# Bab 15 Vektor

4.7 Vektor, SPM Praktis (Kertas 2)

Soalan 1:

Rajah di atas menunjukkan segi tiga OAB. Garis lurus AP bersilang dengan garis lurus OQpada titik R.
Diberi bahawa
(a)    Ungkapakan dalam sebutan
(i) $\stackrel{\to }{AP}$
(ii) $\stackrel{\to }{OQ}$
(b)   (i) Diberi bahawa
(ii) Diberi bahawa
(c)    Dengan menggunakan  cari nilai bagi h dan k.

Penyelesaian:
(a)(i)
$\begin{array}{l}\stackrel{\to }{AP}=\stackrel{\to }{AO}+\stackrel{\to }{OP}\\ \stackrel{\to }{AP}=-\stackrel{\to }{OA}+\stackrel{\to }{OP}\\ \stackrel{\to }{AP}=-8\underset{˜}{a}+4\underset{˜}{b}\end{array}$

(a)(ii)
$\begin{array}{l}\stackrel{\to }{OQ}=\stackrel{\to }{OA}+\stackrel{\to }{AQ}\\ \stackrel{\to }{OQ}=8\underset{˜}{a}+\frac{1}{4}\stackrel{\to }{AB}\\ \stackrel{\to }{OQ}=8\underset{˜}{a}+\frac{1}{4}\left(\stackrel{\to }{AO}+\stackrel{\to }{OB}\right)\\ \stackrel{\to }{OQ}=8\underset{˜}{a}+\frac{1}{4}\left(-8\underset{˜}{a}+4\stackrel{\to }{OP}\right)\\ \stackrel{\to }{OQ}=8\underset{˜}{a}+\frac{1}{4}\left(-8\underset{˜}{a}+4\left(4\underset{˜}{b}\right)\right)\\ \stackrel{\to }{OQ}=8\underset{˜}{a}-2\underset{˜}{a}+4\underset{˜}{b}\\ \stackrel{\to }{OQ}=6\underset{˜}{a}+4\underset{˜}{b}\end{array}$

(b)(i)
$\begin{array}{l}\stackrel{\to }{AR}=h\stackrel{\to }{AP}\\ \stackrel{\to }{AR}=h\left(-8\underset{˜}{a}+4\underset{˜}{b}\right)\\ \stackrel{\to }{AR}=-8h\underset{˜}{a}+4h\underset{˜}{b}\end{array}$

(b)(ii)
$\begin{array}{l}\stackrel{\to }{RQ}=k\stackrel{\to }{OQ}\\ \stackrel{\to }{RQ}=k\left(6\underset{˜}{a}+4\underset{˜}{b}\right)\\ \stackrel{\to }{RQ}=6k\underset{˜}{a}+4k\underset{˜}{b}\end{array}$

(c)
$\begin{array}{l}\stackrel{\to }{AQ}=\stackrel{\to }{AR}+\stackrel{\to }{RQ}\\ \stackrel{\to }{AQ}=-8h\underset{˜}{a}+4h\underset{˜}{b}+\left(6k\underset{˜}{a}+4k\underset{˜}{b}\right)\\ \stackrel{\to }{AO}+\stackrel{\to }{OQ}=-8h\underset{˜}{a}+4h\underset{˜}{b}+6k\underset{˜}{a}+4k\underset{˜}{b}\\ -8\underset{˜}{a}+6\underset{˜}{a}+4\underset{˜}{b}=-8h\underset{˜}{a}+4h\underset{˜}{b}+6k\underset{˜}{a}+4k\underset{˜}{b}\\ -2\underset{˜}{a}+4\underset{˜}{b}=-8h\underset{˜}{a}+4h\underset{˜}{b}+6k\underset{˜}{a}+4k\underset{˜}{b}\end{array}$

–2 = –8h + 6k
–1 = –4h + 3k   → (1)

4 = 4h + 4k
1 = h + k
k= 1 – h   → (2)

Gantikan (2) ke dalam (1),
–1 = –4h + 3 (1 – h)
–1 = –4h + 3 – 3h
–4 = –7h