G25 - Vector arguments and proofs

Vectors can prove a geometric statement for every figure satisfying the given conditions. You will describe positions and ratios, distinguish parallelism from collinearity, find unknown scalars, and build general midpoint and diagonal proofs. The important final step is to explain what the vector equality means geometrically.

Position vectors and division in a ratio (Higher)

Why does joining the midpoints of two sides of any triangle always give a line parallel to the third side, however lopsided the triangle is? A numerical drawing can suggest the result. A vector route will prove it for every triangle.

You already know

$\overrightarrow{AB}=\overrightarrow{OB}-\overrightarrow{OA}$ . Reversing an arrow changes its sign, and multiplying by a fraction takes that fraction of its displacement.

Choose a fixed origin $O$ . If $\overrightarrow{OA}=\mathbf a$ and $\overrightarrow{OB}=\mathbf b$ , the route from $A$ to $B$ is $\mathbf b-\mathbf a$ . A point halfway along this route has position $\mathbf a+\frac12(\mathbf b-\mathbf a)$ .

Position vector

The position vector of a point is its displacement from a fixed origin. For example, $\overrightarrow{OA}$ locates $A$ relative to $O$ . It differs from $\overrightarrow{AB}$ , which starts at $A$ .

If $AP:PB=2:3$ , the complete segment $AB$ has five parts, so $AP$ is $2/5$ of $AB$ . Start at $A$ and add that fraction of the whole displacement. This route method is not on the formula sheet: learn it.

Worked example 1: a midpoint position

Question. $\overrightarrow{OA}=\mathbf a$ , $\overrightarrow{OB}=\mathbf b$ , and $M$ is the midpoint of $AB$ . Express $\overrightarrow{OM}$ in terms of $\mathbf a,\mathbf b$ .

Plan. Travel to $A$ , then halfway along $AB$ .

\begin{aligned} \overrightarrow{AB}&=\mathbf b-\mathbf a && \text{finish minus start} \\ \overrightarrow{OM}&=\mathbf a+\tfrac12(\mathbf b-\mathbf a) && \text{route through }A \\ &=\tfrac12\mathbf a+\tfrac12\mathbf b && \text{collect coefficients} \end{aligned}

Why add $\mathbf a$ ? Half the displacement from $A$ is not itself a position from $O$ .

Answer. $\overrightarrow{OM}=\frac12(\mathbf a+\mathbf b)$ .

Check. If $\mathbf a=\begin{pmatrix}2\\1\end{pmatrix}$ , $\mathbf b=\begin{pmatrix}8\\7\end{pmatrix}$ , this gives $\begin{pmatrix}5\\4\end{pmatrix}$ , halfway between the endpoint coordinates.

Change only the division ratio, from $1:1$ to $2:3$ . The endpoint vectors remain general.

Worked example 2: unequal division

Question. With the same endpoint position vectors, $AP:PB=2:3$ . Find $\overrightarrow{OP}$ .

Plan. Add $2/5$ of the displacement $AB$ to the position of $A$ .

\begin{aligned} \overrightarrow{AP}&=\tfrac25(\mathbf b-\mathbf a) && \text{two of five equal parts} \\ \overrightarrow{OP}&=\mathbf a+\tfrac25(\mathbf b-\mathbf a) && \text{route through }A \\ &=\tfrac35\mathbf a+\tfrac25\mathbf b && \text{expand and collect} \end{aligned}

Why is the coefficient of $\mathbf b$ $2/5$ ? It records the fraction travelled from $A$ towards $B$ .

Answer. $\overrightarrow{OP}=\frac35\mathbf a+\frac25\mathbf b$ .

Check. With endpoints $(2,1)$ and $(8,7)$ , the position is $(4.4,3.4)$ ; its displacement from $A$ is $2/5$ of $(6,6)$ .

More generally, $\mathbf a+t(\mathbf b-\mathbf a)$ describes the line through distinct points $A,B$ . At $t=0$ you are at $A$ ; at $t=1$ , at $B$ . Between zero and one gives the segment, while other values give its extensions. This expression is not on the formula sheet: derive it from the directed-route rule.

Watch out

Wrong: $AP:PB=2:3$ , so $\overrightarrow{AP}=\frac23\overrightarrow{AB}$ . The three parts belong to $PB$ , not the whole segment. Correct: divide by $2+3$ , giving $\frac25\overrightarrow{AB}$ , then add $\mathbf a$ to get a position vector.

Change just the ratio from example 2.

Parallel vectors and collinear points (Higher)

Scalar multiples give a direction relationship. To prove points lie on one line, you also need their positions to connect that relationship.

Parallel directions and collinear points

Non-zero vectors that are scalar multiples have parallel directions, possibly opposite. Points are collinear when they lie on one straight line. If $\overrightarrow{AP}=k\overrightarrow{AB}$ and $A\ne B$ , both routes start at $A$ , so $P$ lies on the line $AB$ .

In contrast, $\overrightarrow{CD}=3\overrightarrow{AB}$ alone does not put all four points on one line. For example, the horizontal arrows from $(0,0)$ to $(2,0)$ , and from $(0,1)$ to $(6,1)$ , satisfy that relation but lie on separate parallel lines. The scalar-multiple test is not on the formula sheet: learn it.

Worked example 3: beyond an endpoint

Question. $\overrightarrow{AB}=2\mathbf a+4\mathbf b\ne\mathbf0$ , and $\overrightarrow{AP}=3\mathbf a+6\mathbf b$ . Show that $A,B,P$ are collinear and locate $P$ .

Plan. Look for one multiplier that changes both coefficients together.

\begin{aligned} \tfrac32\overrightarrow{AB}&=\tfrac32(2\mathbf a+4\mathbf b) && \text{test a common multiplier} \\ &=3\mathbf a+6\mathbf b=\overrightarrow{AP} && \text{both coefficients match} \end{aligned}

Why does this prove collinearity? Both arrows start at $A$ , so the parallel directions belong to the same line.

Answer. $A,B,P$ are collinear. Since $3/2>1$ , $P$ is beyond $B$ on the ray from $A$ .

Check. $AP$ is $1.5$ times $AB$ , so it passes $B$ before reaching $P$ .

Change only the given vector $AP$ . The reference vector $AB$ stays fixed, but the multiplier is now negative.

Worked example 4: the opposite ray

Question. $\overrightarrow{AB}=2\mathbf a+4\mathbf b\ne\mathbf0$ , and $\overrightarrow{AP}=-\mathbf a-2\mathbf b$ . Show that $A,B,P$ are collinear and locate $P$ .

Plan. Factor $AP$ as a multiple of $AB$ .

\begin{aligned} \overrightarrow{AP}&=-\tfrac12(2\mathbf a+4\mathbf b) && \text{factor both coefficients} \\ &=-\tfrac12\overrightarrow{AB} && \text{identify the reference vector} \end{aligned}

Why does the sign matter? A negative factor reverses the direction from the common start $A$ .

Answer. The points are collinear, in the order $P,A,B$ . Point $P$ is half as far from $A$ as $B$ is, on the opposite ray.

Check. Multiplying $2$ and $4$ by $-1/2$ gives $-1$ and $-2$ , so both components agree.

A factor between zero and one instead puts the point between the endpoints. The zero vector has no unique direction; its being a multiple of every vector cannot by itself establish a line or non-trivial parallelism.

Watch out

Wrong: the coefficient of $\mathbf a$ is doubled, so the vectors are parallel. Both coefficients must use the same multiplier; $2\mathbf a+\mathbf b$ is not twice $\mathbf a+\mathbf b$ when $\mathbf a,\mathbf b$ are non-parallel. Correct: compare the complete vectors, then state the common point for a collinearity proof.

Change just $AP$ once more to locate an interior point.

Finding unknown scalars from collinearity (Higher)

With two non-parallel, non-zero vectors, equal vector expressions must have equal coefficients of each vector.

Why? If $r\mathbf a+s\mathbf b=u\mathbf a+v\mathbf b$ , rearranging gives $(r-u)\mathbf a=(v-s)\mathbf b$ . Non-parallel vectors cannot be non-zero multiples of each other, so both coefficients must be zero. Hence $r=u$ and $s=v$ . This coefficient-comparison principle is not on the formula sheet: learn it.

Worked example 5: comparing coefficients

Question. $\mathbf a,\mathbf b$ are non-parallel. $\overrightarrow{OP}=2\mathbf a+3\mathbf b$ , $\overrightarrow{OQ}=k\mathbf a+9\mathbf b$ , and $O,P,Q$ are collinear. Find $k$ .

Plan. Write one complete vector as $t$ times the other, then compare coefficients.

\begin{aligned} \overrightarrow{OQ}&=t\overrightarrow{OP} && \text{common-start collinearity} \\ k\mathbf a+9\mathbf b&=2t\mathbf a+3t\mathbf b && \text{expand} \\ 9&=3t\quad\Rightarrow\quad t=3 && \text{compare }\mathbf b \\ k&=2t=6 && \text{compare }\mathbf a \end{aligned}

Why introduce $t$ ? The unknown coefficient $k$ is not necessarily the multiplier of the whole vector.

Answer. $k=6$ .

Check. $6\mathbf a+9\mathbf b=3(2\mathbf a+3\mathbf b)$ .

Change only the $\mathbf b$ -coefficient in $OQ$ , from $9$ to $-6$ .

Worked example 6: a negative route multiplier

Question. $\overrightarrow{OP}=2\mathbf a+3\mathbf b$ , $\overrightarrow{OQ}=k\mathbf a-6\mathbf b$ , with non-parallel $\mathbf a,\mathbf b$ and collinear $O,P,Q$ . Find $k$ .

Plan. Use $OQ=tOP$ and compare the two coefficients.

\begin{aligned} k\mathbf a-6\mathbf b&=2t\mathbf a+3t\mathbf b && \text{scalar relation} \\ -6&=3t\quad\Rightarrow\quad t=-2 && \text{compare }\mathbf b \\ k&=2t=-4 && \text{compare }\mathbf a \end{aligned}

Why keep the negative sign? $Q$ is now on the opposite ray from $P$ .

Answer. $k=-4$ .

Check. $-4\mathbf a-6\mathbf b=-2(2\mathbf a+3\mathbf b)$ .

Often the useful vectors are not supplied directly: form routes from position vectors first. In the diagram, $M$ is the midpoint of $AB$ , while $Q$ is beyond $A$ on $OA$ .

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Papers 1, 2 and 3

G25 - Vector arguments and proofs

Position vectors and division in a ratio (Higher)

Parallel vectors and collinear points (Higher)

Finding unknown scalars from collinearity (Higher)

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