Ultimate Guide to Vectors and Geometric Proofs

Introduction

Euclidean geometry (using angles, congruent triangles, and circle properties) can be beautiful, but it often relies on drawing complex auxiliary lines and relying on visual intuition. In HSC Mathematics Extension 2, we replace visual intuition with pure algebra. Using Vectors and Geometric Proofs, you can prove ancient theorems about medians, altitudes, and perpendicular bisectors without ever needing to draw a perfectly scaled diagram.

Executive Summary

This guide covers the application of vectors to geometry:

• Dot Product Properties: Using commutativity and distributivity to expand vector expressions algebraically.
• The Cauchy-Schwarz Inequality: Proving and applying the fundamental inequality $|\mathbf{u} \cdot \mathbf{v}| \le |\mathbf{u}| |\mathbf{v}|$.
• Geometric Definitions: Understanding medians, altitudes, and bisectors in a vector context.
• Geometric Proofs: Using the dot product (where $\mathbf{a} \cdot \mathbf{b} = 0$ means perpendicular) to prove theorems about shapes in 2D and 3D.

What is this about?

Vector proofs rely almost entirely on the Scalar (Dot) Product. The geometric definition of the dot product is: $\mathbf{a} \cdot \mathbf{b} = |\mathbf{a}| |\mathbf{b}| \cos\theta$

This definition gives us two immense powers:

1. Perpendicularity: If two non-zero vectors are perpendicular ($\theta = 90^\circ$), then $\cos(90^\circ) = 0$. Therefore, their dot product is exactly $0$. If you need to prove two lines meet at a right angle, you just prove their dot product is zero.
2. Magnitudes: If you dot a vector with itself, $\theta = 0^\circ$, and $\cos(0^\circ) = 1$. Therefore, $\mathbf{a} \cdot \mathbf{a} = |\mathbf{a}| |\mathbf{a}| = |\mathbf{a}|^2$. If you need to prove two lengths are equal, you can compare their dot-product squares!

Main Content

1. Vector Algebraic Properties

Because vectors are not standard numbers, you cannot assume normal algebra applies without proof. The syllabus requires you to know these properties hold:

• Commutativity: $\mathbf{a} \cdot \mathbf{b} = \mathbf{b} \cdot \mathbf{a}$
• Distributivity: $\mathbf{a} \cdot (\mathbf{b} + \mathbf{c}) = \mathbf{a} \cdot \mathbf{b} + \mathbf{a} \cdot \mathbf{c}$

This means you can expand brackets almost exactly like normal algebra! Example: Expand $(\mathbf{a} + \mathbf{b}) \cdot (\mathbf{a} + \mathbf{b})$. $= \mathbf{a} \cdot \mathbf{a} + \mathbf{a} \cdot \mathbf{b} + \mathbf{b} \cdot \mathbf{a} + \mathbf{b} \cdot \mathbf{b}$ $= |\mathbf{a}|^2 + 2(\mathbf{a} \cdot \mathbf{b}) + |\mathbf{b}|^2$

(Notice how similar this is to $(a+b)^2 = a^2 + 2ab + b^2$!)

2. The Cauchy-Schwarz Inequality

The Cauchy-Schwarz inequality states that for any two vectors $\mathbf{u}$ and $\mathbf{v}$: $|\mathbf{u} \cdot \mathbf{v}| \le |\mathbf{u}| |\mathbf{v}|$

The Proof: We know the geometric definition: $\mathbf{u} \cdot \mathbf{v} = |\mathbf{u}| |\mathbf{v}| \cos\theta$. Take the absolute value of both sides: $|\mathbf{u} \cdot \mathbf{v}| = ||\mathbf{u}| |\mathbf{v}| \cos\theta|$ Since magnitudes are already positive: $|\mathbf{u} \cdot \mathbf{v}| = |\mathbf{u}| |\mathbf{v}| |\cos\theta|$

We know from trigonometry that the cosine graph is bounded between $-1$ and $1$. Therefore, the absolute value $|\cos\theta|$ must be $\le 1$. Because $|\cos\theta| \le 1$, if we remove it from the equation, the right side must get larger or stay the same: $|\mathbf{u} \cdot \mathbf{v}| \le |\mathbf{u}| |\mathbf{v}|$

3. Triangle Definitions

To prove theorems, you must know what these terms mean as vectors:

• Median: A line from a vertex to the midpoint of the opposite side. If $M$ is the midpoint of $\vec{BC}$, then $\vec{OM} = \frac{1}{2}(\vec{OB} + \vec{OC})$.
• Altitude: A line from a vertex perpendicular to the opposite side. If altitude $AD$ is drawn to $BC$, then $\vec{AD} \cdot \vec{BC} = 0$.
• Perpendicular Bisector: A line that passes through the midpoint of a side and is perpendicular to it.

Simple Worked Example

Question: Prove using vectors that the diagonals of a rhombus intersect at right angles.

Solution: Step 1: Set up the vectors. Let the rhombus be $OABC$, with the origin at $O$. Let $\vec{OA} = \mathbf{a}$ and $\vec{OC} = \mathbf{c}$. Because $OABC$ is a parallelogram, the vector $\vec{OB}$ (the diagonal) is $\mathbf{a} + \mathbf{c}$. The other diagonal is $\vec{AC}$, which is $\vec{OC} - \vec{OA} = \mathbf{c} - \mathbf{a}$.

Step 2: Define the property of a rhombus. A rhombus is a parallelogram where all four sides are equal in length. Therefore, the length of side $OA$ equals the length of side $OC$. $|\mathbf{a}| = |\mathbf{c}|$ Squaring this gives: $|\mathbf{a}|^2 = |\mathbf{c}|^2$.

Step 3: Prove the diagonals are perpendicular. To prove they intersect at right angles, we must prove their dot product equals zero. Diagonal 1: $\vec{OB} = \mathbf{a} + \mathbf{c}$ Diagonal 2: $\vec{AC} = \mathbf{c} - \mathbf{a}$

Calculate the dot product: $\vec{OB} \cdot \vec{AC} = (\mathbf{a} + \mathbf{c}) \cdot (\mathbf{c} - \mathbf{a})$ Expand using distributivity (Difference of Two Squares): $= \mathbf{a} \cdot \mathbf{c} - \mathbf{a} \cdot \mathbf{a} + \mathbf{c} \cdot \mathbf{c} - \mathbf{c} \cdot \mathbf{a}$ Since dot product is commutative, $\mathbf{a} \cdot \mathbf{c}$ and $-\mathbf{c} \cdot \mathbf{a}$ cancel out. $= \mathbf{c} \cdot \mathbf{c} - \mathbf{a} \cdot \mathbf{a}$ $= |\mathbf{c}|^2 - |\mathbf{a}|^2$

Step 4: Use the rhombus property. From Step 2, we know $|\mathbf{a}|^2 = |\mathbf{c}|^2$. Therefore: $= |\mathbf{c}|^2 - |\mathbf{c}|^2 = 0$

Conclusion: Since the dot product of the diagonals is zero ($\vec{OB} \cdot \vec{AC} = 0$), the diagonals must intersect at right angles.

mini-FAQ page

Q: Can I just draw a diagram and use normal geometry angles? A: No! If the HSC question says "Prove using vector methods", you will receive zero marks for a standard Euclidean geometry proof, even if it is completely flawless. You must use $\mathbf{a}$, $\mathbf{b}$, and dot products.

Q: What is the triangle inequality? A: A corollary to Cauchy-Schwarz is the Triangle Inequality: $|\mathbf{u} + \mathbf{v}| \le |\mathbf{u}| + |\mathbf{v}|$. This states the geometric fact that the shortest path between two points is a straight line, not a detour through a third point.

Common mistakes to avoid

• Writing vectors as scalars: Writing $a \cdot b$ instead of $\mathbf{a} \cdot \mathbf{b}$ will lose you marks. You must use tildes underneath (like $\underset{\sim}{a}$) in handwriting, or bold font in typing, to distinguish vectors from scalar numbers.
• Multiplying vectors without dots: Writing $\mathbf{a}\mathbf{b}$ is mathematically meaningless in this syllabus. It must be $\mathbf{a} \cdot \mathbf{b}$ (the scalar product). There is no "normal" multiplication for two vectors!

Practice on Vu's Maths Hub

Vector proofs require you to translate English geometry shapes into algebra.

• Practice proving properties of parallelograms, kites, and circles in the HSC Vectors Booklet.
• Sharpen your dot product algebraic expansions at Vu's Maths Hub.

Further Readings

• You have conquered the Vectors syllabus!
• Explore more HSC math resources and full worked solutions at Vu's Maths Hub.

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Want to master 3D Vectors and guarantee top marks in HSC Mathematics Extension 2? Visit Vu's Maths Hub for in-depth booklets, rigorous worked solutions, and expert advice to help you ace your exams!