Vectors in 2D and 3D – 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: $\overrightarrow{AN}=\overrightarrow{AO}+\overrightarrow{ON}=-\mathbf{a}+\tfrac23(\mathbf{a}+\mathbf{c})$.

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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 $\overrightarrow{CB}$.

7CoreCalculator4 marks

(a) Given $\mathbf{a}=2\mathbf{i}-3\mathbf{j}+6\mathbf{k}$, find $|\mathbf{a}|$ and hence a unit vector in the direction of $\mathbf{a}$. [2]

(b) The points $A$ and $B$ have coordinates $(1,2,-1)$ and $(4,-2,11)$. Find the distance $AB$. [2]

(a) $|\mathbf{a}|=\sqrt{4+9+36}=7$B1
$\tfrac17(2\mathbf{i}-3\mathbf{j}+6\mathbf{k})$ (follow through their $|\mathbf{a}|$)B1
(b) Finds $\overrightarrow{AB}=3\mathbf{i}-4\mathbf{j}+12\mathbf{k}$ (or $B-A$ with at least two correct components) and uses PythagorasM1
$13$A1
Answer: (a) $7$; $\tfrac17(2\mathbf{i}-3\mathbf{j}+6\mathbf{k})$ (b) $13$
Examiner tip: Square the negative components properly: $(-3)^{2}=+9$.

8StandardCalculator7 marks

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

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

(b) Find the position vector of the midpoint $M$ of $AB$. [2]

The point $C$ has position vector $p\mathbf{i}+5\mathbf{j}+q\mathbf{k}$, where $p$ and $q$ are constants. Given that $A$, $B$ and $C$ lie on a straight line,

(a) $\overrightarrow{AB}=2\mathbf{i}-4\mathbf{j}+4\mathbf{k}$B1
$|\overrightarrow{AB}|=6$B1
(b) $\overrightarrow{OM}=\overrightarrow{OA}+\tfrac12\overrightarrow{AB}$ or $\tfrac12(\overrightarrow{OA}+\overrightarrow{OB})$M1
$4\mathbf{i}-\mathbf{j}$ (accept $4\mathbf{i}-\mathbf{j}+0\mathbf{k}$)A1
(c) Sets $\overrightarrow{AC}=\lambda\overrightarrow{AB}$: $(p-3)\mathbf{i}+4\mathbf{j}+(q+2)\mathbf{k}=\lambda(2\mathbf{i}-4\mathbf{j}+4\mathbf{k})$ (or equivalent, e.g. $\overrightarrow{BC}=\mu\overrightarrow{AB}$, or $\overrightarrow{OC}=\overrightarrow{OA}+\lambda\overrightarrow{AB}$)M1
Uses the $\mathbf{j}$ component to find the scalar ($\lambda=-1$, or $\mu=-2$ for $\overrightarrow{BC}$)M1
$p=1,\ q=-6$A1
Answer: (a) $2\mathbf{i}-4\mathbf{j}+4\mathbf{k}$, $6$ (b) $4\mathbf{i}-\mathbf{j}$ (c) $p=1,\ q=-6$
Examiner tip: Collinear means $\overrightarrow{AC}$ is a scalar multiple of $\overrightarrow{AB}$ – find the scalar from the component you know.

9HardCalculator7 marks

The points $A(2,1,-3)$, $B(5,-1,0)$ and $C(4,3,2)$ are three vertices of the parallelogram $ABCD$.

(a) Find the coordinates of $D$. [2]

(b) Find the size of angle $ABC$, giving your answer in degrees to one decimal place. [3]

(a) Uses $\overrightarrow{AD}=\overrightarrow{BC}$ (or $\overrightarrow{OD}=\overrightarrow{OA}+\overrightarrow{OC}-\overrightarrow{OB}$)M1
$D(1,5,-1)$A1
(b) Finds all three lengths: $AB=\sqrt{22}$, $BC=\sqrt{21}$, $AC=\sqrt{33}$M1
Uses the cosine rule: $\cos ABC=\dfrac{22+21-33}{2\sqrt{22}\sqrt{21}}$M1
$76.6^\circ$A1
(c) Area $=AB\times BC\times\sin ABC$ (or $2\times\tfrac12ab\sin C$)M1
$20.9$ (exact $\sqrt{437}$)A1
Answer: (a) $(1,5,-1)$ (b) $76.6^\circ$ (c) $20.9$
Examiner tip: The parallelogram is twice the triangle – using $\tfrac12ab\sin C$ alone halves your answer.

10StretchCalculator7 marks

The diagram shows triangle $OAB$ with $\overrightarrow{OA}=\mathbf{a}$ and $\overrightarrow{OB}=\mathbf{b}$, where $\mathbf{a}$ and $\mathbf{b}$ are not parallel. $M$ is the midpoint of $AB$ and $N$ is the point on $OA$ such that $ON:NA=2:1$. The lines $OM$ and $BN$ intersect at $X$.

(a) Find $\overrightarrow{OX}$ in terms of $\mathbf{a}$ and $\mathbf{b}$. [6]

(b) Hence find the ratio $OX:XM$. [1]

(a) $\overrightarrow{OM}=\tfrac12(\mathbf{a}+\mathbf{b})$ so $\overrightarrow{OX}=\lambda\cdot\tfrac12(\mathbf{a}+\mathbf{b})$M1
$\overrightarrow{BN}=\tfrac23\mathbf{a}-\mathbf{b}$M1
Second expression: $\overrightarrow{OX}=\mathbf{b}+\mu\left(\tfrac23\mathbf{a}-\mathbf{b}\right)=\tfrac23\mu\mathbf{a}+(1-\mu)\mathbf{b}$M1
Equates coefficients of $\mathbf{a}$ and $\mathbf{b}$ (valid as $\mathbf{a},\mathbf{b}$ not parallel): $\tfrac{\lambda}{2}=\tfrac23\mu$, $\tfrac{\lambda}{2}=1-\mu$ and solvesM1
$\mu=\tfrac35$ or $\lambda=\tfrac45$A1
$\overrightarrow{OX}=\tfrac25\mathbf{a}+\tfrac25\mathbf{b}$A1
(b) $OX:XM=4:1$B1
Answer: (a) $\tfrac25\mathbf{a}+\tfrac25\mathbf{b}$ (b) $4:1$
Examiner tip: Use two different parameters for the two lines – using $\lambda$ for both forces a wrong answer.

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