HG25 - Geometric arguments with vectors

Vector routes

In a non-collinear triangle $OAB$, let $M,N$ be the midpoints of $OA,OB$. Why must $MN$ be parallel to $AB$, whatever the triangle's shape? Vector routes will give a general argument.

A position vector records the displacement from a chosen origin $O$ to a point. Write $\overrightarrow{OA}=\mathbf a$, $\overrightarrow{OB}=\mathbf b$. In a non-collinear triangle, $\mathbf a,\mathbf b$ are nonzero and non-parallel; they provide two independent directions for describing the diagram. Throughout this lesson, non-parallel vectors are nonzero and neither is a scalar multiple of the other.

Adding vectors joins routes head to tail. Reversing an arrow changes its sign. For any points $X,Y$, travel from $X$ back to the origin, then to $Y$:

Write the start and destination on each intermediate route; this makes sign errors visible.

[DIAGRAM: asset_name: Directed vector routes; asset_slug: vector-proof-routes; caption: Route through known edges, reverse arrows when necessary, and use the correct fraction of a whole side.; description: Non-collinear triangle OAB with OA a OB b, P dividing AB in ratio AP to PB 1 to 2 and OP highlighted. Beside it a parallelogram OABC with OA a OC b, M midpoint of AB and CM highlighted. All arrows point in the stated directions.]

A parallelogram supplies equal directed opposite sides; a midpoint supplies a half-side vector.

Routes give the same displacement when their starts and ends agree, even if their intermediate points differ.

Use a different internal ratio and reverse the result.

Division and parallelism

For a point $P$ internally dividing $AB$ in ratio $AP:PB=m:n$, with $m,n>0$, the whole side has $m+n$ parts. Therefore $AP$ is $m/(m+n)$ of $AB$, not $m/n$ of it.

If $\overrightarrow{OA}=\mathbf a$, $\overrightarrow{OB}=\mathbf b$ and $t=m/(m+n)$, routing through $A$ gives

The positive coefficients add to $1$, consistent with $P$ lying between $A$ and $B$.

For nonzero vectors $\mathbf u,\mathbf v$, an equation $\mathbf v=k\mathbf u$, with nonzero scalar $k$, proves parallel directions. Positive $k$ preserves direction; negative $k$ reverses it. Their supporting lines are parallel or coincident. The zero vector has no direction, so a zero-vector equation cannot establish a line's direction.

Matching fractions along the two sides produce a parallel segment. Different fractions generally do not. When $\mathbf a,\mathbf b$ are non-parallel, two expressions in them are equal only if their respective coefficients match: otherwise rearrangement would make one a scalar multiple of the other.