
Three-dimensional space is described by three mutually perpendicular axes: the \(x\)-, \(y\)- and \(z\)-axes. A point is located by an ordered triple \((x,y,z)\), and plotting it means moving \(x\) along the \(x\)-axis, \(y\) along the \(y\)-axis and \(z\) up the \(z\)-axis.
The unit vectors \(\mathbf{i}\), \(\mathbf{j}\) and \(\mathbf{k}\) each have length \(1\) and point along the positive \(x\)-, \(y\)- and \(z\)-axes. Because they are perpendicular, every vector in space is a combination \(x\mathbf{i}+y\mathbf{j}+z\mathbf{k}\) — its Cartesian (component) form, the same information as the column vector \(\begin{pmatrix} x \\ y \\ z \end{pmatrix}\).
The position vector of \(P(x,y,z)\) is \(\overrightarrow{OP}=x\mathbf{i}+y\mathbf{j}+z\mathbf{k}\), running from the origin \(O\) to \(P\). Its magnitude \(|\overrightarrow{OP}|\) is the straight-line distance of \(P\) from the origin.
A unit vector \(\hat{\mathbf{a}}\) points the same way as \(\mathbf{a}\) but has length \(1\). The altitude angle \(\theta\) is the angle the vector makes with the horizontal \(xy\)-plane, found from the vertical (\(z\)) component and the magnitude.
For a vector \(\mathbf{a}=x\mathbf{i}+y\mathbf{j}+z\mathbf{k}\), the magnitude (length) is the space-diagonal:
The unit vector in the direction of \(\mathbf{a}\) divides each component by the magnitude:
The altitude angle \(\theta\) (measured up from the horizontal \(xy\)-plane) comes from the vertical component over the magnitude:
Working with a 3D vector
- Read off the components: write the vector as \(x\mathbf{i}+y\mathbf{j}+z\mathbf{k}\), or as the column \(\begin{pmatrix} x \\ y \\ z \end{pmatrix}\), keeping the sign of each component.
- Magnitude: square all three components, add, and take one square root, \(|\mathbf{a}|=\sqrt{x^{2}+y^{2}+z^{2}}\).
- Unit vector: divide each component by the magnitude, \(\hat{\mathbf{a}}=\dfrac{\mathbf{a}}{|\mathbf{a}|}\).
- Altitude angle: use \(\sin\theta=\dfrac{z}{|\mathbf{a}|}\) and take the inverse sine for the angle above the horizontal plane.
Each coordinate is the coefficient of the matching unit vector:
| \(\overrightarrow{OP}\) | \(=\) | \(x\mathbf{i}+y\mathbf{j}+z\mathbf{k}\) |
| \(=\) | \(4\mathbf{i}+(-1)\mathbf{j}+8\mathbf{k}\) | |
| \(=\) | \(4\mathbf{i}-\mathbf{j}+8\mathbf{k}\) |
\(\overrightarrow{OP}=4\mathbf{i}-\mathbf{j}+8\mathbf{k}\).
Square each of the three components, add, then take the square root:
| \(|\mathbf{v}|\) | \(=\) | \(\sqrt{3^2+4^2+12^2}\) |
| \(=\) | \(\sqrt{9+16+144}\) | |
| \(=\) | \(\sqrt{169}\) | |
| \(=\) | \(13\) |
\(|\mathbf{v}|=13\).
First find the magnitude:
| \(|\mathbf{a}|\) | \(=\) | \(\sqrt{1^2+8^2+4^2}\) |
| \(=\) | \(\sqrt{1+64+16}\) | |
| \(=\) | \(\sqrt{81}\) | |
| \(=\) | \(9\) |
Then divide each component by the magnitude:
| \(\hat{\mathbf{a}}\) | \(=\) | \(\dfrac{1}{9}(\mathbf{i}+8\mathbf{j}+4\mathbf{k})\) |
| \(=\) | \(\dfrac{1}{9}\mathbf{i}+\dfrac{8}{9}\mathbf{j}+\dfrac{4}{9}\mathbf{k}\) |
\(\hat{\mathbf{a}}=\dfrac{1}{9}\mathbf{i}+\dfrac{8}{9}\mathbf{j}+\dfrac{4}{9}\mathbf{k}\).
Find the magnitude first:
| \(|\mathbf{v}|\) | \(=\) | \(\sqrt{3^2+4^2+12^2}\) |
| \(=\) | \(\sqrt{169}\) | |
| \(=\) | \(13\) |
Then use the vertical component over the magnitude:
| \(\sin\theta\) | \(=\) | \(\dfrac{z}{|\mathbf{v}|}\) |
| \(=\) | \(\dfrac{12}{13}\) | |
| \(\theta\) | \(=\) | \(\sin^{-1}\!\left(\dfrac{12}{13}\right)\) |
| \(=\) | \(67.4^\circ\) | |
| \(\approx\) | \(67^\circ\) |
\(\theta\approx 67^\circ\).
Common pitfalls
Frequently asked questions
What are the unit vectors i, j and k?
\(\mathbf{i}\), \(\mathbf{j}\) and \(\mathbf{k}\) are vectors of length \(1\) pointing along the positive \(x\)-, \(y\)- and \(z\)-axes. They are mutually perpendicular, so any vector in space is \(x\mathbf{i}+y\mathbf{j}+z\mathbf{k}\).
How do you find the magnitude of a 3D vector?
Square each of the three components, add the squares, then take the square root: for \(\mathbf{a}=x\mathbf{i}+y\mathbf{j}+z\mathbf{k}\), \(|\mathbf{a}|=\sqrt{x^{2}+y^{2}+z^{2}}\).
What is a position vector in three dimensions?
The position vector of \(P(x,y,z)\) is \(\overrightarrow{OP}=x\mathbf{i}+y\mathbf{j}+z\mathbf{k}\), running from the origin to \(P\). Its magnitude is the straight-line distance of \(P\) from the origin.
How do you find a unit vector in three dimensions?
Divide the vector by its magnitude: \(\hat{\mathbf{a}}=\dfrac{\mathbf{a}}{|\mathbf{a}|}\). Every component is divided by \(|\mathbf{a}|\), and the result always has length \(1\).
What is the altitude angle of a vector?
It is the angle \(\theta\) the vector makes with the horizontal \(xy\)-plane, found from the vertical component and the magnitude: \(\sin\theta=\dfrac{z}{|\mathbf{a}|}\).
Is column vector notation the same as i, j, k form?
Yes. \(\begin{pmatrix} x \\ y \\ z \end{pmatrix}\) holds exactly the same information as \(x\mathbf{i}+y\mathbf{j}+z\mathbf{k}\); the top entry is the \(\mathbf{i}\) part, the middle the \(\mathbf{j}\) part and the bottom the \(\mathbf{k}\) part.