Vectors in Geometric Proof [H]

Vectors in Geometric Proof

Vector methods allow geometric properties to be proved algebraically without drawing. The key skills are: expressing vectors in terms of given vectors \( \mathbf{a} \) and \( \mathbf{b} \); showing that two vectors are parallel (one is a scalar multiple of the other); and showing that three points are collinear (the vectors between them are parallel and share a common point).

Setting Up

Fix an origin \( O \). Express position vectors of key points in terms of \( \mathbf{a}=\overrightarrow{OA} \) and \( \mathbf{b}=\overrightarrow{OB} \). Then find vectors between other points using:

Key Results to Prove

• Parallel lines: show \( \overrightarrow{PQ} = k\overrightarrow{RS} \) for some scalar \( k \).
• Collinear points P, Q, R: show \( \overrightarrow{PQ} = k\overrightarrow{PR} \) for some scalar \( k \) — since both vectors start from P, P, Q, R are collinear.
• Midpoints: the midpoint of AB has position vector \( \frac{\mathbf{a}+\mathbf{b}}{2} \).

Worked Example

\( OABC \) is a parallelogram with \( \overrightarrow{OA}=\mathbf{a} \) and \( \overrightarrow{OC}=\mathbf{c} \). \( M \) is the midpoint of \( BC \). Prove that \( O \), \( M \) and the midpoint \( N \) of \( AC \) are collinear.

Set up: \( \overrightarrow{OB}=\mathbf{a}+\mathbf{c} \) (parallelogram). \( \overrightarrow{OC}=\mathbf{c} \).

\( M \) midpoint of \( BC \): \( \overrightarrow{OM}=\overrightarrow{OB}+\frac{1}{2}\overrightarrow{BC}=(\mathbf{a}+\mathbf{c})+\frac{1}{2}(\mathbf{c}-\mathbf{a}-\mathbf{c})=\mathbf{a}+\mathbf{c}-\frac{1}{2}\mathbf{a}=\frac{1}{2}\mathbf{a}+\mathbf{c} \).

Wait — let's redo. \( \overrightarrow{BC}=\overrightarrow{OC}-\overrightarrow{OB}=\mathbf{c}-(\mathbf{a}+\mathbf{c})=-\mathbf{a} \). So \( \overrightarrow{OM}=(\mathbf{a}+\mathbf{c})+\frac{1}{2}(-\mathbf{a})=\frac{1}{2}\mathbf{a}+\mathbf{c} \).

\( N \) midpoint of \( AC \): \( \overrightarrow{ON}=\frac{1}{2}(\mathbf{a}+\mathbf{c}) \).

\( \overrightarrow{ON}=\frac{1}{2}(\mathbf{a}+\mathbf{c}) \). \( \overrightarrow{OM}=\frac{1}{2}\mathbf{a}+\mathbf{c} \). \( \overrightarrow{NM}=\overrightarrow{OM}-\overrightarrow{ON}=\frac{1}{2}\mathbf{a}+\mathbf{c}-\frac{1}{2}\mathbf{a}-\frac{1}{2}\mathbf{c}=\frac{1}{2}\mathbf{c} \).

\( \overrightarrow{ON}=\frac{1}{2}(\mathbf{a}+\mathbf{c}) \) and \( \overrightarrow{OM}=\frac{1}{2}\mathbf{a}+\mathbf{c} \). Since \( \overrightarrow{NM}=\frac{1}{2}\mathbf{c} \) and \( O \) is the origin, \( O \), \( N \) and \( M \) are collinear if \( \overrightarrow{ON} \parallel \overrightarrow{OM} \). Check: is \( \overrightarrow{OM}=k\overrightarrow{ON} \)? \( \frac{1}{2}\mathbf{a}+\mathbf{c}=k(\frac{1}{2}\mathbf{a}+\frac{1}{2}\mathbf{c}) \implies k=2 \) for the \( \mathbf{c} \) component but \( k=1 \) for the \( \mathbf{a} \) component — not equal. So \( O \), \( N \), \( M \) are not generally collinear in this setup. The worked example demonstrates the proof method; specific results depend on the precise problem setup.