CIE Vector (2018-2020 Solution)

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1. $(\mathrm{CIE} 0606 / 2018 / \mathrm{w} / 11 / \mathrm{q} 10)$
Particle $A$ is at the point with position vector $\left(\begin{array}{r}2 \\ -5\end{array}\right)$ at time $t=0$ and moves with a speed of $10 \mathrm{~ms}^{-1}$ in the same direction as $\left(\begin{array}{l}3 \\ 4\end{array}\right)$
(i) Given that $A$ is at the point with position vector $\left(\begin{array}{c}38 \\ a\end{array}\right)$ when $t=6 \mathrm{~s}$, find the value of the constant $a$. [3]
Particle $B$ is at the point with position vector $\left(\begin{array}{l}16 \\ 37\end{array}\right)$ at time $t=0$ and moves with velocity $\left(\begin{array}{l}4 \\ 2\end{array}\right) \mathrm{ms}^{-1}$ [3]
(ii) Write down, in terms of $t$, the position vector of $B$ at time $t \mathrm{~s}$$[1]$
(iii) Verify that particles $A$ and $B$ collide.[4]
(iv) Write down the position vector of the point of collision.$[1]$


Solution 1


Velocity of particle $A=k\left(\begin{array}{l}3 \\ 4\end{array}\right)=\left(\begin{array}{l}2 k \\ 4 k\end{array}\right)$

$\therefore$ speed of particle $A=\left|\left(\begin{array}{c}3 k \\ 4 k\end{array}\right)\right|=\sqrt{(3 k)^{2}+(4 k)^{2}}=\sqrt{9 k^{2}+16 k^{2}}=5 k$, for $k>0$.

Thus $5 k=10, k=2$.

Hence velocity vector of particle $A=2\left(\begin{array}{l}3 \\ 4\end{array}\right)=\left(\begin{array}{l}6 \\ 8\end{array}\right)$

(i) Position vector of $A$ after $t$ sec is

$$\left(\begin{array}{l}x \\y\end{array}\right)=\left(\begin{array}{c}2 \\-5\end{array}\right)+t\left(\begin{array}{l}6 \\8\end{array}\right)=\left(\begin{array}{c}2+6 t \\-5+8 t\end{array}\right)$$

when $t=6, \quad\left(\begin{array}{l}x \\ y\end{array}\right)=\left(\begin{array}{c}2+6(6) \\ -5+8(6)\end{array}\right)=\left(\begin{array}{c}38 \\ 43\end{array}\right)$

(ii) Position vector of $B$ after $t \sec$ is

$$\left(\begin{array}{l}x \\y\end{array}\right)=\left(\begin{array}{l}16 \\37\end{array}\right)+t\left(\begin{array}{l}4 \\2\end{array}\right)=\left(\begin{array}{c}16+4 t \\37+2 t\end{array}\right)$$

(iii) Particles $A$ and B collide,

$$\begin{aligned}&\left(\begin{array}{r}2+6 t \\-5+8 t\end{array}\right)=\left(\begin{array}{l}16+4 t \\37+2 t\end{array}\right) \\&2+6 t=16+4 t \Rightarrow 2 t=14 \Rightarrow t=7 \\&-5+8 t=37+2 t \Rightarrow 6 t=42 \Rightarrow t=7\end{aligned}$$

Thus after $t=7 \mathrm{sec}$, two particles have the same position vector.

(1v) Position vector at $t=7$. is

$$\left(\begin{array}{c}2+6(7) \\-5+8(7)\end{array}\right)=\left(\begin{array}{c}2+42 \\-5+56\end{array}\right)=\left(\begin{array}{l}44 \\51\end{array}\right)$$


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2. $(\mathrm{CIE} 0606 / 2018 / \mathrm{w} / 12 / \mathrm{q} 7)$
(a) The vector $v$ has a magnitude of 39 units and is in the same direction as $\left(\begin{array}{r}-12 \\ 5\end{array}\right)$. Write $\mathrm{v}$ in the form $\left(\begin{array}{l}a \\ b\end{array}\right)$, where $a$ and $b$ are constants. [2]
(b) Vectors $\mathbf{p}$ and $q$ are such that $p=\left(\begin{array}{c}r+s \\ r+6\end{array}\right)$ and $q=\left(\begin{array}{c}5 r+1 \\ 2 s-1\end{array}\right)$, where $r$ and $s$ are constants. Given that $2 \mathbf{p}+3 \mathbf{q}=\left(\begin{array}{l}0 \\ 0\end{array}\right)$, find the value of $r$ and of $s$ [4]

Solution 2


2(a)  $v=k\left(\begin{array}{c}-12 \\ 5\end{array}\right)=\left(\begin{array}{c}-12 k \\ 5 k\end{array}\right), k>0$

since $|v|=39$, $3 q=\sqrt{(-12 k)^{2}+(5 k)^{2}}=\sqrt{144 k^{2}+25 k^{2}}=13 k, k>0 .$ $\therefore k=3$ $\therefore v=3\left(\begin{array}{c}-12 \\ 5\end{array}\right)=\left(\begin{array}{l}-36 \\ 15\end{array}\right)$

(b)$\begin{aligned}\left(\begin{array}{l}0 \\0\end{array}\right) &=2 \vec{p}+3 \vec{q} \\&=2\left(\begin{array}{l}r+5 \\r+6\end{array}\right)+3\left(\begin{array}{c}5 r+1 \\2 s-1\end{array}\right)=\left(\begin{array}{c}2 r+2 s+15 r+3 \\2 r+12+6 s-3\end{array}\right)=\left(\begin{array}{c}17 r+2 s+3 \\2 r+6 s+9\end{array}\right)\end{aligned}$

$\begin{array}{rll}\therefore \quad 17 r+2s+3&=0 \quad&\cdots(1)\\2 r+6 s+9&=0&\cdots(2)\\(1) \times 3: 51 r+65+9&=0 &\dots (3)\\(3)-(2): 49 r &=0 \\ r &=0 \\ s &=-3 / 2 \end{array}$

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3. (CIE $0606 / 2018 / \mathrm{w} / 13 / \mathrm{q} 7)$
The diagram shows a quadrilateral $O A B C .$ The point $D$ lies on $O B$ such that $\overrightarrow{O D}=2 \overrightarrow{D B}$ and $\overrightarrow{A D}=m \overrightarrow{A C}$, where $m$ is a scalar quantity. $$\overrightarrow{O A}=\mathrm{a} \quad \overrightarrow{O B}=\mathbf{b} \quad \overrightarrow{O C}=\mathrm{c}$$
(i) Find $\overrightarrow{A D}$ in terms of $m$, a and $\mathbf{c}$. $ [1]$
(ii) Find $\overrightarrow{A D}$ in terms of a and $\mathbf{b}$. $ [2]$
(iii) Given that $15 \mathrm{a}=16 \mathrm{~b}-9 \mathrm{c}$, find the value of $m$. [3]

Solution 3


$$ \begin{aligned} (i) &\overrightarrow{A C}=\overrightarrow{A O}+\overrightarrow{O C}=-\overrightarrow{O A}+\overrightarrow{O C}=-\vec{a}+\vec{c} \\ &\overrightarrow{A D}=m \vec{AC}=m(-\vec{a}+\vec{c})=-m \vec{a}+\vec{m} \vec{c} \end{aligned} $$ (ii) since $\overrightarrow{O D}=2 \overrightarrow{D B}, \overrightarrow{O D}=\frac{2}{3} \overrightarrow{O B}=\frac{2}{3} \vec{b}$ Thun $\overrightarrow{A D}=\overrightarrow{A O}+\overrightarrow{O D}=-\overrightarrow{O A}+\overrightarrow{O D}=-\vec{a}+\frac{2}{3} \vec{b}$ $$ \begin{aligned} (iii) 15 \vec{a} &=16 \vec{b}-9 \vec{c} \\ -m \vec{a}+m \vec{c} &=-\vec{a}+\frac{2}{3} \bar{b} \quad\text{ (by(i) and (ii)) }\\ m \vec{i} &=m \vec{a}-\vec{a}+\frac{2}{3} \vec{b} \\ \vec{c} &=\vec{a}-\frac{1}{m} \vec{a}+\frac{2}{3 m} \bar{b} \end{aligned} $$ Thus $\begin{aligned} 15 \vec{a} &=16 \vec{b}-9\left(\bar{a}-\frac{1}{m} \vec{a}+\frac{2}{3 m} \bar{b}\right) \\ 0 &=16-\frac{18}{3 m} \\ \frac{18}{3 m} &=16 \\ m &=\frac{3}{8} \end{aligned}$


4. (CIF 0606/2019/s/11/95)
A particle $P$ is moving with a velocity of $20 \mathrm{~ms}^{-1}$ in the same direction as $\left(\begin{array}{l}3 \\ 4\end{array}\right)$.
(i) Find the velocity vector of $P$.
At time $t=0 \mathrm{~s}, P$ has position vector $\left(\begin{array}{l}1 \\ 2\end{array}\right)$ relative to a fixed point $O$
(ii) Write down the position vector of $P$ after $t \mathrm{~s}$.
A particle $Q$ has position vector $\left(\begin{array}{l}17 \\ 18\end{array}\right)$ relative to $O$ at time $t=0 \mathrm{~s}$ and has a velocity vector $\left(\begin{array}{c}8 \\ 12\end{array}\right) \mathrm{ms}^{-1}$
(iii) Given that $P$ and $Q$ collide, find the value of $t$ when they collide and the position vector of the point of collision.



Solution 4


(i) veloity of $P=k\left(\begin{array}{l}3 \\ 4\end{array}\right), k>0$ sime $\left|k\left(\begin{array}{l}3 \\ 4\end{array}\right)\right|=20, k \sqrt{3^{2}+4^{2}}=20$. Thus $k=\frac{20}{5}=4$ veloity veetor of $p=4\left(\begin{array}{l}3 \\ 4\end{array}\right)=\left(\begin{array}{c}12 \\ 16\end{array}\right)$
(ii) Thus position veetor of $P$ after $t$ sec is $$ \begin{aligned} \left(\begin{array}{l} x \\ y \end{array}\right) &=\left(\begin{array}{l} 1 \\ 2 \end{array}\right)+t\left(\begin{array}{l} 12 \\ 16 \end{array}\right) \\ &=\left(\begin{array}{l} 1+12 t \\ 2+16 t \end{array}\right) \end{aligned} $$ (iii) Position vactor of $Q$ after t see is $$ \begin{aligned} \left(\begin{array}{l} x \\ y \end{array}\right) &=\left(\begin{array}{l} 17 \\ 18 \end{array}\right)+t\left(\begin{array}{l} 8 \\ 12 \end{array}\right) \\ &=\left(\begin{array}{c} 17+8 t \\ 18+12 t \end{array}\right) \end{aligned} $$ Since $P$ and $Q$ collide ofter $t$ seconds, $$ \left(\begin{array}{c} 1+12 t \\ 2+16 t \end{array}\right)=\left(\begin{array}{l} 17+8 t \\ 18+12 t \end{array}\right) $$ Hence
$\begin{aligned} 1+12 t &=17+8 t, 2+16 t=18+12 t . \\ 4 t &=16 \quad, \quad 4 t=16 \\ t &=4 \quad, \quad t=4 \end{aligned}$
$\therefore t=4$, postion veetor of the point of collision is $\left(\begin{array}{l}1+12 \times 4 \\ 2+16 \times 4\end{array}\right)=\left(\begin{array}{l}49 \\ 66\end{array}\right)$


Solution 5




$$ \alpha=90^{\circ}-55=35^{\circ}=\beta $$ By the law of sine, $\frac{\sin \theta}{120}=\frac{\sin 35}{650}$ $$ \theta=6.08 $$ $\therefore 55^{\circ}-6.08^{\circ}=48.92^{\circ}$
(i) direction of plane $=048.9^{\circ}$
(ii) $r=18 \dot{0}-(\theta+\beta)=180^{\circ}-\dot{6} .08-35^{\circ}=138.92^{\circ}$
By the law of cosine, $$ \begin{aligned} |\overrightarrow{A B}|^{2} &=650^{2}+120^{2}-2 \times 650 \times 120 \times \cos (138.92) \\ &=554491 \\ \therefore|\overrightarrow{A B}| &=744.64 \end{aligned} $$
The speed of plane $=744.64 \mathrm{~km} / \mathrm{h}$.
Thus time from $A$ to $B=\frac{1250}{744.64}=1.68$ hours.


Solution 6


$$ \beta=180^{\circ}-50^{\circ}=130^{\circ} $$ By law of sine, $\frac{\sin \theta}{120}=\frac{\sin 130}{600}$
$\therefore \quad \theta=8.81$
(i) Bearing of plane $=50^{\circ}-8.81=041.2^{\circ}$
(ii) $r=41.19$, By law of cosine, $$ \begin{aligned} A B^{2} &=600^{2}+120^{2}-2 \times 600 \times 120 \cos (41.19) \\ &=266035 \\ A B &=515.78 \end{aligned} $$
speed of plane $=515.78 \mathrm{~km} / \mathrm{h}$.
$\therefore$ time from $A$ to $B=\frac{2500}{515.78}=4.85$ hours.

Solution 7


(a) $|5 \hat{i}-15 \hat{j}|=\sqrt{5^{2}+15^{2}}=\sqrt{25+225}=\sqrt{250}=5 \sqrt{10}$
$\therefore$ unit vector $=\frac{1}{5 \sqrt{10}}(5 \hat{i}-15 \hat{j})$
(b) $$ \begin{aligned} &\overrightarrow{A B}=\overrightarrow{O B}-\overrightarrow{O A}=\left(\begin{array}{c} 12 \\ 7 \end{array}\right)-\left(\begin{array}{c} 3 \\ -5 \end{array}\right)=\left(\begin{array}{l} 9 \\ 12 \end{array}\right) \\ &\overrightarrow{A C}=\frac{2}{3} \overrightarrow{A B}=\frac{2}{3}\left(\begin{array}{c} 9 \\ 12 \end{array}\right)=\left(\begin{array}{l} 6 \\ 8 \end{array}\right) \end{aligned} $$ (i) $\overrightarrow{O C}=\overrightarrow{O A}+\overrightarrow{A C}=\left(\begin{array}{c}3 \\ -5\end{array}\right)+\left(\begin{array}{l}6 \\ 8\end{array}\right)=\left(\begin{array}{l}9 \\ 3\end{array}\right)$
(ii) Since $O D: O B=1: \lambda, \overrightarrow{O D}=\frac{1}{\lambda} \overrightarrow{O B}$ $$ \begin{aligned} \overrightarrow{D C} &=\overrightarrow{D O}+\overrightarrow{O C} \\ \left(\begin{array}{l} 6 \\ 1.25 \end{array}\right)=-\frac{1}{\lambda}\left(\begin{array}{l} 12 \\ 7 \end{array}\right)+\left(\begin{array}{l} 9 \\ 3 \end{array}\right) \\ =\left(\begin{array}{l} -\frac{12}{\lambda}+9 \\ -\frac{7}{\lambda}+3 \end{array}\right) \\ 6=-\frac{12}{\lambda}+9,1,25=\frac{-7}{\lambda}+3 \\ \lambda=4=4=4 \\ \therefore \lambda=4 \end{aligned} $$

Solution 8


(i) $\overrightarrow{O A}=2 \hat{i}+12 \hat{j}, \overrightarrow{O B}=6 \hat{i}-4 \hat{j}$ $$ \begin{aligned} \overrightarrow{A B} &=\overrightarrow{A O}+\overrightarrow{O B} \cdot \\ &=-\overrightarrow{O A}+\overrightarrow{O B} \\ &=-(2 \hat{i}+12 \hat{j})+(6 \hat{i}-4 \hat{j}) \\ &=4 \hat{i}-16 \hat{j} \end{aligned} $$ (ii) Since $\frac{A C}{C B}=\frac{1}{3}, \overrightarrow{A C}=\frac{1}{4} \overrightarrow{A B}=\frac{1}{4}(\hat{i}-16 \hat{j})=\hat{i}-4 \hat{j}$. $$ \begin{aligned} &\overrightarrow{O C}=\overrightarrow{O A}+\overrightarrow{A C}=(2 \hat{j}+12 \hat{\hat{j}})+(\hat{i}-4 \hat{j})=3 \hat{i}+8 \hat{j} \\ &|\overrightarrow{O C}|=\sqrt{3^{2}+8^{2}}=\sqrt{9+64}=\sqrt{73} \end{aligned} $$ $\therefore$ unit vector in the direction $\overrightarrow{O C}=\frac{1}{\sqrt{7^{3}}}(3 \hat{i}+8 \hat{j})$
(iii) Since $O D: D A=1: \lambda, \overrightarrow{A D}=\frac{-\lambda}{1+\lambda} \overrightarrow{O A}=\frac{-\lambda}{1+\lambda}(2 \hat{i}+12 \hat{j})$

Solution 9


(i) $\overrightarrow{P Q}=\overrightarrow{O Q}-\overrightarrow{O P}=\vec{q}-\vec{p}$
Since $P$ is the mid point of $O A, \overrightarrow{O A}=2 \vec{p}$.
$\overrightarrow{Q A}=\overrightarrow{O A}-\overrightarrow{O Q}=2 \vec{p}-\vec{q}$
since $\overrightarrow{O Q}=\frac{1}{4} \overrightarrow{O B}, \overrightarrow{O B}=4 \vec{q} .$
Thus $\overrightarrow{P B}=\overrightarrow{O B}-\overrightarrow{O P}=4 \vec{q}-\vec{p}$. \begin{aligned} &\text { (ii) }\\ &\vec{P R}=\lambda \overrightarrow{P B}=\lambda(4 \vec{q}-\vec{p})=4 \lambda \vec{q}-\lambda \vec{p}\\ &\overrightarrow{Q R}=\mu \overrightarrow{Q A}=\mu(2 \vec{p}-\vec{q})=2 \mu \vec{p}-\mu \vec{q}\\ &\overrightarrow{P Q}=\overrightarrow{P R}+\overrightarrow{R Q}=4 \vec{\lambda} q-\lambda \vec{p}+(-2 \mu \vec{p}+\mu \vec{q})\\ &=(4 \lambda+\mu) \vec{q}-(\lambda+2 \mu) \vec{p}\\ &\operatorname{since} \overrightarrow{P Q}=\vec{q}-\vec{p}, \end{aligned} $4 \lambda+\mu=1 \quad \cdots(1)$
$\lambda+2 \mu=1 \quad \ldots(2)$
$(1) \times 2: 8 \lambda+2 \mu=2 \quad \cdots(3)$
(3) $-(2): \quad 7 \lambda=1$ $$ \lambda=\frac{1}{7}, \mu=\frac{3}{7} $$

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