Vectors (2D) – A-level Maths exam questions with mark scheme

1CoreCalculator6 marks

The vectors $\mathbf{a}$ and $\mathbf{b}$ are given by $\mathbf{a}=3\mathbf{i}-4\mathbf{j}$ and $\mathbf{b}=-\mathbf{i}+2\mathbf{j}$.

(a) Find $2\mathbf{a}-3\mathbf{b}$. [2]

(b) Find a unit vector in the direction of $\mathbf{a}$. [2]

(a) for $2(3\mathbf{i}-4\mathbf{j})-3(-\mathbf{i}+2\mathbf{j})$ with at least one component correctM1
(a) $9\mathbf{i}-14\mathbf{j}$A1
(b) for $|\mathbf{a}|=\sqrt{3^{2}+4^{2}}=5$M1
(b) $\tfrac35\mathbf{i}-\tfrac45\mathbf{j}$A1
(c) for $\tan\theta=\tfrac43$ (or $\tfrac34$ with correct subtraction from $90^\circ$)M1
(c) $53.1^\circ$ below the positive $\mathbf{i}$ direction (accept $-53.1^\circ$ or $306.9^\circ$)A1
Answer: (a) $9\mathbf{i}-14\mathbf{j}$ (b) $\tfrac35\mathbf{i}-\tfrac45\mathbf{j}$ (c) $53.1^\circ$ below $\mathbf{i}$
Examiner tip: Sketch the vector before finding angles – $\mathbf{a}$ points down and right, so the angle is below the $\mathbf{i}$ direction.

2StandardCalculator4 marks

Relative to a fixed origin $O$, the points $A$, $B$ and $C$ have position vectors $$\overrightarrow{OA}=2\mathbf{i}-\mathbf{j},\quad \overrightarrow{OB}=5\mathbf{i}+3\mathbf{j},\quad \overrightarrow{OC}=11\mathbf{i}+11\mathbf{j}.$$

(a) Show that $A$, $B$ and $C$ lie on a straight line. [3]

(b) Write down the ratio $AB:BC$. [1]

(a) for finding two of $\overrightarrow{AB},\ \overrightarrow{BC},\ \overrightarrow{AC}$ by subtracting position vectorsM1
(a) two correct vectors, e.g. $\overrightarrow{AB}=3\mathbf{i}+4\mathbf{j}$ and $\overrightarrow{BC}=6\mathbf{i}+8\mathbf{j}$ (or $\overrightarrow{AC}=9\mathbf{i}+12\mathbf{j}$)A1
(a) one vector shown to be a scalar multiple of the other (e.g. $\overrightarrow{BC}=2\overrightarrow{AB}$ or $\overrightarrow{AC}=3\overrightarrow{AB}$), so parallel, and a common point stated, so $A$, $B$, $C$ are collinearA1
(b) $1:2$ (accept any equivalent ratio, e.g. $5:10$ or $\sqrt{25}:\sqrt{100}$)B1
Answer: (a) $\overrightarrow{BC}=2\overrightarrow{AB}$ with common point $B$ (b) $1:2$
Examiner tip: 'Parallel' alone isn't enough for collinear – you must say the vectors share a common point.

3HardCalculator6 marks

The diagram shows a parallelogram $OABC$ with $\overrightarrow{OA}=\mathbf{a}$ and $\overrightarrow{OC}=\mathbf{c}$.

The point $M$ is the midpoint of $CB$. The point $N$ lies on $OB$ such that $ON:NB=2:1$.

(a) Find, in terms of $\mathbf{a}$ and $\mathbf{c}$, (i) $\overrightarrow{OB}$ (ii) $\overrightarrow{AM}$. [2]

(b) Prove that $A$, $N$ and $M$ are collinear. [4]

(a)(i) $\overrightarrow{OB}=\mathbf{a}+\mathbf{c}$B1
(a)(ii) $\overrightarrow{AM}=\mathbf{c}-\tfrac12\mathbf{a}$ (accept $-\tfrac12\mathbf{a}+\mathbf{c}$ or any equivalent)B1
(b) Uses $\overrightarrow{ON}=\tfrac23\overrightarrow{OB}=\tfrac23(\mathbf{a}+\mathbf{c})$ (or $\overrightarrow{BN}=-\tfrac13(\mathbf{a}+\mathbf{c})$) to find a vector between two of $A$, $N$, $M$ involving $N$M1
(b) $\overrightarrow{AN}=\tfrac23\mathbf{c}-\tfrac13\mathbf{a}$ (or $\overrightarrow{NM}=\tfrac13\mathbf{c}-\tfrac16\mathbf{a}$, or the negative of either, correctly simplified)A1
(b) Expresses one vector as a scalar multiple of another, e.g. $\overrightarrow{AM}=\tfrac32\overrightarrow{AN}$ or $\overrightarrow{AN}=2\overrightarrow{NM}$ or $\overrightarrow{AM}=3\overrightarrow{NM}$M1
(b) Complete conclusion: vectors parallel and share a common point (e.g. $A$, or $N$) stated, so $A$, $N$, $M$ are collinear (A0 if 'common point' is omitted)A1
Answer: (a) $\mathbf{a}+\mathbf{c}$, $\ \mathbf{c}-\tfrac12\mathbf{a}$ (b) $\overrightarrow{AM}=\tfrac32\overrightarrow{AN}$, common point $A$
Examiner tip: Route every vector through points you know: $\overrightarrow{AN}=\overrightarrow{AO}+\overrightarrow{ON}=-\mathbf{a}+\tfrac23(\mathbf{a}+\mathbf{c})$.

4StandardCalculator4 marks

The vector $\mathbf{p}=2k\mathbf{i}+(k-3)\mathbf{j}$, where $k$ is a constant, has magnitude $\sqrt{65}$.

Find the possible values of $k$. [4]

for $(2k)^{2}+(k-3)^{2}=65$M1
$5k^{2}-6k-56=0$A1
for solving their 3-term quadratic: $(5k+14)(k-4)=0$M1
$k=4$ and $k=-\tfrac{14}{5}$A1
Answer: $k=4,\ k=-\tfrac{14}{5}$
Examiner tip: Square the whole component: $(2k)^{2}=4k^{2}$, not $2k^{2}$.

5HardCalculator6 marks

Relative to a fixed origin $O$, the points $A$ and $B$ have position vectors $4\mathbf{i}+2\mathbf{j}$ and $\mathbf{i}+6\mathbf{j}$ respectively.

(a) Find $\overrightarrow{AB}$ and $|\overrightarrow{AB}|$. [2]

(b) Find the size of angle $AOB$, giving your answer in degrees to 1 decimal place. [2]

(a) $\overrightarrow{AB}=-3\mathbf{i}+4\mathbf{j}$B1
(a) $|\overrightarrow{AB}|=5$B1
(b) for a complete method for angle $AOB$, e.g. cosine rule with $|\overrightarrow{OA}|=\sqrt{20}$, $|\overrightarrow{OB}|=\sqrt{37}$ and their $|\overrightarrow{AB}|$: $\cos AOB=\dfrac{20+37-25}{2\sqrt{20}\sqrt{37}}$, or $\tan^{-1}6-\tan^{-1}\tfrac12$ (angles each vector makes with $\mathbf{i}$)M1
(b) $54.0^\circ$ (awrt)A1
(c) for $\tfrac12\sqrt{20}\sqrt{37}\sin(\text{their angle})$ or another valid method (e.g. enclosing rectangle $4\times6$ minus triangles, or $\tfrac12|4\times6-2\times1|$)M1
(c) $11$ (accept awrt $11.0$ from a rounded angle)A1
Answer: (a) $-3\mathbf{i}+4\mathbf{j}$, $5$ (b) $54.0^\circ$ (c) $11$
Examiner tip: $\overrightarrow{AB}=\overrightarrow{OB}-\overrightarrow{OA}$ – 'end minus start'; reversing it changes the vector (though not its length).

6StretchCalculator7 marks

Relative to a fixed origin $O$, the points $A$, $B$ and $C$ have coordinates $(1,2)$, $(7,5)$ and $(9,11)$ respectively.

(a) Find the position vector of the point $D$ such that $ABCD$ is a parallelogram. [2]

(b) The point $P$ lies on $AC$ such that $AP:PC=1:3$. Find the position vector of $P$. [2]

(a) for $\overrightarrow{OD}=\overrightarrow{OA}+\overrightarrow{BC}$ (or $\overrightarrow{OC}+\overrightarrow{BA}$)M1
(a) $3\mathbf{i}+8\mathbf{j}$A1
(b) for $\overrightarrow{OP}=\overrightarrow{OA}+\tfrac14\overrightarrow{AC}$ with $\overrightarrow{AC}=8\mathbf{i}+9\mathbf{j}$M1
(b) $3\mathbf{i}+\tfrac{17}{4}\mathbf{j}$A1
(c) for a complete method for the area of triangle $ABC$ (or $ABD$) or of the parallelogram directly, e.g. enclosing $8\times9$ rectangle minus right-angled triangles $9,\ 6,\ 36$ and the $2\times3$ rectangle at $(9,2)$ for triangle $ABC$; or $\tfrac12|AB||BC|\sin B$; or base $\times$ perpendicular heightM1
(c) a correct intermediate value: triangle area $=15$ (or, for the direct method, $72-42$)A1
(c) parallelogram area $=30$A1
Answer: (a) $3\mathbf{i}+8\mathbf{j}$ (b) $3\mathbf{i}+\tfrac{17}{4}\mathbf{j}$ (c) $30$
Examiner tip: Go round the parallelogram in order: $ABCD$ means $\overrightarrow{AD}=\overrightarrow{BC}$, not $\overrightarrow{CB}$.

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