Curriculum

Mathematics

Unit 2

0/14

Unit 5

0/12

Text lesson

The 3D Coordinate System and Vectors in Space

Mathematics

The 3D Coordinate System and Vectors in Space

Overview

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 $\hat{i}$ , $\hat{j}$ , $\hat{k}$

│ ├── 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

The 3D Coordinate System and Vectors in Space · 1 / 15

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 $R^{3}$ .

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).

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

Comparison aspect	Main idea	Keywords
Example 1	Locate $A (- 2, 3, 2)$ and $C (4, 0, - 1)$ .	Locate
•	Solution	Solution
For A	plot (−2 3) in the xy-plane then move 2 units in the positive z direction	plotplanethenmove
For C	the y-coordinate is 0 so C lies in the xz-plane. Take 4 on the x-axis then move 1 unit in the negative z direction	coordinateliesplaneTake

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 $(0, 0, c)$ .

The 3D Coordinate System and Vectors in Space · 2 / 15

3-1 (b) Distance between two points

Rule

For A(x1,y1,z1) and B(x2,y2,z2)AB=(x2−x1)2+(y2−y1)2+(z2−z1)2

It comes from applying Pythagoras’ theorem twicefirst in the horizontal plane, then with the vertical difference.

The 3D Coordinate System: a room corner with three axes places the point (2, 3, 6) in the three-dimensional coordinate system and vectors: its distance from the origin is 7 m.

✍️ 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 $9 0^{\circ}$ .
Example 3 Show that $A (4, 4, 0)$ , $B (4, 0, 4)$ , $C (0, 4, 4)$ form an equilateral triangle and find its area.
Solution:
$A B = \sqrt{0 + 16 + 16} = 4 \sqrt{2}$ , $B C = \sqrt{16 + 16 + 0} = 4 \sqrt{2}$ , $A C = \sqrt{16 + 0 + 16} = 4 \sqrt{2}$ .
All sides are equal, so it is equilateral.
Area $= \frac{\sqrt{3}}{4} {(4 \sqrt{2})}^{2} = \frac{\sqrt{3}}{4} \times 32 = 8 \sqrt{3}$ square units.
Example 4 Find $k$ if the triangle with vertices $(7, 1, 3)$ , $(5, 3, k)$ , $(3, 5, 3)$ is equilateral.
Solution:
Let the points be $P$ , $Q$ , $R$ . $P R^{2} = 16 + 16 + 0 = 32$ .
$P Q^{2} = 4 + 4 + (k - 3)^{2}$ and $Q R^{2} = 4 + 4 + (k - 3)^{2}$ , so $P Q = Q R$ for every $k$ (the triangle is always isosceles).
Equilateral requires $8 + (k - 3)^{2} = 32$ , so $(k - 3)^{2} = 24$ and $k = 3 \pm 2 \sqrt{6}$ .

The 3D Coordinate System and Vectors in Space · 3 / 15

3-1 (c) The midpoint of a segment

Rule

The midpoint of AB is M=(x1+x22, y1+y22, z1+z22)

✍️ Worked examples

Example 5 Find the midpoint of $A B$ for $A (1, - 3, 2)$ and $B (4, - 1, 4)$ .
Solution:
$M = (\frac{1 + 4}{2}, \frac{- 3 - 1}{2}, \frac{2 + 4}{2}) = (\frac{5}{2}, - 2, 3)$ .
Example 6 $C (2, 2, 6)$ is the midpoint of $A B$ and $A = (1, - 4, 0)$ . Find $B$ .
Solution:
$\frac{1 + x}{2} = 2$ , $\frac{- 4 + y}{2} = 2$ , $\frac{0 + z}{2} = 6$ .
So $x = 3$ , $y = 8$ , $z = 12$ and $B = (3, 8, 12)$ .
Shortcut the end point is B=2C−A.
Example 7 $B (- 1, 4, 2)$ is the midpoint of $A C$ with $A (1, 0, 2)$ . Find $C$ .
Solution:
$C = 2 B - A = (- 2 - 1, 8 - 0, 4 - 2) = (- 3, 8, 2)$ .
A common error is to take $B - A$ or to halve instead of doubling.

The 3D Coordinate System and Vectors in Space · 4 / 15

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 A⃗ is its length ∥A⃗∥=Ax2+Ay2+Az2
Example for A⃗=(2,3,1) the norm is 14 and for B⃗=(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.

The 3D Coordinate System and Vectors in Space · 5 / 15

3-2 (b) Operations on vectors

Addition

For A⃗=(Ax,Ay,Az) and B⃗=(Bx,By,Bz)A⃗+B⃗=(Ax+Bx, Ay+By, Az+Bz)

Propertiesclosure, commutative (A⃗+B⃗=B⃗+A⃗), associative, the zero vector 0⃗=(0,0,0) is the identity, and −A⃗=(−Ax,−Ay,−Az) is the additive inverse.

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

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

$\vec{A} = \vec{B}$ 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 $k = - 3$ .
Example 10 Find $l$ and $m$ if $(l - 4, m^{2} - 3, 1) = (5, 1, 1)$ .
Solution:
$l - 4 = 5$ gives $l = 9$ .
$m^{2} - 3 = 1$ gives $m = \pm 2$ .

The 3D Coordinate System and Vectors in Space · 6 / 15

3-2 (c) Unit vectors and the basis i^, j^, 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 A⃗=Axi^+Ayj^+Azk^

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.

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 A⃗=(−2,2,1) ∥A⃗∥=3, so u^A=(−23,23,13).
A vector of norm $r$ in the direction of $\vec{A}$ is $r {\hat{u}}_{A}$ ; the opposite direction uses $- {\hat{u}}_{A}$ .

The 3D Coordinate System and Vectors in Space · 7 / 15

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:
$\vec{A B} = (4 + 2, 0 - 3, 2 - 1) = (6, - 3, 1)$ .
$\vec{B A} = (- 6, 3, - 1)$ .
Example 13 If $A (1, 1, - 2)$ and $\vec{A B} = (4, - 1, 2)$ , find $B$ .
Solution:
$B = A + \vec{A B} = (5, 0, 0)$ .
Example 14 Find the unit vector in the direction of $\vec{A B}$ for $A (- 2, 2, 1)$ and $B (3, 1, - 2)$ .
Solution:
$\vec{A B} = (5, - 1, - 3)$ , with norm $\sqrt{25 + 1 + 9} = \sqrt{35}$ .
${\hat{u}}_{A B} = \frac{1}{\sqrt{35}} (5, - 1, - 3)$ .

The 3D Coordinate System and Vectors in Space · 8 / 15

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 13, 23, 23 (their squares add to 1+4+49=1).
Angles cos⁡−113≈70.5∘, and cos⁡−123≈48.2∘ twice.
Example 16 A force of 10 N makes angles of $6 0^{\circ}$ with $O x$ and $4 5^{\circ}$ with $O y$ and an acute angle with $O z$ . Write it in component form.
Solution:
$\cos^{2} 6 0^{\circ} + \cos^{2} 4 5^{\circ} + \cos^{2} θ_{z} = 1$ gives $\frac{1}{4} + \frac{1}{2} + \cos^{2} θ_{z} = 1$ , so $\cos θ_{z} = \frac{1}{2}$ and $θ_{z} = 6 0^{\circ}$ .
$\vec{F} = 10 (\frac{1}{2}, \frac{\sqrt{2}}{2}, \frac{1}{2}) = (5, 5 \sqrt{2}, 5) \approx (5, 7.07, 5)$ N.
Example 17 A vector of norm 5 makes equal angles with the three positive axes. Write it in component form.
Solution:
Equal angles give $3 \cos^{2} θ = 1$ , so $\cos θ = \frac{1}{\sqrt{3}}$ and $θ \approx 54. 7^{\circ}$ .
$\vec{A} = 5 (\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}) = \frac{5}{\sqrt{3}} (1, 1, 1) \approx (2.89, 2.89, 2.89)$ .
Example 18 Resolve a vector of norm 10 making $4 0^{\circ}$ with $O z$ , whose projection on the $x y$ -plane makes $7 0^{\circ}$ with $O x$ .
Solution:
$A_{z} = 10 \cos 4 0^{\circ} \approx 7.66$ and the projection on the $x y$ -plane has length $10 \sin 4 0^{\circ} \approx 6.43$ .
$A_{x} = 6.43 \cos 7 0^{\circ} \approx 2.20$ and $A_{y} = 6.43 \sin 7 0^{\circ} \approx 6.04$ .
$\vec{A} \approx 2.20 \hat{i} + 6.04 \hat{j} + 7.66 \hat{k}$ .
Direction angles cos⁡−12.2010≈77.3∘, cos⁡−16.0410≈52.8∘, 40∘.

The 3D Coordinate System and Vectors in Space · 9 / 15

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 A⃗=(k,0,0) ∥A⃗∥=∣k∣.
If the norm is known, e.g. $\vec{A} = (- 2, k, 1)$ with $∥ \vec{A} ∥ = 3$ , then $4 + k^{2} + 1 = 9$ , so $k = \pm 2$ .

The 3D Coordinate System and Vectors in Space · 10 / 15

📌 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 = (Ax, Ay, Az) with magnitude |A| = √(Ax² + Ay² + Az²).
Direction angles (α, β, γ) made with positive X, Y, Z axes yield direction cosines: cos α = Ax/|A|, cos β = Ay/|A|, cos γ = Az/|A|.

Comparison aspect	Main idea	Keywords
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.	magnitudesatisfyingcos²
Standard Equation of a Sphere (معادلة الكرة في الفراغ)	A sphere centered at C(a 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 -N) and r = √(L² + M² + N² – D)	centeredradiusGeneralform

The 3D Coordinate System and Vectors in Space · 11 / 15

Comparison aspect	Main idea	Keywords
Why cos² α + cos² β + cos² γ = 1	Because unit vector u = (Ax/\|A\|) i + (Ay/\|A\|) j + (Az/\|A\|) k = (cos α) i + (cos β) j + (cos γ) k and its magnitude \|u\|² = cos² α + cos² β + cos² γ = 1² = 1	Becauseunitvectormagnitude
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	axisderivedPythagoreantheorem
Sphere tangent to the coordinate planes	If a sphere touches all three positive coordinate planes its center is C(r r) and its equation is (x – r)² + (y – r)² + (z – r)² = r²	touchesthreepositivecenter

Target Geometric Entity	Geometric Formula / Equation	Significance in Secondary 3 Exam
Distance to XY-Plane	d = \|c\|	Depends solely on the absolute z-coordinate
Distance to YZ-Plane	d = \|a\|	Depends solely on the absolute x-coordinate
Distance to XZ-Plane	d = \|b\|	Depends solely on the absolute y-coordinate
Distance to X-Axis	d = √(b² + c²)	Pythagorean norm of y and z components
Distance to Y-Axis	d = √(a² + c²)	Pythagorean norm of x and z components
Distance to Z-Axis	d = √(a² + b²)	Pythagorean norm of x and y components

💡 High-Yield Exam Traps & Key Takeaways:

Comparison aspect	Main idea	Keywords
Exam Trap	Don’t confuse direction ratios (a c) with direction cosines (l n). Direction ratios can be scaled by any non-zero constant k but direction cosines must have a sum of squares equal to 1	confusedirectionratioscosines
Quick Identity	sin² α + sin² β + sin² γ = (1 – cos² α) + (1 – cos² β) + (1 – cos² γ) = 3 – 1 = 2	sin²cos²

The 3D Coordinate System and Vectors in Space · 15 / 15

Curriculum

Mathematics

Unit 1

Unit 2

Unit 3

Unit 4

Unit 5

Unit 6

Unit 7

The 3D Coordinate System and Vectors in Space

🌳 Concept tree

The 3D Coordinate System and Vectors in Space

│ ├── 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

│ ├── The fundamental unit vectors

│ └── Unit vector

├── 3-2 (d) The vector between two points

│ └── Rule

├── Definitions

└── Useful facts

Why three axes

Plotting a point

✍️ Worked examples

Rule

✍️ Worked examples

Rule

The midpoint of AB is M=(x1+x22, y1+y22, z1+z22)

✍️ Worked examples

Definitions

Addition

✍️ Worked examples

✍️ Worked examples

The fundamental unit vectors

Every vector can be written A⃗=Axi+Ayj+Azk^

✍️ Worked examples

Rule

✍️ Worked examples

✍️ Worked examples

Well done! You finished this lesson

Every vector can be written A⃗=Axi^+Ayj^+Azk^