9.03a - Geometric proof with vectors

Writing directed journeys

To travel from $A$ to $B$, you can go directly or go back through a convenient point $O$. Both journeys have the same displacement, even if their travelled distances differ.

Write $\overrightarrow{OA}=\mathbf a$ and $\overrightarrow{OB}=\mathbf b$. Going $A\to O\to B$ gives $-\mathbf a+\mathbf b$, so $\overrightarrow{AB}=\mathbf b-\mathbf a$.

[DIAGRAM: asset_name: Direct and indirect directed journeys; asset_slug: directed-vector-paths; caption: Going back along OA contributes negative a before the move b.; description: A general triangle OAB with directed arrows OA labelled a, OB labelled b and AB labelled b minus a. A second panel reverses the direct arrow to BA and labels it a minus b. The sides are not parallel or equal and O is not on AB.]

A position vector means a vector from the chosen reference point to a point. Changing the reference point changes the position vectors, but not the displacement between the two points. A closed journey returns to its starting point, so its vector sum is $\mathbf0$.

Change only the direction of the requested journey. Every displacement must reverse, including both terms.

Letters such as $\mathbf a$ and $\mathbf b$ describe whole vectors, not lengths. Do not turn a subtraction into an unsigned distance.

Proving lines parallel

Two nonzero vectors are parallel exactly when one is a nonzero scalar multiple of the other.

If $\overrightarrow{MN}=k\overrightarrow{AB}$, multiplying by $k$ changes length and possibly reverses direction, but it does not introduce a new line direction. Require $k\ne0$ and $A\ne B$, $M\ne N$: the zero vector has no direction. For two distinct supporting lines this proves they are parallel. If they share a point, the same calculation instead places them on one line.

Consider a non-degenerate triangle $OAB$, so its side vectors $\mathbf a=\overrightarrow{OA}$ and $\mathbf b=\overrightarrow{OB}$ are not parallel. Mark $M$ on $OA$ and $N$ on $OB$ at the same fraction of the way from $O$.

[DIAGRAM: asset_name: Equal fractions on triangle sides give a parallel segment; asset_slug: parallel-vector-segments; caption: The same fraction multiplies both side vectors and therefore the connecting vector.; description: Two exact non-degenerate triangles OAB. In the first M and N are midpoints of OA and OB so MN is one half of directed AB. In the second M and N are one third of the way from O on their sides so MN is one third of AB. All vector arrows follow M to N and A to B.]

Change only the common fraction along the two sides. Both coefficients must change together.

A negative scalar would also establish parallel directions, with the arrows reversed. A statement about lengths alone would miss that directed relationship.

Proving points collinear

To prove three distinct points lie on one straight line, compare two directed vectors with a common starting point.

Positive $k$ places $B$ and $C$ on the same ray from $A$. Negative $k$ puts them on opposite rays. If $0<k<1$, $C$ lies between $A$ and $B$; if $k>1$, $B$ lies between $A$ and $C$. Values $k=0$ or $1$ make two points coincide, so exclude them when the question specifies three distinct points.