8.7.3 Vektor, SPM Praktis (Kertas 2)

Soalan 5:
Rajah di bawah menunjukkan sebuah segi tiga KLM.


$\begin{array}{l}\text{Diberi }KP:PL=1:2,LR:RM=2:1,\overrightarrow{KP}=2\underset{˜}{x},\overrightarrow{KM}=3\underset{˜}{y}.\\ \text{(a) Ungkapkan dalam sebutan }\underset{˜}{x}\text{ dan }\underset{˜}{y},\\ \text{(i) }\overrightarrow{MP}\\ \text{(ii) }\overrightarrow{MR}\\ \\ \text{(b) Diberi }\underset{˜}{x}=2\underset{˜}{i}\text{ dan }\underset{˜}{y}=-\underset{˜}{i}+4\underset{˜}{j},\text{ cari }\overrightarrow{\left|MR\right|}.\\ \\ \text{(c) Diberi }\overrightarrow{MQ}=h\overrightarrow{MP}\text{ dan }\overrightarrow{QR}=n\overrightarrow{KR},\text{ dengan keadaan }h\text{ dan }n\text{ ialah pemalar, }\\ \text{ cari nilai }h\text{ dan nilai }n\text{.}\end{array}$


Penyelesaian:
(a)(i)
$\begin{array}{l}\overrightarrow{MP}=\overrightarrow{MK}+\overrightarrow{KP}\\ =-3\underset{˜}{y}+2\underset{˜}{x}\\ =2\underset{˜}{x}-3\underset{˜}{y}\end{array}$

(a)(ii)
$\begin{array}{l}\overrightarrow{MR}=\frac{1}{3}\overrightarrow{ML}\\ =\frac{1}{3}\left(\overrightarrow{MK}+\overrightarrow{KL}\right)\\ =\frac{1}{3}\left(-3\underset{˜}{y}+6\underset{˜}{x}\right)\\ =2\underset{˜}{x}-\underset{˜}{y}\end{array}$

(b)
$\begin{array}{l}\overrightarrow{MR}=2\left(2\underset{˜}{i}\right)-\left(-\underset{˜}{i}+4\underset{˜}{j}\right)\\ =4\underset{˜}{i}+\underset{˜}{i}-4\underset{˜}{j}\\ =5\underset{˜}{i}-4\underset{˜}{j}\\ \left|\overrightarrow{MR}\right|=\sqrt{5^2+{\left(-4\right)}^2}\\ =\sqrt{41}\text{ unit}\end{array}$

(c)
$\begin{array}{l}\overrightarrow{MQ}+\overrightarrow{QR}=\overrightarrow{MR}\\ h\overrightarrow{MP}+n\overrightarrow{KR}=\overrightarrow{MR}\\ h\left(2\underset{˜}{x}-3\underset{˜}{y}\right)+n\left(\overrightarrow{KM}+\overrightarrow{MR}\right)=2\underset{˜}{x}-\underset{˜}{y}\\ h\left(2\underset{˜}{x}-3\underset{˜}{y}\right)+n\left(3\underset{˜}{y}+2\underset{˜}{x}-\underset{˜}{y}\right)=2\underset{˜}{x}-\underset{˜}{y}\\ 2h\underset{˜}{x}-3h\underset{˜}{y}+2n\underset{˜}{x}+2n\underset{˜}{y}=2\underset{˜}{x}-\underset{˜}{y}\\ \left(2h+2n\right)\underset{˜}{x}+\left(-3h+2n\right)\underset{˜}{y}=2\underset{˜}{x}-\underset{˜}{y}\\ 2h+2n=\mathrm{2\dots \dots \dots .}(1)\\ -3h+2n=-\mathrm{1\dots \dots \dots .}(2)\\ \left(1\right)-\left(2\right):5h=3\\ h=\frac{3}{5}\\ \text{Dari }\left(1\right):h+n=1\\ \frac{3}{5}+n=1\\ n=1-\frac{3}{5}\\ n=\frac{2}{5}\end{array}$

Soalan 6:
Diagram below shows a trapezium OABC and point D lies on AC.


$\begin{array}{l}\text{Diberi bahawa }\overrightarrow{OC}=18\underset{˜}{b},\overrightarrow{OA}=6\underset{˜}{a}\text{ dan }\overrightarrow{OC}=2\overrightarrow{AB}.\\ \text{(a) Ungkapkan dalam sebutan }\underset{˜}{a}\text{ dan }\underset{˜}{b},\\ \text{(i) }\overrightarrow{AC}\\ \text{(ii) }\overrightarrow{OB}\\ \\ \text{(b) Diberi bahawa }\overrightarrow{AD}=k\overrightarrow{AC},\text{ dengan keadaan }k\text{ ialah pemalar}\text{.}\\ \text{Cari nilai }k\text{ jika titik-titik }O\text{, }D\text{ dan }B\text{ adalah segaris}\text{.}\end{array}$

Penyelesaian:
(a)(i)
$\begin{array}{l}\overrightarrow{AC}=\overrightarrow{AO}+\overrightarrow{OC}\\ =-6\underset{˜}{a}+18\underset{˜}{b}\\ =18\underset{˜}{b}-6\underset{˜}{a}\end{array}$


(a)(ii)
$\begin{array}{l}\overrightarrow{OC}=2\overrightarrow{AB}\\ 18\underset{˜}{b}=2\left(\overrightarrow{AO}+\overrightarrow{OB}\right)\\ 18\underset{˜}{b}=2\left(-6\underset{˜}{a}+\overrightarrow{OB}\right)\\ 18\underset{˜}{b}=-12\underset{˜}{a}+2\overrightarrow{OB}\\ \overrightarrow{OB}=6\underset{˜}{a}+9\underset{˜}{b}\end{array}$


(b)
$\begin{array}{l}\overrightarrow{OD}=h\overrightarrow{OB}\\ =h\left(6\underset{˜}{a}+9\underset{˜}{b}\right)\\ =6h\underset{˜}{a}+9h\underset{˜}{b}\\ \\ \overrightarrow{AD}=\overrightarrow{OD}-\overrightarrow{OA}\\ =6h\underset{˜}{a}+9h\underset{˜}{b}-6\underset{˜}{a}\\ =\underset{˜}{a}\left(6h-6\right)+9h\underset{˜}{b}\\ \\ \overrightarrow{AD}=k\overrightarrow{AC}\\ \underset{˜}{a}\left(6h-6\right)+9h\underset{˜}{b}=k\left(18\underset{˜}{b}-6\underset{˜}{a}\right)\\ \underset{˜}{a}\left(6h-6\right)+9h\underset{˜}{b}=-6k\underset{˜}{a}+18k\underset{˜}{b}\\ 6h-6=-6k\\ h-1=-k\\ h=1-k\mathrm{\dots \dots \dots .}\left(1\right)\\ \\ 9h=18k\\ h=2k\\ \text{Dari }\left(1\right),\\ 1-k=2k\\ 3k=1\\ k=\frac{1}{3}\end{array}$