3D vectors & proof by contradiction – A-level Maths exam questions with mark scheme

1CoreCalculator4 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: $(-3)^{2}=+9$.

2StandardCalculator4 marks

Prove by contradiction that $\sqrt3$ is irrational.

[You may assume that if $n^{2}$ is a multiple of 3 then $n$ is a multiple of 3.] [4]

Assumes $\sqrt3$ is rational: $\sqrt3=\dfrac{a}{b}$ where $a,b$ are integers with no common factors ($b\neq0$) (allow 'in its simplest form', 'in lowest terms' or 'HCF$(a,b)=1$')B1
Squares and rearranges: $a^{2}=3b^{2}$, so $a^{2}$ is a multiple of 3 and hence $a$ is a multiple of 3M1
Writes $a=3k$: $9k^{2}=3b^{2}\Rightarrow b^{2}=3k^{2}$, so $b$ is also a multiple of 3M1
States that $a$ and $b$ have a common factor 3, contradicting the assumption, so $\sqrt3$ is irrationalA1
Answer: Proof as in mark scheme
Examiner tip: State 'no common factors' in your assumption – without it there is nothing to contradict.

3HardCalculator5 marks

Prove by contradiction that there are infinitely many prime numbers. [5]

Assumes there are finitely many primes and lists them all: $p_{1},p_{2},\ldots,p_{n}$B1
Considers $N=p_{1}p_{2}\cdots p_{n}+1$M1
States that $N$ leaves remainder 1 when divided by each of $p_{1},\ldots,p_{n}$, so none of them is a factor of $N$M1
Deduces that $N$ is either prime itself (and, being larger than every $p_i$, not in the list) or has a prime factor not in the list (allow: $N>1$ so has a prime factor, but no $p_i$ divides $N$)A1
Contradiction with the list containing all primes, so there are infinitely many primesA1
Answer: Proof as in mark scheme
Examiner tip: $N$ need not itself be prime (e.g. $2\cdot3\cdot5\cdot7\cdot11\cdot13+1=30031=59\times509$) – say 'prime or has a new prime factor'.

4CoreCalculator3 marks

Prove by contradiction that there are no integers $a$ and $b$ such that $$6a+9b=1$$ [3]

Assumes there exist integers $a$ and $b$ with $6a+9b=1$B1
Factorises: $3(2a+3b)=1\Rightarrow 2a+3b=\tfrac13$ (or states that $6a+9b=3(2a+3b)$ is a multiple of 3)M1
$2a+3b$ is an integer but $\tfrac13$ is not (or: 1 is not a multiple of 3) – contradiction, so no such integers existA1
Answer: Proof as in mark scheme
Examiner tip: End with a clear statement of the contradiction and the conclusion – the final mark is for the logic, not the algebra.

5StandardCalculator7 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.

6HardCalculator7 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.

7StretchCalculator7 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.

Still losing marks on this topic?

Book a free introductory session and we'll find exactly which step is costing you.

Book a free intro