# 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 $a$.  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}$$ (iii) Verify that particles A and B collide. (iv) Write down the position vector of the point of collision. 2. (\mathrm{CIE} 0606 / 2018 / \mathrm{w} / 12 / \mathrm{q} 7) Solution (a) The vector v has a magnitude of 39 units and is in the same direction as \left(\begin{array}{r}-12 \\ 5\end{array}\right). Write \mathrm{v} in the form \left(\begin{array}{l}a \\ b\end{array}\right), where a and b are constants. (b) Vectors \mathbf{p} and q are such that p=\left(\begin{array}{c}r+s \\ r+6\end{array}\right) and q=\left(\begin{array}{c}5 r+1 \\ 2 s-1\end{array}\right), where r and s are constants. Given that 2 \mathbf{p}+3 \mathbf{q}=\left(\begin{array}{l}0 \\ 0\end{array}\right), find the value of r and of s 3. (CIE 0606 / 2018 / \mathrm{w} / 13 / \mathrm{q} 7) Solution The diagram shows a quadrilateral O A B C . The point D lies on O B such that \overrightarrow{O D}=2 \overrightarrow{D B} and \overrightarrow{A D}=m \overrightarrow{A C}, where m is a scalar quantity.$$\overrightarrow{O A}=\mathrm{a} \quad \overrightarrow{O B}=\mathbf{b} \quad \overrightarrow{O C}=\mathrm{c}$$(i) Find \overrightarrow{A D} in terms of m, a and \mathbf{c}. (ii) Find \overrightarrow{A D} in terms of a and \mathbf{b}. (iii) Given that 15 \mathrm{a}=16 \mathrm{~b}-9 \mathrm{c}, find the value of m. 4. (CIF 0606/2019/s/11/95) Solution A particle P is moving with a velocity of 20 \mathrm{~ms}^{-1} in the same direction as \left(\begin{array}{l}3 \\ 4\end{array}\right). (i) Find the velocity vector of P. At time t=0 \mathrm{~s}, P has position vector \left(\begin{array}{l}1 \\ 2\end{array}\right) relative to a fixed point O (ii) Write down the position vector of P after t \mathrm{~s}. A particle Q has position vector \left(\begin{array}{l}17 \\ 18\end{array}\right) relative to O at time t=0 \mathrm{~s} and has a velocity vector \left(\begin{array}{c}8 \\ 12\end{array}\right) \mathrm{ms}^{-1} (iii) Given that P and Q collide, find the value of t when they collide and the position vector of the point of collision. 5. (\mathrm{CIE} 0606 / 2019 / \mathrm{s} / 12 / \mathrm{q} 7) Solution A pilot wishes to fly his plane from a point A to a point B on a bearing of 055^{\circ}. There is a wind blowing at 120 \mathrm{~km} \mathrm{~h}^{-1} from the west. The plane can fly at 650 \mathrm{~km} \mathrm{~h}^{-1} in still air. (i) Find the direction in which the pilot must fly his plane in order to reach B. (ii) Given that the distance between A and B is 1250 \mathrm{~km}, find the time it will take the pilot to fly from A to B$$$

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 $B$.

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 $\mu$.$$

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.$$

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

(c) Given that $\overrightarrow{O Y}=h \overrightarrow{O X}$, find $\overrightarrow{A Y}$ in terms of $\mathbf{a}, \mathbf{b}$ and $h .$$ (d) Given that \overrightarrow{A Y}=m \overrightarrow{A C}, find the value of h and of m. 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}. (b) Find \overrightarrow{Q S} in terms of \mathbf{a} and \mathbf{b}. (c) Given that \overrightarrow{Q X}=\mu \overrightarrow{Q S}, find \overrightarrow{O X} in terms of \mathbf{a}, \mathbf{b} and \mu. (d) Given that \overrightarrow{O X}=\lambda \overrightarrow{O R}, find \overrightarrow{O X} in terms of \mathrm{a}, \mathbf{b} and \lambda. (e) Find the value of \lambda and of \mu. (f) Find the value of \frac{Q X}{X S}. (g) Find the value of \frac{O R}{O X}. 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}. (iii) Hence find the value of \lambda and of \mu. (iv) Find the ratio \frac{A X}{X D}. 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. 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. (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. (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. 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}$$$

(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 $\mathbf{b}$.$$

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

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

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 $\mathbf{b}-4 \mathrm{a}$.$$

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 $\overrightarrow{OC.}$ 

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 $A$. 

(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.$$

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

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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