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# 8.7.5 Vectors, SPM Practice (Question 9)

Question 9:
Diagram 2 shows a triangle EFG. [3 marks]
(b) If line GR is extended to point K such that
[3 marks]

Solution:
(a)(i)
$\begin{array}{l}\stackrel{\to }{GR}=\stackrel{\to }{GE}+\stackrel{\to }{ER}\\ =-6\underset{˜}{y}+4\underset{˜}{x}\end{array}$

(a)(ii)
$\begin{array}{l}\stackrel{\to }{GT}=\frac{1}{3}\stackrel{\to }{GF}\\ \text{}=\frac{1}{3}\left(\stackrel{\to }{GR}+\stackrel{\to }{RF}\right)\\ \text{}=\frac{1}{3}\left(-6\underset{˜}{y}+4\underset{˜}{x}+8\underset{˜}{x}\right)←\overline{)\stackrel{\to }{RF}=2\stackrel{\to }{ER}}\\ \text{}=-2\underset{˜}{y}+4\underset{˜}{x}\end{array}$

(b)
$\begin{array}{l}\stackrel{\to }{GK}=h\stackrel{\to }{GR}\\ \stackrel{\to }{GE}+\stackrel{\to }{EK}=h\stackrel{\to }{GR}\\ -6\underset{˜}{y}+\left(6\underset{˜}{x}-3\underset{˜}{y}\right)=h\left(-6\underset{˜}{y}+4\underset{˜}{x}\right)\\ -9\underset{˜}{y}+6\underset{˜}{x}=-6h\underset{˜}{y}+4h\underset{˜}{x}\\ \text{Compare,}\\ -6h=-9\\ h=\frac{3}{2}\end{array}$