Study Guide

Vectors: basics, operations, dot product

IB Mathematics: Applications and Interpretation HLΒ· Topic 3.7 VectorsΒ· 6 min read

1. Vector Basicsβ˜…β˜…β˜†β˜†β˜†β± 15 min

πŸ“˜ Definition

Scalar vs Vector

A scalar is a quantity with only magnitude (size). A vector is a quantity with both magnitude and direction.

Example:

Distance (5 km) is a scalar; displacement (5 km north) is a vector.

For a vector between two points (A(x_1,y_1,z_1)) and (B(x_2,y_2,z_2)), the components of (\vec{AB}) are calculated by subtracting the coordinates of the start point (A) from the end point (B).

ABβƒ—=(x2βˆ’x1y2βˆ’y1z2βˆ’z1),∣vβƒ—βˆ£=v12+v22+v32\vec{AB} = \begin{pmatrix} x_2 - x_1 \\ y_2 - y_1 \\ z_2 - z_1 \end{pmatrix}, \quad |\vec{v}| = \sqrt{v_1^2 + v_2^2 + v_3^2}
πŸ“ Worked Example

Find (\vec{AB}) and its magnitude, where (A = (1, 2, -3)) and (B = (4, 0, 1)).

  1. 1

    Subtract start point coordinates from end point coordinates to get components:

  2. 2
    x:4βˆ’1=3,y:0βˆ’2=βˆ’2,z:1βˆ’(βˆ’3)=4ABβƒ—=(3βˆ’24)x: 4 - 1 = 3, \quad y: 0 - 2 = -2, \quad z: 1 - (-3) = 4 \\ \vec{AB} = \begin{pmatrix} 3 \\ -2 \\ 4 \end{pmatrix}
  3. 3

    Calculate magnitude using the Pythagorean formula:

  4. 4
    ∣ABβƒ—βˆ£=32+(βˆ’2)2+42=9+4+16=29β‰ˆ5.39|\vec{AB}| = \sqrt{3^2 + (-2)^2 + 4^2} = \sqrt{9 + 4 + 16} = \sqrt{29} \approx 5.39

Exam tip:

Always remember: end point minus start point for vectors between two points.

2. Vector Addition and Scalar Multiplicationβ˜…β˜…β˜†β˜†β˜†β± 20 min

All basic vector operations are done component-wise: you work on each x, y, z component separately to get the result.

  • Addition: Add corresponding components: (\vec{a} + \vec{b} = (a_1 + b_1, a_2 + b_2, a_3 + b_3))

  • Scalar multiplication: Multiply every component by the scalar: (k\vec{v} = (kv_1, kv_2, kv_3))

  • Negative of a vector: Reverse direction: (-\vec{v} = (-v_1, -v_2, -v_3))

πŸ“ Worked Example

Given (\vec{a} = \begin{pmatrix} 2 \ -1 \end{pmatrix}) and (\vec{b} = \begin{pmatrix} -3 \ 4 \end{pmatrix}), find (3\vec{a} - 2\vec{b}).

  1. 1

    First calculate each scalar multiple:

  2. 2
    3aβƒ—=3(2βˆ’1)=(6βˆ’3),2bβƒ—=2(βˆ’34)=(βˆ’68)3\vec{a} = 3\begin{pmatrix} 2 \\ -1 \end{pmatrix} = \begin{pmatrix} 6 \\ -3 \end{pmatrix}, \quad 2\vec{b} = 2\begin{pmatrix} -3 \\ 4 \end{pmatrix} = \begin{pmatrix} -6 \\ 8 \end{pmatrix}
  3. 3

    Subtract component-wise:

  4. 4
    3aβƒ—βˆ’2bβƒ—=(6βˆ’(βˆ’6)βˆ’3βˆ’8)=(12βˆ’11)3\vec{a} - 2\vec{b} = \begin{pmatrix} 6 - (-6) \\ -3 - 8 \end{pmatrix} = \begin{pmatrix} 12 \\ -11 \end{pmatrix}
βœ“ Quick check

What is the magnitude of (3\vec{a} - 2\vec{b}) from the example above?

  1. Select the correct answer:

    • (5)

    • (\sqrt{265} \approx 16.3)

    • (1)

    Reveal answer
    1 β€”

    You must square each component before adding: (12^2 + (-11)^2 = 144 + 121 = 265), so magnitude is (\sqrt{265}).

3. Dot Product and Applicationsβ˜…β˜…β˜…β˜†β˜†β± 25 min

πŸ“˜ Definition

Dot Product (Scalar Product)

\(\vec{a} \cdot \vec{b}\)

For two vectors (\vec{a} = (a_1,a_2,a_3)) and (\vec{b} = (b_1,b_2,b_3)), the dot product is a scalar calculated by summing the product of corresponding components. It is linked to the angle (\theta) between the vectors.

aβƒ—β‹…bβƒ—=a1b1+a2b2+a3b3=∣aβƒ—βˆ£βˆ£bβƒ—βˆ£cos⁑θ\vec{a} \cdot \vec{b} = a_1b_1 + a_2b_2 + a_3b_3 = |\vec{a}||\vec{b}|\cos\theta

The most important geometric property of the dot product: two non-zero vectors are perpendicular if and only if their dot product is zero (since (\cos 90^\circ = 0)). We can rearrange the formula to find the angle between any two vectors:

cos⁑θ=aβƒ—β‹…bβƒ—βˆ£aβƒ—βˆ£βˆ£bβƒ—βˆ£\cos\theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|}
πŸ“ Worked Example

Find the angle between (\vec{v} = \begin{pmatrix} 2 \ 1 \ -1 \end{pmatrix}) and (\vec{w} = \begin{pmatrix} 3 \ -2 \ 1 \end{pmatrix}), correct to 1 decimal place.

  1. 1

    Step 1: Calculate the dot product:

  2. 2
    vβƒ—β‹…wβƒ—=(2)(3)+(1)(βˆ’2)+(βˆ’1)(1)=6βˆ’2βˆ’1=3\vec{v} \cdot \vec{w} = (2)(3) + (1)(-2) + (-1)(1) = 6 - 2 - 1 = 3
  3. 3

    Step 2: Calculate the magnitude of each vector:

  4. 4
    ∣vβƒ—βˆ£=22+12+(βˆ’1)2=6β‰ˆ2.449,∣wβƒ—βˆ£=32+(βˆ’2)2+12=14β‰ˆ3.742|\vec{v}| = \sqrt{2^2 + 1^2 + (-1)^2} = \sqrt{6} \approx 2.449, \quad |\vec{w}| = \sqrt{3^2 + (-2)^2 + 1^2} = \sqrt{14} \approx 3.742
  5. 5

    Step 3: Calculate (\cos\theta) and find the angle:

  6. 6
    cos⁑θ=3614β‰ˆ0.327β€…β€ŠβŸΉβ€…β€ŠΞΈ=arccos⁑(0.327)β‰ˆ70.9∘\cos\theta = \frac{3}{\sqrt{6}\sqrt{14}} \approx 0.327 \implies \theta = \arccos(0.327) \approx 70.9^\circ

Exam tip:

Check if the dot product is zero first before calculating magnitudes, this saves time for perpendicularity questions.

4. Common Pitfalls

Wrong move:

Subtract end point coordinates from start point when finding (\vec{AB})

Why:

This gives the vector pointing in the opposite direction, with all components negated, leading to wrong results for angles and magnitudes

Correct move:

Always use: (\vec{AB} = \text{end point } B - \text{start point } A)

Wrong move:

Add components before multiplying when calculating the dot product

Why:

This confuses vector addition with dot product calculation, leading to an incorrect scalar result

Correct move:

Multiply each pair of corresponding components first, then sum the products

Wrong move:

Leave the answer as (\cos\theta) when asked for the angle between two vectors

Why:

Examiners expect the actual angle value, not the cosine of the angle, so you will lose marks

Correct move:

Always calculate the inverse cosine (arccos) of your value to get the final angle

Wrong move:

Calculate magnitude as the sum of components, skipping squares and the square root

Why:

This confuses magnitude calculation with vector addition, leading to a completely wrong result

Correct move:

Square each component, sum the squares, then take the positive square root

Wrong move:

Claim two vectors are perpendicular just because their dot product is zero, even if one is the zero vector

Why:

The zero vector has no defined direction, so perpendicularity does not apply

Correct move:

Confirm both vectors are non-zero before concluding perpendicularity from a zero dot product

5. Quick Reference Cheatsheet

Concept

Formula

Vector (\vec{AB}) (A to B)

(\begin{pmatrix} x_2-x_1 \ y_2-y_1 \ z_2-z_1 \end{pmatrix})

Magnitude of (\vec{v} = (v_1,v_2,v_3))

(|\vec{v}| = \sqrt{v_1^2 + v_2^2 + v_3^2})

Vector addition

(\vec{a} + \vec{b} = (a_1+b_1, a_2+b_2, a_3+b_3))

Scalar multiplication

(k\vec{v} = (kv_1, kv_2, kv_3))

Dot product

(\vec{a} \cdot \vec{b} = \sum a_i b_i = |a||b|\cos\theta)

Angle between two vectors

(\cos\theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|})

Perpendicular vectors

(\vec{a} \cdot \vec{b} = 0) (non-zero vectors)

When this came up on past exams

AI-estimated based on syllabus patterns β€” cross-check with official past papers for accuracy. Use only as revision-focus signals.

  • 2025 Β· 1

    Dot product angle calculation

  • 2024 Β· 2

    3D vector between points

What's Next

Now that you have mastered the basics of vectors, vector operations, and the dot product, you can apply these core skills to working with lines and planes in 2D and 3D space, the next key topic in IB AI HL geometry. A solid understanding of these concepts is essential for finding intersections of lines, calculating distances from points to lines or planes, and solving applied geometric problems in fields like navigation, engineering, and computer graphics. These vector basics also form the foundation for kinematics problems involving velocity and displacement vectors that you will encounter later in the course.