Vectors in Two Dimensions

🟢 Lite — Quick Review (1h–1d)

Rapid summary for last-minute revision before your exam.

Vectors in Two Dimensions — Quick Facts

A vector is a quantity that has both magnitude (size/length) and direction. A scalar has only magnitude. Speed is scalar; velocity is vector.

Key formulas to memorise right now:

• Magnitude: $|\vec{a}| = \sqrt{x^2 + y^2}$
• Direction: $\tan \theta = \frac{y}{x}$
• Dot product: $\vec{a} \cdot \vec{b} = x_1x_2 + y_1y_2 = |\vec{a}||\vec{b}|\cos\theta$
• Unit vectors: $\⃗{i} = (1, 0)$ along x-axis, $\⃗{j} = (0, 1)$ along y-axis
• Scalar multiplication: $k\⃗{a}$ stretches by factor $k$, reverses direction if $k < 0$

⚡ Exam tip: JAMB often asks you to find the angle between two vectors using the dot product formula. Memorise $\cos\theta = \frac{\⃗{a} \cdot \⃗{b}}{|\⃗{a}||\⃗{b}|}$. Units vectors $\⃗{i}$ and $\⃗{j}$ are perpendicular — their dot product is zero.

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🟡 Standard — Regular Study (2d–2mo)

Standard content for students with a few days to months.

What Is a Vector?

A vector in two dimensions is a directed line segment — an arrow with a specific length (magnitude) and pointing in a specific direction. It lives in the 2D Cartesian plane and is written using notation like $\⃗{a}$, $\⃗{AB}$, or component form $(x, y)$.

A scalar is just a number — it has magnitude only, no direction. Mass, temperature, speed (not velocity), and time are scalars. Distance is scalar; displacement is vector.

Key distinction: Two vectors are equal if they have the same magnitude and the same direction. They can start at different points and still be equal (free vectors). Position vectors, however, are anchored to the origin.

Vector Representation

A vector $\⃗{v}$ in the Cartesian plane is written as: $$\⃗{v} = x\⃗{i} + y\⃗{j}$$

where:

• $x$ is the x-component (horizontal displacement)
• $y$ is the y-component (vertical displacement)
• $\⃗{i}$ is the unit vector along the x-axis: $\⃗{i} = (1, 0)$
• $\⃗{j}$ is the unit vector along the y-axis: $\⃗{j} = (0, 1)$

The magnitude (length) of $\⃗{v}$ is: $$|\vec{v}| = \sqrt{x^2 + y^2}$$

The direction — the angle $\⃗{v}$ makes with the positive x-axis — is: $$\tan \theta = \frac{y}{x} \quad \Rightarrow \quad \theta = \tan^{-1}\left(\frac{y}{x}\right)$$

Important: Always check which quadrant the vector is in. $\ tan^{-1}$ gives an angle in the first quadrant; adjust by $\pm 180°$ if the vector lies in QII, QIII, or QIV. JAMB MCQs often test this trap.

Notation reminders:

• $\⃗{AB}$ means the vector from point A to point B
• $|\⃗{AB}|$ is the magnitude (length) of $\⃗{AB}$
• $\⃗{a} = \vec{a}$

Types of Vectors

Type	Description
Equal vectors	Same magnitude and same direction (even if at different positions)
Parallel vectors	Point in the same or exactly opposite direction; $\⃗{a} = k\⃗{b}$ for some scalar $k > 0$
Antiparallel vectors	Point in exactly opposite directions; $\⃗{a} = k\⃗{b}$ with $k < 0$
Zero vector	$\⃗{0} = (0, 0)$; no magnitude, no direction — it’s the identity for addition
Unit vector	Magnitude equals 1; used to indicate direction. $\⃗{i} = (1, 0)$ and $\⃗{j} = (0, 1)$ are the standard unit vectors

A unit vector in the direction of $\⃗{a}$ is: $$\hat{\vec{a}} = \frac{\⃗{a}}{|\⃗{a}|}$$

Vector Addition

Vectors are added component-wise: $$\⃗{a} + \⃗{b} = (a_x + b_x,\ a_y + b_y)$$

Two laws govern vector addition:

1. Triangle Law: Place the tail of $\⃗{b}$ at the head of $\⃗{a}$. The resultant vector goes from the tail of $\⃗{a}$ to the head of $\⃗{b}$. The three vectors form a triangle.

2. Parallelogram Law: Place both vectors tail-to-tail (at the same origin). Complete the parallelogram. The resultant is the diagonal from the common tail to the opposite corner.

Head-to-tail method is the same idea as the triangle law — it’s the most common method in JAMB questions. Draw $\⃗{a}$, then draw $\⃗{b}$ starting from the tip of $\⃗{a}$. The closing vector from the start of $\⃗{a}$ to the end of $\⃗{b}$ is $\⃗{a} + \⃗{b}$.

Multiplication by a Scalar

Multiplying a vector by a scalar $k$: $$k\⃗{a} = (ka_x,\ ka_y)$$

This scales the magnitude by $|k|$ and:

• Leaves direction unchanged if $k > 0$
• Reverses direction if $k < 0$

For example, if $\⃗{a} = (3, 4)$, then $2\⃗{a} = (6, 8)$ and $-\⃗{a} = (-3, -4)$.

Position Vectors

A position vector of point $P(x, y)$ is the vector from the origin $O(0, 0)$ to $P$: $$\⃗{OP} = x\⃗{i} + y\⃗{j} = (x, y)$$

The position vector of point $A(x_1, y_1)$ to point $B(x_2, y_2)$ is: $$\⃗{AB} = (x_2 - x_1)\\⃗{i} + (y_2 - y_1)\\⃗{j} = (x_2 - x_1,\ y_2 - y_1)$$

This is also the displacement vector from A to B. Note: $\⃗{AB} = -\⃗{BA}$.

Dot Product (Scalar Product)

The dot product (also called the scalar product) of two vectors produces a scalar, not a vector: $$\⃗{a} \cdot \⃗{b} = |\⃗{a}||\⃗{b}|\cos\theta = x_1x_2 + y_1y_2$$

Where $\theta$ is the angle between $\⃗{a}$ and $\⃗{b}$.

Key properties:

• If $\⃗{a} \cdot \⃗{b} = 0$, the vectors are perpendicular (orthogonal)
• If $\⃗{a} \cdot \⃗{b} = |\⃗{a}||\⃗{b}|$, they are in the same direction
• If $\⃗{a} \cdot \⃗{b} = -|\⃗{a}||\⃗{b}|$, they are in opposite directions
• $\⃗{i} \cdot \⃗{i} = 1$, $\⃗{j} \cdot \⃗{j} = 1$, $\⃗{i} \cdot \⃗{j} = 0$ (very useful!)

Finding the Angle Between Two Vectors

From the dot product formula: $$\cos\theta = \frac{\⃗{a} \cdot \⃗{b}}{|\⃗{a}||\⃗{b}|}$$

Then find $\theta = \cos^{-1}\left(\frac{\⃗{a} \cdot \⃗{b}}{|\⃗{a}||\⃗{b}|}\right)$.

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🔴 Extended — Deep Study (3mo+)

Comprehensive coverage for students on a longer study timeline.

Detailed Vector Operations

Component Form Deep Dive

When a vector $\⃗{v}$ makes an angle $\theta$ with the positive x-axis, its components are: $$x = |\⃗{v}|\cos\theta, \quad y = |\⃗{v}|\sin\theta$$

Conversely: $$\⃗{v} = (|\⃗{v}|\cos\theta)\⃗{i} + (|\⃗{v}|\sin\theta)\⃗{j}$$

JAMB frequently asks: Given magnitude and direction, find components. Given components, find magnitude and direction.

Worked Example 1 — Finding Magnitude and Direction

Question: Find the magnitude and direction of $\⃗{a} = (3, 4)$.

Solution:

• Magnitude: $|\⃗{a}| = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5$
• Direction: $\tan\theta = \frac{4}{3} = 1.333…$
• $\theta = \tan^{-1}(1.333…) = 53.13°$

$\⃗{a}$ lies in the first quadrant, so no adjustment needed. Answer: magnitude = 5, direction = 53.13°.

Worked Example 2 — Vector Addition

Question: If $\⃗{a} = (2, 3)$ and $\⃗{b} = (-1, 5)$, find $\⃗{a} + \⃗{b}$.

Solution: $$\⃗{a} + \⃗{b} = (2 + (-1),\ 3 + 5) = (1,\ 8)$$

Using the triangle law: draw $\⃗{a}$, then draw $\⃗{b}$ from its head. The closing vector is $(1, 8)$.

Worked Example 3 — Dot Product

Question: Find $\⃗{a} \cdot \⃗{b}$ where $\⃗{a} = (4, -3)$ and $\⃗{b} = (2, 5)$.

Solution: $$\⃗{a} \cdot \⃗{b} = (4)(2) + (-3)(5) = 8 - 15 = -7$$

This negative dot product tells us the angle between them is obtuse (> 90°).

Worked Example 4 — Angle Between Two Vectors (JAMB Classic)

Question: Find the angle between $\⃗{a} = (1, 2)$ and $\⃗{b} = (3, -1)$. (JAMB 2023 style)

Solution:

Step 1: Find the dot product. $$\⃗{a} \cdot \⃗{b} = (1)(3) + (2)(-1) = 3 - 2 = 1$$

Step 2: Find magnitudes. $$|\⃗{a}| = \sqrt{1^2 + 2^2} = \sqrt{5}, \quad |\⃗{b}| = \sqrt{3^2 + (-1)^2} = \sqrt{10}$$

Step 3: Apply the formula. $$\cos\theta = \frac{\⃗{a} \cdot \⃗{b}}{|\⃗{a}||\⃗{b}|} = \frac{1}{\sqrt{5} \cdot \sqrt{10}} = \frac{1}{\sqrt{50}} = \frac{1}{5\sqrt{2}}$$

$$\cos\theta = \frac{1}{5\sqrt{2}} \approx 0.1414$$

$$\theta = \cos^{-1}(0.1414) \approx 81.9°$$

Worked Example 5 — Position Vector

Question: Find the position vector of point $B(5, -2)$ relative to point $A(1, 4)$.

Solution: $$\⃗{AB} = (5 - 1)\⃗{i} + (-2 - 4)\⃗{j} = 4\⃗{i} - 6\⃗{j} = (4, -6)$$

The magnitude is $|\⃗{AB}| = \sqrt{4^2 + (-6)^2} = \sqrt{16 + 36} = \sqrt{52} = 2\sqrt{13}$.

Worked Example 6 — Scalar Multiplication and Parallelism

Question: Show that $\⃗{a} = (6, 9)$ and $\⃗{b} = (2, 3)$ are parallel.

Solution: $$\⃗{a} = k\⃗{b} \Rightarrow (6, 9) = k(2, 3)$$

$6 = 2k \Rightarrow k = 3$ $9 = 3k \Rightarrow k = 3$ ✓

Since the same scalar $k = 3 > 0$ works, $\⃗{a}$ and $\⃗{b}$ are parallel and point in the same direction.

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📊 JAMB Topic Weight & Exam Pattern

Detail	Value
Topic weight (Mathematics)	Medium-High
Questions from Vectors (est.)	2–3 per UTME
Common question types	Find magnitude, find angle between vectors, dot product, components
Trigonometric connection	$\sin/\cos$ values; always check quadrant for direction

Vectors connect closely with trigonometry (resolving into components) and coordinate geometry (position vectors). Questions sometimes combine all three topics in a single MCQ.

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⚠️ Common Mistakes to Avoid

1. Confusing speed (scalar) and velocity (vector) — JAMB uses both in options to trap you. Read carefully.
2. Wrong quadrant for direction — $\tan^{-1}(y/x)$ always gives an acute angle. If the vector is in QII or QIII, add 180°. If in QIV, add 360° or report as a negative angle.
3. Forgetting the unit vectors in component form — always express your answer as $x\⃗{i} + y\⃗{j}$ or $(x, y)$, not just $x$ and $y$.
4. Using the wrong dot product formula — the component formula $x_1x_2 + y_1y_2$ and the geometric formula $|\⃗{a}||\⃗{b}|\cos\theta$ give the same result. Know both.
5. Sign errors in position vectors — $\⃗{AB} = (x_2 - x_1, y_2 - y_1)$. Swapping the order gives the negative vector.
6. Assuming perpendicular equals zero dot product is automatic — verify: $x_1x_2 + y_1y_2 = 0$ confirms perpendicular.
7. Mixing up the triangle and parallelogram laws — triangle law is far more commonly tested on JAMB.

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📝 Quick-Reference Formula Sheet

$$\text{Magnitude: } |\⃗{v}| = \sqrt{x^2 + y^2}$$ $$\text{Direction: } \tan\theta = \frac{y}{x}$$ $$\text{Unit vector: } \hat{\⃗{v}} = \frac{\⃗{v}}{|\⃗{v}|}$$ $$\text{Dot product: } \⃗{a} \cdot \⃗{b} = x_1x_2 + y_1y_2 = |\⃗{a}||\⃗{b}|\cos\theta$$ $$\cos\theta = \frac{\⃗{a} \cdot \⃗{b}}{|\⃗{a}||\⃗{b}|}$$ $$\⃗{AB} = (x_2 - x_1,\ y_2 - y_1)$$ $$\⃗{i} \cdot \⃗{j} = 0 \quad (perpendicular unit vectors)$$ $$\⃗{a} + \⃗{b} = (a_x + b_x,\ a_y + b_y)$$ $$k\⃗{a} = (ka_x,\ ka_y)$$

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📋 Study Priority & Order

Recommended order for mastering this topic:

1. First: Scalar vs vector — get this distinction clear (it costs marks in JAMB if you slip)
2. Second: Magnitude formula and unit vectors — foundation for everything else
3. Third: Components in $\⃗{i}$ and $\⃗{j}$ form — very common in JAMB
4. Fourth: Vector addition (triangle/head-to-tail) — graphical understanding helps
5. Fifth: Position vectors — builds on components
6. Sixth: Dot product — the most powerful tool; connects magnitude and angle
7. Seventh: Angle between vectors — the formula to memorise last; it’s the climax of the topic

Spend the most time on dot product and angle calculations — they appear in nearly every JAMB paper for this topic.

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🔗 Related Topics

Topic	How it connects to Vectors
math-17 — Statics of a Particle	Vector addition in equilibrium problems
math-13 — Trigonometry	Resolving vectors into $\sin/\cos$ components
math-10 — Coordinate Geometry	Position vectors and the Cartesian plane
math-09 — Quadratic Equations	Vector magnitude involves $\sqrt{x^2 + y^2}$ — squaring and solving

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