IB Mathematics AI AHL Definition and calculation of the product MAI Study Notes- New Syllabus

IB Mathematics AI AHL Definition and calculation of the product MAI Study Notes

LEARNING OBJECTIVE

Definition and calculation of the scalar and vector product of two vectors.

Key Concepts:

The Scalar Product
The Vector Product
Components of vectors.

Scalar Product (Dot Product)

Definition

The scalar product, also known as the dot product, is an algebraic operation that takes two vectors and returns a scalar quantity (a real number). It is a measure of how much two vectors align with each other.

Algebraic Formula

If $ \vec{a} = \langle a_1, a_2, a_3 \rangle $ and $ \vec{b} = \langle b_1, b_2, b_3 \rangle $, then: $ \vec{a} \cdot \vec{b} = a_1b_1 + a_2b_2 + a_3b_3 $

Geometric Interpretation

The dot product is also defined as:

$ \vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}|\cos\theta $

Where:

$ |\vec{a}| $, $ |\vec{b}| $ are the magnitudes of the vectors
$ \theta $ is the angle between the vectors

Properties of Scalar Product

Commutative: $ \vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{a} $
Distributive over addition: $ \vec{a} \cdot (\vec{b} + \vec{c}) = \vec{a} \cdot \vec{b} + \vec{a} \cdot \vec{c} $
Scalar multiplication: $ (k\vec{a}) \cdot \vec{b} = k(\vec{a} \cdot \vec{b}) $
Orthogonality: If $ \vec{a} \cdot \vec{b} = 0 $, then $ \vec{a} $ and $ \vec{b} $ are perpendicular.
Self dot product: $ \vec{a} \cdot \vec{a} = |\vec{a}|^2 $

Applications

Finding the angle between two vectors
Determining orthogonality (perpendicularity)
Work done: $ W = \vec{F} \cdot \vec{d} $

Example:

Let $ \vec{a} = \langle 3, 4 \rangle $, $ \vec{b} = \langle 2, -1 \rangle $.

Find:The scalar product $ \vec{a} \cdot \vec{b} $

▶️ Show Solution

Scalar Product:

$ \vec{a} \cdot \vec{b} = 3 \cdot 2 + 4 \cdot (-1) = 6 – 4 = 2 $

Angle Between Two Vectors

Concept

The angle between two vectors is the measure of the smallest angle $ \theta \in [0^\circ, 180^\circ] $ formed when the vectors are placed tail-to-tail. It tells us how much one vector must be rotated to align with the other.

Formula

If $ \vec{a} $ and $ \vec{b} $ are two vectors, the angle $ \theta $ between them is given by:

$ \cos\theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|} \quad \Rightarrow \quad \theta = \cos^{-1} \left( \frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|} \right) $

Key Requirements

Vectors must be in the same dimension (2D or 3D)
Dot product $ \vec{a} \cdot \vec{b} $ gives scalar projection
Magnitudes $ |\vec{a}| $, $ |\vec{b}| $ are required

Interpretation

If $ \theta = 0^\circ $, vectors point in the same direction.
If $ \theta = 90^\circ $, vectors are perpendicular ($ \vec{a} \cdot \vec{b} = 0 $).
If $ \theta = 180^\circ $, vectors point in opposite directions.

Example:

Let $ \vec{a} = \langle 4, -2 \rangle $, $ \vec{b} = \langle 1, 2 \rangle $. Find the angle between the vectors.

Find the dot product:

▶️Answer/Explanation

Solution:

$ \vec{a} \cdot \vec{b} = 4 \cdot 1 + (-2) \cdot 2 = 4 – 4 = 0 $

Find magnitudes:

$ |\vec{a}| = \sqrt{4^2 + (-2)^2} = \sqrt{16 + 4} = \sqrt{20} $ $ |\vec{b}| = \sqrt{1^2 + 2^2} = \sqrt{1 + 4} = \sqrt{5} $

Apply the formula:

$ \cos\theta = \frac{0}{\sqrt{20} \cdot \sqrt{5}} = 0 \Rightarrow \theta = \cos^{-1}(0) = 90^\circ $

Final Answer: $ \theta = 90^\circ $

Vector Product (Cross Product)

Definition

The vector product, also known as the cross product, is a binary operation on two vectors in three-dimensional space. It results in a vector that is perpendicular to both input vectors.

Algebraic Formula

If $ \vec{a} = \langle a_1, a_2, a_3 \rangle $ and $ \vec{b} = \langle b_1, b_2, b_3 \rangle $, then:
$ \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \end{vmatrix} = \langle a_2b_3 – a_3b_2,\; a_3b_1 – a_1b_3,\; a_1b_2 – a_2b_1 \rangle $

Geometric Interpretation

The magnitude of the cross product is given by:

$ |\vec{a} \times \vec{b}| = |\vec{a}||\vec{b}|\sin\theta $

Where:

$ \theta $ is the angle between the vectors ( $ 0^\circ \leq \theta \leq 180^\circ $ )
Resulting vector is perpendicular to both $ \vec{a} $ and $ \vec{b} $
Direction follows the right-hand rule

Properties of Cross Product

Not commutative: $ \vec{a} \times \vec{b} = -(\vec{b} \times \vec{a}) $
Distributive over addition: $ \vec{a} \times (\vec{b} + \vec{c}) = \vec{a} \times \vec{b} + \vec{a} \times \vec{c} $
Scalar multiplication: $ (k\vec{a}) \times \vec{b} = k(\vec{a} \times \vec{b}) $
Zero vector: If $ \vec{a} $ is parallel to $ \vec{b} $, then $ \vec{a} \times \vec{b} = \vec{0} $
Magnitude represents area: $ |\vec{a} \times \vec{b}| $ is the area of the parallelogram formed by $ \vec{a} $ and $ \vec{b} $

Applications

Finding a vector perpendicular to two given vectors
Computing torque: $ \vec{\tau} = \vec{r} \times \vec{F} $
Finding area of parallelograms and triangles

Example:

Let $ \vec{a} = \langle 1, 2, 3 \rangle $, $ \vec{b} = \langle 4, 5, 6 \rangle $.

Find: The cross product $ \vec{a} \times \vec{b} $

▶️ Show Solution

Cross Product:

$ \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 3 \\ 4 & 5 & 6 \end{vmatrix} = \hat{i}(2 \cdot 6 – 3 \cdot 5) – \hat{j}(1 \cdot 6 – 3 \cdot 4) + \hat{k}(1 \cdot 5 – 2 \cdot 4) $

$ = \hat{i}(12 – 15) – \hat{j}(6 – 12) + \hat{k}(5 – 8) = \langle -3, 6, -3 \rangle $

Example

Let

$\vec{u} = \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix}$ and $\vec{v} = \begin{pmatrix} 4 \\ 5 \\ 6 \end{pmatrix}$.

a) Find $\vec{u} \cdot \vec{v}$.
b) Find $\vec{u} \times \vec{v}$ and $\vec{v} \times \vec{u}$.
c) Verify that $\vec{u} \times \vec{v}$ is perpendicular to both $\vec{u}$ and $\vec{v}$.

▶️Answer/Explanation

Solution:

a) Dot product:
$\vec{u} \cdot \vec{v} = (1)(4) + (2)(5) + (3)(6) = 4 + 10 + 18 = 32$

b) Cross product:
$\vec{u} \times \vec{v} = \begin{pmatrix} (2)(6) – (5)(3) \\ (3)(4) – (6)(1) \\ (1)(5) – (4)(2) \end{pmatrix} = \begin{pmatrix} -3 \\ 6 \\ -3 \end{pmatrix}$
$\vec{v} \times \vec{u} = -\vec{u} \times \vec{v} = \begin{pmatrix} 3 \\ -6 \\ 3 \end{pmatrix}$

c) Perpendicularity check:
$(\vec{u} \times \vec{v}) \cdot \vec{u} = (-3)(1) + (6)(2) + (-3)(3) = -3 + 12 – 9 = 0$
$(\vec{u} \times \vec{v}) \cdot \vec{v} = (-3)(4) + (6)(5) + (-3)(6) = -12 + 30 – 18 = 0$
Hence, $\vec{u} \times \vec{v}$ is perpendicular to both $\vec{u}$ and $\vec{v}$.

◆ THE MAGNITUDE OF $\vec{u} \times \vec{v}$

Geometric Interpretation:

The magnitude of $\vec{u} \times \vec{v}$ equals the area of the parallelogram formed by $\vec{u}$ and $\vec{v}$.
The area of the triangle formed by $\vec{u}$ and $\vec{v}$ is half of this:
$\text{Area of triangle} = \frac{1}{2} |\vec{u} \times \vec{v}|$

From the geometric definition:

$|\vec{u} \times \vec{v}| = |\vec{u}| |\vec{v}| \sin \theta$

Example

For

$\vec{u} = \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix}$ and $\vec{v} = \begin{pmatrix} 4 \\ 5 \\ 6 \end{pmatrix}$, we found $\vec{u} \times \vec{v} = \begin{pmatrix} -3 \\ 6 \\ -3 \end{pmatrix}$.

Area of parallelogram:
Area of triangle:

▶️Answer/Explanation

Solution:

Area of parallelogram:
$|\vec{u} \times \vec{v}| = \sqrt{(-3)^2 + 6^2 + (-3)^2} = \sqrt{54} \approx 7.35$
Area of triangle:
$\frac{1}{2} \times 7.35 \approx 3.67$

Example

Find the area of the triangle determined by points

$A(1, 1, 1)$, $B(1, 3, 1)$, and $C(-3, 3, 4)$.

Also Find

$\overrightarrow{AB}$

▶️Answer/Explanation

Solution:

Vectors:
$\overrightarrow{AB} = \begin{pmatrix} 0 \\ 2 \\ 0 \end{pmatrix}, \quad \overrightarrow{AC} = \begin{pmatrix} -4 \\ 2 \\ 3 \end{pmatrix}$

Cross product:
$\overrightarrow{AB} \times \overrightarrow{AC} = \begin{pmatrix} 6 \\ 0 \\ 8 \end{pmatrix}$

Area of triangle:
$\frac{1}{2} |\overrightarrow{AB} \times \overrightarrow{AC}| = \frac{1}{2} \sqrt{6^2 + 0^2 + 8^2} = \frac{1}{2} \times 10 = 5$

In vector problems, we often break a vector into components along specific directions.
Suppose a force represented by vector a is applied, but we’re only interested in how much of that force acts in the direction of another vector b.

Only part of a contributes in the direction of b — this part is called the component of a in the direction of b.

Component of a in direction of b:

$
\text{component} = \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|} = |\vec{a}| \cos \theta
$

Where:

$\vec{a} \cdot \vec{b}$ is the dot product of vectors a and b
$|\vec{b}|$ is the magnitude of vector b
$\theta$ is the angle between vectors a and b

PERPENDICULAR COMPONENTS

In some situations (especially in Physics), we are interested in how much of vector a acts perpendicular to vector b.

This is called the perpendicular component of a to b, and it’s calculated as:

Perpendicular component of a to b:

$
\text{component}_\perp = \frac{|\vec{a} \times \vec{b}|}{|\vec{b}|} = |\vec{a}| \sin \theta
$

Where:

$\vec{a} \times \vec{b}$ is the magnitude of the cross product
$\theta$ is again the angle between vectors a and b

This tells you how much of vector a is acting in a direction that is orthogonal (at 90°) to b.

Example:

Let:

$ \vec{a} = 3\mathbf{i} + 4\mathbf{j}, \quad \vec{b} = 5\mathbf{i} + 5\mathbf{j} $

Find:

The component of a in the direction of b
The component of a perpendicular to b

▶️Answer/Explanation

Magnitudes

$ |\vec{a}| = \sqrt{3^2 + 4^2} = \sqrt{25} = 5 $
$ |\vec{b}| = \sqrt{5^2 + 5^2} = \sqrt{50} = 5\sqrt{2} $

Dot Product for Parallel Component

$ \vec{a} \cdot \vec{b} = (3)(5) + (4)(5) = 35 $
$ \text{Component of } \vec{a} \text{ in direction of } \vec{b} = \frac{35}{5\sqrt{2}} = \frac{7}{\sqrt{2}} \approx 4.95 $

Parallel component magnitude ≈ 4.95

Cross Product for Perpendicular Component

$ |\vec{a} \times \vec{b}| = |(3)(5) – (4)(5)| = |15 – 20| = 5 $
$ \text{Perpendicular component} = \frac{5}{5\sqrt{2}} = \frac{1}{\sqrt{2}} \approx 0.71 $

Perpendicular component magnitude ≈ 0.71

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IB Mathematics AI AHL Definition and calculation of the product MAI Study Notes- New Syllabus

IB Mathematics AI AHL Definition and calculation of the product MAI Study Notes

SCALAR (DOT) PRODUCT

ANGLE BETWEEN VECTORS

VECTOR CROSS PRODUCT

THE MAGNITUDE OF \(\vec{u} \times \vec{v}\)

COMPONENTS OF VECTORS

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