4.7.8 Vektor, SPM Praktis (Kertas 2)

Soalan 8:
Rajah di bawah menunjukkan sisi empat OPQR. Garis lurus PR bersilang dengan garis lurus OQ di titik S.

$\begin{array}{l}\text{Diberi bahawa }\overrightarrow{OP}=7\underset{˜}{x},\overrightarrow{OR}=5\underset{˜}{y},PS:SR=3:1\text{ dan }\overrightarrow{OR}\text{ selari dengan }\overrightarrow{PQ}.\\ \text{(a) Ungkapkan dalam sebutan }\underset{˜}{x}\text{ dan }\underset{˜}{y},\\ \text{(i) }\overrightarrow{PR}\\ \text{(ii) }\overrightarrow{OS}\\ \\ \text{(b) Guna }\overrightarrow{PQ}=m\overrightarrow{OR}\text{ dan }\overrightarrow{SQ}=n\overrightarrow{OS},\text{ dengan keadaan }m\text{ dan }n\text{ adalah pemalar,}\\ \text{ Cari nilai }m\text{ dan nilai }n.\\ \\ \text{(c) Diberi bahawa }\left|\underset{˜}{y}\right|=4\text{ unit dan luas }ORS{\text{ ialah 50 cm}}^2\text{, cari jarak tegak}\\ \text{ dari titik }S\text{ ke }OR\text{.}\end{array}$


Penyelesaian:
(a)(i)
$\begin{array}{l}\overrightarrow{PR}=\overrightarrow{PO}+\overrightarrow{OR}\\ =-7\underset{˜}{x}+5\underset{˜}{y}\end{array}$


(a)(ii)
$\begin{array}{l}\overrightarrow{OS}=\overrightarrow{OP}+\overrightarrow{PS}\\ =7\underset{˜}{x}+\frac{3}{4}\overrightarrow{PR}\\ =7\underset{˜}{x}+\frac{3}{4}\left(-7\underset{˜}{x}+5\underset{˜}{y}\right)\\ =7\underset{˜}{x}-\frac{21}{4}\underset{˜}{x}+\frac{15}{4}\underset{˜}{y}\\ =\frac{7}{4}\underset{˜}{x}+\frac{15}{4}\underset{˜}{y}\end{array}$


(b)
$\begin{array}{l}\overrightarrow{PS}=\overrightarrow{PQ}-\overrightarrow{SQ}\\ \frac{3}{4}\overrightarrow{PR}=m\overrightarrow{OR}-n\overrightarrow{OS}\\ \frac{3}{4}\left(-7\underset{˜}{x}+5\underset{˜}{y}\right)=m\left(5\underset{˜}{y}\right)-n\left(\frac{7}{4}\underset{˜}{x}+\frac{15}{4}\underset{˜}{y}\right)\\ -\frac{21}{4}\underset{˜}{x}+\frac{15}{4}\underset{˜}{y}=5m\underset{˜}{y}-\frac{7}{4}n\underset{˜}{x}-\frac{15}{4}m\underset{˜}{y}\\ -\frac{21}{4}\underset{˜}{x}+\frac{15}{4}\underset{˜}{y}=-\frac{7}{4}n\underset{˜}{x}+\frac{5}{4}m\underset{˜}{y}\\ -21\underset{˜}{x}+15\underset{˜}{y}=-7n\underset{˜}{x}+5m\underset{˜}{y}\\ 7n=21\\ n=3\\ 5m=15\\ m=3\end{array}$


(c)
$\begin{array}{l}\text{Luas }\Delta ORS=50\\ \frac{1}{2}\times \left(5\underset{˜}{y}\right)\times t=50\\ \frac{1}{2}\times 5\left(4\right)\times t=50\\ 10t=50\\ t=5\\ \therefore \text{ Jarak tegak dari titik }S\text{ ke }OR=5\text{ unit}\text{.}\end{array}$