IB Mathematics AI – Topic 3

Vector Notation:

• Bold lowercase: \(\mathbf{a}\), \(\mathbf{v}\)
• Arrow notation: \(\vec{a}\), \(\overrightarrow{AB}\)
• Column vector: \(\begin{pmatrix} x \\ y \end{pmatrix}\) in 2D, \(\begin{pmatrix} x \\ y \\ z \end{pmatrix}\) in 3D

Position Vectors:

Position vector of point A relative to origin O:

\[ \overrightarrow{OA} = \mathbf{a} = \begin{pmatrix} x \\ y \\ z \end{pmatrix} \]

Displacement Vectors:

Vector from point A to point B:

\[ \overrightarrow{AB} = \mathbf{b} - \mathbf{a} = \begin{pmatrix} x_B - x_A \\ y_B - y_A \\ z_B - z_A \end{pmatrix} \]

Magnitude (Length):

\[ |\mathbf{v}| = \sqrt{x^2 + y^2 + z^2} \]

Unit Vector (Direction):

A vector with magnitude 1 in the direction of \(\mathbf{v}\):

\[ \hat{\mathbf{v}} = \frac{\mathbf{v}}{|\mathbf{v}|} \]

Vector Operations:

Addition: \(\mathbf{a} + \mathbf{b} = \begin{pmatrix} a_1 + b_1 \\ a_2 + b_2 \\ a_3 + b_3 \end{pmatrix}\)

Subtraction: \(\mathbf{a} - \mathbf{b} = \begin{pmatrix} a_1 - b_1 \\ a_2 - b_2 \\ a_3 - b_3 \end{pmatrix}\)

Scalar multiplication: \(k\mathbf{a} = \begin{pmatrix} ka_1 \\ ka_2 \\ ka_3 \end{pmatrix}\)

⚠️ Common Pitfalls & Tips:

• Vectors have both magnitude AND direction (not just coordinates)
• Displacement \(\overrightarrow{AB} = \mathbf{b} - \mathbf{a}\), not \(\mathbf{a} - \mathbf{b}\)
• Don't confuse position vectors with coordinates
• Unit vectors always have magnitude 1

Definition:

The scalar product (dot product) of two vectors produces a scalar (number):

\[ \mathbf{a} \cdot \mathbf{b} = |\mathbf{a}||\mathbf{b}|\cos\theta \]

where θ is the angle between the vectors

Component Form (in 3D):

\[ \mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2 + a_3b_3 \]

Multiply corresponding components and add

Properties:

• Commutative: \(\mathbf{a} \cdot \mathbf{b} = \mathbf{b} \cdot \mathbf{a}\)
• If \(\mathbf{a} \cdot \mathbf{b} = 0\): Vectors are perpendicular
• If \(\mathbf{a} \cdot \mathbf{b} > 0\): Angle is acute (< 90°)
• If \(\mathbf{a} \cdot \mathbf{b} < 0\): Angle is obtuse (> 90°)
• \(\mathbf{a} \cdot \mathbf{a} = |\mathbf{a}|^2\)

⚠️ Common Pitfalls & Tips:

• Dot product gives a NUMBER, not a vector
• Perpendicular vectors have dot product = 0
• Use component form for calculation, then formula for angle
• Don't forget to take cos⁻¹ when finding angle

Formula for Angle Between Two Vectors:

\[ \cos\theta = \frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{a}||\mathbf{b}|} \]

Steps to Find Angle:

1. Calculate scalar product: \(\mathbf{a} \cdot \mathbf{b}\)
2. Calculate magnitudes: \(|\mathbf{a}|\) and \(|\mathbf{b}|\)
3. Substitute into formula: \(\cos\theta = \frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{a}||\mathbf{b}|}\)
4. Find angle: \(\theta = \cos^{-1}\left(\frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{a}||\mathbf{b}|}\right)\)

Angle Between Two Lines:

For lines with direction vectors \(\mathbf{d}_1\) and \(\mathbf{d}_2\):

The acute angle between lines is:

\[ \theta = \cos^{-1}\left(\frac{|\mathbf{d}_1 \cdot \mathbf{d}_2|}{|\mathbf{d}_1||\mathbf{d}_2|}\right) \]

Use absolute value to ensure acute angle

⚠️ Common Pitfalls & Tips:

• Check calculator mode (degrees or radians)
• For acute angle between lines, use absolute value in formula
• Angle between vectors is always 0° ≤ θ ≤ 180°
• Use GDC for calculations to avoid errors

Solution:

Step 1: Calculate scalar product

\[ \mathbf{a} \cdot \mathbf{b} = (2)(1) + (3)(-1) + (1)(2) = 2 - 3 + 2 = 1 \]

Step 2: Calculate magnitude of a

\[ |\mathbf{a}| = \sqrt{2^2 + 3^2 + 1^2} = \sqrt{4 + 9 + 1} = \sqrt{14} \]

Step 3: Calculate magnitude of b

\[ |\mathbf{b}| = \sqrt{1^2 + (-1)^2 + 2^2} = \sqrt{1 + 1 + 4} = \sqrt{6} \]

Step 4: Find angle

\[ \cos\theta = \frac{1}{\sqrt{14} \times \sqrt{6}} = \frac{1}{\sqrt{84}} \]

\[ \theta = \cos^{-1}\left(\frac{1}{\sqrt{84}}\right) \approx 83.7° \]

Answer: θ = 83.7° (or 1.46 radians)

Definition:

The vector product (cross product) of two vectors produces a VECTOR perpendicular to both:

\[ \mathbf{a} \times \mathbf{b} = \begin{pmatrix} a_2b_3 - a_3b_2 \\ a_3b_1 - a_1b_3 \\ a_1b_2 - a_2b_1 \end{pmatrix} \]

In IB formula booklet

Magnitude:

\[ |\mathbf{a} \times \mathbf{b}| = |\mathbf{a}||\mathbf{b}|\sin\theta \]

Geometric Interpretation:

\(|\mathbf{a} \times \mathbf{b}|\) equals the area of parallelogram formed by \(\mathbf{a}\) and \(\mathbf{b}\)

\[ \text{Area of parallelogram} = |\mathbf{a} \times \mathbf{b}| \]

\[ \text{Area of triangle} = \frac{1}{2}|\mathbf{a} \times \mathbf{b}| \]

Properties:

• NOT commutative: \(\mathbf{a} \times \mathbf{b} = -(\mathbf{b} \times \mathbf{a})\)
• If parallel: \(\mathbf{a} \times \mathbf{b} = \mathbf{0}\)
• Result is perpendicular to both vectors
• Use right-hand rule for direction

⚠️ Common Pitfalls & Tips:

• Cross product gives a VECTOR, not a number
• Order matters: \(\mathbf{a} \times \mathbf{b} \neq \mathbf{b} \times \mathbf{a}\)
• Parallel vectors have cross product = zero vector
• Use formula booklet - easy to make sign errors

Vector Equation:

\[ \mathbf{r} = \mathbf{a} + \lambda\mathbf{b} \]

where:

• \(\mathbf{r}\): Position vector of general point on line
• \(\mathbf{a}\): Position vector of known point on line
• \(\mathbf{b}\): Direction vector (parallel to line)
• \(\lambda\): Parameter (scalar, can be any real number)

Component Form (in 3D):

\[ \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} a_1 \\ a_2 \\ a_3 \end{pmatrix} + \lambda\begin{pmatrix} b_1 \\ b_2 \\ b_3 \end{pmatrix} \]

Cartesian Form:

From vector equation, eliminate λ:

\[ \frac{x - a_1}{b_1} = \frac{y - a_2}{b_2} = \frac{z - a_3}{b_3} \]

Finding Line Through Two Points:

Given points A and B:

1. Choose one point as \(\mathbf{a}\) (usually A)
2. Direction vector: \(\mathbf{b} = \overrightarrow{AB} = \mathbf{b} - \mathbf{a}\)
3. Write equation: \(\mathbf{r} = \mathbf{a} + \lambda(\mathbf{b} - \mathbf{a})\)

⚠️ Common Pitfalls & Tips:

• Any scalar multiple of direction vector works (e.g., 2b, -b)
• Different forms of same line can look different
• λ can take ANY real value (not just 0 ≤ λ ≤ 1)
• Check point is actually on line by substituting

Solution:

Step 1: Write position vectors

\[ \mathbf{a} = \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix}, \quad \mathbf{b} = \begin{pmatrix} 4 \\ 0 \\ -1 \end{pmatrix} \]

Step 2: Find direction vector

\[ \overrightarrow{AB} = \mathbf{b} - \mathbf{a} = \begin{pmatrix} 4-1 \\ 0-2 \\ -1-3 \end{pmatrix} = \begin{pmatrix} 3 \\ -2 \\ -4 \end{pmatrix} \]

Step 3: Write vector equation

Using point A and direction vector:

\[ \mathbf{r} = \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix} + \lambda\begin{pmatrix} 3 \\ -2 \\ -4 \end{pmatrix} \]

Alternative form (using point B):

\[ \mathbf{r} = \begin{pmatrix} 4 \\ 0 \\ -1 \end{pmatrix} + \mu\begin{pmatrix} 3 \\ -2 \\ -4 \end{pmatrix} \]

Answer: \(\mathbf{r} = \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix} + \lambda\begin{pmatrix} 3 \\ -2 \\ -4 \end{pmatrix}\)

Key Quantities:

1. Position Vector:

\(\mathbf{r}(t)\) - location of object at time t

2. Displacement:

Change in position:

\[ \Delta\mathbf{r} = \mathbf{r}(t_2) - \mathbf{r}(t_1) \]

3. Velocity Vector:

Rate of change of position:

\[ \mathbf{v}(t) = \frac{d\mathbf{r}}{dt} \]

Speed: \(|\mathbf{v}(t)|\) (magnitude of velocity)

4. Acceleration Vector:

Rate of change of velocity:

\[ \mathbf{a}(t) = \frac{d\mathbf{v}}{dt} = \frac{d^2\mathbf{r}}{dt^2} \]

Constant Velocity Motion:

\[ \mathbf{r}(t) = \mathbf{r}_0 + t\mathbf{v} \]

where \(\mathbf{r}_0\) is initial position, \(\mathbf{v}\) is constant velocity

Distance Travelled:

\[ \text{Distance} = \int_{t_1}^{t_2} |\mathbf{v}(t)| \, dt \]

⚠️ Common Pitfalls & Tips:

• Displacement is a vector; distance is a scalar
• Speed = |velocity|, not just velocity
• Differentiate position to get velocity, then again for acceleration
• Units: check if time is in seconds, minutes, etc.

Key Formulas

• Magnitude: \(|\mathbf{v}| = \sqrt{x^2+y^2+z^2}\)
• Dot: \(\mathbf{a} \cdot \mathbf{b} = a_1b_1+a_2b_2+a_3b_3\)
• Angle: \(\cos\theta = \frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{a}||\mathbf{b}|}\)

Line Equation

• \(\mathbf{r} = \mathbf{a} + \lambda\mathbf{b}\)
• a = point on line
• b = direction vector

Products

• Dot product → scalar
• Cross product → vector
• Perpendicular: dot = 0

Kinematics

• \(\mathbf{v} = \frac{d\mathbf{r}}{dt}\)
• \(\mathbf{a} = \frac{d\mathbf{v}}{dt}\)
• Speed = \(|\mathbf{v}|\)