8.7.4 Vektor, SPM Praktis (Kertas 2) - Matematik Tambahan SPM

Soalan 7:
Rajah di bawah menunjukkan sisi empat OPBC. Garis lurus AC bersilang dengan garis lurus PQ di titik B.

\begin{array}{l} Diberi bahawa \vec{O P} = \underset{˜}{a}, \vec{O Q} = \underset{˜}{b}, \vec{O A} = 4 \vec{A P}, \vec{O C} = 3 \vec{O Q}, \vec{P B} = h \vec{P Q} dan \\ \vec{A B} = k \vec{A C} . \\ (a) Ungkapkan \vec{O B} dalam sebutan h, \underset{˜}{a} dan \underset{˜}{b} . \\ (b) Ungkapkan \vec{O B} dalam sebutan k, \underset{˜}{a} dan \underset{˜}{b} . \\ (c)(i) Cari nilai h dan nilai k . \\ (ii) Seterusnya, nyatakan \vec{O B} dalam sebutan \underset{˜}{a} dan \underset{˜}{b} . \end{array}

Penyelesaian:
(a)

\begin{array}{l} \vec{O B} = \vec{O P} + \vec{P B} \\ = \underset{˜}{a} + h \vec{P Q} \\ = \underset{˜}{a} + h (\vec{P O} + \vec{O Q}) \\ = \underset{˜}{a} + h (- \underset{˜}{a} + \underset{˜}{b}) \\ = \underset{˜}{a} - h \underset{˜}{a} + h \underset{˜}{b} \\ \vec{O B} = (1 - h) \underset{˜}{a} + h \underset{˜}{b} \end{array}

(b)

\begin{array}{l} \vec{O B} = \vec{O P} + \vec{P B} \\ = \underset{˜}{a} + \vec{P A} + \vec{A B} \\ = \underset{˜}{a} + (- \frac{1}{5} \vec{O P}) + k \vec{A C} \\ = \underset{˜}{a} + (- \frac{1}{5} \underset{˜}{a}) + k (\vec{A O} + \vec{O C}) \\ = \frac{4}{5} \underset{˜}{a} + k (- \frac{4}{5} \vec{O P} + 3 \vec{O Q}) \\ = \frac{4}{5} \underset{˜}{a} + k (- \frac{4}{5} \underset{˜}{a} + 3 \underset{˜}{b}) \\ = \frac{4}{5} \underset{˜}{a} - \frac{4}{5} k \underset{˜}{a} + 3 k \underset{˜}{b} \\ \vec{O B} = \frac{4}{5} (1 - k) \underset{˜}{a} + 3 k \underset{˜}{b} \end{array}

(c)(i)

\begin{array}{l} (1 - h) \underset{˜}{a} + h \underset{˜}{b} = \frac{4}{5} (1 - k) \underset{˜}{a} + 3 k \underset{˜}{b} \\ 1 - h = \frac{4}{5} - \frac{4}{5} k ………. (1) \\ h = 3 k ………. (2) \\ Gantikan (2) ke dalam (1) \\ 1 - 3 k = \frac{4}{5} - \frac{4}{5} k \\ 5 - 15 k = 4 - 4 k \\ 11 k = 1 \\ k = \frac{1}{11} \\ Gantikan k = \frac{1}{11} ke dalam (2) \\ h = 3 (\frac{1}{11}) \\ = \frac{3}{11} \end{array}

(c)(ii)

\begin{array}{l} \vec{O B} = (1 - h) \underset{˜}{a} + h \underset{˜}{b} \\ apabila h = \frac{3}{11} \\ = (1 - \frac{3}{11}) \underset{˜}{a} + (\frac{3}{11}) \underset{˜}{b} \\ = \frac{8}{11} \underset{˜}{a} + \frac{3}{11} \underset{˜}{b} \end{array}

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

\begin{array}{l} Diberi bahawa \vec{O P} = 7 \underset{˜}{x}, \vec{O R} = 5 \underset{˜}{y}, P S : S R = 3 : 1 dan \vec{O R} selari dengan \vec{P Q} . \\ (a) Ungkapkan dalam sebutan \underset{˜}{x} dan \underset{˜}{y}, \\ (i) \vec{P R} \\ (ii) \vec{O S} \\ (b) Guna \vec{P Q} = m \vec{O R} dan \vec{S Q} = n \vec{O S}, dengan keadaan m dan n adalah pemalar, \\ Cari nilai m dan nilai n . \\ (c) Diberi bahawa | \underset{˜}{y} | = 4 unit dan luas O R S {ialah 50 cm}^{2}, cari jarak tegak \\ dari titik S ke O R . \end{array}

Penyelesaian:
(a)(i)

\begin{array}{l} \vec{P R} = \vec{P O} + \vec{O R} \\ = - 7 \underset{˜}{x} + 5 \underset{˜}{y} \end{array}

(a)(ii)

\begin{array}{l} \vec{O S} = \vec{O P} + \vec{P S} \\ = 7 \underset{˜}{x} + \frac{3}{4} \vec{P R} \\ = 7 \underset{˜}{x} + \frac{3}{4} (- 7 \underset{˜}{x} + 5 \underset{˜}{y}) \\ = 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} \vec{P S} = \vec{P Q} - \vec{S Q} \\ \frac{3}{4} \vec{P R} = m \vec{O R} - n \vec{O S} \\ \frac{3}{4} (- 7 \underset{˜}{x} + 5 \underset{˜}{y}) = m (5 \underset{˜}{y}) - n (\frac{7}{4} \underset{˜}{x} + \frac{15}{4} \underset{˜}{y}) \\ - \frac{21}{4} \underset{˜}{x} + \frac{15}{4} \underset{˜}{y} = 5 m \underset{˜}{y} - \frac{7}{4} n \underset{˜}{x} - \frac{15}{4} n \underset{˜}{y} \\ - \frac{21}{4} \underset{˜}{x} + \frac{15}{4} \underset{˜}{y} = - \frac{7}{4} n \underset{˜}{x} + 5 m \underset{˜}{y} - \frac{15}{4} n \underset{˜}{y} \\ - \frac{7}{4} n = - \frac{21}{4} \\ 7 n = 21 \\ n = 3 \\ 5 m - \frac{15}{4} n = \frac{15}{4} \\ 5 m - \frac{15}{4} (3) = \frac{15}{4} \\ 5 m - \frac{45}{4} = \frac{15}{4} \\ 5 m = 15 \\ m = 3 \end{array}

(c)

\begin{array}{l} Luas Δ O R S = 50 \\ \frac{1}{2} \times (5 \underset{˜}{y}) \times t = 50 \\ \frac{1}{2} \times 5 (4) \times t = 50 \\ 10 t = 50 \\ t = 5 \\ ∴ Jarak tegak dari titik S ke O R = 5 unit . \end{array}