IGCSE, Edexcel, IB and AP Science Courses | Dr Dabour Hub Academy

Overview

🎯 Learning outcomes

Establish the 3D Cartesian Coordinate System $(x, y, z)$ and distance/midpoint formulas.
Calculate Dot Product (Scalar) and Cross Product (Vector) and determine angles between vectors.
Derive the vector, parametric, and Cartesian forms of straight line equations in space.
Formulate vector and general scalar equations of a plane in 3D space ( $A x + B y + C z + D = 0$ ).

🌳 Concept tree

Lesson 1→Three-Dimensional Cartesian Coordinates and Vector Operations

├── Key terms

├── Laws and key relationships

├── Geometry of 3D Vectors and Scalar/Cross Product Physics
├── Equations of Straight Lines in Three Dimensions
├── Equations of Planes and Perpendicular Distance Formulas

├── Worked examples

├── Exam tips

└── Real-world applications

Lesson 1: Three-Dimensional Cartesian Coordinates and Vector Operations · 1 / 15

Key terms

Comparison aspect	Main idea	Keywords
D Coordinate System	A spatial coordinate frame defined by three mutually perpendicular axes ( $x, y, z$ ) intersecting at the origin $(0, 0, 0)$ , governed by the right-hand rule.	spatialframedefinedthree
Dot (Scalar) Product	An algebraic operation between two vectors yielding a scalar: $\vec{A} \cdot \vec{B} = ∣ ∣ \vec{A} ∣ ∣ ∣ ∣ \vec{B} ∣ ∣ \cos θ = A_{x} B_{x} + A_{y} B_{y} + A_{z} B_{z}$ .	algebraicoperationbetweenvectors
Cross (Vector) Product	An algebraic operation yielding a vector perpendicular to both vectors: $\vec{A} \times \vec{B} = ∣ ∣ \vec{A} ∣ ∣ ∣ ∣ \vec{B} ∣ ∣ \sin θ \hat{n}$ , evaluated via a $3 \times 3$ determinant.	algebraicoperationyieldingperpendicular
Direction Vector of a Line (d⃗)	A non-zero vector (a c) that is parallel to the straight line in space	zeroparallelstraightspace
Normal Vector to a Plane (n⃗)	A non-zero vector (A C) perpendicular to every line lying entirely within the plane	zeroperpendiculareveryline

Lesson 1: Three-Dimensional Cartesian Coordinates and Vector Operations · 2 / 15

Laws and key relationships

Vector Norm & Distance Formula[ Formula:

∣ ∣ \vec{A} ∣ ∣ = \sqrt{A_{x}^{2} + A_{y}^{2} + A_{z}^{2}}

| Distance

A B = \sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2} + (z_{2} - z_{1})^{2}}

. ]

Orthogonality and Parallelism Conditions[ Formula: Orthogonal:

\vec{A} \cdot \vec{B} = 0

. Parallel:

\vec{A} \times \vec{B} = \vec{0} ⟺ \frac{A_{x}}{B_{x}} = \frac{A_{y}}{B_{y}} = \frac{A_{z}}{B_{z}}

. ]

Straight Line Equations in Space[ Formula: Vector form:

\vec{r} = {\vec{r}}_{0} + t \vec{d}

. Parametric:

x = x_{0} + a t, y = y_{0} + b t, z = z_{0} + c t

.Cartesian (Symmetric):

\frac{x - x_{0}}{a} = \frac{y - y_{0}}{b} = \frac{z - z_{0}}{c}

. ]

Plane Equation in Space[ Formula: Vector form:

\vec{n} \cdot \vec{r} = \vec{n} \cdot {\vec{r}}_{0}

. General scalar:

A (x - x_{0}) + B (y - y_{0}) + C (z - z_{0}) = 0 ⟹ A x + B y + C z + D = 0

. ]

Lesson 1: Three-Dimensional Cartesian Coordinates and Vector Operations · 3 / 15

Geometry of 3D Vectors and Scalar/Cross Product Physics

Three-Dimensional Cartesian Coordinates and Vector: two cables from a room corner to two points show vectors in three-dimensional space: both of length 3 and perpendicular, dot product zero.

In three dimensions, any vector is represented by components $\vec{A} = A_{x} \hat{i} + A_{y} \hat{j} + A_{z} \hat{k}$ .
Direction cosines satisfy the identity $\cos^{2} α + \cos^{2} β + \cos^{2} γ = 1$ .- Dot Product: $\vec{A} \cdot \vec{B} = A_{x} B_{x} + A_{y} B_{y} + A_{z} B_{z}$ .
It calculates the angle between vectors $\cos θ = \frac{\vec{A} \cdot \vec{B}}{∣ ∣ \vec{A} ∣ ∣ ∣ ∣ \vec{B} ∣ ∣}$ and the scalar projection of $\vec{A}$ onto $\vec{B}$ ( $= \vec{A} \cdot {\hat{u}}_{B}$ ).
Two non-zero vectors are perpendicular if and only if $\vec{A} \cdot \vec{B} = 0$ .- Cross Product: $\vec{A} \times \vec{B} = ∣ \begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ A_{x} & A_{y} & A_{z} \\ B_{x} & B_{y} & B_{z} \end{array} ∣$ .
Its magnitude $∣ ∣ \vec{A} \times \vec{B} ∣ ∣$ geometrically represents the AREA of the parallelogram spanned by $\vec{A}$ and $\vec{B}$ .
The triangle area is half of this magnitude.

Lesson 1: Three-Dimensional Cartesian Coordinates and Vector Operations · 4 / 15

Equations of Straight Lines in Three Dimensions

A line in 3D is uniquely fixed by a point $P_{0} (x_{0}, y_{0}, z_{0})$ and a direction vector $\vec{d} = (a, b, c)$ .
The position vector of any point on the line is $\vec{r} = {\vec{r}}_{0} + t \vec{d}$ , where $t \in R$ is the scalar parameter. Expressing each coordinate gives the parametric equations $x = x_{0} + a t, y = y_{0} + b t, z = z_{0} + c t$ .
Solving for t yields the symmetric Cartesian form x−x0a=y−y0b=z−z0c.
If any direction number is zero (e.g., $c = 0$ ), the line lies in a plane parallel to the xy-plane, written as $\frac{x - x_{0}}{a} = \frac{y - y_{0}}{b}, z = z_{0}$ .

Lesson 1: Three-Dimensional Cartesian Coordinates and Vector Operations · 5 / 15

Equations of Planes and Perpendicular Distance Formulas

A plane in space is defined by a point $P_{0} (x_{0}, y_{0}, z_{0})$ and a normal vector $\vec{n} = (A, B, C)$ perpendicular to the plane.
For any point $P (x, y, z)$ on the plane, the vector $\vec{P_{0} P}$ is orthogonal to $\vec{n}$ , giving $\vec{n} \cdot (\vec{r} - {\vec{r}}_{0}) = 0 ⟹ A x + B y + C z + D = 0$ , where $D = - (A x_{0} + B y_{0} + C z_{0})$ .
The perpendicular length $L$ from an external point $(x_{1}, y_{1}, z_{1})$ to the plane is given by the elegant formula: $L = \frac{∣ A x_{1} + B y_{1} + C z_{1} + D ∣}{\sqrt{A^{2} + B^{2} + C^{2}}}$ .

Lesson 1: Three-Dimensional Cartesian Coordinates and Vector Operations · 6 / 15

Worked examples

✍️ Worked examples→Worked Problem 2 – Line-Plane Intersection and Perpendicular Distance

Problem Statement Problem: Given the plane 2x−y+2z−6=0 and the point P(3,4,−1):1) Find the perpendicular distance from point P to the plane.2) Find the Cartesian equation of the straight line passing through P perpendicular to the plane. Detailed Solution:1) Normal to plane: n⃗=(2,−1,2).
$A = 2, B = - 1, C = 2, D = - 6$ .Perpendicular distance: $L = \frac{∣ A x_{1} + B y_{1} + C z_{1} + D ∣}{\sqrt{A^{2} + B^{2} + C^{2}}} = \frac{∣ 2 (3) + (- 1) (4) + 2 (- 1) - 6 ∣}{\sqrt{2^{2} + (- 1)^{2} + 2^{2}}} = \frac{∣ 6 - 4 - 2 - 6 ∣}{\sqrt{4 + 1 + 4}} = \frac{∣ - 6 ∣}{\sqrt{9}} = \frac{6}{3} = 2 units of length$ .2) Since the line is perpendicular to the plane, its direction vector $\vec{d}$ is parallel to the normal vector $\vec{n}$ : $\vec{d} = \vec{n} = (2, - 1, 2)$ .The line passes through $P (3, 4, - 1)$ :Cartesian symmetric equation: $\frac{x - 3}{2} = \frac{y - 4}{- 1} = \frac{z + 1}{2}$ .

Lesson 1: Three-Dimensional Cartesian Coordinates and Vector Operations · 7 / 15

Exam tips

Angle Between Two PlanesThe angle between two planes is the angle between their normal vectors:

\cos θ = \frac{∣ {\vec{n}}_{1} \cdot {\vec{n}}_{2} ∣}{∣ ∣ {\vec{n}}_{1} ∣ ∣ ∣ ∣ {\vec{n}}_{2} ∣ ∣}

.

Line Parallel to PlaneIf a line is parallel to a plane, its direction vector

\vec{d}

is PERPENDICULAR to the plane’s normal:

\vec{d} \cdot \vec{n} = 0

.

Direction Cosines Identity

\cos^{2} α + \cos^{2} β + \cos^{2} γ = 1

. Consequently,

\sin^{2} α + \sin^{2} β + \sin^{2} γ = 2

.

Lesson 1: Three-Dimensional Cartesian Coordinates and Vector Operations · 8 / 15

Real-world applications

Comparison aspect	Main idea	Keywords
Robotics & Computer Graphics	3D transformation matrices and vector cross products govern the rendering of 3D computer animation and robotic arm inverse kinematics	transformationmatricesvectorcross
GPS & Satellite Positioning	Triangulating radio signals from four orbital GPS satellites to compute the precise latitude longitude and elevation of a mobile user	Triangulatingradiosignalsfour
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Key Scientific Tables & Anatomical Diagrams Extracted from Official Curricula

💡 Core Diagrams & Reference Data→All data, reference comparison tables, and anatomical figures have been accurately digitized from official curriculum textbooks.

🔬 solid_geometry_vectors_and_planes

Scientific details and precise observational breakdown:

(1) » (r2) (2) 2 (-r)

3) ¥ (= (4yc9 & (L-+)

( »d (Sy rat \rtl or

(1) Find in two different methods :

(1) ¥ (2r?-3r45) (2) % G-2r+r’)

r=1 ts

If you know that→pant) and ye r= onmetent’ , use the properti

ie) r=1 r=

of the summation notation = to find the value of each of the following :

n 6

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Part 11

📖

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📌 Comprehensive Scientific Terminology Dictionary:

3D Coordinate System:
A spatial coordinate frame defined by three mutually perpendicular axes (x, y, z) intersecting at the origin (0, 0, 0), governed by the right-hand rule.
Dot (Scalar) Product:
An algebraic operation between two vectors yielding a scalar A⃗ · B⃗ = ||A⃗|| ||B⃗|| cos θ = Ax Bx + Ay By + Az Bz.
Cross (Vector) Product:
An algebraic operation yielding a vector perpendicular to both vectors: A⃗ × B⃗ = ||A⃗|| ||B⃗|| sin θ n̂, evaluated via a 3 × 3 determinant.
Direction Vector of a Line (d⃗):
A non-zero vector $(a, b, c)$ that is parallel to the straight line in space.
Normal Vector to a Plane (n⃗):
A non-zero vector $(A, B, C)$ perpendicular to every line lying entirely within the plane.

⚖️ Governing Scientific Laws, Theorems & Mathematical Models:
Vector Norm & Distance Formula:
[ Formula ||A⃗|| = √(Ax2 + Ay2 + Az2) | Distance AB = √((x2-x1)2 + (y2-y1)2 + (z2-z1)2). ]
Orthogonality and Parallelism Conditions:
[ Formula Orthogonal: A⃗ · B⃗ = 0. Parallel: A⃗ × B⃗ = 0⃗ ⇔ Ax / Bx = Ay / By = Az / Bz. ]
Straight Line Equations in Space:
[ Formula Vector form: r⃗ = r⃗0 + t d⃗. Parametric: $x = x0 + at, y = y0 + bt, z = z0 + ct$.
Cartesian (Symmetric) (x – x0) / a = (y – y0) / b = (z – z0) / c. ]
Plane Equation in Space:
[ Formula Vector form: n⃗ · r⃗ = n⃗ · r⃗0. General scalar: A(x – x0) + B(y – y0) + C(z – z0) = 0 ⇒ Ax + By + Cz + D = 0. ]

📌 Exhaustive In:
Depth Pedagogical Breakdown:
◄ I. Geometry of 3D Vectors and Scalar/Cross Product Physics
In three dimensions, any vector is represented by components A⃗ = Ax î + Ay ĵ + Az k̂.
Direction cosines satisfy the identity cos 2 α + cos 2 β + cos 2 γ = 1.:
Dot Product A⃗ · B⃗ = Ax Bx + Ay By + Az Bz. It calculates the angle between vectors cos θ = (A⃗ · B⃗) / (||A⃗|| ||B⃗||) and the scalar projection of $\vec{A}$ onto $\vec{B}$ (= A⃗ · ûB). Two non-zero vectors are perpendicular if and only if A⃗ · B⃗ = 0.
– Cross Product A⃗ × B⃗ = î amp; ĵ amp; k̂ Ax amp; Ay amp; Az Bx amp; By amp; Bz. Its magnitude ||A⃗ × B⃗|| geometrically represents the AREA of the parallelogram spanned by $\vec{A}$ and $\vec{B}$. The triangle area is half of this magnitude.

Figure (solid_geometry_vectors_and_planes.png):
Figure (8) Spatial 3D coordinate planes, vector projections, and normal vector to a plane
◄ II. Equations of Straight Lines in Three Dimensions
A line in 3D is uniquely fixed by a point $P0(x0, y0, z0)$ and a direction vector $\vec{d} = (a, b, c)$.
The position vector of any point on the line is r⃗ = r⃗0 + t d⃗, where t ∈ ℝ is the scalar parameter.
Expressing each coordinate gives the parametric equations $x = x0 + at, y = y0 + bt, z = z0 + ct$.
Solving for t yields the symmetric Cartesian form:
(x – x0) / a = (y – y0) / b = (z – z0) / c.
If any direction number is zero (e.g., c = 0), the line lies in a plane parallel to the xy-plane, written as (x – x0) / a = (y – y0) / b, z = z0.

Lesson 1: Three-Dimensional Cartesian Coordinates and Vector Operations · 12 / 15

⚖️ Governing Physical Laws & Mathematical Relations:

A plane in space is defined by a point $P0(x0, y0, z0)$ and a normal vector $\vec{n} = (A, B, C)$ perpendicular to the plane.
For any point P(x, y, z) on the plane, the vector $\vec{P0 P}$ is orthogonal to $\vec{n}$, giving n⃗ · (r⃗ – r⃗0) = 0 ⇒ Ax + By + Cz + D = 0, where D = -(Ax0 + By0 + Cz0).
The perpendicular length L from an external point $(x₁, y₁, z₁)$ to the plane is given by the elegant formula: L = (|A x₁ + B y₁ + C z₁ + D|) / (√(A² + B² + C²)).
🧮 Exemplary Step-by-Step Worked Problems & Applied Exercises:
Worked Example 1 Worked Problem 2 – Line-Plane Intersection and Perpendicular Distance

Problem Statement Problem: Given the plane 2x – y + 2z – 6 = 0 and the point P(3, 4, -1):
1) Find the perpendicular distance from point P to the plane.
2) Find the Cartesian equation of the straight line passing through P perpendicular to the plane.
Detailed Solution:
1) Normal to plane n⃗ = (2, -1, 2). A = 2, B = -1, C = 2, D = -6.
Perpendicular distance:
L = (|A x₁ + B y₁ + C z₁ + D|) / (√(A² + B² + C²)) = (|2(3) + (-1)(4) + 2(-1) – 6|) / (√(2² + (-1)² + 2²)) = (|6 – 4 – 2 – 6|) / (√(4 + 1 + 4)) = (|-6|) / √9 = 6 / 3 = 2 units of length.
2) Since the line is perpendicular to the plane, its direction vector d⃗ is parallel to the normal vector n⃗:
d⃗ = n⃗ = (2, -1, 2).
The line passes through P(3, 4, -1):
Cartesian symmetric equation (x – 3) / 2 = (y – 4) / (-1) = (z + 1) / 2.

Comparison aspect	Main idea	Keywords
•	💡 Master Teacher Insights & Critical Exam Cautions	MasterTeacherInsightsCritical
Angle Between Two Planes	The angle between two planes is the angle between their normal vectors: cos θ = (\|n⃗1 · n⃗2\|) / (\|\|n⃗1\|\| \|\|n⃗2\|\|)	normalvectors
Line Parallel to Plane	If a line is parallel to a plane its direction vector d⃗ is PERPENDICULAR to the plane's normal: d⃗ · n⃗ = 0	directionvectorPERPENDICULARnormal
Direction Cosines Identity	cos 2 α + cos 2 β + cos 2 γ = 1. Consequently sin 2 α + sin 2 β + sin 2 γ = 2	Consequently
•	🌍 Practical Real-World Industrial & Societal Connections	PracticalRealWorldIndustrial
✔ Robotics & Computer Graphics	3D transformation matrices and vector cross products govern the rendering of 3D computer animation and robotic arm inverse kinematics	transformationmatricesvectorcross

✔ GPS & Satellite Positioning→Triangulating radio signals from four orbital GPS satellites to compute the precise latitude, longitude, and elevation of a mobile user.

Lesson 1: Three-Dimensional Cartesian Coordinates and Vector Operations · 13 / 15

🔍 In-Depth Explanations, Deductions & Multi-Book Insights:

★ Comprehensive Study & Detailed Pedagogical Guide ★
Comprehensive Explanation Three-Dimensional Cartesian Coordinates, Vectors, Dot and Cross Products, and Planes in Space

| | 2026/2027 |

🎯 Targeted Learning Outcomes & Pedagogical Objectives:
• Establish the 3D Cartesian Coordinate System $(x, y, z)$ and distance/midpoint formulas.
• Calculate Dot Product (Scalar) and Cross Product (Vector) and determine angles between vectors.
• Derive the vector, parametric, and Cartesian forms of straight line equations in space.
• Formulate vector and general scalar equations of a plane in 3D space (Ax + By + Cz + D = 0).

Lesson 1: Three-Dimensional Cartesian Coordinates and Vector Operations · 14 / 15

Lesson summary

★ High-Yield Summary, Mind Maps & Comparison Tables ★
🗺️ Core Concept Map & Knowledge Architecture:
✔ 3D Vectors Norm ||A⃗|| = √(x2+y2+z2) | Direction Cosines cos 2α + cos 2β + cos 2γ = 1.
✔ Products Dot Product (scalar, A⃗ · B⃗ = 0 ⇔ ⊥) | Cross Product (vector, area of parallelogram).
✔ Planes Ax + By + Cz + D = 0 | Distance L = (|Ax1+By1+Cz1+D|) / (√(A2+B2+C2)).
📊 Classification & Comparative Summary Tables:
Table Comparison Between Line and Plane Equations in Three Dimensions

Geometric Entity	Defining Spatial Elements	Vector Form	Cartesian General Form
Straight Line	Point P₀(x₀, y₀, z₀) and direction vector d⃗ = (a, b, c)	r⃗ = r⃗₀ + t d⃗	(x – x₀) / a = (y – y₀) / b = (z – z₀) / c
Plane	Point P₀(x₀, y₀, z₀) and normal vector n⃗ = (A, B, C)	n⃗ · r⃗ = n⃗ · r⃗₀	Ax + By + Cz + D = 0

💡 Exam Flashcard Capsules & High-Yield Secrets:
★ Distance from (x₁, y₁, z₁) to xy-plane is |z₁|, to yz-plane is |x₁|, and to xz-plane is |y₁|.
★ Distance from point to the x-axis is √(y₁² + z₁²).
★ Parallel planes have identical normal vectors A1/A2 = B1/B2 = C1/C2.
★ Two non-parallel lines that do not intersect in 3D space are called SKEW LINES.

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Curriculum

Mathematics

Unit 1

Unit 2

Unit 3

Unit 4

Unit 5

Unit 6

Unit 7

Lesson 1: Three-Dimensional Cartesian Coordinates and Vector Operations