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Geometry & measures · Vectors

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Vector proofs

What a vector equation actually proves — a scalar multiple gives parallel and nothing more, a multiple plus a shared point gives collinear — and how to write the conclusion in the form a mark scheme rewards.

Geometry & measures · Vectors

Vector proofs

What a vector equation actually proves — a scalar multiple gives parallel and nothing more, a multiple plus a shared point gives collinear — and how to write the conclusion in the form a mark scheme rewards.

Why it works

A vector holds only two pieces of information: how far, and in which direction. It carries nothing at all about where. XY→\overrightarrow{XY} is the journey from XX to YY; it does not know where on the page XX happens to sit. Everything a vector proof can and cannot do falls out of that one sentence.

Fact 1 — a scalar multiple proves parallel, and only parallel. If XY→=k PQ→\overrightarrow{XY} = k\,\overrightarrow{PQ} for some number kk, then the journey X→YX \to Y runs in the same direction as P→QP \to Q (or exactly the reverse, when kk is negative) and is kk times as long. Same direction ⇒\Rightarrow the two segments are parallel. That is the whole of the deduction. It says nothing whatsoever about where those segments are.

Make it concrete. Take PQ→=(32)\overrightarrow{PQ} = \binom{3}{2} and XY→=(64)=2(32)\overrightarrow{XY} = \binom{6}{4} = 2\binom{3}{2}, so XY→=2 PQ→\overrightarrow{XY} = 2\,\overrightarrow{PQ}. Now put P=(0,0)P = (0,0), Q=(3,2)Q = (3,2), X=(0.5, 4.5)X = (0.5,\,4.5), Y=(6.5, 8.5)Y = (6.5,\,8.5):PQXYPQXYFour points sitting on two different parallel lines. The algebra is word for word the algebra you would write in a collinearity proof — and these points are not collinear. So "one vector is a multiple of the other" can never, by itself, prove that points lie on one straight line.

Fact 2 — a multiple plus a shared point proves collinear. Suppose BC→=k AB→\overrightarrow{BC} = k\,\overrightarrow{AB}. Both of those vectors involve BB. Stand at BB. The direction of AB→\overrightarrow{AB} picks out one straight line through BB, and AA lies on it. Because BC→\overrightarrow{BC} points along that same direction, CC lies on it too — and there is only one line through BB in a given direction. So AA, BB and CC all sit on that single line, which is exactly what collinear means.

The shared point is doing all the work, and it is the half students leave out. Compare two calculations:
  • A=(1,1)A = (1,1), B=(3,2)B = (3,2), C=(7,4)C = (7,4). Then
AB→=(21)\overrightarrow{AB} = \binom{2}{1} and BC→=(42)=2 AB→\overrightarrow{BC} = \binom{4}{2} = 2\,\overrightarrow{AB}, and the two vectors share BB — so AA, BB, CC are collinear. ✓
  • A=(1,1)A = (1,1), B=(3,2)B = (3,2), D=(10,0)D = (10,0), E=(12,1)E = (12,1). Then
DE→=(21)=1×AB→\overrightarrow{DE} = \binom{2}{1} = 1 \times \overrightarrow{AB} — an even tidier multiple — yet AA, BB, DD, EE are nowhere near one line. ✗

Identical algebra, opposite geometry. The only difference is the shared point. So the sentence a mark scheme is waiting for has two clauses: *parallel, because one is a scalar multiple of the other, and they have the point BB in common*.

Two other shapes of conclusion.

A parallelogram. A quadrilateral is a parallelogram when one pair of opposite sides is equal in length and parallel. In vector language that is a single equation: for PQRSPQRS, PQ→=SR→\overrightarrow{PQ} = \overrightarrow{SR}. Equality is stronger than being a multiple — it pins down the length as well as the direction — and that extra strength is precisely the "equal in length" half. Watch the direction: going round the shape, PQ→\overrightarrow{PQ} and RS→\overrightarrow{RS} run in opposite senses, so what usually drops out is RS→=−PQ→\overrightarrow{RS} = -\overrightarrow{PQ}. The minus sign is not a failure. It says the two sides are the same length and parallel; deducing "so it isn't a parallelogram" from it is the standard slip.

A ratio. If AP→=34AB→\overrightarrow{AP} = \tfrac{3}{4}\overrightarrow{AB}, then PP is three-quarters of the way from AA to BB. The rest of the journey is PB→=AB→−AP→=14AB→\overrightarrow{PB} = \overrightarrow{AB} - \overrightarrow{AP} = \tfrac{1}{4}\overrightarrow{AB}, so AP:PB=34:14=3:1.AP : PB = \tfrac{3}{4} : \tfrac{1}{4} = 3 : 1. A fraction compares a part with the whole; a ratio compares the two parts. Writing 3:43:4 is the reflex to resist.

Writing it so it scores. On these questions the final mark is almost never for algebra. The method mark buys the vector expression; the last mark buys a sentence in English naming what has been shown and why. So "BC→=52AB→\overrightarrow{BC} = \tfrac{5}{2}\overrightarrow{AB}" followed by a full stop earns the method and stops there. What earns the rest is: "BC→\overrightarrow{BC} is a scalar multiple of AB→\overrightarrow{AB}, so ABAB and BCBC are parallel; they share the point BB, so AA, BB and CC are collinear." A claim, a reason, and — for collinearity — the common point.

One check before you write that sentence: the multiple must be a plain number. If your comparison still contains an a\mathbf{a} or a b\mathbf{b}, you have not finished. Factorise until what sits outside the bracket is a number and what sits inside it is identical to the other vector.