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Geometry & measures · Vectors

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Vectors on a shape

Expressing any journey around a figure in terms of two given vectors — why every route gives the same answer, why going against an arrow flips the sign, and why a point splitting a side in the ratio 2:3 sits two fifths along, not two thirds.

Geometry & measures · Vectors

Vectors on a shape

Expressing any journey around a figure in terms of two given vectors — why every route gives the same answer, why going against an arrow flips the sign, and why a point splitting a side in the ratio 2:3 sits two fifths along, not two thirds.

Why it works

A vector is a journey, not a place. It records two things and only two things — how far, and in which direction. It records nothing about where the journey started. That one fact is the engine of this whole topic, because it means the same vector can appear in several different places on the same diagram. In a parallelogram OABCOABC with OA→=a\overrightarrow{OA} = \mathbf{a}, the opposite side CBCB is the same length as OAOA and points the same way, so CB→=a\overrightarrow{CB} = \mathbf{a} as well. Not "something like a\mathbf{a}" — the very same vector, written with the very same letter.

Journeys join up. Walk from AA to BB, then from BB to CC, and you have gone from AA to CC:

AC→=AB→+BC→.\overrightarrow{AC} = \overrightarrow{AB} + \overrightarrow{BC}.

The middle letters match and drop out, which is a handy way to check a chain is legal. It works for any number of legs: AD→=AB→+BC→+CD→\overrightarrow{AD} = \overrightarrow{AB} + \overrightarrow{BC} + \overrightarrow{CD}.

Any route will do. Every route from XX to YY gives the same vector, because every route finishes in the same place. So when a question asks for a side you have no arrow along, you are free to hunt around the figure for a chain of legs you do know — through the origin, around the outside, along a diagonal — and the answer comes out the same either way. Two routes disagreeing is a signal that one of them contains a slip, so a second route is a free check.

Going against an arrow flips the sign. If AB→\overrightarrow{AB} takes you from AA to BB, then the return trip undoes it exactly: AB→+BA→=AA→=0\overrightarrow{AB} + \overrightarrow{BA} = \overrightarrow{AA} = \mathbf{0}, so BA→=−AB→\overrightarrow{BA} = -\overrightarrow{AB}. Same length, opposite direction.

Put those together and you get the single most-asked question in the topic. Given OA→=a\overrightarrow{OA} = \mathbf{a} and OB→=b\overrightarrow{OB} = \mathbf{b}, find AB→\overrightarrow{AB}. There is no arrow from AA to BB, so travel via OO. The first leg runs backwards along a\mathbf{a}, so it is −a-\mathbf{a}:

AB→=AO→+OB→=−a+b.\overrightarrow{AB} = \overrightarrow{AO} + \overrightarrow{OB} = -\mathbf{a} + \mathbf{b}.

That minus sign is not a rule to memorise. It is there because the first leg of the journey goes against the arrow. Watch it fail with numbers: take a=(14)\mathbf{a} = \binom{1}{4} and b=(52)\mathbf{b} = \binom{5}{2}, so AA is at (1,4)(1, 4) and BB is at (5,2)(5, 2). The journey from AA to BB is (4−2)\binom{4}{-2}, and −a+b=(−1+5−4+2)=(4−2)-\mathbf{a} + \mathbf{b} = \binom{-1 + 5}{-4 + 2} = \binom{4}{-2}. ✓ Whereas a+b=(66)\mathbf{a} + \mathbf{b} = \binom{6}{6} — nowhere near. The sign is doing real work.

Position vectors. Pick one point as the origin OO. Then every point on the figure has an address: the position vector of PP is OP→\overrightarrow{OP}, the journey from OO to PP. With addresses a\mathbf{a} and b\mathbf{b} for AA and BB, the rule above reads destination minus start: AB→=b−a\overrightarrow{AB} = \mathbf{b} - \mathbf{a}. Same thing, tidier.

Midpoints. If MM is the midpoint of ABAB, get to AA first and then go half of the way along ABAB:

OM→=a+12(b−a)=12a+12b.\overrightarrow{OM} = \mathbf{a} + \tfrac{1}{2}(\mathbf{b} - \mathbf{a}) = \tfrac{1}{2}\mathbf{a} + \tfrac{1}{2}\mathbf{b}.

Add the two ends and halve — exactly the coordinate midpoint, in vector clothing.

The ratio trap, which costs more marks than anything else here. Suppose MM lies on ABAB with AM:MB=2:3AM:MB = 2:3. The tempting move is AM→=23AB→\overrightarrow{AM} = \tfrac{2}{3}\overrightarrow{AB}, and it is wrong. A ratio of 2:32:3 chops ABAB into 2+3=52 + 3 = 5 equal parts and puts MM after two of them, so MM is two fifths of the way along.

Make it concrete. Say ABAB is 1010 cm. Splitting it 2:32:3 gives AM=4AM = 4 cm and MB=6MB = 6 cm, and 44 out of 1010 is 25\tfrac{2}{5}. ✓ Now try the tempting version: 23\tfrac{2}{3} of 1010 cm is 6.6˙6.\dot{6} cm, which leaves MB=3.3˙MB = 3.\dot{3} cm — a ratio of 2:12:1, not the 2:32:3 you were given. So the fraction is always your part over the total number of parts, never part over part. Then

OM→=OA→+25AB→=a+25(b−a)=35a+25b.\overrightarrow{OM} = \overrightarrow{OA} + \tfrac{2}{5}\overrightarrow{AB} = \mathbf{a} + \tfrac{2}{5}(\mathbf{b} - \mathbf{a}) = \tfrac{3}{5}\mathbf{a} + \tfrac{2}{5}\mathbf{b}.

There is a quick sanity check on the last line: for any point on the line ABAB, the two coefficients add to 11, and 35+25=1\tfrac{3}{5} + \tfrac{2}{5} = 1. ✓ It catches arithmetic slips (it will not catch a wrong ratio, since 13+23=1\tfrac{1}{3} + \tfrac{2}{3} = 1 too), so use it as a tidy-up, not a proof.

Sides that are multiples. Trapezia and other figures often give one side as a multiple of another: "DCDC is parallel to ABAB and DC=3ABDC = 3AB". Parallel and three times as long, pointing the same way round the shape, means DC→=3a\overrightarrow{DC} = 3\mathbf{a}. If the two sides pointed in opposite directions it would be −3a-3\mathbf{a} — so read the letter order off the shape before you commit to a sign.