Using Vectors in Geometry Problems
CSEC Mathematics: The Power of Vector Geometry
Essential Understanding: Vectors are not just for magnitude and direction—they're powerful tools for solving complex geometry problems. Learn to navigate through points, prove parallelism and collinearity, and master the vector approach to geometric proofs.
1. The Vector Journey (Pathfinding)
The most important rule in vector geometry is that you can get from one point to another by any valid route.
The Route Rule
To find vector \(\vec{AC}\), you don't need a direct line. You can go via any intermediate point:
This is the foundation of all vector pathfinding. The path from \(A\) to \(C\) is the sum of the paths from \(A\) to \(B\) and \(B\) to \(C\).
Reverse Movement
If you move against the arrow of a known vector \(\mathbf{a}\), you must use \(-\mathbf{a}\).
Example: If \(\vec{AB} = \mathbf{a}\), then \(\vec{BA} = -\mathbf{a}\).
This is crucial when navigating backwards along a known path.
Multiple Routes
There are often multiple valid routes between two points. All correct routes will give the same final vector.
Example: In triangle \(ABC\):
\(\vec{AC} = \vec{AB} + \vec{BC}\)
Also: \(\vec{AC} = \vec{AD} + \vec{DC}\) if \(D\) is another point.
2. The Midpoint and Ratio Constants
CSEC problems often introduce a point \(M\) that divides a line segment in a given ratio.
Ratio Division Formulas
Midpoint: If \(M\) is the midpoint of \(AB\), then \(\vec{AM} = \frac{1}{2}\vec{AB}\)
General Ratio: If \(P\) divides \(AB\) in the ratio \(m:n\), then:
\(\vec{AP} = \frac{m}{m+n}\vec{AB}\) and \(\vec{PB} = \frac{n}{m+n}\vec{AB}\)
Key Strategy: Always express the small segment as a fraction of the whole line before trying to find its position vector.
3. Parallelism and Collinearity
These two concepts are the "proof" based objectives of the CSEC syllabus.
A. Proving Parallel Vectors
Two vectors \(\mathbf{u}\) and \(\mathbf{v}\) are parallel if one is a scalar multiple of the other.
Example: If \(\vec{PQ} = \mathbf{a} + 2\mathbf{b}\) and \(\vec{XY} = 3\mathbf{a} + 6\mathbf{b}\), then \(\vec{XY} = 3(\vec{PQ})\).
Therefore, \(PQ\) is parallel to \(XY\).
B. Proving Collinear Points
Points are collinear if they all lie on the same straight line. In the exam, you must prove two things:
- Parallelism: Show that the vector between the first two points is a scalar multiple of the vector between the next two.
- Common Point: Explicitly state that the two vectors share a point.
The CSEC "Magic Sentence"
"Since \(\vec{AB} = k\vec{BC}\), the vectors are parallel. Since they share the common point \(B\), the points \(A, B,\) and \(C\) are collinear (lie on a straight line)."
Remember: You must mention BOTH conditions to get full marks!
4. Interactive "Path Builder" Lab
Build Your Vector Path
Objective: Construct the correct path from \(O\) to \(M\) by selecting vector steps. See how complex vectors are built from simple components.
Challenge: How do you get from \(O\) to \(M\)?
Base vectors: \(\mathbf{a} = \vec{OA}\), \(\mathbf{b} = \vec{OB}\)
\(M\) is the midpoint of diagonal \(AC\)
\(\vec{OM} = \)
5. Common Geometric Shapes in Vectors
Vectors are frequently applied to two specific shapes in CSEC exams:
The Parallelogram
In parallelogram \(OABC\):
- \(\vec{OA} = \mathbf{a}\)
- \(\vec{OC} = \mathbf{c}\)
- \(\vec{CB} = \mathbf{a}\) (opposite sides equal and parallel)
- \(\vec{AB} = \mathbf{c}\) (opposite sides equal and parallel)
Key Property: \(\vec{OB} = \vec{OA} + \vec{AB} = \mathbf{a} + \mathbf{c}\)
The Trapezium
In trapezium \(PQRS\) with \(PQ \parallel SR\):
- \(\vec{PQ} = k\vec{SR}\) (parallel sides are scalar multiples)
- The scalar \(k\) is the ratio of the parallel sides
- Opposite sides are NOT equal (unlike parallelogram)
Key Property: One pair of opposite sides are parallel but not necessarily equal.
6. CSEC Exam Mastery Tips
Avoid These Common Mistakes
Simplify Your Fractions
- Always keep your path expressions in the simplest form.
- \(\frac{2}{4}\vec{AB}\) should be written as \(\frac{1}{2}\vec{AB}\).
- Examiners deduct marks for unsimplified fractions.
State the "Common Point"
- You will lose a mark in a collinearity proof if you forget to mention the shared point.
- Always explicitly state: "Since they share the common point \(B\)..."
- This is a specific CSEC marking requirement.
Draw Your Own Arrows
- If the exam diagram doesn't have arrows, draw them in immediately.
- This prevents sign errors when navigating against vector directions.
- Use different colors for different vectors if possible.
7. Worked Example: The Collinearity Proof
Problem: In a diagram, \(\vec{OP} = 2\mathbf{a}\) and \(\vec{OQ} = 6\mathbf{a}\). Point \(R\) lies on \(OQ\). Show that \(O, P,\) and \(Q\) are collinear.
Find \(\vec{OP}\): Given as \(2\mathbf{a}\).
Find \(\vec{PQ}\): Using pathfinding:
\[ \vec{PQ} = \vec{PO} + \vec{OQ} = -2\mathbf{a} + 6\mathbf{a} = 4\mathbf{a} \]
Compare vectors:
\[ \vec{PQ} = 4\mathbf{a} = 2(2\mathbf{a}) = 2\vec{OP} \]
So \(\vec{PQ} = 2\vec{OP}\), meaning \(\vec{PQ}\) is parallel to \(\vec{OP}\).
Conclusion:
"Since \(\vec{PQ} = 2\vec{OP}\), the vectors are parallel. Since they share the common point \(P\), the points \(O, P,\) and \(Q\) are collinear (lie on a straight line)."
8. Practice Mission: "The Midpoint Diagonal"
Step 1: Find \(\vec{AC}\):
\[ \vec{AC} = \vec{AO} + \vec{OC} = -\mathbf{a} + \mathbf{c} \]
Step 2: Since \(M\) is the midpoint:
\[ \vec{AM} = \frac{1}{2}\vec{AC} = \frac{1}{2}(-\mathbf{a} + \mathbf{c}) \]
Step 3: Find \(\vec{OM}\):
\[ \vec{OM} = \vec{OA} + \vec{AM} = \mathbf{a} + \frac{1}{2}(-\mathbf{a} + \mathbf{c}) \]
\[ = \mathbf{a} - \frac{1}{2}\mathbf{a} + \frac{1}{2}\mathbf{c} = \frac{1}{2}\mathbf{a} + \frac{1}{2}\mathbf{c} \]
Final Answer: \(\vec{OM} = \frac{1}{2}\mathbf{a} + \frac{1}{2}\mathbf{c}\)
Step 1: Find \(\vec{OB}\):
In parallelogram \(OABC\), \(\vec{OB} = \vec{OA} + \vec{AB} = \mathbf{a} + \mathbf{c}\)
Step 2: Midpoint of \(OB\):
\[ \vec{O}(\text{midpoint of }OB) = \frac{1}{2}\vec{OB} = \frac{1}{2}(\mathbf{a} + \mathbf{c}) = \frac{1}{2}\mathbf{a} + \frac{1}{2}\mathbf{c} \]
Step 3: Compare with midpoint of \(AC\):
From Question 1: \(\vec{OM} = \frac{1}{2}\mathbf{a} + \frac{1}{2}\mathbf{c}\)
Conclusion: Both diagonals have the same midpoint.
Key Insight: In any parallelogram, the diagonals bisect each other.
Step 1: Express \(\vec{ST}\) using pathfinding:
\[ \vec{ST} = \vec{SR} + \vec{RT} \]
Step 2: Use midpoint properties:
Since \(S\) is midpoint of \(PR\): \(\vec{SR} = \frac{1}{2}\vec{PR}\)
Since \(T\) is midpoint of \(QR\): \(\vec{RT} = \frac{1}{2}\vec{RQ} = -\frac{1}{2}\vec{QR}\)
Step 3: Substitute and simplify:
\[ \vec{ST} = \frac{1}{2}\vec{PR} - \frac{1}{2}\vec{QR} = \frac{1}{2}(\vec{PR} - \vec{QR}) \]
Step 4: Note that \(\vec{PR} - \vec{QR} = \vec{PQ}\) (from triangle rule)
\[ \vec{ST} = \frac{1}{2}\vec{PQ} \]
Conclusion: Since \(\vec{ST} = \frac{1}{2}\vec{PQ}\), \(ST\) is parallel to \(PQ\) and half its length.
Vector Geometry Rules
Pathfinding Rule
\(\vec{AC} = \vec{AB} + \vec{BC}\)
Any valid route works
Reverse Movement
\(\vec{BA} = -\vec{AB}\)
Moving against arrow = negative
Midpoint Formula
\(\vec{OM} = \frac{1}{2}(\vec{OA} + \vec{OB})\)
Average of endpoints
Collinearity Proof
1. Show parallel (\(\vec{AB} = k\vec{BC}\))
2. State common point