# 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)$$\quad$$\quad$$\quad$$\quad
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}

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