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GCSE / IGCSE Maths – Vectors and Proof

Vector geometry, algebraic proof, geometric proof (Higher).

12 questions  ·  48 marks  ·  answers included  ·  GCSE/IGCSE Higher
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This worksheet covers:

  • Column vectors and vector geometry
  • Algebraic proof
  • Geometric proof at Higher tier

Non-calculator throughout; allow about 60 minutes and set out every proof in full.

Questions

Section A – Skills

Non-calculator.
1

$\mathbf{a} = \begin{pmatrix} 3 \\ -1 \end{pmatrix}$ and $\mathbf{b} = \begin{pmatrix} -2 \\ 4 \end{pmatrix}$.(a) Write $2\mathbf{a} + \mathbf{b}$ as a column vector.(b) Write $\mathbf{a} – 3\mathbf{b}$ as a column vector.(c) Find the magnitude of $\mathbf{a} – 3\mathbf{b}$, giving your answer in the form $k\sqrt{10}$.

Show answer
(a) $\begin{pmatrix} 4 \\ 2 \end{pmatrix}$(b) $\begin{pmatrix} 9 \\ -13 \end{pmatrix}$(c) $\sqrt{81 + 169} = \sqrt{250} = 5\sqrt{10}$
[5]
2

In triangle $OAB$, $\overrightarrow{OA} = \mathbf{a}$ and $\overrightarrow{OB} = \mathbf{b}$. $M$ is the midpoint of $AB$.(a) Find $\overrightarrow{AB}$ in terms of $\mathbf{a}$ and $\mathbf{b}$.(b) Show that $\overrightarrow{OM} = \frac{1}{2}(\mathbf{a} + \mathbf{b})$.

Show answer
(a) $\mathbf{b} – \mathbf{a}$(b) $\overrightarrow{OM} = \mathbf{a} + \frac{1}{2}(\mathbf{b} – \mathbf{a}) = \frac{1}{2}\mathbf{a} + \frac{1}{2}\mathbf{b} = \frac{1}{2}(\mathbf{a} + \mathbf{b})$
[3]
3

Prove that the sum of any three consecutive integers is a multiple of $3$.

Show answer
$n + (n+1) + (n+2) = 3n + 3 = 3(n+1)$, which is a multiple of $3$
[2]
4

Prove that $(n+3)^2 – (n-3)^2$ is a multiple of $12$ for all integers $n$.

Show answer
$(n^2 + 6n + 9) – (n^2 – 6n + 9) = 12n$, a multiple of $12$
[3]
5

Prove that the product of two odd numbers is always odd.

Show answer
$(2m+1)(2n+1) = 4mn + 2m + 2n + 1 = 2(2mn + m + n) + 1$, which is odd
[3]

Section B – Exam-style

6

In triangle $OAB$, $\overrightarrow{OA} = \mathbf{a}$ and $\overrightarrow{OB} = \mathbf{b}$. Point $C$ lies on $OA$ with $OC : CA = 2 : 1$ and point $D$ lies on $AB$ with $AD : DB = 1 : 2$.(a) Find $\overrightarrow{CD}$ in terms of $\mathbf{a}$ and $\mathbf{b}$, simplifying your answer.(b) What does your answer tell you about the lines $CD$ and $OB$?

Show answer
(a) $\overrightarrow{CD} = \frac{1}{3}\mathbf{a} + \frac{1}{3}(\mathbf{b} – \mathbf{a}) = \frac{1}{3}\mathbf{b}$(b) $CD$ is parallel to $OB$ and $CD = \frac{1}{3}OB$
[5]
7

$AB$ is a diameter of a circle with centre $O$, and $C$ is a point on the circumference. Prove that angle $ACB = 90^\circ$. You may use the fact that $OA = OB = OC$.

Show answer
Triangles $OAC$ and $OBC$ are isosceles (radii). Let angle $OAC = x$ so angle $OCA = x$; let angle $OBC = y$ so angle $OCB = y$. Angles in triangle $ABC$: $2x + 2y = 180^\circ$, so $x + y = 90^\circ$, i.e. angle $ACB = 90^\circ$
[4]
8

Prove that $n^2 – n$ is even for every integer $n$.

Show answer
$n^2 – n = n(n – 1)$, the product of two consecutive integers; one of them is even, so the product is even
[3]
9

Prove that the difference between the squares of any two consecutive odd numbers is a multiple of $8$.

Show answer
$(2n+3)^2 – (2n+1)^2 = (4n^2 + 12n + 9) – (4n^2 + 4n + 1) = 8n + 8 = 8(n + 1)$
[4]

Section C – Challenge

10

$\overrightarrow{OA} = 6\mathbf{a}$ and $\overrightarrow{OB} = 6\mathbf{b}$. $M$ is the point on $OA$ with $OM : MA = 1 : 2$ and $N$ is the point on $AB$ with $AN : NB = 2 : 1$.(a) Find $\overrightarrow{ON}$ in terms of $\mathbf{a}$ and $\mathbf{b}$.(b) Show that $MN$ is parallel to $OB$ and find the ratio $MN : OB$.

Show answer
(a) $\overrightarrow{AB} = 6\mathbf{b} – 6\mathbf{a}$, $\overrightarrow{ON} = 6\mathbf{a} + \frac{2}{3}(6\mathbf{b} – 6\mathbf{a}) = 2\mathbf{a} + 4\mathbf{b}$(b) $\overrightarrow{MN} = (2\mathbf{a} + 4\mathbf{b}) – 2\mathbf{a} = 4\mathbf{b} = \frac{2}{3}\overrightarrow{OB}$, so parallel; $MN : OB = 2 : 3$
[6]
11

$P$, $Q$ and $R$ have position vectors $\begin{pmatrix} 2 \\ 1 \end{pmatrix}$, $\begin{pmatrix} 5 \\ 7 \end{pmatrix}$ and $\begin{pmatrix} k \\ 11 \end{pmatrix}$. Find the value of $k$ for which $P$, $Q$ and $R$ lie on a straight line.

Show answer
$\overrightarrow{PQ} = \begin{pmatrix} 3 \\ 6 \end{pmatrix}$, $\overrightarrow{PR} = \begin{pmatrix} k-2 \\ 10 \end{pmatrix}$; parallel when $\frac{k-2}{3} = \frac{10}{6}$, so $k = 7$
[4]
12

(a) A two-digit number has tens digit $a$ and units digit $b$. Prove that the difference between the number and the number formed by reversing its digits is always a multiple of $9$.(b) Prove that $x^2 – 6x + 10$ is positive for all real values of $x$.

Show answer
(a) $(10a + b) – (10b + a) = 9a – 9b = 9(a – b)$(b) $x^2 – 6x + 10 = (x – 3)^2 + 1 \ge 1 > 0$ since $(x-3)^2 \ge 0$
[6]

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