Bab 15 Vektor

Posted on July 6, 2016 by user

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

O P = \frac{1}{4} O B, A Q = \frac{1}{4} A B, \vec{O P} = 4 \underset{˜}{b} dan \vec{O A} = 8 \underset{˜}{a} .

(a) Ungkapakan dalam sebutan

\underset{˜}{a} dan \underset{˜}{b} :

(i)

\vec{A P}

(ii)

\vec{O Q}

(b) (i) Diberi bahawa

\vec{A R} = h \vec{A P}, nyatakan \vec{A R} dalam sebutan h, \underset{˜}{a} dan \underset{˜}{b} .

(ii) Diberi bahawa

\vec{R Q} = k \vec{O Q}, nyatakan \vec{A R} dalam sebutan k, \underset{˜}{a} dan \underset{˜}{b} .

\vec{A Q} = \vec{A R} + \vec{R Q},

cari nilai bagi h dan k.

Penyelesaian:

(a)(i)

\begin{array}{l} \vec{A P} = \vec{A O} + \vec{O P} \\ \vec{A P} = - \vec{O A} + \vec{O P} \\ \vec{A P} = - 8 \underset{˜}{a} + 4 \underset{˜}{b} \end{array}

(a)(ii)

\begin{array}{l} \vec{O Q} = \vec{O A} + \vec{A Q} \\ \vec{O Q} = 8 \underset{˜}{a} + \frac{1}{4} \vec{A B} \\ \vec{O Q} = 8 \underset{˜}{a} + \frac{1}{4} (\vec{A O} + \vec{O B}) \\ \vec{O Q} = 8 \underset{˜}{a} + \frac{1}{4} (- 8 \underset{˜}{a} + 4 \vec{O P}) \\ \vec{O Q} = 8 \underset{˜}{a} + \frac{1}{4} (- 8 \underset{˜}{a} + 4 (4 \underset{˜}{b})) \\ \vec{O Q} = 8 \underset{˜}{a} - 2 \underset{˜}{a} + 4 \underset{˜}{b} \\ \vec{O Q} = 6 \underset{˜}{a} + 4 \underset{˜}{b} \end{array}

(b)(i)

\begin{array}{l} \vec{A R} = h \vec{A P} \\ \vec{A R} = h (- 8 \underset{˜}{a} + 4 \underset{˜}{b}) \\ \vec{A R} = - 8 h \underset{˜}{a} + 4 h \underset{˜}{b} \end{array}

(b)(ii)

\begin{array}{l} \vec{R Q} = k \vec{O Q} \\ \vec{R Q} = k (6 \underset{˜}{a} + 4 \underset{˜}{b}) \\ \vec{R Q} = 6 k \underset{˜}{a} + 4 k \underset{˜}{b} \end{array}

(c)

\begin{array}{l} \vec{A Q} = \vec{A R} + \vec{R Q} \\ \vec{A Q} = - 8 h \underset{˜}{a} + 4 h \underset{˜}{b} + (6 k \underset{˜}{a} + 4 k \underset{˜}{b}) \\ \vec{A O} + \vec{O Q} = - 8 h \underset{˜}{a} + 4 h \underset{˜}{b} + 6 k \underset{˜}{a} + 4 k \underset{˜}{b} \\ - 8 \underset{˜}{a} + 6 \underset{˜}{a} + 4 \underset{˜}{b} = - 8 h \underset{˜}{a} + 4 h \underset{˜}{b} + 6 k \underset{˜}{a} + 4 k \underset{˜}{b} \\ - 2 \underset{˜}{a} + 4 \underset{˜}{b} = - 8 h \underset{˜}{a} + 4 h \underset{˜}{b} + 6 k \underset{˜}{a} + 4 k \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

\begin{array}{l} h = \frac{4}{7} \\ Daripada (2), \\ k = 1 - \frac{4}{7} = \frac{3}{7} \end{array}