4.7.4 Vektor, SPM Praktis (Kertas 2)

Soalan 4:
Rajah di bawah menunjukkan sisi empat ABCD. Garis lurus AC bersilang dengan garis lurus BD di titik E.


$\begin{array}{l}\text{Diberi bahawa }BE:ED=2:3,\overrightarrow{AB}=10\underset{˜}{x},\overrightarrow{AD}=25\underset{˜}{y}\text{ dan }\overrightarrow{BC}=-\underset{˜}{x}+15\underset{˜}{y}.\\ \text{(a) Ungkapkan dalam sebutan }\underset{˜}{x}\text{ dan }\underset{˜}{y},\\ \text{(i) }\overrightarrow{BD}\\ \text{(ii) }\overrightarrow{AE}\\ \\ \text{(b) Cari nisbah }AE:EC.\end{array}$


Penyelesaian:
(a)(i)
$\begin{array}{l}\overrightarrow{BD}=\overrightarrow{BA}+\overrightarrow{AD}\\ =\overrightarrow{AD}-\overrightarrow{AB}\\ =25\underset{˜}{y}-10\underset{˜}{x}\end{array}$

(a)(ii)
$\begin{array}{l}\overrightarrow{AE}=\overrightarrow{AB}+\overrightarrow{BE}\\ =\overrightarrow{AB}+\frac{2}{5}\overrightarrow{BD}\\ =10\underset{˜}{x}+\frac{2}{5}\left(25\underset{˜}{y}-10\underset{˜}{x}\right)\\ =10\underset{˜}{x}+\frac{2}{5}\left(25\underset{˜}{y}-10\underset{˜}{x}\right)\\ =10\underset{˜}{x}+10\underset{˜}{y}-4\underset{˜}{x}\\ =6\underset{˜}{x}+10\underset{˜}{y}\\ =2\left(3\underset{˜}{x}+5\underset{˜}{y}\right)\end{array}$

(b)
$\begin{array}{l}\overrightarrow{EC}=\overrightarrow{EB}+\overrightarrow{BC}\\ =\overrightarrow{BC}-\overrightarrow{BE}\\ =\overrightarrow{BC}-\frac{2}{3}\overrightarrow{ED}\\ =\overrightarrow{BC}-\frac{2}{3}\left(\overrightarrow{EA}+\overrightarrow{AD}\right)\\ =-\underset{˜}{x}+15\underset{˜}{y}-\frac{2}{3}\left(-6\underset{˜}{x}-10\underset{˜}{y}+25\underset{˜}{y}\right)\\ =-\underset{˜}{x}+15\underset{˜}{y}-\frac{2}{3}\left(-6\underset{˜}{x}+15\underset{˜}{y}\right)\\ =-\underset{˜}{x}+15\underset{˜}{y}+4\underset{˜}{x}-10\underset{˜}{y}\\ =3\underset{˜}{x}+5\underset{˜}{y}\\ \frac{AE}{EC}=\frac{2\left(3\underset{˜}{x}+5\underset{˜}{y}\right)}{1\left(3\underset{˜}{x}+5\underset{˜}{y}\right)}\\ AE:EC=2:1\end{array}$