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.

MAI HL and SL Notes – All topics

◆ GEOMETRIC DEFINITION

For vectors $\vec{u}$, $\vec{v}$ with angle $\theta$ between them:

$
\vec{u} \cdot \vec{v} = |\vec{u}||\vec{v}|\cos\theta.
$

If $|\vec{u}| = 5$, $|\vec{v}| = 4$, $\theta = 60^\circ$, then $\vec{u} \cdot \vec{v} = 10$.

Special Cases:

$\theta = 90^\circ \Rightarrow \vec{u} \cdot \vec{v} = 0$ (perpendicular).
$\theta = 0^\circ \Rightarrow \vec{u} \cdot \vec{v} = |\vec{u}||\vec{v}|$ (parallel, same direction).
$\theta = 180^\circ \Rightarrow \vec{u} \cdot \vec{v} = -|\vec{u}||\vec{v}|$ (parallel, opposite direction).

◆ ALGEBRAIC DEFINITION

For $\vec{u} = \begin{pmatrix} a_1 \\ b_1 \end{pmatrix}$, $\vec{v} = \begin{pmatrix} a_2 \\ b_2 \end{pmatrix}$:

$\vec{u} \cdot \vec{v} = a_1a_2 + b_1b_2.$

Example

For

$\begin{pmatrix} 2 \\ 3 \end{pmatrix} \cdot \begin{pmatrix} 5 \\ 4 \end{pmatrix} =?$.

Find Dot Product of given vectors.

▶️Answer/Explanation

Solution:

$\begin{pmatrix} 2 \\ 3 \end{pmatrix} \cdot \begin{pmatrix} 5 \\ 4 \end{pmatrix} = 22$.

◆ ANGLE BETWEEN VECTORS

$\cos\theta = \frac{\vec{u} \cdot \vec{v}}{|\vec{u}||\vec{v}|}.$

Example

For

$\vec{u} = \begin{pmatrix} 3 \\ 4 \end{pmatrix}$, $\vec{v} = \begin{pmatrix} 1 \\ -2 \end{pmatrix}$

Find Angle between given vectors.

▶️Answer/Explanation

Solution:

$\cos\theta = \frac{-5}{5\sqrt{5}} = -\frac{1}{\sqrt{5}} \Rightarrow \theta \approx 116.6^\circ.$

◆ PERPENDICULAR AND PARALLEL VECTORS

Perpendicular: $\vec{u} \cdot \vec{v} = 0$.
Parallel: $\vec{u} = k\vec{v}$ for some scalar $k$.

Example

Find $x$ such that $\vec{v} = \begin{pmatrix} x \\ -6 \end{pmatrix}$ is

1. Perpendicular to $\vec{u} = $

2. Parallel to $\vec{u}$:

▶️Answer/Explanation

Solution:

1. Perpendicular to $\vec{u} = \begin{pmatrix} 3 \\ 4 \end{pmatrix}$:
$3x – 24 = 0 \Rightarrow x = 8.$
2. Parallel to $\vec{u}$:
$\frac{x}{3} = \frac{-6}{4} \Rightarrow x = -\frac{9}{2}.$

◆ PROPERTY

$|\vec{u}|^2 = \vec{u}^2$
$|\vec{u} + \vec{v}| = |\vec{u} – \vec{v}| \Rightarrow \vec{u} \cdot \vec{v} = 0 \quad (\text{perpendicular vectors}).$

◆ 3D VECTORS

For $\vec{u} = \begin{pmatrix} a_1 \\ b_1 \\ c_1 \end{pmatrix}$, $\vec{v} = \begin{pmatrix} a_2 \\ b_2 \\ c_2 \end{pmatrix}$:
$\vec{u} \cdot \vec{v} = a_1a_2 + b_1b_2 + c_1c_2.$

Example: $\begin{pmatrix} 4 \\ 2 \\ -1 \end{pmatrix} \cdot \begin{pmatrix} 5 \\ -3 \\ 14 \end{pmatrix} = 0$ (perpendicular).

◆ THE GEOMETRIC DEFINITION

Let $\vec{u}$ and $\vec{v}$ be two vectors, and let $\theta$ be the angle between them (where $0 \leq \theta \leq \pi$).

The cross product (or vector product) of $\vec{u}$ and $\vec{v}$ is defined as a vector given by:

$\vec{u} \times \vec{v} = (|\vec{u}| |\vec{v}| \sin \theta) \vec{n}$
where $\vec{n}$ is the unit vector perpendicular to both $\vec{u}$ and $\vec{v}$ and follows the “right-hand rule” (or “screw rule”):

If you place a screw at the common starting point of $\vec{u}$ and $\vec{v}$ and rotate it from $\vec{u}$ to $\vec{v}$, the screw will move in the direction of $\vec{n}$.

Thus, $\vec{u} \times \vec{v}$ is a new vector:

Perpendicular to both $\vec{u}$ and $\vec{v}$ (and hence to the plane they determine).
With magnitude $|\vec{u}| |\vec{v}| \sin \theta$.
With direction given by $\vec{n}$.

Note:

The commutative law does not hold for the cross product. Instead:

$\vec{u} \times \vec{v} = -\vec{v} \times \vec{u}$
If $\vec{u}$ and $\vec{v}$ are parallel ($\theta = 0$ or $\pi$), then $\sin \theta = 0$, and $\vec{u} \times \vec{v} = \vec{0}$.

◆ THE ALGEBRAIC DEFINITION

Let $\vec{u} = \begin{pmatrix} a_1 \\ b_1 \\ c_1 \end{pmatrix}$ and $\vec{v} = \begin{pmatrix} a_2 \\ b_2 \\ c_2 \end{pmatrix}$ be two vectors. The cross product is given by:

$\vec{u} \times \vec{v} = \begin{pmatrix} b_1 c_2 – b_2 c_1 \\ c_1 a_2 – c_2 a_1 \\ a_1 b_2 – a_2 b_1 \end{pmatrix}$

Memory Aid:

For the first component, ignore the first row of $\vec{u}$ and $\vec{v}$ and compute:

$\begin{vmatrix} b_1 & c_1 \\ b_2 & c_2 \end{vmatrix} = b_1 c_2 – b_2 c_1$

For the second component, ignore the second row and compute:

$\begin{vmatrix} c_1 & a_1 \\ c_2 & a_2 \end{vmatrix} = c_1 a_2 – c_2 a_1$

For the third component, ignore the third row and compute:

$\begin{vmatrix} a_1 & b_1 \\ a_2 & b_2 \end{vmatrix} = a_1 b_2 – a_2 b_1$

Alternative Notation (Determinant Form):

$\vec{u} \times \vec{v} = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \end{vmatrix}$

where

$\vec{i} = \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}$, $\vec{j} = \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix}$, $\vec{k} = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}$.

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 $\vec{v}$.

Example

Let

$\vec{u} = \begin{pmatrix} 3 \\ 2 \\ 0 \end{pmatrix}$ and $\vec{v} = \begin{pmatrix} 1 \\ 4 \\ 0 \end{pmatrix}$.

a) Find $\vec{u} \times \vec{v}$ using the algebraic definition.
b) Find the angle $\theta$ between $\vec{u}$ and $\vec{v}$.
c) Find the unit vector $\vec{n}$.
d) Find $\vec{u} \times \vec{v}$ using the geometric definition.
e) Verify perpendicularity.

▶️Answer/Explanation

Solution:

a) Cross product:
$\vec{u} \times \vec{v} = \begin{pmatrix} (2)(0) – (4)(0) \\ (0)(1) – (0)(3) \\ (3)(4) – (1)(2) \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 10 \end{pmatrix}$

b) Angle $\theta$:
$\cos \theta = \frac{\vec{u} \cdot \vec{v}}{|\vec{u}| |\vec{v}|} = \frac{11}{\sqrt{13} \sqrt{17}} \approx 0.74 \implies \theta \approx 42.27^\circ$

c) Unit vector $\vec{n}$:
Since $\vec{u}$ and $\vec{v}$ lie in the $xy$-plane, $\vec{n}$ is parallel to the $z$-axis. From the cross product result:
$\vec{n} = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}$

d) Geometric definition:
$\vec{u} \times \vec{v} = (|\vec{u}| |\vec{v}| \sin \theta) \vec{n} = (\sqrt{13} \sqrt{17} \sin 42.27^\circ) \vec{n} \approx 10 \vec{n} = \begin{pmatrix} 0 \\ 0 \\ 10 \end{pmatrix}$

e) Perpendicularity:
Clearly, $\vec{u} \times \vec{v}$ is parallel to $\vec{n}$ and thus perpendicular to both $\vec{u}$ and $\vec{v}$.

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

From the geometric definition:

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

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:
$\text{Area of triangle} = \frac{1}{2} |\vec{u} \times \vec{v}|$

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 \(\vec{u} \times \vec{v}\)

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