Hard GCSE Vector geometry and proof Questions

Write down the two vectors the question gives you

Every vector in the figure has to be written in terms of $\mathbf{a}$ and $\mathbf{b}$, so start from the two the question hands you: $\overrightarrow{OA} = \mathbf{a}$ and $\overrightarrow{OB} = \mathbf{b}$.

Write down the position vector of C

The question gives $\overrightarrow{OC}$ directly, so the position vector of $C$ is $2\mathbf{a} + 3\mathbf{b}$.

Write down the position vector of P, the midpoint of OA

The midpoint of $OA$ has position vector the average of the two ends, so $\overrightarrow{OP} = \frac{1}{2}\mathbf{a}$. Halving is exact — keep it as a fraction.

Write down the position vector of Q, the midpoint of AB

The midpoint of $AB$ has position vector the average of the two ends, so $\overrightarrow{OQ} = \frac{1}{2}\mathbf{a} + \frac{1}{2}\mathbf{b}$. Halving is exact — keep it as a fraction.

Write down the position vector of R, the midpoint of BC

The midpoint of $BC$ has position vector the average of the two ends, so $\overrightarrow{OR} = \mathbf{a} + 2\mathbf{b}$. Halving is exact — keep it as a fraction.

Write down the position vector of S, the midpoint of CO

The midpoint of $CO$ has position vector the average of the two ends, so $\overrightarrow{OS} = \mathbf{a} + \frac{3}{2}\mathbf{b}$. Halving is exact — keep it as a fraction.

Work out the first vector in terms of a and b

Going from $P$ to $Q$ through $O$ gives $\overrightarrow{PQ} = \frac{1}{2}\mathbf{b}$.

Work out the second vector in terms of a and b

The same journey for $O$ to $B$ gives $\overrightarrow{OB} = \mathbf{b}$.

Compare the two and read off the scalar

Every coefficient of $\overrightarrow{PQ}$ is $\frac{1}{2}$ times the matching coefficient of $\overrightarrow{OB}$, and the same multiplier works for both $\mathbf{a}$ and $\mathbf{b}$ — which is exactly what $k = \frac{1}{2}$ means.

State the triangle law that every one of these questions rests on

Any vector between two points of the figure can be routed through $O$: go backwards along the first position vector and forwards along the second. That single move is what turns a picture into algebra.

Note that reversing a vector reverses its sign

$\overrightarrow{PQ}$ and $\overrightarrow{QP}$ have the same length and opposite directions, so one is minus the other. Writing the letters the wrong way round is the single most common way to lose the marks here.

Recall what a midpoint does to a vector

A midpoint halves the vector along the line, so its position vector is the average of the two ends. Every half stays an exact fraction — never round.

Recall how a point that divides a line in a given ratio is found

The ratio $m:n$ cuts the line into $m+n$ equal parts, so the point is $\frac{m}{m+n}$ of the way along — not $\frac{m}{n}$ of the way along. That slip is the classic error in this topic.

State the test for two vectors being parallel

Two non-zero vectors are parallel exactly when one is a scalar multiple of the other. The scalar itself is the scale factor between their lengths, and a negative scalar means they point in opposite directions.

State the answer

So $k = \frac{1}{2}$: $\overrightarrow{PQ}$ is $\frac{1}{2}$ times $\overrightarrow{OB}$, which also proves that $PQ$ is parallel to $OB$.

Write down the two vectors the question gives you

Every vector in the figure has to be written in terms of $\mathbf{a}$ and $\mathbf{b}$, so start from the two the question hands you: $\overrightarrow{OA} = \mathbf{a}$ and $\overrightarrow{OB} = \mathbf{b}$.

Write down the position vector of M, the midpoint of OB

The midpoint of $OB$ has position vector the average of the two ends, so $\overrightarrow{OM} = \frac{1}{2}\mathbf{b}$. Halving is exact — keep it as a fraction.

Write down the position vector of N, the midpoint of OA

The midpoint of $OA$ has position vector the average of the two ends, so $\overrightarrow{ON} = \frac{1}{2}\mathbf{a}$. Halving is exact — keep it as a fraction.

Find where AM meets BN

A point on $AM$ is $\overrightarrow{OA} + s\,\overrightarrow{AM}$ and a point on $BN$ is $\overrightarrow{OB} + t\,\overrightarrow{BN}$. Because $\mathbf{a}$ and $\mathbf{b}$ are not parallel, the $\mathbf{a}$ parts and the $\mathbf{b}$ parts must match separately, which gives two equations and $s = \frac{2}{3}$. Hence $\overrightarrow{OG} = \frac{1}{3}\mathbf{a} + \frac{1}{3}\mathbf{b}$.

Work out the first vector in terms of a and b

Going from $A$ to $G$ through $O$ gives $\overrightarrow{AG} = -\frac{2}{3}\mathbf{a} + \frac{1}{3}\mathbf{b}$.

Work out the second vector in terms of a and b

The same journey for $A$ to $M$ gives $\overrightarrow{AM} = -\mathbf{a} + \frac{1}{2}\mathbf{b}$.

Compare the two and read off the scalar

Every coefficient of $\overrightarrow{AG}$ is $\frac{2}{3}$ times the matching coefficient of $\overrightarrow{AM}$, and the same multiplier works for both $\mathbf{a}$ and $\mathbf{b}$ — which is exactly what $k = \frac{2}{3}$ means.

State the triangle law that every one of these questions rests on

Any vector between two points of the figure can be routed through $O$: go backwards along the first position vector and forwards along the second. That single move is what turns a picture into algebra.

Note that reversing a vector reverses its sign

$\overrightarrow{AG}$ and $\overrightarrow{GA}$ have the same length and opposite directions, so one is minus the other. Writing the letters the wrong way round is the single most common way to lose the marks here.

Recall what a midpoint does to a vector

A midpoint halves the vector along the line, so its position vector is the average of the two ends. Every half stays an exact fraction — never round.

Recall how a point that divides a line in a given ratio is found

The ratio $m:n$ cuts the line into $m+n$ equal parts, so the point is $\frac{m}{m+n}$ of the way along — not $\frac{m}{n}$ of the way along. That slip is the classic error in this topic.

State the test for two vectors being parallel

Two non-zero vectors are parallel exactly when one is a scalar multiple of the other. The scalar itself is the scale factor between their lengths, and a negative scalar means they point in opposite directions.

State the test for three points being collinear

Parallel is not enough on its own — two parallel vectors can sit on different lines. It is the shared point that upgrades parallel to collinear, so the proof must say both things.

Say why a and b must not be parallel

$\mathbf{a}$ and $\mathbf{b}$ are not parallel, so the only way a combination of them can vanish is for both coefficients to vanish. That is what lets you compare the $\mathbf{a}$ parts and the $\mathbf{b}$ parts separately, and it is the hidden engine of every vector proof.

State the answer

So $k = \frac{2}{3}$: $\overrightarrow{AG}$ is $\frac{2}{3}$ times $\overrightarrow{AM}$, which also proves that $AG$ is parallel to $AM$.

Write down the two vectors the question gives you

Every vector in the figure has to be written in terms of $\mathbf{a}$ and $\mathbf{b}$, so start from the two the question hands you: $\overrightarrow{OA} = \mathbf{a}$ and $\overrightarrow{OB} = \mathbf{b}$.

Write down the position vector of C, the fourth vertex of the parallelogram OACB

In a parallelogram $OACB$ the diagonals $OC$ and $AB$ bisect each other, so the two diagonals have the same midpoint and $\overrightarrow{OC} = \overrightarrow{OA} + \overrightarrow{OB}$. That gives $\overrightarrow{OC} = \mathbf{a} + \mathbf{b}$.

Write down the position vector of P, which divides OC in the ratio 1 to 2

$P$ is $\frac{1}{3}$ of the way from $O$ to $C$, because the whole of $OC$ is $1 + 2 = 3$ parts. That gives $\overrightarrow{OP} = \frac{1}{3}\mathbf{a} + \frac{1}{3}\mathbf{b}$.

Write down the position vector of M, the midpoint of OB

The midpoint of $OB$ has position vector the average of the two ends, so $\overrightarrow{OM} = \frac{1}{2}\mathbf{b}$. Halving is exact — keep it as a fraction.

Work out the first of the two vectors

$\overrightarrow{AP} = -\frac{2}{3}\mathbf{a} + \frac{1}{3}\mathbf{b}$, the first part of the line.

Work out the second of the two vectors

$\overrightarrow{PM} = -\frac{1}{3}\mathbf{a} + \frac{1}{6}\mathbf{b}$, the second part of the line.

Write one as a multiple of the other and read off the ratio

The two vectors are parallel and point the same way, so $P$ lies on $AM$ and the lengths are in the ratio of the multipliers. Clearing the fraction turns $1:\frac{1}{2}$ into $2:1$.

State the triangle law that every one of these questions rests on

Any vector between two points of the figure can be routed through $O$: go backwards along the first position vector and forwards along the second. That single move is what turns a picture into algebra.

Note that reversing a vector reverses its sign

$\overrightarrow{AP}$ and $\overrightarrow{PA}$ have the same length and opposite directions, so one is minus the other. Writing the letters the wrong way round is the single most common way to lose the marks here.

Recall what a midpoint does to a vector

A midpoint halves the vector along the line, so its position vector is the average of the two ends. Every half stays an exact fraction — never round.

Recall how a point that divides a line in a given ratio is found

The ratio $m:n$ cuts the line into $m+n$ equal parts, so the point is $\frac{m}{m+n}$ of the way along — not $\frac{m}{n}$ of the way along. That slip is the classic error in this topic.

State the test for two vectors being parallel

Two non-zero vectors are parallel exactly when one is a scalar multiple of the other. The scalar itself is the scale factor between their lengths, and a negative scalar means they point in opposite directions.

State the test for three points being collinear

Parallel is not enough on its own — two parallel vectors can sit on different lines. It is the shared point that upgrades parallel to collinear, so the proof must say both things.

Say why a and b must not be parallel

$\mathbf{a}$ and $\mathbf{b}$ are not parallel, so the only way a combination of them can vanish is for both coefficients to vanish. That is what lets you compare the $\mathbf{a}$ parts and the $\mathbf{b}$ parts separately, and it is the hidden engine of every vector proof.

State the answer

So $AP:PM = 2:1$. The ratio is a ratio of lengths along one straight line, so it is read straight off the scalar that links the two vectors.

Write down the two vectors the question gives you

Every vector in the figure has to be written in terms of $\mathbf{a}$ and $\mathbf{b}$, so start from the two the question hands you: $\overrightarrow{OA} = \mathbf{a}$ and $\overrightarrow{OB} = \mathbf{b}$.

Write down the position vector of M, the midpoint of OB

The midpoint of $OB$ has position vector the average of the two ends, so $\overrightarrow{OM} = \frac{1}{2}\mathbf{b}$. Halving is exact — keep it as a fraction.

Write down the position vector of N, the midpoint of OA

The midpoint of $OA$ has position vector the average of the two ends, so $\overrightarrow{ON} = \frac{1}{2}\mathbf{a}$. Halving is exact — keep it as a fraction.

Write down the position vector of L, the midpoint of AB

The midpoint of $AB$ has position vector the average of the two ends, so $\overrightarrow{OL} = \frac{1}{2}\mathbf{a} + \frac{1}{2}\mathbf{b}$. Halving is exact — keep it as a fraction.

Find where AM meets OL

A point on $AM$ is $\overrightarrow{OA} + s\,\overrightarrow{AM}$ and a point on $OL$ is $t\,\overrightarrow{OL}$. Because $\mathbf{a}$ and $\mathbf{b}$ are not parallel, the $\mathbf{a}$ parts and the $\mathbf{b}$ parts must match separately, which gives two equations and $s = \frac{2}{3}$. Hence $\overrightarrow{OG} = \frac{1}{3}\mathbf{a} + \frac{1}{3}\mathbf{b}$.

Work out the first of the two vectors

$\overrightarrow{OG} = \frac{1}{3}\mathbf{a} + \frac{1}{3}\mathbf{b}$, the first part of the line.

Work out the second of the two vectors

$\overrightarrow{GL} = \frac{1}{6}\mathbf{a} + \frac{1}{6}\mathbf{b}$, the second part of the line.

Write one as a multiple of the other and read off the ratio

The two vectors are parallel and point the same way, so $G$ lies on $OL$ and the lengths are in the ratio of the multipliers. Clearing the fraction turns $1:\frac{1}{2}$ into $2:1$.

State the triangle law that every one of these questions rests on

Any vector between two points of the figure can be routed through $O$: go backwards along the first position vector and forwards along the second. That single move is what turns a picture into algebra.

Note that reversing a vector reverses its sign

$\overrightarrow{OG}$ and $\overrightarrow{GO}$ have the same length and opposite directions, so one is minus the other. Writing the letters the wrong way round is the single most common way to lose the marks here.

Recall what a midpoint does to a vector

A midpoint halves the vector along the line, so its position vector is the average of the two ends. Every half stays an exact fraction — never round.

Recall how a point that divides a line in a given ratio is found

The ratio $m:n$ cuts the line into $m+n$ equal parts, so the point is $\frac{m}{m+n}$ of the way along — not $\frac{m}{n}$ of the way along. That slip is the classic error in this topic.

State the test for two vectors being parallel

Two non-zero vectors are parallel exactly when one is a scalar multiple of the other. The scalar itself is the scale factor between their lengths, and a negative scalar means they point in opposite directions.

State the test for three points being collinear

Parallel is not enough on its own — two parallel vectors can sit on different lines. It is the shared point that upgrades parallel to collinear, so the proof must say both things.

State the answer

So $OG:GL = 2:1$. The ratio is a ratio of lengths along one straight line, so it is read straight off the scalar that links the two vectors.

Write down the two vectors the question gives you

Every vector in the figure has to be written in terms of $\mathbf{a}$ and $\mathbf{b}$, so start from the two the question hands you: $\overrightarrow{OA} = \mathbf{a}$ and $\overrightarrow{OB} = \mathbf{b}$.

Write down the position vector of M, the midpoint of OB

The midpoint of $OB$ has position vector the average of the two ends, so $\overrightarrow{OM} = \frac{1}{2}\mathbf{b}$. Halving is exact — keep it as a fraction.

Write down the position vector of N, the midpoint of OA

The midpoint of $OA$ has position vector the average of the two ends, so $\overrightarrow{ON} = \frac{1}{2}\mathbf{a}$. Halving is exact — keep it as a fraction.

Find where AM meets BN

A point on $AM$ is $\overrightarrow{OA} + s\,\overrightarrow{AM}$ and a point on $BN$ is $\overrightarrow{OB} + t\,\overrightarrow{BN}$. Because $\mathbf{a}$ and $\mathbf{b}$ are not parallel, the $\mathbf{a}$ parts and the $\mathbf{b}$ parts must match separately, which gives two equations and $s = \frac{2}{3}$. Hence $\overrightarrow{OG} = \frac{1}{3}\mathbf{a} + \frac{1}{3}\mathbf{b}$.

Work out the first of the two vectors

$\overrightarrow{BG} = \frac{1}{3}\mathbf{a} - \frac{2}{3}\mathbf{b}$, the first part of the line.

Work out the second of the two vectors

$\overrightarrow{GN} = \frac{1}{6}\mathbf{a} - \frac{1}{3}\mathbf{b}$, the second part of the line.

Write one as a multiple of the other and read off the ratio

The two vectors are parallel and point the same way, so $G$ lies on $BN$ and the lengths are in the ratio of the multipliers. Clearing the fraction turns $1:\frac{1}{2}$ into $2:1$.

State the triangle law that every one of these questions rests on

Any vector between two points of the figure can be routed through $O$: go backwards along the first position vector and forwards along the second. That single move is what turns a picture into algebra.

Note that reversing a vector reverses its sign

$\overrightarrow{BG}$ and $\overrightarrow{GB}$ have the same length and opposite directions, so one is minus the other. Writing the letters the wrong way round is the single most common way to lose the marks here.

Recall what a midpoint does to a vector

A midpoint halves the vector along the line, so its position vector is the average of the two ends. Every half stays an exact fraction — never round.

Recall how a point that divides a line in a given ratio is found

The ratio $m:n$ cuts the line into $m+n$ equal parts, so the point is $\frac{m}{m+n}$ of the way along — not $\frac{m}{n}$ of the way along. That slip is the classic error in this topic.

State the test for two vectors being parallel

Two non-zero vectors are parallel exactly when one is a scalar multiple of the other. The scalar itself is the scale factor between their lengths, and a negative scalar means they point in opposite directions.

State the test for three points being collinear

Parallel is not enough on its own — two parallel vectors can sit on different lines. It is the shared point that upgrades parallel to collinear, so the proof must say both things.

Say why a and b must not be parallel

$\mathbf{a}$ and $\mathbf{b}$ are not parallel, so the only way a combination of them can vanish is for both coefficients to vanish. That is what lets you compare the $\mathbf{a}$ parts and the $\mathbf{b}$ parts separately, and it is the hidden engine of every vector proof.

State the answer

So $BG:GN = 2:1$. The ratio is a ratio of lengths along one straight line, so it is read straight off the scalar that links the two vectors.