A vector has both size and direction. In Year 1 you worked with vectors in a flat plane (see Year 1 Vectors). The real world has three dimensions, so in Year 2 every vector gets a third component, $\mathbf{k}$, pointing along a $z$-axis at right angles to the other two.
Magnitude and distance in 3-D · geometric problems in 3-D · forces as 3-D vectors
| Word | Meaning | Example |
|---|---|---|
| Scalar | A quantity with size only. | Mass $3$ kg, speed $5$ m s$^{-1}$ |
| Vector | A quantity with size and direction. | $2\mathbf{i} - \mathbf{j} + 4\mathbf{k}$ |
| $\mathbf{i}$, $\mathbf{j}$, $\mathbf{k}$ | Unit vectors along the $x$, $y$, $z$ axes. | $\mathbf{k} = \begin{pmatrix}0\\0\\1\end{pmatrix}$ |
| Magnitude | The length of a vector, written $|\mathbf{a}|$. | $|2\mathbf{i} - 3\mathbf{j} + 6\mathbf{k}| = 7$ |
| Unit vector | A vector of magnitude $1$, written $\hat{\mathbf{a}}$. | $\tfrac{1}{3}(\mathbf{i} + 2\mathbf{j} + 2\mathbf{k})$ |
| Position vector | The vector from the origin to a point. | $A(1, 4, -2)$ has $\mathbf{a} = \mathbf{i} + 4\mathbf{j} - 2\mathbf{k}$ |
| Displacement | The vector from one point to another. | $\overrightarrow{AB} = \mathbf{b} - \mathbf{a}$ |
| Parallel | One vector is a scalar multiple of the other. | $(1, 2, 3)$ and $(-2, -4, -6)$ |
| Collinear | Points lying on one straight line. | $\overrightarrow{AC} = \lambda\overrightarrow{AB}$ |
| Resultant | The sum of several vectors (often forces). | $\mathbf{R} = \mathbf{F}_1 + \mathbf{F}_2$ |
$A$ is $(3, -1, 2)$ and $B$ is $(5, 3, -2)$. Find $\overrightarrow{AB}$, the distance $AB$, a unit vector in the direction of $\overrightarrow{AB}$, and the midpoint of $AB$.
The Year 2 part of DfE content area J (Vectors), together with the Year 1 pages it builds on.
- JVectors in Three Dimensions$\mathbf{i}$, $\mathbf{j}$, $\mathbf{k}$ and column vectors, magnitude, distance, position vectors, 3-D geometry and forces.
- Yr 1Vectors in Two DimensionsThe 2-D foundations: notation, magnitude and direction, and arithmetic.
- Yr 1Vector GeometryProving geometric results with vectors in 2-D — the same methods are used in 3-D.
Components
$a_1\mathbf{i} + a_2\mathbf{j} + a_3\mathbf{k} = \begin{pmatrix}a_1\\a_2\\a_3\end{pmatrix}$
Magnitude
$\sqrt{a_1^2 + a_2^2 + a_3^2}$
Displacement
$\overrightarrow{AB} = \mathbf{b} - \mathbf{a}$
Unit vector
Divide by the magnitude.
Parallel
$\mathbf{a} = \lambda\mathbf{b}$
Collinear
Parallel plus a shared point.
Midpoint
$\tfrac{1}{2}(\mathbf{a} + \mathbf{b})$
Forces
Add components; $\mathbf{R} = m\mathbf{a}$.
Equilibrium
Each component of the resultant is $0$.
Ten questions drawing on the whole Vectors topic. Work each one through before opening the solution.
$\mathbf{a} = 3\mathbf{i} - \mathbf{j} + 2\mathbf{k}$ and $\mathbf{b} = \mathbf{i} + 4\mathbf{j} - \mathbf{k}$. Find $2\mathbf{a} - 3\mathbf{b}$.
▶ Show solution
$2\mathbf{a} = 6\mathbf{i} - 2\mathbf{j} + 4\mathbf{k}$; $3\mathbf{b} = 3\mathbf{i} + 12\mathbf{j} - 3\mathbf{k}$.
$2\mathbf{a} - 3\mathbf{b} = 3\mathbf{i} - 14\mathbf{j} + 7\mathbf{k}$
Find $|2\mathbf{i} + 3\mathbf{j} - 6\mathbf{k}|$.
▶ Show solution
$\sqrt{4 + 9 + 36} = \sqrt{49} = 7$
Write the position vector of the point $P(-1, 4, 0)$ in $\mathbf{i}$, $\mathbf{j}$, $\mathbf{k}$ form and as a column vector.
▶ Show solution
$\overrightarrow{OP} = -\mathbf{i} + 4\mathbf{j} = \begin{pmatrix} -1 \\ 4 \\ 0 \end{pmatrix}$ (the $\mathbf{k}$ component is zero).
Determine whether each pair of vectors is parallel: (a) $(2, -4, 6)$ and $(-3, 6, -9)$; (b) $(1, 2, 3)$ and $(2, 4, 5)$.
▶ Show solution
(a) $(-3, 6, -9) = -\tfrac{3}{2}(2, -4, 6)$, so they are parallel (opposite directions).
(b) The first two components double but $3 \times 2 = 6 \neq 5$, so they are not parallel.
Find the distance between $(1, 2, 3)$ and $(3, -1, 9)$.
▶ Show solution
Difference $(2, -3, 6)$. Distance $= \sqrt{4 + 9 + 36} = 7$.
Find a unit vector in the direction of $6\mathbf{i} - 2\mathbf{j} + 3\mathbf{k}$.
▶ Show solution
Magnitude $\sqrt{36 + 4 + 9} = 7$.
$\tfrac{1}{7}(6\mathbf{i} - 2\mathbf{j} + 3\mathbf{k}) = \tfrac{6}{7}\mathbf{i} - \tfrac{2}{7}\mathbf{j} + \tfrac{3}{7}\mathbf{k}$
$A(0, 1, 2)$, $B(1, 3, 5)$ and $C(3, q, r)$ are collinear. Find $q$ and $r$.
▶ Show solution
$\overrightarrow{AB} = (1, 2, 3)$, $\overrightarrow{AC} = (3, q - 1, r - 2)$.
The $x$-components give $\overrightarrow{AC} = 3\overrightarrow{AB}$, so $q - 1 = 6$ and $r - 2 = 9$.
$q = 7$, $r = 11$.
$M$ is the midpoint of $A(4, -2, 7)$ and $B(-2, 6, 1)$. Find the exact distance $OM$, where $O$ is the origin.
▶ Show solution
$\mathbf{m} = \tfrac{1}{2}(2, 4, 8) = (1, 2, 4)$.
$OM = \sqrt{1 + 4 + 16} = \sqrt{21}$
A particle is in equilibrium under the forces $(p\mathbf{i} + 2\mathbf{j} - \mathbf{k})$ N, $(3\mathbf{i} + q\mathbf{j} + 4\mathbf{k})$ N and $(-5\mathbf{i} - 6\mathbf{j} + r\mathbf{k})$ N. Find $p$, $q$ and $r$.
▶ Show solution
Equilibrium: each component of the resultant is zero.
$\mathbf{i}$: $p + 3 - 5 = 0 \Rightarrow p = 2$
$\mathbf{j}$: $2 + q - 6 = 0 \Rightarrow q = 4$
$\mathbf{k}$: $-1 + 4 + r = 0 \Rightarrow r = -3$
The points $A(1, 2, 0)$, $B(3, 5, 1)$, $C(4, 6, 4)$ and $D(2, 3, 3)$ form a quadrilateral $ABCD$.
(a) Show that $ABCD$ is a parallelogram. (b) Show that it is not a rhombus. (c) Show that the diagonals $AC$ and $BD$ bisect each other. (d) Find the perimeter of $ABCD$ to 3 significant figures.
▶ Show solution
(a) $\overrightarrow{AB} = (2, 3, 1)$ and $\overrightarrow{DC} = (4 - 2, 6 - 3, 4 - 3) = (2, 3, 1)$.
$\overrightarrow{AB} = \overrightarrow{DC}$, so one pair of opposite sides is equal and parallel: $ABCD$ is a parallelogram.
(b) $\overrightarrow{AD} = (1, 1, 3)$. $|\overrightarrow{AB}| = \sqrt{14}$ and $|\overrightarrow{AD}| = \sqrt{11}$.
Adjacent sides are different lengths, so it is not a rhombus.
(c) Midpoint of $AC$: $\tfrac{1}{2}(5, 8, 4) = (2.5, 4, 2)$. Midpoint of $BD$: $\tfrac{1}{2}(5, 8, 4) = (2.5, 4, 2)$.
The diagonals share their midpoint, so they bisect each other.
(d) Perimeter $= 2\left(\sqrt{14} + \sqrt{11}\right) \approx 14.1$ units.