Soal Vektor

Posted by Unknown Tuesday, 30 April 2013

Diketahui titik A(5, 1, 3), B(2, -1, -1), dan C(4, 2, -4). Besar sudut ABC = ….

Penyelesaian :

Diketahui titik A(5, 1, 3), B(2, -1, -1), dan C(4, 2, -4).
$\inline \fn_cm \overrightarrow{a}=\begin{pmatrix} 5 \\ 1\\3 \end{pmatrix},\overrightarrow{b}=\begin{pmatrix} 2 \\ -1\\-1 \end{pmatrix},dan \overrightarrow{c}=\begin{pmatrix} 4 \\ 2\\-4 \end{pmatrix}$
Misalkan $\inline \fn_cm \theta$ adalah besar sudut ABC, hal ini berarti $\inline \fn_cm \theta$ adalah sudut antara vektor $\inline \fn_cm \overrightarrow{BA}$ dan vektor $\inline \fn_cm \overrightarrow{BC}$
$\inline \fn_cm \overrightarrow{BA}=\overrightarrow{a}-\overrightarrow{b}=\begin{pmatrix} 5 \\ 1\\3 \end{pmatrix}-\begin{pmatrix} 2 \\-1\\-1 \end{pmatrix}=\begin{pmatrix} 3 \\2\\4 \end{pmatrix}$
$\inline \fn_cm \overrightarrow{BC}=\overrightarrow{c}-\overrightarrow{b}=\begin{pmatrix} 4 \\2\\-4 \end{pmatrix}-\begin{pmatrix} 2 \\-1\\-1 \end{pmatrix}=\begin{pmatrix} 2 \\3\\-3 \end{pmatrix}$
sehingga :
$\cos \theta =\frac{\overrightarrow{BA}.\overrightarrow{BC}}{|\overrightarrow{BA}|.\overrightarrow{BC}|}=\frac{(3.2)+(2.3)+(4.-3)}{\sqrt{(3)^2+(2)^2+(4)^2}\: .\: \sqrt{(2)^2+(3)^2+(-3)^2}}$

$=\frac{6+6-12}{\sqrt{9+4+16}.\sqrt{4+9+9}}=\frac{0}{\sqrt{29}.\sqrt{22}}=0$

$\cos \theta =0$ maka

$\theta =\frac{\pi }{2}$
Diketahui vektor $\overrightarrow{a}=4\overrightarrow{i}-2\overrightarrow{j}+2\overrightarrow{k}$ dan vektor $\overrightarrow{b}=2\overrightarrow{i}-6\overrightarrow{j}+4\overrightarrow{k}$ . Proyeksi vektor orthogonal vektor $\overrightarrow{a}$ pada vektor $\overrightarrow{b}$ adalah ….
Penyelesaian :

$\inline \fn_cm \overrightarrow{a}=4\overrightarrow{i}-2\overrightarrow{j}+2\overrightarrow{k}=\begin{pmatrix} 4\\-2\\2 \end{pmatrix}$
$\inline \fn_cm \overrightarrow{b}=2\overrightarrow{i}-6\overrightarrow{j}+4\overrightarrow{k}=\begin{pmatrix} 2\\-6\\4 \end{pmatrix}$
Misalkan $\inline \fn_cm \overrightarrow{c}$ adalah proyeksi vektor orthogonal vektor $\inline \fn_cm \overrightarrow{a}$ pada vektor $\inline \fn_cm \overrightarrow{b}$ maka :
$\inline \fn_cm \overrightarrow{c}=\frac{\overrightarrow{a}.\overrightarrow{b}}{|\overrightarrow{b}|^2}.\overrightarrow{b}$

$\inline \fn_phv \overrightarrow{c}=\frac{(4.2)+(-2.-6)+(2.4)}{\left ( \sqrt{2^2+6^2+4^2} \right )^2}.(2\overrightarrow{i}-6\overrightarrow{j}+4\overrightarrow{k})$

$\inline \fn_phv \overrightarrow{c}=\frac{8+12+8}{2^2+6^2+4^2}.(2\overrightarrow{i}-6\overrightarrow{j}+4\overrightarrow{k})=\frac{28}{56}.(2\overrightarrow{i}-6\overrightarrow{j}+4\overrightarrow{k})$

$\inline \fn_phv \overrightarrow{c}=\frac{1}{2}.(2\overrightarrow{i}-6\overrightarrow{j}+4\overrightarrow{k})=\overrightarrow{i}-3\overrightarrow{j}+2\overrightarrow{k}$

Oleh : Manase Rumahombar (120170118)

No comments

Leave a Reply