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Module 2: Force and Motion

 

PHYS-2325 M2L1 Vectors

"The miracle of the appropriateness of the language of mathematics for the formulation of the laws of physics is a wonderful gift which we neither understand nor deserve."
— Eugene Wigner


Physics describes a world that moves in three dimensions — and most physical quantities that matter, like force, velocity, and displacement, have both a size and a direction. These are vector quantities, and understanding how to work with them mathematically is the foundation of everything in this course. In this lesson, you will build the tools to represent vectors in component form using unit vectors, to add and subtract vectors analytically, and to compute two distinct types of vector products — the dot product and the cross product — each of which appears repeatedly in mechanics, energy, and rotation. The calculus-based approach developed here moves beyond geometric arrows on a page; you will work with vectors expressed as derivatives and integrals of position, velocity, and force fields throughout the semester.

Three-dimensional coordinate axes with vectors decomposed into x, y, and z components, illustrating vector addition by components.
Vectors in 3D space, resolved into component form along each axis.

Course Competencies and Learning Objectives

A ★ indicates that this page contains content related to that LO.

CC2.1 Analyze the components of motion and any applied forces

★ LO2.1.1 Apply Newton's Laws and the formulas of motion to be able to predict the behavior of the motion

★ LO2.1.2 Identify the behavior of a particle of motion

LO2.1.3 Identify the initial conditions of the motion of the particle

LO2.1.4 Apply the equations of motion to describe how the particle will move

LO2.1.5 Determine the uniform circular motion of an object

Required Reading

Click the blue buttons to go to the OpenStax reading assignments. Complete all four sections before watching the videos.

2.1 Scalars and Vectors 2.2 Coordinate Systems and Components of a Vector 2.3 Algebra of Vectors 2.4 Products of Vectors

 

Optional Reading

Explore More

Want to go deeper or see vectors from a different angle? These free resources complement the OpenStax text and may help clarify the concepts covered in this lesson.

Media

Watch each video in order. The pre-lecture videos introduce the core concepts from the reading; the mini-lectures walk through worked examples step by step.

Video 1: Scalars and Vectors

Pre-Lecture: Scalars and Vectors

This pre-lecture video introduces the distinction between scalar and vector quantities and establishes the geometric representation of vectors — magnitude, direction, tail, and head — using real-world examples from physics.

  • Scalar vs. vector quantities — what distinguishes them
  • Representing vectors as directed arrows with magnitude and direction
  • Parallel, antiparallel, and equal vectors
  • Geometric vector addition: parallelogram rule and tail-to-head method

Time: Video pending — to be recorded by instructor

Video 2: Vector Components

Pre-Lecture: Coordinate Systems and Vector Components

This video covers the analytical method of working with vectors by resolving them into scalar components along each coordinate axis, using unit vectors î, ĵ, and k̂ to express vectors in component form in 2D and 3D.

  • Cartesian coordinate system and unit vectors î, ĵ, k̂
  • Scalar components Ax, Ay, Az and vector component form
  • Finding magnitude from components: A = √(Ax2 + Ay2 + Az2)
  • Direction angles and polar coordinate connection

Time: Video pending — to be recorded by instructor

Video 3: Vector Algebra

Pre-Lecture: Algebra of Vectors

This video demonstrates the analytical method of vector addition and subtraction using scalar components — the technique used throughout the rest of the course to solve kinematics and dynamics problems.

  • Adding and subtracting vectors by adding their components
  • Finding the resultant vector of multiple forces or displacements
  • Writing a vector equation and solving for an unknown vector
  • Unit vector of direction: v̂ = v / |v|

Time: Video pending — to be recorded by instructor

Video 4: Adding Vectors by Components

Mini-Lecture: Adding Two Vectors Component by Component

A worked example video walking through the full process of resolving two vectors given in magnitude-angle form into their scalar components, computing their sum, and expressing the resultant in both component form and magnitude-angle form.

  • Identify magnitude and direction angle for each vector
  • Compute Ax = A cos θ and Ay = A sin θ for each
  • Add x-components and y-components separately
  • Find resultant magnitude and direction angle

Time: Video pending — to be recorded by instructor

Video 5: Dot and Cross Products

Mini-Lecture: Dot Product and Cross Product

This video covers both types of vector multiplication. The dot product produces a scalar and is used to find the angle between vectors and to compute work. The cross product produces a vector and is used to find torque and angular momentum.

  • Dot product: A · B = AB cos φ = AxBx + AyBy + AzBz
  • Finding the angle between two vectors using the dot product
  • Cross product: |A × B| = AB sin φ, direction by right-hand rule
  • Cross product in component form using the determinant method

Time: Video pending — to be recorded by instructor

Practice and Apply - Conceptual

Steps to Find a Vector Resultant

Arrange the following steps in the correct order for finding the resultant of two or more vectors using the analytical (component) method.

Order Items

Arrange the following items in the correct order.

  1. Establish a coordinate system and identify the positive x and y directions.
  2. Identify the magnitude and direction angle of each vector.
  3. Resolve each vector into scalar components: Ax = A cos θ and Ay = A sin θ.
  4. Add all x-components to get Rx; add all y-components to get Ry.
  5. Find the magnitude of the resultant: R = √(Rx2 + Ry2).
  6. Find the direction angle of the resultant: θR = tan-1(Ry / Rx), adjusting for quadrant.

Identify the Vector Quantities

From the list below, select all quantities that are vector quantities (they have both magnitude and direction).

Select All

Select all vector quantities from the list.

  • Displacement
  • Velocity
  • Acceleration
  • Force
  • Torque
  • Distance
  • Speed
  • Mass
  • Temperature
  • Time

Match the Vector Notation

Match each vector symbol or expression on the left to its correct meaning or description.

Match Items

Select each set of two matching items.

Items to Match

|A|
The magnitude (length) of vector A — always a positive scalar.
î (i-hat)
Unit vector pointing in the positive x-direction; magnitude = 1, no units.
A · B
The dot product of A and B — a scalar equal to AB cos φ.
A × B
The cross product of A and B — a vector perpendicular to both, with magnitude AB sin φ.
 (A-hat)
The unit vector of direction for A — found by dividing A by its magnitude.
R = A + B
The resultant vector — found by adding corresponding scalar components of A and B.

Key Vocabulary Check

Click each card to flip it and check your understanding of the key terms from this lesson.

What is a scalar quantity? Give two examples from physics.
Answer

A scalar quantity is fully described by a single number and a unit — it has magnitude only, no direction. Examples: mass (5.0 kg), temperature (300 K), distance (12 m), speed (25 m/s), time (4.0 s).

 

What does the dot product A · B equal, and what type of quantity is the result?
Answer

A · B = AB cos φ, where φ is the angle between the two vectors. The result is a scalar. In component form: A · B = AxBx + AyBy + AzBz. The dot product is used to compute work and find angles between vectors.

 

What is a unit vector, and what is its purpose?
Answer

A unit vector has a magnitude of exactly 1 and carries no physical units. Its only purpose is to specify direction. The standard unit vectors î, ĵ, k̂ point in the positive x, y, and z directions respectively. Any vector A can be written as A = Axî + Ayĵ + Azk̂.

 

What does the cross product A × B produce, and how is its direction determined?
Answer

The cross product produces a vector perpendicular to both A and B, with magnitude |A × B| = AB sin φ. Its direction is found using the right-hand rule: point fingers in the direction of A, curl them toward B — your thumb points in the direction of A × B. Note: A × B = −B × A (anticommutative).

 

How do you find the angle between two vectors using the dot product?
Answer

Use the relation cos φ = (A · B) / (AB). Steps: (1) Compute the dot product using components: AxBx + AyBy + AzBz. (2) Find the magnitudes A and B. (3) Divide and take the inverse cosine: φ = cos-1[(A · B) / (AB)].

 

What is the resultant vector, and how is it found analytically?
Answer

The resultant is the vector sum of two or more vectors: R = A + B + ... Analytically, add corresponding scalar components: Rx = Ax + Bx, Ry = Ay + By, Rz = Az + Bz. Then write R = Rxî + Ryĵ + Rzk̂ and compute its magnitude and direction angle.

 

Practice and Apply - Computational

Important: Answer Formatting

Follow these formatting rules when entering answers:

  • Component form: Express vectors as A = Axî + Ayĵ + Azk̂. Always include the unit vector symbols.
  • Magnitude: Report as a positive number with correct SI units (m, m/s, N, etc.). Round to 3 significant figures unless stated otherwise.
  • Direction angles: Report in degrees (°), measured counterclockwise from the positive x-axis. Verify the quadrant matches the sign of your components.
  • Dot product result: A scalar — include units (e.g., N·m, m2) but no vector notation.
  • Cross product result: A vector — express in component form with units.

Practice Problem 1

A displacement vector has a magnitude of 8.50 m and points at an angle of 32.0° above the positive x-axis. (a) Find the scalar x- and y-components of this vector. (b) Express the vector in component form using unit vectors. (c) Find the unit vector of its direction.

Show Solution

A displacement vector has a magnitude of 8.50 m and points at an angle of 32.0° above the positive x-axis. (a) Find the scalar x- and y-components. (b) Express in component form. (c) Find the unit vector of direction.

Given: A = 8.50 m, θ = 32.0°

Part (a) — Scalar components:

Ax = A cos θ = (8.50 m) cos 32.0° = (8.50)(0.848) = 7.21 m

Ay = A sin θ = (8.50 m) sin 32.0° = (8.50)(0.530) = 4.50 m

Part (b) — Component form:

A = 7.21î + 4.50ĵ (m)

Part (c) — Unit vector of direction:

 = A / A = (7.21î + 4.50ĵ) / 8.50 = 0.848î + 0.530ĵ

Check: |Â| = √(0.848² + 0.530²) = √(0.719 + 0.281) = √1.000 = 1.00 ✓

Practice Problem 2

Three displacement vectors are given in component form: A = (4.0î − 3.0ĵ) m, B = (−2.0î + 6.0ĵ − 1.0k̂) m, and C = (1.0î + 0.0ĵ + 5.0k̂) m. Find the resultant R = A + B + C and its magnitude.

Show Solution

Three displacement vectors: A = (4.0î − 3.0ĵ) m, B = (−2.0î + 6.0ĵ − 1.0k̂) m, C = (1.0î + 0.0ĵ + 5.0k̂) m. Find R = A + B + C and its magnitude.

Add corresponding components:

Rx = Ax + Bx + Cx = 4.0 + (−2.0) + 1.0 = 3.0 m

Ry = Ay + By + Cy = (−3.0) + 6.0 + 0.0 = 3.0 m

Rz = Az + Bz + Cz = 0.0 + (−1.0) + 5.0 = 4.0 m

Resultant vector:

R = 3.0î + 3.0ĵ + 4.0k̂ (m)

Magnitude:

R = √(Rx2 + Ry2 + Rz2) = √(3.02 + 3.02 + 4.02)

R = √(9.0 + 9.0 + 16.0) = √34.0 = 5.83 m

Practice Problem 3

Two force vectors are F1 = (6.0î + 4.0ĵ − 2.0k̂) N and F2 = (−3.0î + 2.0ĵ + 5.0k̂) N. (a) Find the dot product F1 · F2. (b) Find the angle between the two force vectors.

Show Solution

F1 = (6.0î + 4.0ĵ − 2.0k̂) N, F2 = (−3.0î + 2.0ĵ + 5.0k̂) N. Find the dot product and angle between them.

Part (a) — Dot product:

F1 · F2 = F1xF2x + F1yF2y + F1zF2z

= (6.0)(−3.0) + (4.0)(2.0) + (−2.0)(5.0)

= −18.0 + 8.0 − 10.0 = −20.0 N2

Part (b) — Angle between the vectors:

F1 = √(6.02 + 4.02 + (−2.0)2) = √(36 + 16 + 4) = √56 = 7.48 N

F2 = √((−3.0)2 + 2.02 + 5.02) = √(9 + 4 + 25) = √38 = 6.16 N

cos φ = (F1 · F2) / (F1 F2) = −20.0 / (7.48 × 6.16) = −20.0 / 46.1 = −0.434

φ = cos-1(−0.434) = 115.7°

The negative dot product confirms the angle is greater than 90°, meaning the force vectors point in generally opposite directions.

Practice Problem 4

A wrench applies a force F = (15.0ĵ) N at a position r = (0.25î) m from a bolt. (a) Find the torque vector τ = r × F. (b) Find the magnitude of the torque. (c) Describe the direction of the torque vector and explain what it means physically.

Show Solution

r = (0.25î) m, F = (15.0ĵ) N. Find τ = r × F, its magnitude, and direction.

Part (a) — Cross product:

τ = r × F = (0.25î) × (15.0ĵ)

= (0.25)(15.0)(î × ĵ)

Since î × ĵ = k̂ (cyclic order in the right-hand coordinate system):

τ = (0.25)(15.0)k̂ = 3.75k̂ N·m

Part (b) — Magnitude:

|τ| = rF sin φ = (0.25 m)(15.0 N) sin 90° = (0.25)(15.0)(1) = 3.75 N·m

(The angle between r and F is 90° because r points in the x-direction and F points in the y-direction.)

Part (c) — Direction and physical meaning:

The torque vector points in the +z direction (out of the xy-plane, toward you). Physically, this means the force applied by the wrench will cause the bolt to rotate counterclockwise when viewed from above (from the +z direction), by the right-hand rule. The magnitude 3.75 N·m is the turning effectiveness of the applied force.

Ready to Move On?

Lesson Checklist

Before moving on to M2L2, confirm you can do each of the following:

  • ☐ Distinguish between scalar and vector quantities and give two examples of each from physics.
  • ☐ Given a vector's magnitude and direction angle, find its scalar components using Ax = A cos θ and Ay = A sin θ.
  • ☐ Express a vector in component form (A = Axî + Ayĵ + Azk̂) and find its magnitude from components.
  • ☐ Find the resultant of multiple vectors by adding their scalar components.
  • ☐ Compute the dot product of two vectors in component form and use it to find the angle between them.
  • ☐ Compute the cross product of two vectors and determine the direction of the result using the right-hand rule.

If any of these feel uncertain, revisit the reading or videos before continuing. Vectors are the language of the rest of this course!