The derivative of $f(x) = 5^{x^2}$ with respect to $x$ is:

B) $2x \cdot 5^{x^2} \cdot \ln 5$

The sum of the first n terms of an arithmetic sequence (S_n) is formulated as:

S_n = n / 2 * [ 2a + (n - 1) * d ] = n / 2 * [ a + l ]

The sum to infinity (S_infinity) of a geometric series exists and converges if and only if the common ratio satisfies:

|r| < 1 (i.e. -1 < r < 1), and S_infinity = a / (1 - r)

كيف يحسب الرقم الهيدروجيني pH للمحاليل؟ How to calculate pH and pOH?

يحسب الرقم الهيدروجيني بالعلاقة: pH = -log[H⁺]، وللمحاليل المائية عند 25°C: pH + pOH = 14. إذا كان pH 7 فالمحلول قاعدي. [الصيغة الرمزية الموزونة: pH = -log[H⁺] | pOH = -log[OH⁻] | pH + pOH = 14 | [H⁺][OH⁻] = 10⁻¹⁴]

$Mathematics cover$

MathematicsEgyptian Curriculum 2026/2027 · lessons, notes and exam-style quizzes

Home/ Courses/ Mathematics/ Secondary 3 Mathematics 2026/2027 — further lessons/ The 3D Coordinate System and Vectors in Space

DABOUR HUB / LESSON 2026-2027

The 3D Coordinate System and Vectors in Space

Q: كيف يحسب الرقم الهيدروجيني pH للمحاليل؟ How to calculate pH and pOH?

يحسب الرقم الهيدروجيني بالعلاقة: pH = -log[H⁺]، وللمحاليل المائية عند 25°C: pH + pOH = 14. إذا كان pH 7 فالمحلول قاعدي. [الصيغة الرمزية الموزونة: pH = -log[H⁺] | pOH = -log[OH⁻] | pH + pOH = 14 | [H⁺][OH⁻] = 10⁻¹⁴]

Course: Mathematics Unit: Secondary 3 Mathematics 2026/2027 — further lessons

🧭 Lesson points — pick a point and study it with its question bank

1. 3-1 (a) The three-dimensional orthogonal coordinate systemExplanation

2. 3-1 (b) Distance between two pointsExplanation

3. 3-1 (c) The midpoint of a segmentExplanation

4. 3-2 (a) Position vectors and componentsExplanation

5. 3-2 (b) Operations on vectorsExplanation

6. 3-2 (c) Unit vectors and the basis i^, j^, k^Explanation

7. 3-2 (d) The vector between two pointsExplanation

8. 3-2 (e) Direction angles and direction cosinesExplanation

This lesson covers lessons 3-1 and 3-2 of the unit "Geometry and Measurement in Two and Three Dimensions".
After it you should be able to:

locate a point in space and find the distance between two points and the midpoint of a segment;
represent a vector by three components and carry out addition, scalar multiplication and equality in space;
find the norm, the unit vector, the direction angles and direction cosines of a vector.

🌳 Concept tree

The 3D Coordinate System and Vectors in Space
├── 3-1 (a) The three-dimensional orthogonal coordinate system
│   ├── Why three axes
│   ├── The right-hand rule
│   ├── The coordinate planes
│   ├── Plotting a point
│   └── Distances to planes and axes
├── 3-1 (b) Distance between two points
│   └── Rule
├── 3-1 (c) The midpoint of a segment
│   └── Rule
├── 3-2 (a) Position vectors and components
│   └── Definitions
├── 3-2 (b) Operations on vectors
│   ├── Addition
│   ├── Multiplication by a real number
│   └── Equality of vectors
├── 3-2 (c) Unit vectors and the basis i^, j^, k^
│   ├── The fundamental unit vectors
│   └── Unit vector
├── 3-2 (d) The vector between two points
│   └── Rule
└── 3-2 (e) Direction angles and direction cosines
    ├── Definitions
    └── Useful facts

01 3-1 (a) The three-dimensional orthogonal coordinate system

Why three axes

On a line, a body's position is one number: its distance from the origin $O$ .
In a plane, two perpendicular axes give the pair $(x, y)$ by projecting the point onto each axis.
In space we need three mutually perpendicular axes $O x$ , $O y$ and $O z$ meeting at the origin $O (0, 0, 0)$ .
A point $A$ is given by the ordered triple $(x, y, z)$ , the projections of $A$ on the three axes. The set of all such triples is $R^{3}$ .

The right-hand rule

Curl the fingers of the right hand from the positive $x$ -axis towards the positive $y$ -axis; the thumb then points along the positive $z$ -axis.
Every system in this course follows this rule (a right-handed system).

The coordinate planes

All points $(x, y, 0)$ lie in the $x y$ -plane, whose equation is $z = 0$ .
All points $(x, 0, z)$ lie in the $x z$ -plane, whose equation is $y = 0$ .
All points $(0, y, z)$ lie in the $y z$ -plane, whose equation is $x = 0$ .
The three planes divide space into eight regions called octants.

Plotting a point

✍️ Worked examples

Example 1: Locate $A (- 2, 3, 2)$ and $C (4, 0, - 1)$ .

Solution:

For $A$ : plot $(- 2, 3)$ in the $x y$ -plane, then move 2 units in the positive $z$ direction.
For $C$ : the $y$ -coordinate is $0$ , so $C$ lies in the $x z$ -plane. Take 4 on the $x$ -axis, then move 1 unit in the negative $z$ direction.

Distances to planes and axes

Distance from $(x, y, z)$ to the $x y$ -plane is $∣ z ∣$ ; to the $x z$ -plane it is $∣ y ∣$ ; to the $y z$ -plane it is $∣ x ∣$ .
Distance to the $x$ -axis is $\sqrt{y^{2} + z^{2}}$ ; to the $y$ -axis it is $\sqrt{x^{2} + z^{2}}$ ; to the $z$ -axis it is $\sqrt{x^{2} + y^{2}}$ .
For example, the distance from $(- 2, 3, 4)$ to the $x$ -axis is $\sqrt{9 + 16} = 5$ .
A point on the $x$ -axis has the form $(a, 0, 0)$ ; on the $y$ -axis $(0, b, 0)$ ; on the $z$ -axis $(0, 0, c)$ .

02 3-1 (b) Distance between two points

Rule

For $A (x_{1}, y_{1}, z_{1})$ and $B (x_{2}, y_{2}, z_{2})$ : $A B = \sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2} + (z_{2} - z_{1})^{2}}$
It comes from applying Pythagoras' theorem twice: first in the horizontal plane, then with the vertical difference.

✍️ Worked examples

Example 2: Show that the triangle $A (2, - 1, 3)$ , $B (- 4, 4, 2)$ , $C (- 2, 5, 1)$ is right-angled at $C$ .

Solution:

$A B^{2} = (- 6)^{2} + 5^{2} + (- 1)^{2} = 36 + 25 + 1 = 62$ .
$B C^{2} = (2)^{2} + (1)^{2} + (- 1)^{2} = 6$ .
$A C^{2} = (4)^{2} + (- 6)^{2} + (2)^{2} = 16 + 36 + 4 = 56$ .
$B C^{2} + A C^{2} = 6 + 56 = 62 = A B^{2}$ , so the angle at $C$ is $9 0^{\circ}$ .

03 3-1 (c) The midpoint of a segment

Rule

The midpoint of $A B$ is $M = (\frac{x_{1} + x_{2}}{2}, \frac{y_{1} + y_{2}}{2}, \frac{z_{1} + z_{2}}{2})$

✍️ Worked examples

Example 5: Find the midpoint of $A B$ for $A (1, - 3, 2)$ and $B (4, - 1, 4)$ .

Solution:

$M = (\frac{1 + 4}{2}, \frac{- 3 - 1}{2}, \frac{2 + 4}{2}) = (\frac{5}{2}, - 2, 3)$ .

04 3-2 (a) Position vectors and components

Definitions

The position vector of $A (A_{x}, A_{y}, A_{z})$ is the directed segment $\vec{O A}$ from the origin to $A$ , written $\vec{A} = (A_{x}, A_{y}, A_{z})$ .
$A_{x}$ , $A_{y}$ , $A_{z}$ are the components of the vector along the three axes.
The norm (magnitude) of $\vec{A}$ is its length: $∥ \vec{A} ∥ = \sqrt{A_{x}^{2} + A_{y}^{2} + A_{z}^{2}}$
Example: for $\vec{A} = (2, 3, 1)$ the norm is $\sqrt{14}$ and for $\vec{B} = (0, 4, - 3)$ it is $5$ .
A vector with one zero component lies in a coordinate plane; for example $\vec{B}$ above lies in the $y z$ -plane.

05 3-2 (b) Operations on vectors

Addition

For $\vec{A} = (A_{x}, A_{y}, A_{z})$ and $\vec{B} = (B_{x}, B_{y}, B_{z})$ : $\vec{A} + \vec{B} = (A_{x} + B_{x}, A_{y} + B_{y}, A_{z} + B_{z})$
Properties: closure, commutative ( $\vec{A} + \vec{B} = \vec{B} + \vec{A}$ ), associative, the zero vector $\vec{0} = (0, 0, 0)$ is the identity, and $- \vec{A} = (- A_{x}, - A_{y}, - A_{z})$ is the additive inverse.

Multiplication by a real number

$k \vec{A} = (k A_{x}, k A_{y}, k A_{z})$ .
If $k > 0$ the direction is unchanged; if $k < 0$ it is reversed; the norm is multiplied by $∣ k ∣$ .
Properties: $(k + l) \vec{A} = k \vec{A} + l \vec{A}$ , $k (\vec{A} + \vec{B}) = k \vec{A} + k \vec{B}$ , $k (l \vec{A}) = (k l) \vec{A}$ .

✍️ Worked examples

Example 8: For $\vec{A} = (- 1, 5, 2)$ and $\vec{B} = (4, - 1, 3)$ find $2 \vec{A} - 3 \vec{B}$ and the vector $\vec{C}$ with $2 \vec{C} + 3 \vec{A} = 2 \vec{B}$ .

Solution:

$2 \vec{A} - 3 \vec{B} = (- 2, 10, 4) - (12, - 3, 9) = (- 14, 13, - 5)$ .
$2 \vec{C} = 2 \vec{B} - 3 \vec{A} = (8, - 2, 6) - (- 3, 15, 6) = (11, - 17, 0)$ .
$\vec{C} = (\frac{11}{2}, - \frac{17}{2}, 0)$ .

Equality of vectors

$\vec{A} = \vec{B}$ if and only if $A_{x} = B_{x}$ , $A_{y} = B_{y}$ and $A_{z} = B_{z}$ .

✍️ Worked examples

Example 9: Find $x$ , $y$ , $k$ if $(2 x + 1, 5, k + 4) = (1, y^{2} - 4, x + 1)$ .

Solution:

$2 x + 1 = 1$ gives $x = 0$ .
$y^{2} - 4 = 5$ gives $y = \pm 3$ .
$k + 4 = x + 1 = 1$ gives $k = - 3$ .

06 3-2 (c) Unit vectors and the basis $\hat{i}$ , $\hat{j}$ , $\hat{k}$

The fundamental unit vectors

$\hat{i} = (1, 0, 0)$ , $\hat{j} = (0, 1, 0)$ and $\hat{k} = (0, 0, 1)$ have norm 1 along the positive $x$ , $y$ and $z$ axes.
Every vector can be written $\vec{A} = A_{x} \hat{i} + A_{y} \hat{j} + A_{z} \hat{k}$
For example $(- 1, 0, 0) = - \hat{i}$ and $(0, 0, - 1) = - \hat{k}$ .

✍️ Worked examples

Example 11: For $\vec{A} = 2 \hat{i} - 3 \hat{j} + \hat{k}$ and $\vec{B} = - \hat{i} - 2 \hat{j}$ find $2 \vec{A} - 3 \vec{B}$ , then compare $∥ \vec{A} + \vec{B} ∥$ with $∥ \vec{A} ∥ + ∥ \vec{B} ∥$ .

Solution:

$2 \vec{A} = 4 \hat{i} - 6 \hat{j} + 2 \hat{k}$ and $3 \vec{B} = - 3 \hat{i} - 6 \hat{j}$ , so $2 \vec{A} - 3 \vec{B} = 7 \hat{i} + 2 \hat{k}$ .
$\vec{A} + \vec{B} = \hat{i} - 5 \hat{j} + \hat{k}$ , with norm $\sqrt{27} = 3 \sqrt{3} \approx 5.20$ .
$∥ \vec{A} ∥ + ∥ \vec{B} ∥ = \sqrt{14} + \sqrt{5} \approx 3.74 + 2.24 = 5.98$ .
So $∥ \vec{A} + \vec{B} ∥ < ∥ \vec{A} ∥ + ∥ \vec{B} ∥$ . In general $∥ \vec{A} + \vec{B} ∥ \leq ∥ \vec{A} ∥ + ∥ \vec{B} ∥$ , with equality only when the vectors point the same way.

Unit vector

A unit vector has norm 1.
The unit vector in the direction of $\vec{A}$ is ${\hat{u}}_{A} = \frac{\vec{A}}{∥ \vec{A} ∥}$
For $\vec{A} = (- 2, 2, 1)$ : $∥ \vec{A} ∥ = 3$ , so ${\hat{u}}_{A} = (- \frac{2}{3}, \frac{2}{3}, \frac{1}{3})$ .
A vector of norm $r$ in the direction of $\vec{A}$ is $r {\hat{u}}_{A}$ ; the opposite direction uses $- {\hat{u}}_{A}$ .

07 3-2 (d) The vector between two points

Rule

$\vec{A B} = \vec{B} - \vec{A}$ , because $\vec{O A} + \vec{A B} = \vec{O B}$ .
Always "end minus start"; also $\vec{B A} = - \vec{A B}$ .
$∥ \vec{A B} ∥$ equals the distance $A B$ .

✍️ Worked examples

Example 12: For $A (- 2, 3, 1)$ and $B (4, 0, 2)$ find $\vec{A B}$ and $\vec{B A}$ .

Solution:

$\vec{A B} = (4 + 2, 0 - 3, 2 - 1) = (6, - 3, 1)$ .
$\vec{B A} = (- 6, 3, - 1)$ .

08 3-2 (e) Direction angles and direction cosines

Definitions

Let $θ_{x}$ , $θ_{y}$ , $θ_{z}$ be the angles a vector $\vec{A}$ makes with the positive $x$ , $y$ , $z$ axes.
They are the direction angles; $\cos θ_{x}$ , $\cos θ_{y}$ , $\cos θ_{z}$ are the direction cosines, often written $l$ , $m$ , $n$ .
The components are $A_{x} = ∥ \vec{A} ∥ \cos θ_{x}$ , $A_{y} = ∥ \vec{A} ∥ \cos θ_{y}$ , $A_{z} = ∥ \vec{A} ∥ \cos θ_{z}$ .
So the unit vector is ${\hat{u}}_{A} = (\cos θ_{x}, \cos θ_{y}, \cos θ_{z})$ , and therefore $\cos^{2} θ_{x} + \cos^{2} θ_{y} + \cos^{2} θ_{z} = 1$
Each direction angle lies between $0^{\circ}$ and $18 0^{\circ}$ .

✍️ Worked examples

Example 15: Find the direction cosines and angles of $\vec{A} = (1, 2, 2)$ .

Solution:

$∥ \vec{A} ∥ = 3$ .
Direction cosines: $\frac{1}{3}$ , $\frac{2}{3}$ , $\frac{2}{3}$ (their squares add to $\frac{1 + 4 + 4}{9} = 1$ ).
Angles: $\cos^{- 1} \frac{1}{3} \approx 70. 5^{\circ}$ , and $\cos^{- 1} \frac{2}{3} \approx 48. 2^{\circ}$ twice.

Useful facts

A vector parallel to the $y z$ -plane has $A_{x} = 0$ .
A vector along the $x$ -axis has the form $(a, 0, 0)$ and norm $∣ a ∣$ .
For $\vec{A} = (k, 0, 0)$ : $∥ \vec{A} ∥ = ∣ k ∣$ .
If the norm is known, e.g. $\vec{A} = (- 2, k, 1)$ with $∥ \vec{A} ∥ = 3$ , then $4 + k^{2} + 1 = 9$ , so $k = \pm 2$ .

📖

Comprehensive Curriculum Encyclopedia & Core Terminology — The 3D Coordinate System and Vectors in Space

📌 Comprehensive Scientific Terminology & Definitions:

3D Cartesian Coordinate Frame (نظام الإحداثيات المتعامد ثلاثي الأبعاد): Defined by three mutually perpendicular axes (X, Y, Z) intersecting at origin O(0,0,0), satisfying the Right-Hand Rule, partitioning space into eight octants.

Coordinate Planes and Projections (المستويات الإحداثية): XY-plane (z = 0); YZ-plane (x = 0); XZ-plane (y = 0). Distance from point P(x, y, z) to XY-plane is |z|; distance to z-axis is √(x² + y²).

Position Vector & Direction Cosines (متجه الموضع وجيوب تمام الاتجاه): Vector A = (A_x, A_y, A_z) with magnitude |A| = √(A_x² + A_y² + A_z²). Direction angles (α, β, γ) made with positive X, Y, Z axes yield direction cosines: cos α = A_x/|A|, cos β = A_y/|A|, cos γ = A_z/|A|.

Unit Vector (متجه الوحدة): A vector of magnitude 1: u_A = A / |A| = (cos α, cos β, cos γ), satisfying cos² α + cos² β + cos² γ = 1.

Standard Equation of a Sphere (معادلة الكرة في الفراغ): A sphere centered at C(a, b, c) with radius r has equation: (x - a)² + (y - b)² + (z - c)² = r²; General form: x² + y² + z² + 2Lx + 2My + 2Nz + D = 0, where C = (-L, -M, -N) and r = √(L² + M² + N² - D).

⚖️ Governing Scientific Laws, Principles & Mathematical Formulas:

('Fundamental Direction Cosines Identity: cos² α + cos² β + cos² γ = 1; Consequently, sin² α + sin² β + sin² γ = 2.',)

('Distance Between Two Points in 3D: AB = √[(x₂ - x₁)² + (y₂ - y₁)² + (z₂ - z₁)²].',)

('Internal Division of a Line Segment: Point P dividing AB in ratio m₁ : m₂ has coordinates x = (m₁x₂ + m₂x₁) / (m₁ + m₂), similarly for y and z.',)

('Condition for Real Sphere: L² + M² + N² - D > 0 (if = 0 it represents a point; if < 0 it represents an imaginary set).',)

🔍 In-Depth Explanations, Deductions & Core Concepts:

Why cos² α + cos² β + cos² γ = 1: Because unit vector u = (A_x/|A|) i + (A_y/|A|) j + (A_z/|A|) k = (cos α) i + (cos β) j + (cos γ) k, and its magnitude |u|² = cos² α + cos² β + cos² γ = 1² = 1.
Distance from P(x, y, z) to the coordinate axes: Distance to x-axis = √(y² + z²); Distance to y-axis = √(x² + z²); Distance to z-axis = √(x² + y²), derived from the Pythagorean theorem in 3D projection.
Sphere tangent to the coordinate planes: If a sphere touches all three positive coordinate planes, its center is C(r, r, r) and its equation is (x - r)² + (y - r)² + (z - r)² = r².