IB Mathematics AI AHL Concept of a vector MAI Study Notes- New Syllabus
IB Mathematics AI AHL Concept of a vector MAI Study Notes
LEARNING OBJECTIVE
- Concept of a vector and a scalar.
Key Concepts:
- Introduction to Vectors
- Position & Displacement Vectors
- Magnitude of a Vector
- IBDP Maths AI SL- IB Style Practice Questions with Answer-Topic Wise-Paper 1
- IBDP Maths AI SL- IB Style Practice Questions with Answer-Topic Wise-Paper 2
- IB DP Maths AI HL- IB Style Practice Questions with Answer-Topic Wise-Paper 1
- IB DP Maths AI HL- IB Style Practice Questions with Answer-Topic Wise-Paper 2
- IB DP Maths AI HL- IB Style Practice Questions with Answer-Topic Wise-Paper 3
Scalars and Vectors
◆ DEFINITION
We distinguish two kinds of quantities in nature:
SCALARS:
Have magnitude only. Examples: age = 28, length = 4 m, temperature = 25°C.
VECTORS:
Have magnitude and direction. Examples: force = 7 N, velocity = 35 m/s towards southeast.
Geometrically, a vector is represented by an arrow and denoted by:
- Single letter: \(\vec{u}\)
- Two letters: \(\overrightarrow{AB}\) (A = tail, B = head)
The length of the arrow gives the vector’s magnitude, denoted by \(|\vec{u}|\) or \(|\overrightarrow{AB}|\).
Example Identify which of the following are scalar or vector quantities:
▶️ Answer/ExplanationSolution:
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Representation of Vectors
◆ DEFINITION
A vector is represented geometrically by a directed line segment — an arrow — showing both magnitude and direction.
Let \(\overrightarrow{AB}\) represent a vector from point \(A\) to point \(B\):
- Tail at point \(A\), Head at point \(B\).
- The length of the arrow represents the vector’s magnitude.
- The arrowhead shows the direction.
You can move a vector around without changing it (as long as direction and magnitude are preserved). Vectors are said to be free vectors.
Equal vectors have the same magnitude and direction, regardless of position
Example Draw a vector \(\vec{AB}\) from point \(A(2, 1)\) to point \(B(5, 4)\). ▶️ Answer/ExplanationSolution:
Vector \(\vec{AB} = \begin{pmatrix} 5 – 2 \\ 4 – 1 \end{pmatrix} = \begin{pmatrix} 3 \\ 3 \end{pmatrix}\) |
Unit Vectors and Base Vectors
◆ DEFINITION
A unit vector is a vector with magnitude 1, used to indicate direction.
In Cartesian coordinates, we use the standard unit vectors:
- \(\hat{i} = \begin{pmatrix} 1 \\ 0 \end{pmatrix}\) → unit vector in the \(x\)-direction
- \(\hat{j} = \begin{pmatrix} 0 \\ 1 \end{pmatrix}\) → unit vector in the \(y\)-direction
- \(\hat{k} = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}\) → unit vector in the \(z\)-direction (3D only)
Any vector can be expressed as a combination of unit vectors:
\(\vec{v} = a\hat{i} + b\hat{j}\) in 2D, or \(\vec{v} = a\hat{i} + b\hat{j} + c\hat{k}\) in 3D.
Unit vectors are useful for breaking a vector into components and for vector projections.
Example Write the vector \(\vec{v} = \begin{pmatrix} 4 \\ -2 \\ 7 \end{pmatrix}\) in terms of \(i, j, k\). ▶️ Answer/ExplanationSolution: \(\vec{v} = 4\mathbf{i} – 2\mathbf{j} + 7\mathbf{k}\) |
Column Vector Representation
◆ DEFINITION
A vector can be written in column form as a matrix with one column and multiple rows, showing its components vertically.
For example:
- 2D vector: \(\vec{v} = \begin{pmatrix} x \\ y \end{pmatrix}\)
- 3D vector: \(\vec{v} = \begin{pmatrix} x \\ y \\ z \end{pmatrix}\)
Column vectors are useful for calculations in linear algebra, vector geometry, and matrix transformations.
The transition between unit vector form and column form is straightforward:
\(\vec{v} = 3\hat{i} + 2\hat{j} \quad \longleftrightarrow \quad \vec{v} = \begin{pmatrix} 3 \\ 2 \end{pmatrix}\)
Example Express \(\vec{v} = 2\mathbf{i} + 5\mathbf{j} – 3\mathbf{k}\) as a column vector. ▶️ Answer/ExplanationSolution: \(\vec{v} = \begin{pmatrix} 2 \\ 5 \\ -3 \end{pmatrix}\) |
The Zero Vector and Negative Vectors
◆ THE ZERO VECTOR
The zero vector, denoted \(\vec{0}\), has all components equal to zero and no direction.
Examples:
\(\vec{0} = \begin{pmatrix} 0 \\ 0 \end{pmatrix} \) in 2D,
\(\vec{0} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} \) in 3D.
Properties:
- Adding the zero vector doesn’t change a vector: \(\vec{v} + \vec{0} = \vec{v}\)
- The zero vector is the result of subtracting a vector from itself: \(\vec{v} – \vec{v} = \vec{0}\)
◆ NEGATIVE VECTORS
The negative of a vector \(-\vec{v}\) has the same magnitude as \(\vec{v}\), but opposite direction.
If \(\vec{v} = \begin{pmatrix} a \\ b \end{pmatrix}\), then \(-\vec{v} = \begin{pmatrix} -a \\ -b \end{pmatrix}\)
Adding a vector and its negative gives the zero vector:
\(\vec{v} + (-\vec{v}) = \vec{0}\)
Example If \(\vec{a} = \begin{pmatrix} 4 \\ -2 \end{pmatrix}\), find \(-\vec{a}\) and \(\vec{a} + (-\vec{a})\). ▶️ Answer/ExplanationSolution:
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Vector Operations
◆ ADDITION & SUBTRACTION
Vectors can be added or subtracted by combining their corresponding components.
If \(\vec{a} = \begin{pmatrix} a_1 \\ a_2 \end{pmatrix}\) and \(\vec{b} = \begin{pmatrix} b_1 \\ b_2 \end{pmatrix}\), then:
- \(\vec{a} + \vec{b} = \begin{pmatrix} a_1 + b_1 \\ a_2 + b_2 \end{pmatrix}\)
- \(\vec{a} – \vec{b} = \begin{pmatrix} a_1 – b_1 \\ a_2 – b_2 \end{pmatrix}\)
Addition can also be visualized geometrically using the tip-to-tail rule.
◆ SCALAR MULTIPLICATION
Multiplying a vector by a scalar stretches or shrinks it in the same or opposite direction:
If \(\vec{v} = \begin{pmatrix} x \\ y \end{pmatrix}\), then \(k\vec{v} = \begin{pmatrix} kx \\ ky \end{pmatrix}\)
Scalar multiplication preserves direction if \(k > 0\) and reverses direction if \(k < 0\).
Effects of Scalar Multiplication:
Stretching: If $|k| > 1$, the vector becomes longer.
Shrinking: If $0 < |k| < 1$, the vector becomes shorter.
Rescaling: Scalar multiplication changes the length of the vector. This process is called rescaling.
Direction preserved: If $k > 0$, the direction of the vector stays the same.
Direction reversed: If $k < 0$, the direction of the vector is reversed.
Example Given \(\vec{a} = \begin{pmatrix} 2 \\ 3 \end{pmatrix}\) and \(\vec{b} = \begin{pmatrix} -1 \\ 4 \end{pmatrix}\), find \(\vec{a} + \vec{b}\) and \(2\vec{b}\). ▶️ Answer/ExplanationSolution:
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Vector Magnitude
◆ DEFINITION
The magnitude (or length) of a vector \(\vec{v} = \begin{pmatrix} x \\ y \end{pmatrix}\) is the distance from the origin to the point \((x, y)\).
It is denoted by \(|\vec{v}|\) and is found using the Pythagorean theorem:
\(|\vec{v}| = \sqrt{x^2 + y^2}\) (in 2D)
\(|\vec{v}| = \sqrt{x^2 + y^2 + z^2}\) (in 3D)
The magnitude is always a non-negative scalar and represents the length of the vector.
Example Find the magnitude of \(\vec{v} = \begin{pmatrix} 5 \\ 12 \end{pmatrix}\) ▶️ Answer/ExplanationSolution: \(|\vec{v}| = \sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13\) |
Position Vectors
◆ DEFINITION
A position vector locates a point in space relative to the origin.
If point \(A\) has coordinates \((x, y)\), then the position vector of \(A\) is:
\(\vec{OA} = \begin{pmatrix} x \\ y \end{pmatrix}\)
Similarly, for 3D: If \(A = (x, y, z)\), then \(\vec{OA} = \begin{pmatrix} x \\ y \\ z \end{pmatrix}\)
To find the vector from point \(A\) to point \(B\), use:
\(\vec{AB} = \vec{OB} – \vec{OA}\)
Example Find the position vector of point \(P(7, -2)\) from the origin. ▶️ Answer/ExplanationSolution: \(\vec{OP} = \begin{pmatrix} 7 \\ -2 \end{pmatrix}\) |
Normalizing Vectors
◆ DEFINITION
To normalize a vector means to convert it into a unit vector (a vector of length 1) that points in the same direction.
Given a non-zero vector \(\vec{v}\), the unit vector in its direction is:
\(\hat{v} = \dfrac{\vec{v}}{|\vec{v}|}\)
This process is called normalization. It is useful when only the direction of a vector is needed, not its magnitude.
Example Let \(\vec{v} = \begin{pmatrix} 6 \\ 8 \end{pmatrix}\). Find the unit vector in the same direction. ▶️ Answer/ExplanationSolution:
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Velocity Vectors (Applied Context)
◆ APPLICATION
A velocity vector shows both the speed and direction of an object’s motion.
If an object moves from point \(A(x_1, y_1)\) to point \(B(x_2, y_2)\) in time \(t\), then its velocity vector is:
\(\vec{v} = \dfrac{\vec{AB}}{t} = \dfrac{1}{t} \begin{pmatrix} x_2 – x_1 \\ y_2 – y_1 \end{pmatrix}\)
This represents displacement per unit time (vector form of velocity).
Example A car moves from point \(A(1, 3)\) to point \(B(9, -1)\) in 4 seconds. Find its velocity vector and speed. ▶️ Answer/ExplanationSolution:
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| 2D | 3D | |
|---|---|---|
| Points | \( A(x_1, y_1),\ B(x_2, y_2) \) | \( A(x_1, y_1, z_1),\ B(x_2, y_2, z_2) \) |
| Mid-point | \( M = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) \) | \( M = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}, \frac{z_1 + z_2}{2} \right) \) |
| Position vectors of A and B | \( \vec{OA} = \begin{pmatrix} x_1 \\ y_1 \end{pmatrix},\ \vec{OB} = \begin{pmatrix} x_2 \\ y_2 \end{pmatrix} \) | \( \vec{OA} = \begin{pmatrix} x_1 \\ y_1 \\ z_1 \end{pmatrix},\ \vec{OB} = \begin{pmatrix} x_2 \\ y_2 \\ z_2 \end{pmatrix} \) |
| Vector \( \vec{AB} \) | \( \vec{AB} = \vec{OB} – \vec{OA} = \begin{pmatrix} x_2 – x_1 \\ y_2 – y_1 \end{pmatrix} \) | \( \vec{AB} = \vec{OB} – \vec{OA} = \begin{pmatrix} x_2 – x_1 \\ y_2 – y_1 \\ z_2 – z_1 \end{pmatrix} \) |
| Distance (A, B) Magnitude \( |\vec{AB}| \) | \( \sqrt{(x_2 – x_1)^2 + (y_2 – y_1)^2} \) | \( \sqrt{(x_2 – x_1)^2 + (y_2 – y_1)^2 + (z_2 – z_1)^2} \) |
