CAIE IGCSE Additional Math 13 Vectors in…ensions

[Maximum number: 7]

The point O is the origin.
Two points P and Q are such that $\overrightarrow{PQ}$ is in the same direction as $-\mathbf{i}+5\mathbf{j}$ .

(a)

The point R is such that $\overrightarrow{OR}$ is in the same direction as $\overrightarrow{PQ}$ and the magnitude of $\overrightarrow{OR}$ is $3\sqrt{26}$ .
Find $\overrightarrow{OR}$ .

[ 3 ]

$[\overrightarrow{OR}=]k(-\mathbf{i}+5\mathbf{j})$ or $k\binom{-1}{5}$
B1 soi
$\sqrt{1^2+5^2}=\sqrt{26}$
B1 for finding the magnitude of $\overrightarrow{PQ}$ ;
allow unsimplified.
$-3\mathbf{i}+15\mathbf{j}$ or $\binom{-3}{15}$
B1
Allow $3(-\mathbf{i}+5\mathbf{j})$ or $3\binom{-1}{5}$ . Do not isw.

(b)

$\overrightarrow{OP}$ is in the same direction as $2\mathbf{i}-3\mathbf{j}$ and $\overrightarrow{OQ}=10\mathbf{i}+6\mathbf{j}$ .
Find $\overrightarrow{OP}$ .

[ 4 ]

$a(2\mathbf{i}-3\mathbf{j})+b(-\mathbf{i}+5\mathbf{j})=10\mathbf{i}+6\mathbf{j}$ oe
2a-b=10 oe, -3a+5b=6 oe
or $\frac{10-2a}{6+3a}=\frac{-1}{5}$ or $\frac{10+b}{6-5b}=\frac{-2}{3}$
M1 dep on previous B1 for equating like vectors.
a=8 or b=6 $16\mathbf{i}-24\mathbf{j}$ oe
Allow $8(2\mathbf{i}-3\mathbf{j})$ or $8\binom{2}{-3}$ .

[Maximum number: 8]

In this question lengths are in centimetres and time, t, is in seconds.
A particle P is moving in a straight line with a speed of 26 in the direction of the vector $\binom{5}{-12}$ .

All through this question allow correctly written vectors in $\mathbf{i},\mathbf{j}$ form;
incorrectly written column vectors do not earn full marks unless the correct form is recovered.

(a)

Find the velocity vector of P.

When t=0, P passes through a point A which has position vector $\binom{3}{6}$ .

[ 2 ]

$\binom{10}{-24}$ oe, isw
B1 for the magnitude of the direction vector is 13 soi, or M1 for $\sqrt{5^2+(-12)^2}$ oe.
A1

(b)

Write down the position vector of P at time t.

At the same time that P passes through A, a particle Q passes through a point B.
The position vector of Q at time t is given by $\binom{8 t-5}{2-25 t}$ .
The distance between P and Q at time t is d.

[ 2 ]

$\binom{3}{6}+t\binom{10}{-24}$ oe, isw
M1 FT for $\binom{3}{6}+t(\text{their }\binom{10}{-24})$ .
A1

(c)

Show that $d^{2}=m t^{2}+n t+r$ , where m, n and r are integers to be found.

[ 3 ]

$\binom{3+10t}{6-24t}-\binom{8t-5}{2-25t}=\binom{8+2t}{4+t}$ oe, soi
Finds $d^2$ or d for the difference of their position vectors.
M1 FT.
$d^2=5t^2+40t+80$ mark final answer

(d)

Hence show that P and Q do not collide.

[ 1 ]

Correct argument including d, e.g. when d=0, t=-4, but $t\geqslant0$ , so no collision;
or d>0 when $t\geqslant0$ , so $d\ne0$ .
B1 dep on part (c) being correct.

[Maximum number: 6]

(a)

Figure for Question (a) — CAIE IGCSE Additional Math

The diagram shows a triangle OAB. The point P lies on AB. The ratio A P: P B is 1: 3.
Given that $\overrightarrow{O A}=\mathbf{a}$ and $\overrightarrow{O B}=\mathbf{b}$ , find an expression for $\overrightarrow{O P}$ in terms of a and b. Simplify your answer.

[ 2 ]

$\frac{3}{4} \mathbf{a}+\frac{1}{4} \mathbf{b}$ or equivalent simplified
expression
B1 for $\mathbf{a}+\frac{1}{4}(\mathbf{b}-\mathbf{a})$ or $\mathbf{b}+\frac{3}{4}(\mathbf{a}-\mathbf{b})$
ое
or for $3(\overrightarrow{O P}-\mathbf{a})=\mathbf{b}-\overrightarrow{O P}$ oe

(b)

Vector q has magnitude $12 \sqrt{5}$ and direction $\binom{6}{-3}$ .

Vector r has magnitude $15 \sqrt{2}$ and direction $\binom{-5}{5}$ .
Find the unit vector in the direction of q+r.

[ 4 ]

$\mathbf q=\binom{24}{-12}$ oe. M1 for $12\sqrt{5}\times \dfrac{1}{\sqrt{6^{2}+(-3)^{2}}}\binom{6}{-3}$ oe, soi.
$\mathbf r=\binom{-15}{15}$ oe. M1 for $15\sqrt{2}\times \dfrac{1}{\sqrt{(-5)^{2}+5^{2}}}\binom{-5}{5}$ oe, soi.
If M0 M0, then SC1 for the unit direction vectors $\dfrac{1}{\sqrt{45}}\binom{6}{-3}$ or better and $\dfrac{1}{\sqrt{50}}\binom{-5}{5}$ or better.
$|\mathbf q+\mathbf r|=\left|\binom{9}{3}\right|=\sqrt{9^{2}+3^{2}}$ . M1, FT their $\mathbf q+\mathbf r$ providing at least M1 previously awarded.
Unit vector in direction $\mathbf q+\mathbf r$ : $\dfrac{1}{\sqrt{90}}\binom{9}{3}$ oe, isw. A1

[Maximum number: 3]

The point O is the origin.
Two points P and Q are such that $\overrightarrow{PQ}$ is in the same direction as $-\mathbf{i}+5\mathbf{j}$ .

The point R is such that $\overrightarrow{OR}$ is in the same direction as $\overrightarrow{PQ}$ and the magnitude of $\overrightarrow{OR}$ is $3\sqrt{26}$ .
Find $\overrightarrow{OR}$ .

$[\overrightarrow{OR}=]k(-\mathbf{i}+5\mathbf{j})$ or $k\binom{-1}{5}$
B1 soi
$\sqrt{1^2+5^2}=\sqrt{26}$
B1 for finding the magnitude of $\overrightarrow{PQ}$ ;
allow unsimplified.
$-3\mathbf{i}+15\mathbf{j}$ or $\binom{-3}{15}$
B1
Allow $3(-\mathbf{i}+5\mathbf{j})$ or $3\binom{-1}{5}$ . Do not isw.

[Maximum number: 4]

The point O is the origin.
Two points P and Q are such that $\overrightarrow{PQ}$ is in the same direction as $-\mathbf{i}+5\mathbf{j}$ .

$\overrightarrow{OP}$ is in the same direction as $2\mathbf{i}-3\mathbf{j}$ and $\overrightarrow{OQ}=10\mathbf{i}+6\mathbf{j}$ .
Find $\overrightarrow{OP}$ .

$a(2\mathbf{i}-3\mathbf{j})+b(-\mathbf{i}+5\mathbf{j})=10\mathbf{i}+6\mathbf{j}$ oe
2a-b=10 oe, -3a+5b=6 oe
or $\frac{10-2a}{6+3a}=\frac{-1}{5}$ or $\frac{10+b}{6-5b}=\frac{-2}{3}$
M1 dep on previous B1 for equating like vectors.
a=8 or b=6 $16\mathbf{i}-24\mathbf{j}$ oe
Allow $8(2\mathbf{i}-3\mathbf{j})$ or $8\binom{2}{-3}$ .

CAIE IGCSE Additional Math 13 Vectors in Two Dimensions Topic Practice

Covered syllabus points

CAIE IGCSE Additional Math 13 Vectors in Two Dimensions Topic Practice

Covered syllabus points

Question 1

Question (a)

Question (b)

Question 2

Question (a)

Question (b)

Question (c)

Question (d)

Question 3

Question (a)

Question (b)

Question 4

Question 5