# Vector Geometry

**** CIE 2020 winter Paper 22 No 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 $\mathbf{a}, \mathbf{b}$ and $\lambda$. [1]

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

(f) Find the value of $\dfrac{Q X}{X S}$. [1]

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

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$\def\va{\mathbf{a}}\def\vb{\mathbf{b}}$

$\begin{array}{llll}&\text { (a) }& \overrightarrow{O Q}&=\overrightarrow{O P}+\overrightarrow{O Q}=2 \vb+\va \\&\text { (b) } &\overrightarrow{Q S}&=\overrightarrow{QP} +\overrightarrow{P O}+\overrightarrow{O S}=-\va-2 \vb+3 \va=2 \va-2 \vb \\&\text { (c) }& \overrightarrow{QX} &=\mu {QS}=\mu(2 \va-2 \vb) \\&\text { (d) } &\overrightarrow{O R}&=\overrightarrow{O S}+\overrightarrow{S R}=3 \va+\vb \\&&\overrightarrow{OX }&=\lambda \overrightarrow{O R}=\lambda(3 \va+\vb) \\&\text { (e) } &\overrightarrow{OX}&=\vec{OP} +\vec{PQ}+\vec{QX} \\&&\therefore \lambda(3 \va+\vb)&=2 \vb+\va+\mu(2 \va-2 \vb) \\&&3 \lambda \va+\lambda \vb&=\va+2 \mu \va+2 \vb-2 \mu \vb \\&&&=(1+2 \mu) \va+(2-2 \mu) \vb\end{array}$

$\begin{array}{rcll}\therefore 3\lambda&=&1+2\mu&\cdots (1)\\ \lambda&=&2-2\mu&\cdots (2)\\(1)+(2): 4\lambda&=& 3\end{array}$

Thus $\lambda=\dfrac{3}{4},\mu=\dfrac{5}{8}.$

(f)  $QX=\mu QS=\dfrac 58 QS.$ Thus $\dfrac{QX}{XS}=\dfrac 53$.

(g) $OX=\lambda OR=\dfrac 34 OR.$ Thus $\dfrac{OR}{OX}=\dfrac 43.$

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