Bab 15 Vektor

Posted on May 17, 2016 by user

4.7 Vektor, SPM Praktis (Kertas 2)

Soalan 2:

Diberi

$\vec{A B} = (\begin{matrix} 10 \\ 14 \end{matrix}), \vec{O B} = (\begin{matrix} 4 \\ 6 \end{matrix}), dan \vec{C D} = (\begin{matrix} m \\ 7 \end{matrix}), carikan$

(a) koordinat A,

(b) vektor unit dalam arah

$\vec{O A}$

Penyelesaian:

(a)

$\begin{array}{l} \vec{A B} = (\begin{matrix} 10 \\ 14 \end{matrix}), \vec{O B} = (\begin{matrix} 4 \\ 6 \end{matrix}) \vec{C D} = (\begin{matrix} m \\ 7 \end{matrix}) \\ \vec{A B} = \vec{A O} + \vec{O B} \\ (\begin{matrix} 10 \\ 14 \end{matrix}) = (\begin{matrix} x \\ y \end{matrix}) + (\begin{matrix} 4 \\ 6 \end{matrix}) \\ (\begin{matrix} x \\ y \end{matrix}) = (\begin{matrix} 10 \\ 14 \end{matrix}) - (\begin{matrix} 4 \\ 6 \end{matrix}) \\ \vec{A O} = (\begin{matrix} 6 \\ 8 \end{matrix}) \\ \vec{O A} = (\begin{matrix} - 6 \\ - 8 \end{matrix}) \\ A = (- 6, - 8) \end{array}$

(b)

$\begin{array}{l} | \vec{O A} | = \sqrt{{(- 6)}^{2} + {(- 8)}^{2}} \\ | \vec{O A} | = \sqrt{100} = 10 \\ the unit vector in the direction of \vec{O A} \\ = \frac{\vec{O A}}{| \vec{O A} |} \\ = \frac{(\begin{matrix} - 6 \\ - 8 \end{matrix})}{10} = \frac{1}{10} (\begin{matrix} - 6 \\ - 8 \end{matrix}) \\ = (\begin{matrix} - \frac{3}{5} \\ - \frac{4}{5} \end{matrix}) \end{array}$

(c)

$\begin{array}{l} Given \vec{C D} parallel \vec{A B} \\ ∴ \vec{C D} = k \vec{A B} \\ (\begin{matrix} m \\ 7 \end{matrix}) = k (\begin{matrix} 10 \\ 14 \end{matrix}) \\ (\begin{matrix} m \\ 7 \end{matrix}) = (\begin{matrix} 10 k \\ 14 k \end{matrix}) \end{array}$

7 = 14k

k= ½

m = 10k = 10 (½) = 5