The Scenario
A survey drone is being used to inspect a building. It starts on the ground at a reference point O. To reach its first inspection point A, it flies 3 m east, 2 m north, and rises 5 m up.
From A, it then moves to a second point B by flying 1 m west, 3 m north, and descending 2 m.
The operator needs to know: where is B relative to the start O? The drone's position involves three independent directions (east, north, up) โ and so does its movement. We need a single, compact mathematical object that captures all three directions at once.
The Question
The drone's position A needs three numbers (east, north, up), and so does its movement AB.
Can you invent a single, compact way to write down a 3D position or movement โ and a rule for adding a position and a movement to get the new position B?
(Hint: think about combining the east, north and up parts separately. What symbol could represent "one unit east"?)
๐ก The Bridge (click to reveal)
We introduce three unit vectors: $\hat{i}$ (one unit east), $\hat{j}$ (one unit north), $\hat{k}$ (one unit up). Then position A is written compactly as:
$$\overrightarrow{OA} = 3\hat{i} + 2\hat{j} + 5\hat{k}$$
The movement $\overrightarrow{AB} = -1\hat{i} + 3\hat{j} - 2\hat{k}$ (west is negative east, down is negative up). Adding position and movement component by component:
$$\overrightarrow{OB} = \overrightarrow{OA} + \overrightarrow{AB} = (3-1)\hat{i} + (2+3)\hat{j} + (5-2)\hat{k} = 2\hat{i} + 5\hat{j} + 3\hat{k}$$
This compact object $x\hat{i} + y\hat{j} + z\hat{k}$ is a 3D vector. Today we learn its component form, magnitude, unit vectors and the rules of vector arithmetic.
A vector is a quantity with both magnitude (size) and direction. In 3D space we describe a vector by its three components along the perpendicular $x$, $y$, $z$ axes, using the unit vectors $\hat{i}$, $\hat{j}$, $\hat{k}$.
A vector $\vec{v}$ with components $x$, $y$, $z$ along the $x$, $y$, $z$ axes is written:
$$\boxed{\;\vec{v} = x\,\hat{i} + y\,\hat{j} + z\,\hat{k}\;}$$Equivalently, as a column vector: $\vec{v} = \begin{pmatrix} x \\ y \\ z \end{pmatrix}$. The three numbers $x$, $y$, $z$ are the components of $\vec{v}$. The unit vectors $\hat{i} = \begin{pmatrix}1\\0\\0\end{pmatrix}$, $\hat{j} = \begin{pmatrix}0\\1\\0\end{pmatrix}$, $\hat{k} = \begin{pmatrix}0\\0\\1\end{pmatrix}$ point one unit along each axis.
The magnitude (length) of $\vec{v} = x\hat{i} + y\hat{j} + z\hat{k}$ is found by applying Pythagoras' theorem twice (once in the $xy$-plane, then with the $z$-component):
$$\boxed{\;|\vec{v}| = \sqrt{x^2 + y^2 + z^2}\;}$$
For a 2D vector $\vec{v} = x\hat{i} + y\hat{j}$ this reduces to $|\vec{v}| = \sqrt{x^2 + y^2}$. Magnitude is always a non-negative scalar (a number, not a vector).
A unit vector has magnitude 1. It points in a direction but carries no "size". To get the unit vector in the direction of any vector $\vec{v}$ (with $\vec{v} \neq \vec{0}$), divide $\vec{v}$ by its magnitude:
$$\boxed{\;\hat{v} = \dfrac{\vec{v}}{|\vec{v}|}\;}$$
The result $\hat{v}$ has length 1 and points the same way as $\vec{v}$. The special unit vectors $\hat{i}, \hat{j}, \hat{k}$ are just the unit vectors along the coordinate axes.
Vectors are added and subtracted component by component. If $\vec{a} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k}$ and $\vec{b} = b_1\hat{i} + b_2\hat{j} + b_3\hat{k}$, then:
$$\vec{a} + \vec{b} = (a_1+b_1)\hat{i} + (a_2+b_2)\hat{j} + (a_3+b_3)\hat{k}, \qquad \vec{a} - \vec{b} = (a_1-b_1)\hat{i} + (a_2-b_2)\hat{j} + (a_3-b_3)\hat{k}$$
The position vector of a point $P(x,y,z)$ is $\overrightarrow{OP} = x\hat{i} + y\hat{j} + z\hat{k}$. The vector from $A$ to $B$ is $\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA}$.
Four rounds, alternating teams: Team A takes Rounds 1 and 3, Team B takes Rounds 2 and 4. Each round gives a 30-second huddle; a correct answer scores 10 points. The live scoreboard stays visible all game.
Five questions, 90 seconds, one attempt each. Click an answer to lock it in; feedback is immediate and the run advances automatically. 10 points per correct answer.
The points A, B and C have coordinates (2, 3), (0, 4) and (โ2, 1).
(i) Write down the position vectors of A and C.
(ii) Write down the vectors of the line segments joining AB and CB.
(iii) What do your answers to parts (i) and (ii) tell you about (a) AB and OC (b) CB and OA?
(iv) Describe the quadrilateral OABC
๐ Reveal Solution
$\overrightarrow{OA} = 2\hat{i} + 3\hat{j}$, โ $\overrightarrow{OC} = -2\hat{i} + \hat{j}$.
$\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA} = (0-2)\hat{i} + (4-3)\hat{j} = -2\hat{i} + \hat{j}$.
$\overrightarrow{CB} = \overrightarrow{OB} - \overrightarrow{OC} = (0-(-2))\hat{i} + (4-1)\hat{j} = 2\hat{i} + 3\hat{j}$.
(a) $\overrightarrow{AB} = -2\hat{i} + \hat{j}$ and $\overrightarrow{OC} = -2\hat{i} + \hat{j}$, so $\overrightarrow{AB} = \overrightarrow{OC}$ โ they are equal (same direction and length).
(b) $\overrightarrow{CB} = 2\hat{i} + 3\hat{j}$ and $\overrightarrow{OA} = 2\hat{i} + 3\hat{j}$, so $\overrightarrow{CB} = \overrightarrow{OA}$ โ they are equal.
Both pairs of opposite sides are equal vectors: $\overrightarrow{AB} = \overrightarrow{OC}$ and $\overrightarrow{CB} = \overrightarrow{OA}$. A quadrilateral whose opposite sides are equal (and parallel) is a parallelogram.
$$\boxed{\text{OABC is a parallelogram.}}$$
The cuboid diagram is referenced from the textbook (Exercise 10B Q10, p. 263). In the diagram, $O$, $A$, $E$, $G$ are the vertices meeting at $O$ with $\overrightarrow{OA}$, $\overrightarrow{OE}$, $\overrightarrow{OG}$ mutually perpendicular edges.
In the cuboid, $\overrightarrow{OA} = \vec{p}$, $\overrightarrow{OE} = \vec{q}$, $\overrightarrow{OG} = \vec{r}$.
(i) Express the following vectors in terms of $\vec{p}, \vec{q}$ and $\vec{r}$. (a) $\overrightarrow{GF}$ (b) $\overrightarrow{CF}$ (c) $\overrightarrow{OB}$ (d) $\overrightarrow{OD}$ (e) $\overrightarrow{OC}$
(ii) The point M divides AD in the ratio 3:2. Find $\overrightarrow{OM}$ in terms of $\vec{p}, \vec{q}$ and $\vec{r}$.
(iii) Use vectors to prove that OC and BG bisect each other
With $\overrightarrow{OA}=\vec{p}$, $\overrightarrow{OE}=\vec{q}$, $\overrightarrow{OG}=\vec{r}$ the three perpendicular edges from $O$, the eight vertices have position vectors: $O=\vec{0}$, $A=\vec{p}$, $E=\vec{q}$, $G=\vec{r}$, $B=\vec{p}+\vec{q}$, $D=\vec{p}+\vec{r}$, $F=\vec{q}+\vec{r}$, $C=\vec{p}+\vec{q}+\vec{r}$. (If your diagram labels $D$ and $F$ the other way round, swap their answers accordingly.)
๐ Reveal Solution
Using $\overrightarrow{XY} = \overrightarrow{OY} - \overrightarrow{OX}$ with the vertex position vectors from the labeling:
(a) $\overrightarrow{GF} = \overrightarrow{OF} - \overrightarrow{OG} = (\vec{q}+\vec{r}) - \vec{r} = \vec{q}$.
(b) $\overrightarrow{CF} = \overrightarrow{OF} - \overrightarrow{OC} = (\vec{q}+\vec{r}) - (\vec{p}+\vec{q}+\vec{r}) = -\vec{p}$.
(c) $\overrightarrow{OB} = \vec{p} + \vec{q}$.
(d) $\overrightarrow{OD} = \vec{p} + \vec{r}$.
(e) $\overrightarrow{OC} = \vec{p} + \vec{q} + \vec{r}$.
$\overrightarrow{OA} = \vec{p}$ and $\overrightarrow{OD} = \vec{p}+\vec{r}$, so $\overrightarrow{AD} = \overrightarrow{OD} - \overrightarrow{OA} = \vec{r}$.
M divides AD in the ratio $AM:MD = 3:2$, so $M$ is $\dfrac{3}{5}$ of the way from A to D:
$$\overrightarrow{OM} = \overrightarrow{OA} + \dfrac{3}{5}\overrightarrow{AD} = \vec{p} + \dfrac{3}{5}\vec{r}.$$
Midpoint of $OC$: โ $\dfrac{\overrightarrow{OO} + \overrightarrow{OC}}{2} = \dfrac{\vec{0} + (\vec{p}+\vec{q}+\vec{r})}{2} = \dfrac{\vec{p}+\vec{q}+\vec{r}}{2}$.
Midpoint of $BG$: โ $\dfrac{\overrightarrow{OB} + \overrightarrow{OG}}{2} = \dfrac{(\vec{p}+\vec{q}) + \vec{r}}{2} = \dfrac{\vec{p}+\vec{q}+\vec{r}}{2}$.
The two midpoints are equal, so the segments $OC$ and $BG$ bisect each other. $\blacksquare$
Attempt each question on paper first, then click Reveal Solution to check your work. Questions are ordered from easiest to hardest.
Find the magnitude of these vectors.
(i) $\begin{pmatrix}1 \\ -2 \\ 3\end{pmatrix}$ โ (ii) $\begin{pmatrix}4 \\ 0 \\ -2\end{pmatrix}$ โ (iii) $2\hat{i} + 4\hat{j} + 2\hat{k}$
(iv) $\hat{i} + \hat{j} - 3\hat{k}$ โ (v) $\begin{pmatrix}6 \\ -2 \\ -3\end{pmatrix}$ โ (vi) $\hat{i} - 2\hat{k}$
๐ก Hint
Use $|\vec{v}| = \sqrt{x^2 + y^2 + z^2}$. Square each component (negatives vanish), add, then take the square root.
๐ Reveal Solution
(i) $\sqrt{1^2+(-2)^2+3^2} = \sqrt{1+4+9} = \boxed{\sqrt{14}}$.
(ii) $\sqrt{4^2+0^2+(-2)^2} = \sqrt{16+0+4} = \sqrt{20} = \boxed{2\sqrt{5}}$.
(iii) $\sqrt{2^2+4^2+2^2} = \sqrt{4+16+4} = \sqrt{24} = \boxed{2\sqrt{6}}$.
(iv) $\sqrt{1^2+1^2+(-3)^2} = \sqrt{1+1+9} = \boxed{\sqrt{11}}$.
(v) $\sqrt{6^2+(-2)^2+(-3)^2} = \sqrt{36+4+9} = \sqrt{49} = \boxed{7}$.
(vi) $\sqrt{1^2+0^2+(-2)^2} = \sqrt{1+0+4} = \boxed{\sqrt{5}}$.
Find unit vectors in the same directions as the following vectors.
(i) $\begin{pmatrix}2 \\ 3\end{pmatrix}$ โ (ii) $3\hat{i} + 4\hat{j}$ โ (iii) $\begin{pmatrix}-2 \\ -2\end{pmatrix}$ โ (iv) $5\hat{i} - 12\hat{j}$
๐ก Hint
The unit vector in the direction of $\vec{v}$ is $\hat{v} = \dfrac{\vec{v}}{|\vec{v}|}$. First find $|\vec{v}| = \sqrt{x^2+y^2}$, then divide each component by it.
๐ Reveal Solution
$|\vec{v}| = \sqrt{4+9} = \sqrt{13}$. So $\hat{v} = \boxed{\dfrac{1}{\sqrt{13}}\begin{pmatrix}2\\3\end{pmatrix} = \dfrac{2}{\sqrt{13}}\hat{i} + \dfrac{3}{\sqrt{13}}\hat{j}}$.
$|\vec{v}| = \sqrt{9+16} = 5$. So $\hat{v} = \boxed{\dfrac{3}{5}\hat{i} + \dfrac{4}{5}\hat{j}}$.
$|\vec{v}| = \sqrt{4+4} = \sqrt{8} = 2\sqrt{2}$. So $\hat{v} = \dfrac{1}{2\sqrt{2}}\begin{pmatrix}-2\\-2\end{pmatrix} = \boxed{-\dfrac{1}{\sqrt{2}}\hat{i} - \dfrac{1}{\sqrt{2}}\hat{j}}$.
$|\vec{v}| = \sqrt{25+144} = \sqrt{169} = 13$. So $\hat{v} = \boxed{\dfrac{5}{13}\hat{i} - \dfrac{12}{13}\hat{j}}$.
Relative to an origin O, the position vectors of the points A, B and C are given by $\overrightarrow{OA} = \begin{pmatrix}2 \\1 \\3\end{pmatrix}$, $\overrightarrow{OB} = \begin{pmatrix}-2 \\4 \\3\end{pmatrix}$ and $\overrightarrow{OC} = \begin{pmatrix}-1 \\2 \\1\end{pmatrix}$. Find the perimeter of triangle ABC
๐ก Hint 1
The sides of the triangle are $\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA}$, $\overrightarrow{BC} = \overrightarrow{OC} - \overrightarrow{OB}$, and $\overrightarrow{CA} = \overrightarrow{OA} - \overrightarrow{OC}$.
๐ก Hint 2
Find the length of each side with $|\vec{v}| = \sqrt{x^2+y^2+z^2}$, then add the three lengths for the perimeter.
๐ Reveal Solution
$\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA} = \begin{pmatrix}-2-2\\4-1\\3-3\end{pmatrix} = \begin{pmatrix}-4\\3\\0\end{pmatrix}$.
$\overrightarrow{BC} = \overrightarrow{OC} - \overrightarrow{OB} = \begin{pmatrix}-1-(-2)\\2-4\\1-3\end{pmatrix} = \begin{pmatrix}1\\-2\\-2\end{pmatrix}$.
$\overrightarrow{CA} = \overrightarrow{OA} - \overrightarrow{OC} = \begin{pmatrix}2-(-1)\\1-2\\3-1\end{pmatrix} = \begin{pmatrix}3\\-1\\2\end{pmatrix}$.
$|\overrightarrow{AB}| = \sqrt{16+9+0} = \sqrt{25} = 5$.
$|\overrightarrow{BC}| = \sqrt{1+4+4} = \sqrt{9} = 3$.
$|\overrightarrow{CA}| = \sqrt{9+1+4} = \sqrt{14}$.
$$\boxed{\text{Perimeter} = 5 + 3 + \sqrt{14} = 8 + \sqrt{14} \approx 11.74}$$
Relative to an origin O, the position vectors of the points A, B, C and D are given by $\overrightarrow{OA} = \begin{pmatrix} 3 \\ 1 \\5 \end{pmatrix}$, $\overrightarrow{OB} = \begin{pmatrix}5 \\5 \\13 \end{pmatrix}$, $\overrightarrow{OC} = \begin{pmatrix}8 \\2 \\7 \end{pmatrix}$, and $\overrightarrow{OD} = \begin{pmatrix}6 \\-2 \\-1 \end{pmatrix}$. Use vectors to prove that ABCD is a parallelogram
๐ก Hint 1
A quadrilateral is a parallelogram if one pair of opposite sides is both parallel and equal in length โ equivalently, if the vectors of one pair of opposite sides are equal.
Try showing $\overrightarrow{AB} = \overrightarrow{DC}$ (or $\overrightarrow{AB} = \overrightarrow{CD}$ depending on direction).
๐ก Hint 2
Compute $\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA}$ and $\overrightarrow{DC} = \overrightarrow{OC} - \overrightarrow{OD}$. If the two vectors are identical, the sides are parallel and equal.
๐ Reveal Solution
$\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA} = \begin{pmatrix}5-3\\5-1\\13-5\end{pmatrix} = \begin{pmatrix}2\\4\\8\end{pmatrix}$.
$\overrightarrow{DC} = \overrightarrow{OC} - \overrightarrow{OD} = \begin{pmatrix}8-6\\2-(-2)\\7-(-1)\end{pmatrix} = \begin{pmatrix}2\\4\\8\end{pmatrix}$.
Since $\overrightarrow{AB} = \overrightarrow{DC}$, the side AB is parallel to DC and equal in length. A quadrilateral with one pair of opposite sides both parallel and equal is a parallelogram.
$\overrightarrow{AD} = \overrightarrow{OD} - \overrightarrow{OA} = \begin{pmatrix}6-3\\-2-1\\-1-5\end{pmatrix} = \begin{pmatrix}3\\-3\\-6\end{pmatrix}$.
$\overrightarrow{BC} = \overrightarrow{OC} - \overrightarrow{OB} = \begin{pmatrix}8-5\\2-5\\7-13\end{pmatrix} = \begin{pmatrix}3\\-3\\-6\end{pmatrix}$.
So $\overrightarrow{AD} = \overrightarrow{BC}$ too โ both pairs of opposite sides are equal. $\blacksquare$
$$\boxed{\text{ABCD is a parallelogram.}}$$
Exit Ticket
Before you leave, tick the objectives you are confident you can achieve. Be honest โ this guides next lesson's review.