🏠 Home › 📚 CSEC Mathematics › → Position Vectors and Notation

Position Vectors and Notation

CSEC Mathematics: The Language of Movement

Essential Understanding: Vectors are the mathematical language of direction and movement. While points tell you "where," vectors tell you "how to get there." Master position and displacement vectors to solve problems involving movement, forces, and geometry in the CSEC exam.

🔑 Key Skill: Column Vector Notation
📈 Exam Focus: \(\vec{AB} = \vec{OB} - \vec{OA}\)
🎯 Problem Solving: Parallel Vectors & Ratio Divisions

1. Points vs. Position Vectors

The foundation of this topic is understanding the shift from a static point \(P(x, y)\) to a Position Vector \(\vec{OP}\).

O

The Origin (\(O\))

In CSEC, position vectors are always relative to the fixed reference point \((0, 0)\).

Key Concept: The position vector \(\vec{OA}\) is an instruction: "Start at the origin and move \(x\) units horizontally and \(y\) units vertically to reach point \(A\)."

⇅

Column Vector Notation

While a point is written horizontally as \((x, y)\), a vector is written as a column matrix:

\[ \vec{OA} = \begin{pmatrix} x \\ y \end{pmatrix} \]

CSEC Requirement: Always use column notation for vectors. \((x, y)\) is for points only!

↔

Visual Comparison

Point \(A\): \(A(3, 2)\) - A location on the coordinate plane.

Position Vector \(\vec{OA}\): \(\begin{pmatrix} 3 \\ 2 \end{pmatrix}\) - Instructions from origin to \(A\).

Same numbers, different meaning!

2. Displacement Vectors (\(\vec{AB}\))

A core CSEC objective is calculating the vector between two points that do not start at the origin.

The Triangle Law

To get from \(A\) to \(B\), you effectively go "backwards" to the origin (\(-\vec{OA}\)) and then "forwards" to \(B\) (\(+\vec{OB}\)).

\[ \vec{AB} = \vec{OB} - \vec{OA} \]

Memory Trick: It is always Destination minus Start.

1
Identify position vectors: Find \(\vec{OA}\) and \(\vec{OB}\) from the coordinates of \(A\) and \(B\).
2
Apply subtraction rule: \(\vec{AB} = \vec{OB} - \vec{OA}\).
3
Subtract component-wise: Subtract x-coordinates, then y-coordinates.
4
Write in column form: Ensure your answer is in \(\begin{pmatrix} x \\ y \end{pmatrix}\) format.

3. Direction and Scalar Multiples

Vectors aren't just about location; they define a specific direction.

↔

Reverse Direction

If \(\vec{AB} = \begin{pmatrix} 4 \\ -2 \end{pmatrix}\), then \(\vec{BA} = \begin{pmatrix} -4 \\ 2 \end{pmatrix}\).

The signs flip because the direction is reversed.

\[ \vec{BA} = -\vec{AB} \]
∥

Parallel Vectors

Two vectors are parallel if one is a scalar multiple of the other.

Example: \(\mathbf{u} = \begin{pmatrix} 2 \\ 3 \end{pmatrix}\) and \(\mathbf{v} = \begin{pmatrix} 4 \\ 6 \end{pmatrix}\) are parallel because \(\mathbf{v} = 2\mathbf{u}\).

Check for parallelism: \(\frac{4}{2} = \frac{6}{3} = 2\) ✓
λ

Scalar Multiplication

Multiplying a vector by a scalar changes its magnitude but not its direction (unless the scalar is negative, which reverses direction).

\(k\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} kx \\ ky \end{pmatrix}\)

4. Interactive "Vector Construction" Lab

🔧

Build Vectors Visually

Objective: Drag points A and B on the grid. Watch how the position vectors \(\vec{OA}\), \(\vec{OB}\) and displacement vector \(\vec{AB}\) change in real-time.

\(\vec{OA}\)

\(\begin{pmatrix} 0 \\ 0 \end{pmatrix}\)

Position vector of A

\(\vec{OB}\)

\(\begin{pmatrix} 0 \\ 0 \end{pmatrix}\)

Position vector of B

\(\vec{AB}\)

\(\begin{pmatrix} 0 \\ 0 \end{pmatrix}\)

Displacement from A to B

Calculation

\(\vec{AB} = \vec{OB} - \vec{OA} = \begin{pmatrix} 0 \\ 0 \end{pmatrix} - \begin{pmatrix} 0 \\ 0 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}\)

5. Midpoints and Ratio Divisions

Advanced vector problems involve finding a specific point along a line segment using ratio notation.

The Midpoint Formula

The position vector of the midpoint \(M\) is the average of the two endpoints:

\[ \vec{OM} = \frac{1}{2}(\vec{OA} + \vec{OB}) \]

Example: If \(A(2, 4)\) and \(B(6, 8)\), then:

\(\vec{OM} = \frac{1}{2}\left(\begin{pmatrix} 2 \\ 4 \end{pmatrix} + \begin{pmatrix} 6 \\ 8 \end{pmatrix}\right) = \frac{1}{2}\begin{pmatrix} 8 \\ 12 \end{pmatrix} = \begin{pmatrix} 4 \\ 6 \end{pmatrix}\)

So \(M\) is at \((4, 6)\).

Sectional Ratio Division

If a point \(P\) divides the line \(AB\) in the ratio \(m:n\), then:

\[ \vec{OP} = \frac{n\vec{OA} + m\vec{OB}}{m+n} \]

Memory Trick: "The other part times the first point, plus this part times the second point, all over the total."

Example: If \(P\) divides \(AB\) in ratio \(2:3\), then \(m=2\), \(n=3\).

6. CSEC Exam Mastery Tips

⚠️

Avoid These Common Mistakes

Avoid Coordinate Confusion

  • Never write a vector as \((x, y)\) in your final answer.
  • CSEC examiners expect the \(\begin{pmatrix} x \\ y \end{pmatrix}\) format.
  • Points use parentheses: \(A(3, 4)\). Vectors use column notation: \(\vec{OA} = \begin{pmatrix} 3 \\ 4 \end{pmatrix}\).

The "Head-to-Tail" Rule

  • When adding vectors visually, place the start of the second vector at the end of the first.
  • This is crucial for understanding vector addition geometrically.
  • The resultant vector goes from the start of the first to the end of the last.

Labeling Vectors

  • Always draw the arrow on your vector lines in diagrams to show the intended direction of travel.
  • Label vectors clearly: \(\vec{AB}\), not just \(AB\).
  • In complex diagrams, use different colors for different vectors.

7. Worked Example: The Path to \(B\)

1

Problem: A point \(A\) has coordinates \((3, -2)\) and point \(B\) has coordinates \((-1, 4)\). Write the position vectors \(\vec{OA}\) and \(\vec{OB}\), and find the displacement vector \(\vec{AB}\).

2

Position Vectors:

\[ \vec{OA} = \begin{pmatrix} 3 \\ -2 \end{pmatrix} \]

\[ \vec{OB} = \begin{pmatrix} -1 \\ 4 \end{pmatrix} \]

3

Displacement Formula:

\[ \vec{AB} = \vec{OB} - \vec{OA} \]

4

Subtraction:

\[ \vec{AB} = \begin{pmatrix} -1 \\ 4 \end{pmatrix} - \begin{pmatrix} 3 \\ -2 \end{pmatrix} \]

5

Result:

\[ \vec{AB} = \begin{pmatrix} -1 - 3 \\ 4 - (-2) \end{pmatrix} = \begin{pmatrix} -4 \\ 6 \end{pmatrix} \]

Final Answer: \(\vec{AB} = \begin{pmatrix} -4 \\ 6 \end{pmatrix}\)

8. Practice Mission: "Vector Navigation"

1
Given points \(P(2, 5)\), \(Q(6, 1)\), \(R(1, 3)\), and \(S(5, -1)\). Find the displacement vectors \(\vec{PQ}\) and \(\vec{RS}\). Are \(\vec{PQ}\) and \(\vec{RS}\) parallel?
\(\vec{PQ} = \begin{pmatrix} 4 \\ -4 \end{pmatrix}\), \(\vec{RS} = \begin{pmatrix} 4 \\ -4 \end{pmatrix}\), Yes
\(\vec{PQ} = \begin{pmatrix} 4 \\ 4 \end{pmatrix}\), \(\vec{RS} = \begin{pmatrix} 4 \\ -4 \end{pmatrix}\), No
\(\vec{PQ} = \begin{pmatrix} 4 \\ -4 \end{pmatrix}\), \(\vec{RS} = \begin{pmatrix} 4 \\ -4 \end{pmatrix}\), Yes
\(\vec{PQ} = \begin{pmatrix} -4 \\ 4 \end{pmatrix}\), \(\vec{RS} = \begin{pmatrix} -4 \\ 4 \end{pmatrix}\), Yes
Solution:
1. \(\vec{PQ} = \vec{OQ} - \vec{OP} = \begin{pmatrix} 6 \\ 1 \end{pmatrix} - \begin{pmatrix} 2 \\ 5 \end{pmatrix} = \begin{pmatrix} 4 \\ -4 \end{pmatrix}\)
2. \(\vec{RS} = \vec{OS} - \vec{OR} = \begin{pmatrix} 5 \\ -1 \end{pmatrix} - \begin{pmatrix} 1 \\ 3 \end{pmatrix} = \begin{pmatrix} 4 \\ -4 \end{pmatrix}\)
3. Since \(\vec{PQ} = \vec{RS}\), they are clearly parallel (in fact, they are equal).
4. Even if they weren't equal, they would be parallel if one was a scalar multiple of the other.
2
Point \(M\) is the midpoint of line segment \(AB\), where \(A(1, 4)\) and \(B(7, 10)\). Find the position vector of \(M\).
\(\begin{pmatrix} 3 \\ 5 \end{pmatrix}\)
\(\begin{pmatrix} 4 \\ 7 \end{pmatrix}\)
\(\begin{pmatrix} 4 \\ 7 \end{pmatrix}\)
\(\begin{pmatrix} 8 \\ 14 \end{pmatrix}\)
Solution:
\[ \vec{OM} = \frac{1}{2}(\vec{OA} + \vec{OB}) = \frac{1}{2}\left(\begin{pmatrix} 1 \\ 4 \end{pmatrix} + \begin{pmatrix} 7 \\ 10 \end{pmatrix}\right) \]
\[ = \frac{1}{2}\begin{pmatrix} 8 \\ 14 \end{pmatrix} = \begin{pmatrix} 4 \\ 7 \end{pmatrix} \]
So \(M\) has coordinates \((4, 7)\).
3
Point \(P\) divides line segment \(AB\) in the ratio \(2:3\), where \(A(1, 1)\) and \(B(6, 11)\). Find the position vector of \(P\).
\(\begin{pmatrix} 2 \\ 3 \end{pmatrix}\)
\(\begin{pmatrix} 3 \\ 5 \end{pmatrix}\)
\(\begin{pmatrix} 3 \\ 5 \end{pmatrix}\)
\(\begin{pmatrix} 4 \\ 7 \end{pmatrix}\)
Solution:
Using the ratio formula with \(m=2\), \(n=3\):
\[ \vec{OP} = \frac{3\vec{OA} + 2\vec{OB}}{2+3} = \frac{3\begin{pmatrix} 1 \\ 1 \end{pmatrix} + 2\begin{pmatrix} 6 \\ 11 \end{pmatrix}}{5} \]
\[ = \frac{\begin{pmatrix} 3 \\ 3 \end{pmatrix} + \begin{pmatrix} 12 \\ 22 \end{pmatrix}}{5} = \frac{\begin{pmatrix} 15 \\ 25 \end{pmatrix}}{5} = \begin{pmatrix} 3 \\ 5 \end{pmatrix} \]
So \(P\) has coordinates \((3, 5)\).

Vector Notation Recap

Position Vector

\(\vec{OA} = \begin{pmatrix} x \\ y \end{pmatrix}\)
From origin \(O\) to point \(A\)

Displacement Vector

\(\vec{AB} = \vec{OB} - \vec{OA}\)
From point \(A\) to point \(B\)

Midpoint Vector

\(\vec{OM} = \frac{1}{2}(\vec{OA} + \vec{OB})\)
Midpoint of \(AB\)

Parallel Vectors

\(\vec{u} = k\vec{v}\)
One is a scalar multiple of the other

Scroll to Top