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Geometry Proofs with Vectors

CSEC Additional Mathematics Essential Knowledge: Vectors provide a powerful algebraic method for proving geometric properties. Instead of relying solely on Euclidean geometry with angles and side lengths, vectors allow us to use algebraic operations to prove collinearity, parallel lines, perpendicularity, and midpoints. This approach is particularly useful for coordinate geometry problems.

Key Concept: A vector is a quantity with both magnitude and direction. In geometry proofs, we use position vectors (vectors from the origin to a point) and displacement vectors (vectors between two points). Vector operations like addition, subtraction, scalar multiplication, and dot products help us establish geometric relationships.

Part 1: Vector Fundamentals for Geometry

Essential Vector Concepts

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Position Vectors and Notation

If point \(A\) has coordinates \((x_1, y_1)\), its position vector from origin \(O\) is:

\[\overrightarrow{OA} = \mathbf{a} = \begin{pmatrix} x_1 \\ y_1 \end{pmatrix} = x_1\mathbf{i} + y_1\mathbf{j}\]

The vector from point \(A\) to point \(B\) is:

\[\overrightarrow{AB} = \mathbf{b} – \mathbf{a} = \overrightarrow{OB} – \overrightarrow{OA}\]

Memory Aid: “End point minus start point” – To go from A to B, take B’s position vector minus A’s position vector.

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Parallel Vectors and Collinearity

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

\[\mathbf{u} \parallel \mathbf{v} \quad \text{if} \quad \mathbf{u} = k\mathbf{v} \quad \text{for some scalar } k\]

Collinearity Test: Points \(A\), \(B\), and \(C\) are collinear if \(\overrightarrow{AB} = k\overrightarrow{AC}\) for some scalar \(k\).

📝 Example 1: Proving Collinearity

Given points \(A(1,2)\), \(B(4,5)\), and \(C(7,8)\), prove they are collinear.

Find vectors: \(\overrightarrow{AB} = \begin{pmatrix} 4-1 \\ 5-2 \end{pmatrix} = \begin{pmatrix} 3 \\ 3 \end{pmatrix}\)

\(\overrightarrow{AC} = \begin{pmatrix} 7-1 \\ 8-2 \end{pmatrix} = \begin{pmatrix} 6 \\ 6 \end{pmatrix}\)

Check for scalar multiple: \(\begin{pmatrix} 6 \\ 6 \end{pmatrix} = 2\begin{pmatrix} 3 \\ 3 \end{pmatrix}\)

Conclusion: Since \(\overrightarrow{AC} = 2\overrightarrow{AB}\), the vectors are parallel and share point A, so \(A\), \(B\), and \(C\) are collinear.

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Section Formula

The position vector of point \(P\) that divides \(AB\) in the ratio \(m:n\) is:

\[\overrightarrow{OP} = \frac{n\mathbf{a} + m\mathbf{b}}{m+n}\]

Special Case – Midpoint: When \(m:n = 1:1\)

\[\overrightarrow{OM} = \frac{\mathbf{a} + \mathbf{b}}{2}\]

Part 2: Proving Geometric Properties

Common Proof Strategies

Strategy 1: Proving Parallel Lines

Approach: Show \(\overrightarrow{AB} = k\overrightarrow{CD}\) for some scalar \(k\)

Example: Prove opposite sides of a parallelogram are parallel

Key Formula: If \(\mathbf{u} = k\mathbf{v}\), then \(\mathbf{u} \parallel \mathbf{v}\)

Strategy 2: Proving Collinearity

Approach: Show \(\overrightarrow{AB} = k\overrightarrow{AC}\)

Example: Prove three points lie on the same straight line

Key Formula: If \(\overrightarrow{AB} = k\overrightarrow{AC}\), then \(A,B,C\) are collinear

Strategy 3: Proving Perpendicularity

Approach: Show \(\mathbf{u} \cdot \mathbf{v} = 0\) using dot product

Example: Prove diagonals of a rhombus are perpendicular

Key Formula: \(\mathbf{u} \cdot \mathbf{v} = |\mathbf{u}||\mathbf{v}|\cos\theta\)

📝 Example 2: Proving a Quadrilateral is a Parallelogram

Prove that quadrilateral \(ABCD\) with vertices \(A(1,2)\), \(B(4,3)\), \(C(5,6)\), and \(D(2,5)\) is a parallelogram.

Find opposite sides: \(\overrightarrow{AB} = \begin{pmatrix} 4-1 \\ 3-2 \end{pmatrix} = \begin{pmatrix} 3 \\ 1 \end{pmatrix}\)

\(\overrightarrow{DC} = \begin{pmatrix} 5-2 \\ 6-5 \end{pmatrix} = \begin{pmatrix} 3 \\ 1 \end{pmatrix}\)

Compare: \(\overrightarrow{AB} = \overrightarrow{DC}\)

Find other opposite sides: \(\overrightarrow{AD} = \begin{pmatrix} 2-1 \\ 5-2 \end{pmatrix} = \begin{pmatrix} 1 \\ 3 \end{pmatrix}\)

\(\overrightarrow{BC} = \begin{pmatrix} 5-4 \\ 6-3 \end{pmatrix} = \begin{pmatrix} 1 \\ 3 \end{pmatrix}\)

Conclusion: Since \(\overrightarrow{AB} = \overrightarrow{DC}\) and \(\overrightarrow{AD} = \overrightarrow{BC}\), both pairs of opposite sides are equal and parallel, so \(ABCD\) is a parallelogram.

📝 Example 3: Using Dot Product for Perpendicularity

Prove that triangle \(ABC\) with vertices \(A(1,1)\), \(B(4,2)\), and \(C(3,5)\) is right-angled at \(B\).

Find vectors meeting at B: \(\overrightarrow{BA} = \begin{pmatrix} 1-4 \\ 1-2 \end{pmatrix} = \begin{pmatrix} -3 \\ -1 \end{pmatrix}\)

\(\overrightarrow{BC} = \begin{pmatrix} 3-4 \\ 5-2 \end{pmatrix} = \begin{pmatrix} -1 \\ 3 \end{pmatrix}\)

Calculate dot product: \(\overrightarrow{BA} \cdot \overrightarrow{BC} = (-3)(-1) + (-1)(3) = 3 – 3 = 0\)

Conclusion: Since the dot product is zero, \(\overrightarrow{BA}\) is perpendicular to \(\overrightarrow{BC}\), so triangle \(ABC\) is right-angled at \(B\).

Part 3: Vector Proofs with Ratios and Midpoints

Division Points and Centroids

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Proving Points Divide Segments in Ratios

To prove point \(P\) divides \(AB\) in ratio \(m:n\), show:

\[\overrightarrow{OP} = \frac{n\mathbf{a} + m\mathbf{b}}{m+n} \quad \text{or} \quad \overrightarrow{AP} = \frac{m}{m+n}\overrightarrow{AB}\]

📝 Example 4: Proving a Point Divides a Line Segment

Prove that point \(P(3,4)\) divides the line segment joining \(A(1,2)\) and \(B(5,6)\) in the ratio 1:1.

Midpoint formula: For midpoint \(M\), \(\overrightarrow{OM} = \frac{\mathbf{a} + \mathbf{b}}{2}\)

Calculate: \(\frac{\mathbf{a} + \mathbf{b}}{2} = \frac{1}{2}\begin{pmatrix} 1+5 \\ 2+6 \end{pmatrix} = \frac{1}{2}\begin{pmatrix} 6 \\ 8 \end{pmatrix} = \begin{pmatrix} 3 \\ 4 \end{pmatrix}\)

Compare: This equals \(\overrightarrow{OP} = \begin{pmatrix} 3 \\ 4 \end{pmatrix}\)

Conclusion: Since \(\overrightarrow{OP} = \frac{\mathbf{a} + \mathbf{b}}{2}\), \(P\) is the midpoint of \(AB\), dividing it in ratio 1:1.

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Centroid of a Triangle

The centroid \(G\) (intersection of medians) of triangle \(ABC\) has position vector:

\[\overrightarrow{OG} = \frac{\mathbf{a} + \mathbf{b} + \mathbf{c}}{3}\]

Proof Strategy: Show that the point dividing median in ratio 2:1 from vertex to midpoint satisfies this formula.

📝 Example 5: CSEC Past Paper Question (Adapted)

The points \(A\), \(B\), and \(C\) have position vectors \(\mathbf{a}\), \(\mathbf{b}\), and \(\mathbf{c}\) respectively. \(M\) is the midpoint of \(BC\).

(a) Find \(\overrightarrow{AM}\) in terms of \(\mathbf{a}\), \(\mathbf{b}\), and \(\mathbf{c}\).

(b) \(G\) is the centroid of triangle \(ABC\) and lies on \(AM\) such that \(AG:GM = 2:1\). Find \(\overrightarrow{OG}\) in terms of \(\mathbf{a}\), \(\mathbf{b}\), and \(\mathbf{c}\).

Part (a): \(\overrightarrow{OM} = \frac{\mathbf{b} + \mathbf{c}}{2}\) (midpoint formula)

\(\overrightarrow{AM} = \overrightarrow{OM} – \overrightarrow{OA} = \frac{\mathbf{b} + \mathbf{c}}{2} – \mathbf{a} = \frac{\mathbf{b} + \mathbf{c} – 2\mathbf{a}}{2}\)

Part (b): Since \(AG:GM = 2:1\), \(G\) divides \(AM\) in ratio 2:1

Using section formula: \(\overrightarrow{OG} = \frac{1\cdot\mathbf{a} + 2\cdot\overrightarrow{OM}}{2+1} = \frac{\mathbf{a} + 2\left(\frac{\mathbf{b} + \mathbf{c}}{2}\right)}{3}\)

Simplify: \(\overrightarrow{OG} = \frac{\mathbf{a} + \mathbf{b} + \mathbf{c}}{3}\)

Part 4: Advanced Vector Proof Techniques

Combining Multiple Concepts

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Proving Special Quadrilaterals

Parallelogram: Show \(\overrightarrow{AB} = \overrightarrow{DC}\) OR \(\overrightarrow{AD} = \overrightarrow{BC}\)

Rectangle: First prove it’s a parallelogram, then show adjacent sides are perpendicular (dot product = 0)

Rhombus: First prove it’s a parallelogram, then show all sides equal in magnitude

Square: Prove it’s both a rectangle and a rhombus

Identify what type of quadrilateral you need to prove

Use vector equality to prove parallel/equal sides

Use dot product for right angles if needed

Use magnitude formula for equal sides if needed

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Proving Concurrency Using Vectors

To prove three lines are concurrent (meet at one point):

Find equations of two lines in vector form
Find their intersection point
Show this point lies on the third line

📝 Example 6: Proving Medians are Concurrent

Prove that the medians of triangle \(ABC\) are concurrent at the centroid \(G\).

Let \(D\), \(E\), \(F\) be midpoints of \(BC\), \(CA\), \(AB\) respectively

Find equations of medians: Median from \(A\): \(\mathbf{r} = \mathbf{a} + t(\overrightarrow{AD})\) where \(\overrightarrow{AD} = \frac{\mathbf{b}+\mathbf{c}}{2} – \mathbf{a}\)

Find intersection of medians from A and B: Solve \(\mathbf{a} + t(\overrightarrow{AD}) = \mathbf{b} + s(\overrightarrow{BE})\)

Solution gives: \(t = s = \frac{2}{3}\), and intersection point \(\mathbf{g} = \frac{\mathbf{a}+\mathbf{b}+\mathbf{c}}{3}\)

Verify on third median: Point with position vector \(\frac{\mathbf{a}+\mathbf{b}+\mathbf{c}}{3}\) lies on median from \(C\) when parameter = \(\frac{2}{3}\)

Conclusion: All three medians pass through \(G\), so they are concurrent.

Part 5: CSEC Past Paper Questions

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Exam-Style Questions with Solutions

📝 CSEC Additional Mathematics 2019 Question 8 (Adapted)

The points \(A\), \(B\), and \(C\) have coordinates \((1,3)\), \((4,1)\), and \((7,5)\) respectively.

(a) Find \(\overrightarrow{AB}\) and \(\overrightarrow{AC}\).

(b) Hence prove that \(A\), \(B\), and \(C\) are collinear.

(a) \(\overrightarrow{AB} = \begin{pmatrix} 4-1 \\ 1-3 \end{pmatrix} = \begin{pmatrix} 3 \\ -2 \end{pmatrix}\)

\(\overrightarrow{AC} = \begin{pmatrix} 7-1 \\ 5-3 \end{pmatrix} = \begin{pmatrix} 6 \\ -4 \end{pmatrix}\)

(b) Since \(\overrightarrow{AC} = 2\begin{pmatrix} 3 \\ -2 \end{pmatrix} = 2\overrightarrow{AB}\), the vectors are parallel and share point \(A\), so \(A\), \(B\), \(C\) are collinear.

(c) Since \(\overrightarrow{AB} = \frac{1}{2}\overrightarrow{AC}\), \(B\) divides \(AC\) in ratio \(1:1\) (it’s actually the midpoint). Wait, let’s check…

Actually, if \(\overrightarrow{AB} = k\overrightarrow{AC}\), then \(AB:AC = k:1\). Here \(k = \frac{1}{2}\), so \(AB:AC = 1:2\), meaning \(B\) divides \(AC\) in ratio \(1:1\)? No, let’s think carefully.

Since \(\overrightarrow{AB} = \frac{1}{2}\overrightarrow{AC}\), \(AB\) is half of \(AC\), so \(B\) is the midpoint. Therefore \(B\) divides \(AC\) in ratio \(1:1\).

📝 CSEC Additional Mathematics 2017 Question 9 (Adapted)

\(OABC\) is a parallelogram with \(\overrightarrow{OA} = \mathbf{a}\) and \(\overrightarrow{OC} = \mathbf{c}\). \(M\) is the midpoint of \(AB\) and \(N\) is the midpoint of \(BC\).

(a) Express \(\overrightarrow{OB}\) and \(\overrightarrow{AC}\) in terms of \(\mathbf{a}\) and \(\mathbf{c}\).

(b) Show that \(\overrightarrow{MN} = \frac{1}{2}\overrightarrow{AC}\).

(a) In parallelogram \(OABC\), \(\overrightarrow{OB} = \overrightarrow{OA} + \overrightarrow{AB} = \mathbf{a} + \mathbf{c}\) (since \(\overrightarrow{AB} = \overrightarrow{OC} = \mathbf{c}\))

\(\overrightarrow{AC} = \overrightarrow{AO} + \overrightarrow{OC} = -\mathbf{a} + \mathbf{c} = \mathbf{c} – \mathbf{a}\)

(b) First find \(\overrightarrow{OM}\): Since \(M\) is midpoint of \(AB\), \(\overrightarrow{OM} = \frac{\overrightarrow{OA} + \overrightarrow{OB}}{2} = \frac{\mathbf{a} + (\mathbf{a}+\mathbf{c})}{2} = \frac{2\mathbf{a}+\mathbf{c}}{2}\)

Find \(\overrightarrow{ON}\): \(N\) is midpoint of \(BC\), \(\overrightarrow{ON} = \frac{\overrightarrow{OB} + \overrightarrow{OC}}{2} = \frac{(\mathbf{a}+\mathbf{c}) + \mathbf{c}}{2} = \frac{\mathbf{a}+2\mathbf{c}}{2}\)

Now \(\overrightarrow{MN} = \overrightarrow{ON} – \overrightarrow{OM} = \frac{\mathbf{a}+2\mathbf{c}}{2} – \frac{2\mathbf{a}+\mathbf{c}}{2} = \frac{-\mathbf{a}+\mathbf{c}}{2} = \frac{1}{2}(\mathbf{c}-\mathbf{a}) = \frac{1}{2}\overrightarrow{AC}\)

(c) Since \(\overrightarrow{MN} = \frac{1}{2}\overrightarrow{AC}\), \(MN\) is parallel to \(AC\) and half its length.

Quiz: Test Your Understanding

Geometry Proofs with Vectors Quiz

Question 1: Points \(P\), \(Q\), and \(R\) have position vectors \(\mathbf{p} = \begin{pmatrix} 1 \\ 3 \end{pmatrix}\), \(\mathbf{q} = \begin{pmatrix} 4 \\ 5 \end{pmatrix}\), and \(\mathbf{r} = \begin{pmatrix} 7 \\ 7 \end{pmatrix}\). Prove that \(P\), \(Q\), and \(R\) are collinear.

Answer:
\(\overrightarrow{PQ} = \mathbf{q} – \mathbf{p} = \begin{pmatrix} 3 \\ 2 \end{pmatrix}\)
\(\overrightarrow{PR} = \mathbf{r} – \mathbf{p} = \begin{pmatrix} 6 \\ 4 \end{pmatrix}\)
Since \(\overrightarrow{PR} = 2\overrightarrow{PQ}\), the vectors are parallel and share point \(P\), so \(P\), \(Q\), \(R\) are collinear.

Question 2: Prove that quadrilateral \(ABCD\) with vertices \(A(0,0)\), \(B(3,1)\), \(C(5,5)\), and \(D(2,4)\) is a parallelogram.

Answer:
\(\overrightarrow{AB} = \begin{pmatrix} 3 \\ 1 \end{pmatrix}\), \(\overrightarrow{DC} = \begin{pmatrix} 5-2 \\ 5-4 \end{pmatrix} = \begin{pmatrix} 3 \\ 1 \end{pmatrix}\)
\(\overrightarrow{AD} = \begin{pmatrix} 2 \\ 4 \end{pmatrix}\), \(\overrightarrow{BC} = \begin{pmatrix} 5-3 \\ 5-1 \end{pmatrix} = \begin{pmatrix} 2 \\ 4 \end{pmatrix}\)
Since \(\overrightarrow{AB} = \overrightarrow{DC}\) and \(\overrightarrow{AD} = \overrightarrow{BC}\), both pairs of opposite sides are equal and parallel, so \(ABCD\) is a parallelogram.

Question 3: Triangle \(ABC\) has vertices \(A(1,2)\), \(B(4,6)\), and \(C(5,0)\). Prove it is a right-angled triangle and find which angle is the right angle.

Answer:
Check dot products:
\(\overrightarrow{AB} = \begin{pmatrix} 3 \\ 4 \end{pmatrix}\), \(\overrightarrow{BC} = \begin{pmatrix} 1 \\ -6 \end{pmatrix}\), \(\overrightarrow{CA} = \begin{pmatrix} -4 \\ 2 \end{pmatrix}\)
\(\overrightarrow{AB} \cdot \overrightarrow{BC} = 3(1) + 4(-6) = 3 – 24 = -21 \neq 0\)
\(\overrightarrow{BC} \cdot \overrightarrow{CA} = 1(-4) + (-6)(2) = -4 – 12 = -16 \neq 0\)
\(\overrightarrow{CA} \cdot \overrightarrow{AB} = (-4)(3) + 2(4) = -12 + 8 = -4 \neq 0\)
None are zero, so let’s check magnitudes and Pythagorean theorem…
Actually, this triangle isn’t right-angled. The correct approach would be to check all three dot products as shown.

Question 4: Points \(A\), \(B\), and \(C\) have position vectors \(\mathbf{a}\), \(\mathbf{b}\), and \(\mathbf{c}\). \(M\) and \(N\) are midpoints of \(AB\) and \(AC\) respectively. Prove that \(\overrightarrow{MN} = \frac{1}{2}\overrightarrow{BC}\).

Answer:
\(\overrightarrow{OM} = \frac{\mathbf{a}+\mathbf{b}}{2}\), \(\overrightarrow{ON} = \frac{\mathbf{a}+\mathbf{c}}{2}\)
\(\overrightarrow{MN} = \overrightarrow{ON} – \overrightarrow{OM} = \frac{\mathbf{a}+\mathbf{c}}{2} – \frac{\mathbf{a}+\mathbf{b}}{2} = \frac{\mathbf{c}-\mathbf{b}}{2} = \frac{1}{2}\overrightarrow{BC}\)
Thus \(MN\) is parallel to \(BC\) and half its length.

Question 5: The position vectors of points \(P\) and \(Q\) are \(\mathbf{p} = 2\mathbf{i} + 3\mathbf{j}\) and \(\mathbf{q} = 5\mathbf{i} – \mathbf{j}\). Point \(R\) divides \(PQ\) in the ratio 2:3. Find the position vector of \(R\).

Answer:
Using section formula with \(m:n = 2:3\):
\(\overrightarrow{OR} = \frac{n\mathbf{p} + m\mathbf{q}}{m+n} = \frac{3(2\mathbf{i}+3\mathbf{j}) + 2(5\mathbf{i}-\mathbf{j})}{5}\)
\(= \frac{6\mathbf{i}+9\mathbf{j} + 10\mathbf{i}-2\mathbf{j}}{5} = \frac{16\mathbf{i}+7\mathbf{j}}{5} = \frac{16}{5}\mathbf{i} + \frac{7}{5}\mathbf{j}\)

🎯 Key Concepts Summary

Position Vector: \(\overrightarrow{OA} = \mathbf{a}\) for point \(A\)
Displacement Vector: \(\overrightarrow{AB} = \mathbf{b} – \mathbf{a}\) (end minus start)
Parallel Vectors: \(\mathbf{u} \parallel \mathbf{v}\) if \(\mathbf{u} = k\mathbf{v}\) for some scalar \(k\)
Collinearity: \(A,B,C\) collinear if \(\overrightarrow{AB} = k\overrightarrow{AC}\)
Dot Product: \(\mathbf{u} \cdot \mathbf{v} = u_1v_1 + u_2v_2 = |\mathbf{u}||\mathbf{v}|\cos\theta\)
Perpendicularity: \(\mathbf{u} \perp \mathbf{v}\) if \(\mathbf{u} \cdot \mathbf{v} = 0\)
Section Formula: Point dividing \(AB\) in ratio \(m:n\) has position vector \(\frac{n\mathbf{a} + m\mathbf{b}}{m+n}\)
Midpoint: \(\frac{\mathbf{a} + \mathbf{b}}{2}\)
Centroid of Triangle: \(\frac{\mathbf{a} + \mathbf{b} + \mathbf{c}}{3}\)
Parallelogram Test: \(\overrightarrow{AB} = \overrightarrow{DC}\) or \(\overrightarrow{AD} = \overrightarrow{BC}\)

CSEC Exam Strategy: When solving vector geometry proofs: (1) Draw a clear diagram, (2) Label all points with their position vectors, (3) Write vectors in terms of given position vectors, (4) Show all steps in logical order, (5) State your conclusion clearly. Remember to use proper vector notation (\(\overrightarrow{AB}\) not just AB) and show when vectors are equal, parallel, or perpendicular.

Common Mistakes to Avoid: 1. Confusing position vectors with displacement vectors
2. Incorrect vector subtraction (\(\overrightarrow{AB} = \mathbf{b} – \mathbf{a}\), not \(\mathbf{a} – \mathbf{b}\))
3. Forgetting that parallel vectors must be scalar multiples, not just same direction
4. Not showing all steps in a proof (examiners need to see your reasoning)
5. Misapplying the section formula (remember: \(m:n\) means \(m\) parts from \(A\) to \(P\), \(n\) parts from \(P\) to \(B\))