Bab 15 Vektor

Posted on July 5, 2016 by user

4.5 Vektor dalam Satah Cartes

1. Vektor unit mempunyai magnitud satu unit.

2. Vektor

$\underset{˜}{i}$ ialah vektor unit ke arah positif paksi-x.

Vektor

$\underset{˜}{j}$ ialah vektor unit ke arah positif paksi-y.

3. Suatu vektor dalam satah Cartes boleh ditulis dalam bentuk

$x \underset{˜}{i} + y \underset{˜}{j} atau vektor lajur (\begin{array}{l} x \\ y \end{array}) .$

4. Magnitud bagi vektor unit ialah

$| \underset{˜}{i} | = | \underset{˜}{j} | = 1.$

5. Magnitud bagi vektor

$\vec{O A} = x \underset{˜}{i} + y \underset{˜}{j} diberi oleh | \vec{O A} | = \sqrt{x^{2} + y^{2}} .$

Contoh 1:

Jika

$\underset{˜}{r} = k \underset{˜}{i} - 8 \underset{˜}{j} dan | \underset{˜}{r} | = 10,$ cari nilai k. Tentukan vektor unit dalam arah

$\underset{˜}{r}$ bagi setiap nilai k.

Penyelesaian:

$\begin{array}{l} Diberi | \underset{˜}{r} | = 10 \\ \sqrt{x^{2} + y^{2}} = 10 \\ \sqrt{k^{2} + {(- 8)}^{2}} = 10 \\ k^{2} + 64 = 100 \\ k = \pm 6 \end{array}$

$\begin{array}{l} Vektor unit, \hat{\underset{˜}{r}} = \frac{x \underset{˜}{i} + y \underset{˜}{j}}{\sqrt{x^{2} + y^{2}}} \\ Apabila k = 6, Apabila k = - 6 \\ \hat{\underset{˜}{r}} = \frac{6 \underset{˜}{i} - 8 \underset{˜}{j}}{10} = \frac{3 \underset{˜}{i} - 4 \underset{˜}{j}}{5} \hat{\underset{˜}{r}} = \frac{- 6 \underset{˜}{i} - 8 \underset{˜}{j}}{10} = \frac{- 3 \underset{˜}{i} - 4 \underset{˜}{j}}{5} \\ \hat{\underset{˜}{r}} = \frac{1}{5} (3 \underset{˜}{i} - 4 \underset{˜}{j}) \hat{\underset{˜}{r}} = \frac{1}{5} (- 3 \underset{˜}{i} - 4 \underset{˜}{j}) \end{array}$

Contoh 2:

Diberi bahawa

$\underset{˜}{a} = (\begin{array}{l} 6 \\ 3 \end{array}) dan \underset{˜}{b} = (\begin{array}{l} 3 \\ 7 \end{array}) .$

(a) Cari

$\underset{˜}{b} - \underset{˜}{a} dan | \underset{˜}{b} - \underset{˜}{a} | .$

(b) Seterusnya, cari vektor unit dalam arah

$\underset{˜}{b} - \underset{˜}{a} .$

Penyelesaian:

(a)

$\begin{array}{l} \underset{˜}{b} - \underset{˜}{a} = (\begin{array}{l} 3 \\ 7 \end{array}) - (\begin{array}{l} 6 \\ 3 \end{array}) \\ = (\begin{array}{l} 3 - 6 \\ 7 - 3 \end{array}) \\ = (\begin{array}{l} - 3 \\ 4 \end{array}) \\ | \underset{˜}{b} - \underset{˜}{a} | = \sqrt{{(- 3)}^{2} + 4^{2}} \\ = \sqrt{9 + 16} \\ = \sqrt{25} = 5 \end{array}$

(b)

$\begin{array}{l} Vektor unit dalam arah \underset{˜}{b} - \underset{˜}{a} \\ = \frac{1}{5} (\begin{array}{l} - 3 \\ 4 \end{array}) \\ = (\begin{array}{l} - \frac{3}{5} \\ \frac{4}{5} \end{array}) \end{array}$