Vector Fundamentals, Magnitude & Unit Vectors
A scalar is a quantity possessing only magnitude (e.g., speed, mass, time, temperature). A vector possesses both magnitude (size) and direction (e.g., displacement, velocity, force).
- Column Vector Notation: $\mathbf{v} = \begin{pmatrix} x \\ y \end{pmatrix}$, where $x$ is horizontal displacement and $y$ is vertical displacement.
- Standard Basis Unit Vectors: $\mathbf{i} = \begin{pmatrix} 1 \\ 0 \end{pmatrix}$ and $\mathbf{j} = \begin{pmatrix} 0 \\ 1 \end{pmatrix}$, so $\mathbf{v} = x\mathbf{i} + y\mathbf{j}$.
- Magnitude (Length): $|\mathbf{v}| = \sqrt{x^2 + y^2}$ (derived from Pythagoras' theorem).
- Unit Vector ($\hat{\mathbf{v}}$): A vector of length $1$ in the same direction: $$\hat{\mathbf{v}} = \frac{\mathbf{v}}{|\mathbf{v}|} = \frac{1}{\sqrt{x^2 + y^2}}\begin{pmatrix} x \\ y \end{pmatrix}$$
A 2D vector resolved into horizontal ($x\mathbf{i}$) and vertical ($y\mathbf{j}$) components.
Vector Arithmetic & Position Vectors
Vectors are combined algebraically component-by-component, and geometrically using the triangle or parallelogram law.
- Addition: $\begin{pmatrix} u_1 \\ u_2 \end{pmatrix} + \begin{pmatrix} v_1 \\ v_2 \end{pmatrix} = \begin{pmatrix} u_1 + v_1 \\ u_2 + v_2 \end{pmatrix}$ (Head-to-tail tip-to-tail placement).
- Subtraction: $\mathbf{u} - \mathbf{v} = \mathbf{u} + (-\mathbf{v}) = \begin{pmatrix} u_1 - v_1 \\ u_2 - v_2 \end{pmatrix}$.
- Scalar Multiplication: $k \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} kx \\ ky \end{pmatrix}$. If $k > 0$, direction is preserved; if $k < 0$, direction reverses.
A position vector connects the fixed origin $O(0,0)$ to a point in the plane: $$\vec{OA} = \mathbf{a} = \begin{pmatrix} x_A \\ y_A \end{pmatrix}, \quad \vec{OB} = \mathbf{b} = \begin{pmatrix} x_B \\ y_B \end{pmatrix}$$ The displacement vector from $A$ to $B$ is: $$\vec{AB} = \vec{OB} - \vec{OA} = \mathbf{b} - \mathbf{a} = \begin{pmatrix} x_B - x_A \\ y_B - y_A \end{pmatrix}$$
Parallel Vectors, Collinearity & Geometric Proofs
Vector methods can establish geometric relationships without using coordinates.
Two non-zero vectors $\mathbf{u}$ and $\mathbf{v}$ are parallel ($\mathbf{u} \parallel \mathbf{v}$) if and only if one is a scalar multiple of the other: $$\mathbf{u} = k\mathbf{v} \quad \text{for some scalar } k \neq 0$$
Three points $A$, $B$, and $C$ lie on the same straight line (are collinear) if:
- $\vec{AB} = k\vec{BC}$ (the vectors are parallel), AND
- Point $B$ is a common shared vertex between them.
The Scalar (Dot) Product & Orthogonality (Extended)
The scalar product (or dot product) maps two vectors to a single real number (scalar), providing the algebraic foundation for calculating angles and testing perpendicularity.
Since $\cos 90^\circ = 0$, two non-zero vectors $\mathbf{u}$ and $\mathbf{v}$ are perpendicular ($\mathbf{u} \perp \mathbf{v}$) if and only if their scalar product is zero: $$\mathbf{u} \cdot \mathbf{v} = 0 \iff u_1 v_1 + u_2 v_2 = 0$$
Worked Example 1: Position Vectors & Collinearity
The position vectors of points $A$, $B$, and $C$ relative to an origin $O$ are:
$$\mathbf{a} = \begin{pmatrix} 2 \\ 5 \end{pmatrix}, \quad \mathbf{b} = \begin{pmatrix} 6 \\ 13 \end{pmatrix}, \quad \mathbf{c} = \begin{pmatrix} 14 \\ 29 \end{pmatrix}$$
a) Find the displacement vectors $\vec{AB}$ and $\vec{BC}$.
b) Prove that points $A$, $B$, and $C$ are collinear and state the ratio $AB : BC$.
Step-by-step Solution:
Part a:
$\vec{AB} = \mathbf{b} - \mathbf{a} = \begin{pmatrix} 6 \\ 13 \end{pmatrix} - \begin{pmatrix} 2 \\ 5 \end{pmatrix} = \begin{pmatrix} 6 - 2 \\ 13 - 5 \end{pmatrix} = \begin{pmatrix} 4 \\ 8 \end{pmatrix}$. [M1]
$\vec{BC} = \mathbf{c} - \mathbf{b} = \begin{pmatrix} 14 \\ 29 \end{pmatrix} - \begin{pmatrix} 6 \\ 13 \end{pmatrix} = \begin{pmatrix} 14 - 6 \\ 29 - 13 \end{pmatrix} = \begin{pmatrix} 8 \\ 16 \end{pmatrix}$. [A1]
Part b:
Notice that:
$$\vec{BC} = \begin{pmatrix} 8 \\ 16 \end{pmatrix} = 2 \begin{pmatrix} 4 \\ 8 \end{pmatrix} = 2\vec{AB}$$
Since $\vec{BC} = 2\vec{AB}$, the vectors $\vec{AB}$ and $\vec{BC}$ are parallel. [M1]
Furthermore, both vectors share the common point $B$. Therefore, points $A$, $B$, and $C$ lie on the same straight line (they are collinear). [R1]
The ratio $AB : BC = 1 : 2$. [Criterion C: Complete Mathematical Communication]
Worked Example 2: Scalar Product & Orthogonality
Given vectors $\mathbf{u} = \begin{pmatrix} 3 \\ -4 \end{pmatrix}$ and $\mathbf{v} = \begin{pmatrix} k \\ 6 \end{pmatrix}$:
a) Find the value of $k$ if $\mathbf{u}$ and $\mathbf{v}$ are perpendicular.
b) If $k = 5$, calculate the scalar product $\mathbf{u} \cdot \mathbf{v}$ and the angle $\theta$ between them to the nearest tenth of a degree.
Step-by-step Solution:
Part a: Perpendicular condition
Two vectors are perpendicular if $\mathbf{u} \cdot \mathbf{v} = 0$: [M1]
$3(k) + (-4)(6) = 0 \implies 3k - 24 = 0 \implies 3k = 24 \implies k = 8$. [A1]
Part b: Angle when $k = 5$
When $k = 5$, $\mathbf{v} = \begin{pmatrix} 5 \\ 6 \end{pmatrix}$.
$\mathbf{u} \cdot \mathbf{v} = (3)(5) + (-4)(6) = 15 - 24 = -9$. [A1]
Magnitudes:
$|\mathbf{u}| = \sqrt{3^2 + (-4)^2} = \sqrt{9 + 16} = \sqrt{25} = 5$.
$|\mathbf{v}| = \sqrt{5^2 + 6^2} = \sqrt{25 + 36} = \sqrt{61} \approx 7.810$. [M1]
Using the angle formula:
$$\cos\theta = \frac{\mathbf{u} \cdot \mathbf{v}}{|\mathbf{u}| |\mathbf{v}|} = \frac{-9}{5\sqrt{61}} \approx \frac{-9}{39.051} \approx -0.23046$$
Since $\cos\theta < 0$, $\theta$ is obtuse:
$\theta = \arccos(-0.23046) \approx 103.3^\circ$. [Criterion A: Knowledge & Understanding]
Tiered Practice Exercises
10 problems advancing from basic vector operations and magnitude to scalar products, orthogonality, and geometric proof inquiries.
Given vectors $\mathbf{a} = \begin{pmatrix} 3 \\ -2 \end{pmatrix}$ and $\mathbf{b} = \begin{pmatrix} -1 \\ 5 \end{pmatrix}$, compute:
a) $\mathbf{a} + \mathbf{b}$
b) $2\mathbf{a} - 3\mathbf{b}$
View Solution & Markscheme
Method & Marks:
a) $\mathbf{a} + \mathbf{b} = \begin{pmatrix} 3 + (-1) \\ -2 + 5 \end{pmatrix} = \begin{pmatrix} 2 \\ 3 \end{pmatrix}$. [A1]
b) $2\mathbf{a} - 3\mathbf{b} = 2\begin{pmatrix} 3 \\ -2 \end{pmatrix} - 3\begin{pmatrix} -1 \\ 5 \end{pmatrix} = \begin{pmatrix} 6 \\ -4 \end{pmatrix} - \begin{pmatrix} -3 \\ 15 \end{pmatrix} = \begin{pmatrix} 6 - (-3) \\ -4 - 15 \end{pmatrix} = \begin{pmatrix} 9 \\ -19 \end{pmatrix}$. [A1]
Find the exact magnitude of the vector $\mathbf{v} = 6\mathbf{i} - 8\mathbf{j}$, and write down the unit vector $\hat{\mathbf{v}}$ in the direction of $\mathbf{v}$.
View Solution & Markscheme
Method & Marks:
Magnitude: $|\mathbf{v}| = \sqrt{6^2 + (-8)^2} = \sqrt{36 + 64} = \sqrt{100} = 10$. [M1][A1]
Unit vector: $\hat{\mathbf{v}} = \frac{\mathbf{v}}{|\mathbf{v}|} = \frac{1}{10}(6\mathbf{i} - 8\mathbf{j}) = \frac{3}{5}\mathbf{i} - \frac{4}{5}\mathbf{j} = \begin{pmatrix} 0.6 \\ -0.8 \end{pmatrix}$. [A1]
Point $P$ has coordinates $(-3, 4)$ and point $Q$ has coordinates $(5, -2)$. Write the position vectors $\vec{OP}$ and $\vec{OQ}$, and calculate the displacement vector $\vec{PQ}$.
View Solution & Markscheme
Method & Marks:
$\vec{OP} = \begin{pmatrix} -3 \\ 4 \end{pmatrix}$, $\vec{OQ} = \begin{pmatrix} 5 \\ -2 \end{pmatrix}$. [A1]
$\vec{PQ} = \vec{OQ} - \vec{OP} = \begin{pmatrix} 5 \\ -2 \end{pmatrix} - \begin{pmatrix} -3 \\ 4 \end{pmatrix} = \begin{pmatrix} 5 - (-3) \\ -2 - 4 \end{pmatrix} = \begin{pmatrix} 8 \\ -6 \end{pmatrix}$. [A1]
Vectors $\mathbf{u} = \begin{pmatrix} 4 \\ 6 \end{pmatrix}$ and $\mathbf{w} = \begin{pmatrix} p \\ -9 \end{pmatrix}$ are parallel. Find the value of $p$ and the scale factor $k$ such that $\mathbf{w} = k\mathbf{u}$.
View Solution & Markscheme
Method & Marks:
Since $\mathbf{w} = k\mathbf{u}$: [M1]
$\begin{pmatrix} p \\ -9 \end{pmatrix} = k\begin{pmatrix} 4 \\ 6 \end{pmatrix} \implies 6k = -9 \implies k = -\frac{9}{6} = -1.5$. [A1]
Then $p = 4k = 4(-1.5) = -6$. [A1]
Calculate the scalar product of $\mathbf{a} = 5\mathbf{i} + 2\mathbf{j}$ and $\mathbf{b} = -3\mathbf{i} + 7\mathbf{j}$. State with justification whether the angle between them is acute, right, or obtuse.
View Solution & Markscheme
Method & Marks:
Scalar product: $\mathbf{a} \cdot \mathbf{b} = (5)(-3) + (2)(7) = -15 + 14 = -1$. [A1]
Since $\mathbf{a} \cdot \mathbf{b} = |\mathbf{a}||\mathbf{b}|\cos\theta$ and magnitudes are positive, $\cos\theta = \frac{-1}{|\mathbf{a}||\mathbf{b}|} < 0$. [M1]
Since $\cos\theta < 0$, the angle $\theta$ between the vectors is obtuse ($90^\circ < \theta \le 180^\circ$). [R1]
Determine the value of $m$ such that the vectors $\mathbf{p} = \begin{pmatrix} m + 2 \\ 5 \end{pmatrix}$ and $\mathbf{q} = \begin{pmatrix} 3 \\ -4 \end{pmatrix}$ are perpendicular.
View Solution & Markscheme
Method & Marks:
For perpendicular vectors, $\mathbf{p} \cdot \mathbf{q} = 0$: [M1]
$3(m + 2) + 5(-4) = 0$
$3m + 6 - 20 = 0 \implies 3m - 14 = 0 \implies 3m = 14 \implies m = \frac{14}{3}$. [A1]
Find the angle between the vectors $\mathbf{u} = \begin{pmatrix} 1 \\ 2 \end{pmatrix}$ and $\mathbf{v} = \begin{pmatrix} 3 \\ 1 \end{pmatrix}$ to the nearest $0.1^\circ$.
View Solution & Markscheme
Method & Marks:
$\mathbf{u} \cdot \mathbf{v} = (1)(3) + (2)(1) = 3 + 2 = 5$. [M1]
$|\mathbf{u}| = \sqrt{1^2 + 2^2} = \sqrt{5}$.
$|\mathbf{v}| = \sqrt{3^2 + 1^2} = \sqrt{10}$. [M1]
$\cos\theta = \frac{5}{\sqrt{5}\sqrt{10}} = \frac{5}{\sqrt{50}} = \frac{5}{5\sqrt{2}} = \frac{1}{\sqrt{2}}$.
$\theta = \arccos\left(\frac{1}{\sqrt{2}}\right) = 45^\circ$. [A1]
In $\triangle OAB$, let $\vec{OA} = \mathbf{a}$ and $\vec{OB} = \mathbf{b}$. Point $M$ is the midpoint of $AB$, and point $P$ lies on $OM$ such that $OP : PM = 2 : 1$. Express $\vec{OP}$ in terms of $\mathbf{a}$ and $\mathbf{b}$. What significant geometric point in the triangle does $P$ represent?
View Solution & Markscheme
Method & Marks:
Since $M$ is the midpoint of $AB$: $\vec{OM} = \frac{1}{2}(\mathbf{a} + \mathbf{b})$. [M1]
Since $OP : PM = 2 : 1$, $P$ is $\frac{2}{3}$ of the way along $OM$:
$\vec{OP} = \frac{2}{3}\vec{OM} = \frac{2}{3}\left(\frac{1}{2}(\mathbf{a} + \mathbf{b})\right) = \frac{1}{3}(\mathbf{a} + \mathbf{b})$. [A1]
Point $P$ is the centroid (center of mass) of $\triangle OAB$, where the medians intersect in a $2:1$ ratio. [R1]
Use vector dot products to prove that the diagonals of a rhombus are perpendicular to each other.
View Solution & Markscheme
Method & Marks:
Let the rhombus $OACB$ have adjacent sides $\vec{OA} = \mathbf{a}$ and $\vec{OB} = \mathbf{b}$. [M1]
Because all sides of a rhombus are equal in length: $|\mathbf{a}| = |\mathbf{b}|$. [M1]
The two diagonals are $\vec{OC} = \mathbf{a} + \mathbf{b}$ and $\vec{BA} = \mathbf{a} - \mathbf{b}$.
Compute their dot product: [M1]
$\vec{OC} \cdot \vec{BA} = (\mathbf{a} + \mathbf{b}) \cdot (\mathbf{a} - \mathbf{b}) = \mathbf{a} \cdot \mathbf{a} - \mathbf{a} \cdot \mathbf{b} + \mathbf{b} \cdot \mathbf{a} - \mathbf{b} \cdot \mathbf{b} = |\mathbf{a}|^2 - |\mathbf{b}|^2$.
Since $|\mathbf{a}| = |\mathbf{b}|$, $|\mathbf{a}|^2 - |\mathbf{b}|^2 = 0$.
Since $\vec{OC} \cdot \vec{BA} = 0$, the diagonals are perpendicular. $\blacksquare$ [A1][R1]
Criterion B & D Navigational Modelling: A rescue boat departs port $O$ at $t = 0$ with constant velocity $\mathbf{v}_b = \begin{pmatrix} 15 \\ 20 \end{pmatrix}\text{ km/h}$. A stranded liferaft is initially located at position $\mathbf{r}_0 = \begin{pmatrix} 60 \\ 20 \end{pmatrix}\text{ km}$ and is drifting with ocean current velocity $\mathbf{v}_c = \begin{pmatrix} 3 \\ 4 \end{pmatrix}\text{ km/h}$.
a) Write the position vector of the rescue boat $\mathbf{r}_b(t)$ and liferaft $\mathbf{r}_L(t)$ at time $t$ hours.
b) Determine the time $t$ when the rescue boat intercepts the liferaft and the coordinates of the intercept point.
View Solution & Markscheme
Method & Marks:
Part a: $\mathbf{r}_b(t) = t\begin{pmatrix} 15 \\ 20 \end{pmatrix} = \begin{pmatrix} 15t \\ 20t \end{pmatrix}$.
$\mathbf{r}_L(t) = \begin{pmatrix} 60 \\ 20 \end{pmatrix} + t\begin{pmatrix} 3 \\ 4 \end{pmatrix} = \begin{pmatrix} 60 + 3t \\ 20 + 4t \end{pmatrix}$. [M1]
Part b: Intercept condition $\mathbf{r}_b(t) = \mathbf{r}_L(t)$:
$x$-components: $15t = 60 + 3t \implies 12t = 60 \implies t = 5\text{ hours}$. [M1]
Check $y$-components: $20(5) = 100$, and $20 + 4(5) = 40 \ne 100$.
Model analysis: With the boat's initial course, the paths do not meet simultaneously ($t = 5$ for $x$, $16t = 20 \implies t = 1.25$ for $y$).
To intercept, the relative displacement $\mathbf{r}_L(t) - \mathbf{r}_b(0)$ must match relative velocity $\mathbf{v}_{\text{rel}} = \mathbf{v}_b - \mathbf{v}_c = \begin{pmatrix} 12 \\ 16 \end{pmatrix}$.
Notice that $\begin{pmatrix} 60 \\ 20 \end{pmatrix} = t\begin{pmatrix} 12 \\ 16 \end{pmatrix}$ requires equal times: the raft position is $(60, 20)$ whereas $(12, 16)$ is in direction $3:4$. [M1][R1: Model evaluation]
2D Vectors & Scalar Product Reference Sheet
Column Vector
$$\mathbf{v} = \begin{pmatrix} x \\ y \end{pmatrix} = x\mathbf{i} + y\mathbf{j}$$
Cartesian component format.Magnitude
$$|\mathbf{v}| = \sqrt{x^2 + y^2}$$
Length of vector via Pythagoras.Unit Vector
$$\hat{\mathbf{v}} = \frac{\mathbf{v}}{|\mathbf{v}|}$$
Length 1 in direction of $\mathbf{v}$.Displacement Vector
$$\vec{AB} = \mathbf{b} - \mathbf{a}$$
Head minus tail position rule.Parallel Vectors
$$\mathbf{u} = k\mathbf{v}$$
Scalar multiple criterion.Scalar (Dot) Product
$$\mathbf{u} \cdot \mathbf{v} = u_1 v_1 + u_2 v_2$$
Component multiplication.Angle Between Vectors
$$\cos\theta = \frac{\mathbf{u} \cdot \mathbf{v}}{|\mathbf{u}| |\mathbf{v}|}$$
$0^\circ \le \theta \le 180^\circ$.Perpendicularity Test
$$\mathbf{u} \perp \mathbf{v} \iff \mathbf{u} \cdot \mathbf{v} = 0$$
Zero dot product condition.Top 5 Common MYP Vector Pitfalls
- Writing a Fraction Line in Column Vectors: A column vector is $\begin{pmatrix} x \\ y \end{pmatrix}$, NOT $\begin{pmatrix} \frac{x}{y} \end{pmatrix}$. Never write a fraction line between vector components!
- Confusing Head and Tail in Displacement: To find $\vec{AB}$, remember: $\vec{AB} = \vec{OB} - \vec{OA} = \mathbf{b} - \mathbf{a}$ ("end minus start"). Writing $\mathbf{a} - \mathbf{b}$ gives $\vec{BA}$, which points in the opposite direction.
- Assuming Dot Product Produces a Vector: The scalar (dot) product $\mathbf{u} \cdot \mathbf{v} = u_1 v_1 + u_2 v_2$ produces a single number (scalar), not a vector.
- Forgetting that Collinearity Requires a Common Point: Proving $\vec{AB} = k\vec{CD}$ only proves that line segments $AB$ and $CD$ are parallel. To prove points $A, B, C$ are collinear, you must show $\vec{AB} = k\vec{BC}$ and explicitly state that $B$ is a shared point.
- Wrong Quadrant Angle Signs: When computing magnitude, remember $(-y)^2 = +y^2$. Also, when computing direction angle with negative components, check which quadrant the vector lies in before concluding the angle.
Self-Grading Mini Test
5 quiz questions on 2D vectors, magnitude, dot product, and perpendicularity.
What is the magnitude of the vector $\mathbf{v} = \begin{pmatrix} -5 \\ 12 \end{pmatrix}$?
Explanation: $|\mathbf{v}| = \sqrt{(-5)^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13$.
If $\vec{OA} = \begin{pmatrix} 1 \\ 4 \end{pmatrix}$ and $\vec{OB} = \begin{pmatrix} 6 \\ -1 \end{pmatrix}$, what is $\vec{AB}$?
Explanation: $\vec{AB} = \vec{OB} - \vec{OA} = \begin{pmatrix} 6 - 1 \\ -1 - 4 \end{pmatrix} = \begin{pmatrix} 5 \\ -5 \end{pmatrix}$.
What is the scalar product of $\mathbf{u} = \begin{pmatrix} 4 \\ -3 \end{pmatrix}$ and $\mathbf{v} = \begin{pmatrix} 2 \\ 5 \end{pmatrix}$?
Explanation: $\mathbf{u} \cdot \mathbf{v} = (4)(2) + (-3)(5) = 8 - 15 = -7$.
Which of the following vectors is perpendicular to $\mathbf{a} = \begin{pmatrix} 3 \\ 2 \end{pmatrix}$?
Explanation: $(3)(-2) + (2)(3) = -6 + 6 = 0$. Since their dot product is zero, $\begin{pmatrix} -2 \\ 3 \end{pmatrix}$ is perpendicular to $\mathbf{a}$.
If $\mathbf{u} = \begin{pmatrix} 2 \\ -5 \end{pmatrix}$ and $\mathbf{w} = \begin{pmatrix} -6 \\ k \end{pmatrix}$ are parallel, what is the value of $k$?
Explanation: The scale factor is $\frac{-6}{2} = -3$. Therefore $k = -5 \times (-3) = 15$.