Vectors in Geometrical Proofs

Problem Statement

Proving geometrical theorems using traditional Euclidean methods (congruent triangles, alternate angles) can be lengthy and confusing. In HSC Mathematics Extension 2, we introduce Vector Proofs. By assigning vectors to the sides of shapes, we can prove properties of triangles, quadrilaterals, and circles using pure algebra.

The key to vector proofs is understanding a few fundamental principles:

1. Vector Addition: $\vec{AB} + \vec{BC} = \vec{AC}$.
2. Parallel Vectors: If $\mathbf{a} = k\mathbf{b}$ (where $k$ is a scalar), the vectors are parallel.
3. The Dot Product: If $\mathbf{a} \cdot \mathbf{b} = 0$, the vectors are perpendicular.

Consider the classic geometry theorem: "The diagonals of a rhombus are perpendicular to each other."

Let $OABC$ be a rhombus. Let the position vectors of $A$ and $C$ relative to the origin $O$ be $\mathbf{a}$ and $\mathbf{c}$, respectively.

(a) Write down the position vector of $B$ in terms of $\mathbf{a}$ and $\mathbf{c}$.

(b) Find the vectors representing the two diagonals, $\vec{OB}$ and $\vec{AC}$, in terms of $\mathbf{a}$ and $\mathbf{c}$.

(c) Using the dot product, prove that the diagonals of the rhombus intersect at right angles.

Hints

• Part (a): A rhombus is a type of parallelogram. Therefore, the vector $\vec{CB}$ is equivalent to the vector $\vec{OA}$. Use vector addition: $\vec{OB} = \vec{OC} + \vec{CB}$.
• Part (b): $\vec{OB}$ is simply the position vector of $B$ you found in part (a). To find the vector from $A$ to $C$ ($\vec{AC}$), use the rule "destination minus origin": $\vec{AC} = \mathbf{c} - \mathbf{a}$.
• Part (c): Calculate the dot product $\vec{OB} \cdot \vec{AC}$. Expand the expression algebraically. Remember the defining property of a rhombus: all four sides are equal in length. This means the magnitude of vector $\mathbf{a}$ equals the magnitude of vector $\mathbf{c}$ ($|\mathbf{a}| = |\mathbf{c}|$), which also means $\mathbf{a} \cdot \mathbf{a} = \mathbf{c} \cdot \mathbf{c}$. If the dot product evaluates to $0$, the diagonals are perpendicular.

Solutions

Part (a): Finding Position Vector B

1. Let the origin $O$ be a vertex of the rhombus.
2. The vector from $O$ to $A$ is $\mathbf{a}$. The vector from $O$ to $C$ is $\mathbf{c}$.
3. Because $OABC$ is a rhombus (and thus a parallelogram), the side $CB$ is parallel and equal in length to $OA$. Therefore, the vector $\vec{CB} = \vec{OA} = \mathbf{a}$.
4. To find the position vector of $B$ (which is $\vec{OB}$), we start at $O$, go to $C$, and then go to $B$: $\vec{OB} = \vec{OC} + \vec{CB}$ $\vec{OB} = \mathbf{c} + \mathbf{a} = \mathbf{a} + \mathbf{c}$

Part (b): Vectors of the Diagonals

1. The first diagonal is $\vec{OB}$. From part (a), this is: $\vec{OB} = \mathbf{a} + \mathbf{c}$
2. The second diagonal goes from $A$ to $C$, denoted as $\vec{AC}$.
3. Using the position vector rule ($\text{final} - \text{initial}$): $\vec{AC} = \vec{OC} - \vec{OA} = \mathbf{c} - \mathbf{a}$

Part (c): Proving Perpendicularity

1. To prove two vectors are perpendicular, we must show their dot product is zero. We need to evaluate $\vec{OB} \cdot \vec{AC}$.
2. Substitute the expressions from part (b): $\vec{OB} \cdot \vec{AC} = (\mathbf{a} + \mathbf{c}) \cdot (\mathbf{c} - \mathbf{a})$
3. Rearrange for clarity before expanding (using commutativity of vector addition): $(\mathbf{c} + \mathbf{a}) \cdot (\mathbf{c} - \mathbf{a})$
4. Expand using the difference of two squares rule for dot products: $\mathbf{c} \cdot \mathbf{c} - \mathbf{c} \cdot \mathbf{a} + \mathbf{a} \cdot \mathbf{c} - \mathbf{a} \cdot \mathbf{a}$ Since the dot product is commutative ($\mathbf{c} \cdot \mathbf{a} = \mathbf{a} \cdot \mathbf{c}$), the middle terms cancel out: $\mathbf{c} \cdot \mathbf{c} - \mathbf{a} \cdot \mathbf{a}$
5. Recall that $\mathbf{v} \cdot \mathbf{v} = |\mathbf{v}|^2$. So: $|\mathbf{c}|^2 - |\mathbf{a}|^2$
6. Here is where the specific geometry of the rhombus comes in. By definition, all sides of a rhombus are equal in length. Therefore, the length of side $OA$ equals the length of side $OC$. $|\mathbf{a}| = |\mathbf{c}|$ $|\mathbf{a}|^2 = |\mathbf{c}|^2$
7. Substitute this back into our dot product equation: $|\mathbf{c}|^2 - |\mathbf{c}|^2 = 0$
8. Since the dot product of the diagonals is $0$, the diagonals $\vec{OB}$ and $\vec{AC}$ are perpendicular.

Takeaways

• Algebraic Geometry: Vector proofs allow you to treat geometric shapes purely as algebraic expressions. Expanding dot products often leads directly to the proof without needing to draw complex construction lines.
• The Difference of Two Squares: The identity $(\mathbf{u} + \mathbf{v}) \cdot (\mathbf{u} - \mathbf{v}) = |\mathbf{u}|^2 - |\mathbf{v}|^2$ is incredibly common in circle and quadrilateral vector proofs.
• Definitions Matter: A proof usually relies on one specific geometric property to evaluate to zero. In this case, it was the fact that a rhombus has equal side lengths ($|\mathbf{a}| = |\mathbf{c}|$). For a rectangle proof, it would be that the adjacent sides are perpendicular ($\mathbf{a} \cdot \mathbf{c} = 0$).

Further Readings

• HSC Proofs: https://vumaths.com/booklets/hsc-proofs/
• HSC Vectors: https://vumaths.com/booklets/hsc-vectors/
• HSC Collections: https://vumaths.com/booklets/hsc-collections/

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