Lesson · Vector language
Vectors in the plane
Some configurations have too many points for angle chasing: six points on the sides of a triangle, three centroids, four midpoints. Vectors handle them by refusing to look at the figure at all - every statement becomes an identity, and identities are settled by bookkeeping.
Displacement, and inserting a point
The vector \(\overrightarrow{XY}\) is the displacement from \(X\) to \(Y\): only its direction and length matter, not where it is drawn. Two rules generate everything:
Theorem
For any three points \(X\), \(Y\), \(Z\),
The first is the triangle rule, and it is what lets you insert any point you like into any vector.
Principle
Choose one point \(X\) as origin and rewrite every vector in the problem as a difference of vectors from \(X\). All the geometry disappears into that one substitution; what remains is regrouping a sum.
Midpoints and centroids
Theorem
If \(M\) is the midpoint of \(AB\) and \(T\) the centroid of triangle \(ABC\), then for every point \(X\)
Taking \(X = T\) in the second identity gives the characterisation \(\overrightarrow{TA} + \overrightarrow{TB} + \overrightarrow{TC} = \vec{0}\).
Read the centroid identity in the form \(\overrightarrow{XA} + \overrightarrow{XB} + \overrightarrow{XC} = 3\,\overrightarrow{XT}\) and you have a translation dictionary: a sum of three vectors to the vertices is three times the vector to the centroid. Most municipal-round vector problems are exactly this sentence, disguised.
Parallelograms and decomposition
- Parallelogram rule. \(ABCD\) is a parallelogram exactly when \(\overrightarrow{AB} = \overrightarrow{DC}\); then the diagonal from \(A\) is \(\overrightarrow{AC} = \overrightarrow{AB} + \overrightarrow{AD}\). So a vector split into two components along given directions is a parallelogram you can point at.
- Unique coefficients. If \(\vec{u}\) and \(\vec{v}\) are not parallel, then \(\alpha \vec{u} + \beta \vec{v} = \alpha' \vec{u} + \beta' \vec{v}\) forces \(\alpha = \alpha'\) and \(\beta = \beta'\). Comparing coefficients is a complete proof, not a heuristic.
- Collinearity and ratios. \(P\) lies on \(AB\) with \(AP : PB = t : (1-t)\) exactly when \(\overrightarrow{XP} = (1-t)\overrightarrow{XA} + t\,\overrightarrow{XB}\) for every \(X\).
Example
Let \(P\), \(Q\), \(R\), \(S\) be the midpoints of the sides \(AB\), \(BC\), \(CD\), \(DA\) of an arbitrary quadrilateral. With any origin \(X\),
and the \(\overrightarrow{XB}\) terms cancel:
The same computation gives \(\overrightarrow{SR} = \tfrac{1}{2} \overrightarrow{AC}\). Equal vectors, so \(PQRS\) is a parallelogram - for every quadrilateral, convex or not.
Warning
A vector proof states equality of vectors, which is direction and length together. Do not conclude \(\overrightarrow{XY} = \overrightarrow{ZW}\) from \(XY = ZW\); and when a problem asks for a length, take lengths only at the very end.
Warm-up
Example
In a quadrilateral \(ABCD\), let \(P\), \(Q\), \(R\), \(S\) be the midpoints of \(AB\), \(BC\), \(CD\), \(DA\), and let \(K\), \(L\) be the midpoints of the diagonals \(AC\) and \(BD\). Prove that the segments \(PR\), \(QS\) and \(KL\) all have the same midpoint.
Only one thing needs computing
Two segments have the same midpoint if and only if the midpoints coincide as points - so compute the midpoint of each segment from a common origin \(X\).
Write the midpoint of \(PR\)
\(\tfrac{1}{2}(\overrightarrow{XP} + \overrightarrow{XR}) = \tfrac{1}{4} (\overrightarrow{XA} + \overrightarrow{XB} + \overrightarrow{XC} + \overrightarrow{XD})\).
Now do the other two
The same four vectors come out for \(QS\) and for \(KL\), in a different order. All three midpoints equal the point \(G\) with \(\overrightarrow{XG} = \tfrac{1}{4} (\overrightarrow{XA} + \overrightarrow{XB} + \overrightarrow{XC} + \overrightarrow{XD})\) - the centroid of the four vertices.