Edexcel IGCSE Math B 8 vectors and transformation geometry

Use this Vectors and Transformations hub to move from vector magnitude and geometry into translations, enlargements, combined transformations and matrices.

Covered syllabus points

8.A Scalars and vectors 8.B Vector notation 8.C Directed line segments 8.D Parallel, unit and position vectors 8.E Vector sums and differences 8.F Vector magnitude 8.G Scalar multiplication of vectors 8.H Resultant vectors 8.I Vector geometry 8.J Transformations 8.K Combined transformations 8.L Vectors and matrices

[Maximum number: 5]

The coordinates of point A are (1,8) and the coordinates of point B are (10,-4).

(a)

Write down $\overrightarrow{AB}$ as a column vector.

[ 1 ]

$\overrightarrow{AB}=\binom{9}{-12}$

Marks: B1.

(b)

The point P lies on the line AB so that AP : PB = 1 : 2.

Find $\overrightarrow{OP}$ as a column vector.

[ 2 ]

$\overrightarrow{OP}=\overrightarrow{OA}+\overrightarrow{AP} =\binom{1}{8}+\frac13\binom{9}{-12} =\binom{4}{4}$

Alternative:
$\overrightarrow{OP}=\overrightarrow{OB}+\overrightarrow{BP} =\binom{10}{-4}-\frac23\binom{9}{-12} =\binom{4}{4}$

Marks: M1, A1.

(c)

Figure for Question (c) — Edexcel IGCSE Math B

Calculate the modulus of A P.

[ 2 ]

$|\overrightarrow{AP}|=\sqrt{3^{2}+(-4)^{2}}=5$
Final answer:
5

Follow through from $\overrightarrow{AP}$ in part (b), or from $\overrightarrow{AP}=\frac13\overrightarrow{AB}$ , is allowed.

[Maximum number: 5]

$21 \overrightarrow{O X}=\binom{2}{5}$ and $\overrightarrow{O Y}=\binom{-2}{7}$

(a)

Express $\overrightarrow{X Y}$ as a column vector.

\overrightarrow{X Y}=(

[ 2 ]

$\overrightarrow{XY}=\overrightarrow{OY}-\overrightarrow{OX}$ $\binom{-2}{7}-\binom{2}{5}=\binom{-4}{2}$

Final answer:
$\binom{-4}{2}$

M1 for a correct vector subtraction method.
A1 for $\binom{-4}{2}$ .

(b)

Calculate $|\overrightarrow{X Y}|$ , giving your answer as a surd.

|\overrightarrow{X Y}|=

[ 2 ]

$|\overrightarrow{XY}|=\sqrt{(-4)^{2}+2^{2}}$ $=\sqrt{20}=2\sqrt{5}$

Final answer:
$2\sqrt{5}$

M1ft: using the magnitude of their $\overrightarrow{XY}$ .
A1: $2\sqrt{5}$ .

(c)

Hence write down a unit vector that is parallel to $\overrightarrow{X Y}$ .

[ 1 ]

One unit vector parallel to $\overrightarrow{XY}$ is:
$\frac{1}{2\sqrt{5}}\binom{-4}{2} =\frac{\sqrt{5}}{5}\binom{-2}{1}$

OR

$\frac{\sqrt{5}}{5}\binom{2}{-1}$

A1: a correct unit vector parallel to $\overrightarrow{XY}$ .

[Maximum number: 7]

Figure 1 shows triangle OAB in which $\overrightarrow{O A}=4 \mathbf{a}$ and $\overrightarrow{O B}=8 \mathbf{b}$
P is the point on OA such that O P: O A=1: 4

(a)

Express in terms of a or bor a and b where appropriate,

[ 2 ]

(i)

$\overrightarrow{A B}$

$\overrightarrow{AB}=8\mathbf{b}-4\mathbf{a}$

B1.

(ii)

$\overrightarrow{PO}$

$\overrightarrow{PO}=-\mathbf{a}$

B1.

(b)

Q is the point on OB such that OQ : OB = 1 : m where m is a constant.
$\overrightarrow{PQ}=\alpha\overrightarrow{AB}$ , where $\alpha$ is a scalar.

Using vectors, find the value of m and the value of $\alpha$ .

[ 3 ]

$\overrightarrow{PQ}=-\mathbf a+\frac{8}{m}\mathbf b$ $\overrightarrow{AB}=8\mathbf b-4\mathbf a$ $-\mathbf a+\frac{8}{m}\mathbf b=\alpha(8\mathbf b-4\mathbf a)$

Comparing coefficients:
$\alpha=\frac14,\quad m=4$

Marks: M1, A1, A1

(c)

R is the point on AB such that AR : AB = 1 : n where n is a constant.

Find and simplify an expression for $\overrightarrow{PR}$ in terms of n, a and b.

[ 2 ]

NB: candidates must use vectors as required by the question.

$\overrightarrow{PR}=\overrightarrow{PA}+\overrightarrow{AR}$ $\overrightarrow{PR}=3\mathbf{a}+\frac{1}{n}(8\mathbf{b}-4\mathbf{a})$ $\overrightarrow{PR}=\left(3-\frac{4}{n}\right)\mathbf{a}+\frac{8}{n}\mathbf{b}$

Equivalent form:
$\overrightarrow{PR}=\frac{(3n-4)\mathbf{a}+8\mathbf{b}}{n}$

M1.
A1.

(d)

Given that PR is parallel to OB,

find the value of n.

[ 2 ]

For $\overrightarrow{PR}$ to be parallel to $\overrightarrow{OB}$ , the coefficient of $\mathbf{a}$ in $\overrightarrow{PR}$ is zero:
$3-\frac{4}{n}=0$ $3=\frac{4}{n}$ $n=\frac{4}{3}$

OR, since $\overrightarrow{PR}=6\mathbf{b}$ ,
$\frac{8}{n}=6$ $n=\frac{4}{3}$

M1.
A1.

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Covered syllabus points

8 Vectors and transformation geometry question 1

Question (a)

Question (b)

Question (c)

8 Vectors and transformation geometry question 2

Question (a)

Question (b)

Question (c)

8 Vectors and transformation geometry question 3

Question (a)

Question (i)

Question (ii)

Question (b)

Question (c)

Question (d)