CIE Vector (2018-2020)

 1. $(\mathrm{CIE} 0606 / 2018 / \mathrm{w} / 11 / \mathrm{q} 10)$

Solution

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 $\left(\begin{array}{l}4 \\ 2\end{array}\right) \mathrm{ms}^{-1}$

(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.

(iv) Write down the position vector of the point of collision.$[1]$


2. $(\mathrm{CIE} 0606 / 2018 / \mathrm{w} / 12 / \mathrm{q} 7)$

Solution

(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.

(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 $s$


3. (CIE $0606 / 2018 / \mathrm{w} / 13 / \mathrm{q} 7)$

Solution

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 $m$.


4. (CIF 0606/2019/s/11/95)

Solution

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.


5. $(\mathrm{CIE} 0606 / 2019 / \mathrm{s} / 12 / \mathrm{q} 7)$

Solution

A pilot wishes to fly his plane from a point $A$ to a point $B$ on a bearing of $055^{\circ}$. There is a wind blowing at $120 \mathrm{~km} \mathrm{~h}^{-1}$ from the west. The plane can fly at $650 \mathrm{~km} \mathrm{~h}^{-1}$ in still air.

(i) Find the direction in which the pilot must fly his plane in order to reach $B$.

(ii) Given that the distance between $A$ and $B$ is $1250 \mathrm{~km}$, find the time it will take the pilot to fly from $A$ to $B$$[4]$


6. $(\mathrm{CIE} 0606 / 2019 / \mathrm{s} / 13 / \mathrm{q} 11)$

Solution

A pilot wishes to fly his plane from a point $A$ to a point $B$. The bearing of $B$ from $A$ is $050^{\circ}$. A wind is blowing from the north at a speed of $120 \mathrm{~km} \mathrm{~h}^{-1}$. The plane can fly at $600 \mathrm{~km} \mathrm{~h}^{-1}$ in still air.

(i) Find the bearing on which the pilot must fly his plane in order to reach $B$.

(ii) Given that the distance from $A$ to $B$ is $2500 \mathrm{~km}$, find the time taken to fly from $A$ to $B$.[4]


7. $(\mathrm{CIE} 0606 / 2019 / \mathrm{s} / 22 / \mathrm{q} 10)$ Solution

(a) Find the unit vector in the direction of $5 \mathbf{i}-15 \mathbf{j}$.

(b) The position vectors of points $A$ and $B$ relative to an origin $O$ are $\left(\begin{array}{r}3 \\ -5\end{array}\right)$ and $\left(\begin{array}{c}12 \\ 7\end{array}\right)$ respectively. The point $C$ lies on $A B$ such that $A C: C B$ is $2: 1$.

(i) Find the position vector of $C$ relative to $O$. The point $D$ lies on $O B$ such that $O D: O B$ is $1: \lambda$ and $\overrightarrow{D C}=\left(\begin{array}{c}6 \\ 1.25\end{array}\right)$

(ii) Find the value of $\lambda$.


8. $(\mathrm{CIE} 0606 / 2019 / \mathrm{m} / 22 / \mathrm{q} 8)$ Solution

Relative to an origin $O$, the position vectors of the points $A$ and $B$ are $2 \mathbf{i}+12 \mathbf{j}$ and $6 \mathbf{i}-4 \mathbf{j}$ respectively.

(i) Write down and simplify an expression for $\overrightarrow{A B}$.

The point $C$ lies on $\overrightarrow{A B}$ such that $A C=C B$ is $1: 3$

(ii) Find the unit vector in the direction of $\overrightarrow{O C}$. 

The point $D$ lies on $\overrightarrow{O A}$ such that $O D: D A$ is $1: \lambda$.

(iii) Find an expression for $\overrightarrow{A D}$ in terms of $\lambda$, $\mathbf{i}$ and $\mathbf{j}$.


9. $(\mathrm{CIE} 0606 / 2019 / \mathrm{s} / 23 / \mathrm{q} 10)$ Solution


The diagram shows a triangle $O A B$. The point $P$ is the midpoint of $O A$ and the point $Q$ lies on $O B$ such that $\overrightarrow{O Q}=\frac{1}{4} \overrightarrow{O B}$. The position vectors of $P$ and $Q$ relative to $O$ are $\mathrm{p}$ and $\mathrm{q}$ respectively.

(i) Find, in terms of $\mathbf{p}$ and $\mathrm{q}$, an expression for each of the vectors $\overrightarrow{P Q}, \overrightarrow{Q A}$ and $\overrightarrow{P B}$.

(ii) Given that $\overrightarrow{P R}=\lambda \overrightarrow{P B}$ and that $\overrightarrow{Q R}=\mu \overrightarrow{Q A}$, find an expression for $\overrightarrow{P Q}$ in terms of $\lambda, \mu, \mathbf{p}$ and $\mathbf{q}$.

(iii) Using your expressions for $\overrightarrow{P Q}$, find the value of $\lambda$ and of $\mu$.$[4]$


10. (CIE $0606 / 2020 / \mathrm{w} / 13 / \mathrm{q} 9)$


The diagram shows the triangle $O A C$. The point $B$ is the midpoint of $O C$. The point $Y$ lies on $A C$ such that $O Y$ intersects $A B$ at the point $X$ where $A X: X B=3: 1$. It is given that $\overrightarrow{O A}=\mathbf{a}$ and $\overrightarrow{O B}=\mathbf{b}$.

(a) Find $\overrightarrow{O X}$ in terms of a and $\mathbf{b}$, giving your answer in its simplest form.$[3]$

(b) Find $\overrightarrow{A C}$ in terms of $\mathbf{a}$ and $\mathbf{b}$.$[1]$

(c) Given that $\overrightarrow{O Y}=h \overrightarrow{O X}$, find $\overrightarrow{A Y}$ in terms of $\mathbf{a}, \mathbf{b}$ and $h .$$[1]$

(d) Given that $\overrightarrow{A Y}=m \overrightarrow{A C}$, find the value of $h$ and of $m$.$[4]$


11. $(\mathrm{CIE} 0606 / 2020 / \mathrm{w} / 22 / \mathrm{q} 9)$


In the diagram $\overrightarrow{O P}=2 \mathbf{b}, \overrightarrow{O S}=3 \mathbf{a}, \overrightarrow{S R}=\mathbf{b}$ and $\overrightarrow{P Q}=\mathbf{a}$. The lines $O R$ and $Q S$ intersect at $X$.

(a) Find $\overrightarrow{O Q}$ in terms of $\mathbf{a}$ and $\mathbf{b}$.$[1]$

(b) Find $\overrightarrow{Q S}$ in terms of $\mathbf{a}$ and $\mathbf{b}$.$[1]$

(c) Given that $\overrightarrow{Q X}=\mu \overrightarrow{Q S}$, find $\overrightarrow{O X}$ in terms of $\mathbf{a}, \mathbf{b}$ and $\mu$.$[1]$

(d) Given that $\overrightarrow{O X}=\lambda \overrightarrow{O R}$, find $\overrightarrow{O X}$ in terms of $\mathrm{a}, \mathbf{b}$ and $\lambda$.$[1]$

(e) Find the value of $\lambda$ and of $\mu$.$[3]$

(f) Find the value of $\frac{Q X}{X S}$.

(g) Find the value of $\frac{O R}{O X}$.$[1]$


12. (CIE 0606/2019/w/23/q9)


The diagram shows points $O, A, B, C, D$ and $X$. The position vectors of $A, B$, and $C$ relative to $O$ are $\overrightarrow{O A}=\mathbf{a}$, $\overrightarrow{O B}=2 \mathbf{b}$ and $\overrightarrow{O C}=3 \mathrm{a}$. The vector $\overrightarrow{C D}=\mathbf{b}$.

(i) Given that $\overrightarrow{A X}=\lambda \overrightarrow{A D}$, find $\overrightarrow{O X}$ in terms of $\lambda$, a and $\mathbf{b}$.

(ii) Given that $\overrightarrow{B X}=\mu \overrightarrow{B C}$, find $\overrightarrow{O X}$ in terms of $\mu, \mathbf{a}$ and $\mathbf{b}$.$[2]$

(iii) Hence find the value of $\lambda$ and of $\mu$.$[4]$

(iv) Find the ratio $\frac{A X}{X D}$.[1]


13. $(\mathrm{CIE} 0606 / 2020 / \mathrm{w} / 11 / \mathrm{q} 6)$

A particle $P$ is initially at the point with position vector $\left(\begin{array}{l}30 \\ 10\end{array}\right)$ and moves with a constant speed of $10 \mathrm{~ms}^{-1}$ in the same direction as $\left(\begin{array}{r}-4 \\ 3\end{array}\right)$

(a) Find the position vector of $P$ after $t \mathrm{~s}$.

As $P$ starts moving, a particle $Q$ starts to move such that its position vector after $t \mathrm{~s}$ is given by $\left(\begin{array}{r}-80 \\ 90\end{array}\right)+t\left(\begin{array}{r}5 \\ 12\end{array}\right)$

(b) Write down the speed of $Q$.

(c) Find the exact distance between $P$ and $Q$ when $t=10$, giving your answer in its simplest surd form.$[3]$


14. (CIE $0606 / 2020 / \mathrm{m} / 12 / \mathrm{q} 8)$

In this question all distances are in $\mathrm{km}$.

A ship $P$ sails from a point $A$, which has position vector $\left(\begin{array}{l}0 \\ 0\end{array}\right)$, with a speed of $52 \mathrm{kmh}^{-1}$ in the direction of $\left(\begin{array}{c}-5 \\ 12\end{array}\right)$

(a) Find the velocity vector of the ship.$[1]$

(b) Write down the position vector of $P$ at a time $t$ hours after leaving $A$.

At the same time that ship $P$ sails from $A$, a ship $Q$ sails from a point $B$, which has position vector $\left(\begin{array}{c}12 \\ 8\end{array}\right)$. with velocity vector $\left(\begin{array}{r}-25 \\ 45\end{array}\right) \mathrm{kmh}^{-1}$

(c) Write down the position vector of $Q$ at a time $t$ hours after leaving $B$.

(d) Using your answers to parts (b) and (c), find the displacement vector $\overrightarrow{P Q}$ at time $t$ hours.$[1]$

(e) Hence show that $P Q=\sqrt{34 t^{2}-168 t+208}$.

(f) Find the value of $t$ when $P$ and $Q$ are first $2 \mathrm{~km}$ apart.$[2]$


15. $(\mathrm{CIE} 0606 / 2020 / \mathrm{s} / 12 / \mathrm{q} 8)$


The diagram shows a triangle $O A B$ such that $\overrightarrow{O A}=\mathrm{a}$ and $\overrightarrow{O B}=\mathbf{b}$. The point $P$ lies on $O A$ such that $O P=\frac{3}{4} O A$. The point $Q$ is the mid-point of $A B$. The lines $O B$ and $P Q$ are extended to meet at the point $R$. Find, in terms of $a$ and $b$,

(a) $\overrightarrow{A B}$$[1]$

(b) $\overline{P Q}$. Give your answer in its simplest form.

It is given that $n \overrightarrow{P Q}=\overrightarrow{Q R}$ and $\overrightarrow{B R}=k \mathbf{b}$, where $n$ and $k$ are positive constants.

(c) Find $\overrightarrow{Q R}$ in terms of $n, \mathbf{a}$ and $\mathbf{b}$.$[1]$

(d) Find $\overrightarrow{Q R}$ in terms of $k, \mathbf{a}$ and $\mathbf{b}$.

(e) Hence find the value of $n$ and of $k$.$[3]$


16. (CIE $0606 / 2020 / \mathrm{s} / 13 / \mathrm{q} 6)$

(a) Given that $\log _{2} x+2 \log _{4} y=8$, find the value of $x y$.

(b) Using the substitution $y=2^{x}$, or otherwise, solve $2^{2 x+1}-2^{x+1}-2^{x}+1=0$.


17. $(\mathrm{CIE} 0606 / 2020 / \mathrm{s} / 21 / \mathrm{q} 5)$

The vectors $\mathbf{a}$ and $\mathbf{b}$ are such that $\mathbf{a}=\alpha \mathbf{i}+\mathbf{j}$ and $\quad \mathbf{b}=12 \mathbf{i}+\beta \mathbf{j}$.

(a) Find the value of each of the constants $\alpha$ and $\beta$ such that $4 \mathrm{a}-\mathbf{b}=(\alpha+3) \mathbf{i}-2 \mathbf{j}$.

(b) Hence find the unit vector in the direction of $\mathbf{b}-4 \mathrm{a}$.$[2]$


18. (CIE $0606 / 2020 / \mathrm{m} / 22 / \mathrm{q} 4)$ $\def\colv#1#2{\left(\begin{array}{c}#1\\#2\end{array}\right)}$

The position vectors of three points, $A, B$ and $C,$ relative to an origin $O,$ are $\colv{-5}{-7},\colv{10}{-4}$ and $\colv xy$  respectively. Given that $\overrightarrow{AC} = 4\overrightarrow{BC,}$ find the unit vector in the direction of $\overrightarrow{OC.}$ [5]


19. (CIE $0606 / 2019 / \mathrm{w} / 21 / \mathrm{q} 11)$

A plane, which can travel at a speed of $300 \mathrm{~km} \mathrm{~h}^{-1}$ in still air, heads due north. The plane is blown off course by a wind so that it travels on a bearing of $010^{\circ}$ at a speed of $280 \mathrm{~km} \mathrm{~h}^{-1}$.

(i) Find the speed of the wind.

(ii) Find the direction of the wind as a bearing correct to the nearest degree.


20. $(\mathrm{CIE} 0606 / 2019 / \mathrm{w} / 22 / \mathrm{q} 8)$

(i) A particle $A$ travels with a speed of $6.5 \mathrm{~ms}^{-1}$ in the direction $-5 \mathrm{i}-12 \mathrm{j}$. Find the velocity, $v_{A}$, of $A$. [2]

(ii) A particle $B$ travels with velocity $v_{B}=12 \hat{i}-9 j$. Find the speed, in $\mathrm{ms}^{-1}$, of $B$.

Particle $A$ starts moving from the point with position vector $20 \mathrm{i}-7 \mathbf{j}$. At the same time particle $B$ starts moving from the point with position vector $-67 \mathbf{i}+11 \mathbf{j}$.

(iii) Find $\boldsymbol{r}_{A}$, the position vector of $A$ after $t$ seconds, and $r_{B}$, the position vector of $B$ after $t$ seconds.$[2]$

(iv) Find the time when the particles collide and the position vector of the point of collision.$[3]$


Answer

1. (i) 43

(ii) $r_{B}=\left(\begin{array}{l}16 \\ 37\end{array}\right)+\left(\begin{array}{l}4 \\ 2\end{array}\right) t$

(iii) $t=7$,

(iv) $\left(\begin{array}{l}44 \\ 51\end{array}\right)$

2. (a) $v=\left(\begin{array}{c}-36 \\ 15\end{array}\right)$

(b) $r=0, s=-\frac{3}{2}$

3. $(i) \overrightarrow{A D}=m(\bar{c}-\bar{a})$

(ii) $\overrightarrow{A D}=\frac{2}{3} \bar{b}-\bar{a}$

(iii) $m=\frac{3}{8}$

4. (i) $v=\left(\begin{array}{l}12 \\ 16\end{array}\right)$

(ii) $r_{p}=\left(\begin{array}{l}1 \\ 2\end{array}\right)+\left(\begin{array}{l}12 \\ 16\end{array}\right) t$

(iii) $t=4,\left(\begin{array}{l}49 \\ 66\end{array}\right)$

5. (i) Bearing $048.9^{\circ}$

(ii) $1.68$ hours

6. (i) Bearing $041.2^{\circ}$ (ii) $t=4.85$

7. (a) $\frac{1}{5 \sqrt{10}}(5 \hat{i}-15 \hat{j})$

(b)(i) $\left(\begin{array}{l}9 \\ 3\end{array}\right)$ (ii) $\lambda=4$

8. (i) $4 \mathrm{i}-16 \mathrm{j}$

(ii) $(3 i+8 j) / \sqrt{73}$

(iii) $\frac{-\lambda}{1+\lambda}(2 i+12 j)$

9. (i) $\overrightarrow{P Q}=\bar{q}-\bar{p}, \overrightarrow{Q A}=2 \bar{p}-\bar{q}, \overrightarrow{P B}=4 \bar{q}-\bar{p}$

(ii) $\overrightarrow{P Q}=\lambda(4 \bar{q}-\bar{p})-\mu(2 \bar{p}-\bar{q})$

(iii) $\lambda=\frac{1}{7}, \mu=\frac{3}{7}$

10. (a) $\overline{\sigma x}=\frac{1}{4} \bar{a}+\frac{3}{4} \bar{b}$

(b) $\overline{A C}=2 \bar{b}-\bar{a}$

(c) $\overline{A Y}=-\bar{a}+h\left(\frac{1}{4} \bar{a}+\frac{3}{4} \bar{b}\right)$

(d) $h=\frac{8}{5}, m=\frac{3}{5}$

11. (a) $\overrightarrow{O Q}=2 \vec{b}+\vec{a}$

(b) $\overrightarrow{Q S}=2 \vec{a}-2 \vec{b}$

(c) $\quad 2 \vec{b}+\vec{a}+\mu(2 \vec{a}-2 \vec{b})$

(d) $\lambda(3 \vec{a}+\vec{b})$

(e) $\lambda=\frac{3}{4}, \mu=\frac{5}{8}$

(f) $\frac{Q X}{X S}=\frac{5}{3}$

(g) $\frac{O R}{O X}=\frac{4}{3}$

12. (i) $\overrightarrow{O X}=\bar{a}+\lambda(2 \bar{a}+\bar{b})$

(ii) $\overline{O X}=2 \bar{b}+\mu(3 \bar{a}-2 \bar{b})$

(iii) $\lambda=\frac{4}{7}, \mu=\frac{5}{7}$ (iv) $\frac{4}{3}$

13. (a) $\left(\begin{array}{c}30 \\ 10\end{array}\right)+\left(\begin{array}{c}-8 \\ 6\end{array}\right) t$

(b) 13 (c) $100 \sqrt{2}$

14. (a) $\left(\begin{array}{c}-20 \\ 48\end{array}\right)$

(b) $\left(\begin{array}{c}-20 \\ 48\end{array}\right) t$

(c) $\left(\begin{array}{c}12 \\ 8\end{array}\right)+\left(\begin{array}{c}-25 \\ 45\end{array}\right) t$

(d) $\left(\begin{array}{c}12 \\ 8\end{array}\right)+$

$\left(\begin{array}{l}-5 \\ -3\end{array}\right) t$ (e) Prove (f) $t=2.15$

15. (a) $\overrightarrow{A B}=\bar{b}-\bar{a}$

(b) $\frac{1}{2} \bar{b}-\frac{1}{4} \bar{a}$

(c) $n\left(\frac{1}{2} \bar{b}-\frac{1}{4} \bar{a}\right)$

(d) $\frac{1}{2}(\vec{b}-\bar{a})+k \bar{b}$

(e) $n=2, k=\frac{1}{2}$

16. (a) $\frac{1}{13}\left(\begin{array}{c}5 \\ -12\end{array}\right)$

(b) $k=-\frac{3}{2}, r=-\frac{7}{10}$

(c) (i) $3 \bar{q}-2 \bar{p}$

(ii) $\overrightarrow{A C}=9 \bar{q}-6 \bar{P}$

(iii) common point and same direction (iv) $1: 2$

17. (a) $\alpha=5, \beta=6$

(b) $\frac{2 j-8 i}{\sqrt{68}}$

18. $\frac{1}{\sqrt{234}}\left(\begin{array}{c}15 \\ -3\end{array}\right)$

19. (i) $v_{w}=54.3$

(ii) Bearing $117^{\circ}$

20. (i) $v_{A}=\frac{1}{2}(-5 \hat{i}-12 \hat{\jmath})$

(ii) 15 .

(iii) $r_{A}=\left(\begin{array}{c}20 \\ -7\end{array}\right)+t\left(\begin{array}{c}-2.5 \\ -6\end{array}\right), r_{B}=\left(\begin{array}{c}-67 \\ 11\end{array}\right)+t\left(\begin{array}{c}12 \\ -9\end{array}\right)$

(iv) $t=6, \quad r=5 \hat{i}-43 \hat{\jmath}$



Solution Group

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

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


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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