Operasi Vektor di Ruang (Lanjutan Materi Vektor di Ruang)

Sebelumnya silahkan lihat di sini

 $\text{A. Operasi Vektor Dalam Ruang}$

Operasi vektor pada dimensi tiga kurang lebih sama dengan operasi pada vektor berdimensi dua.

$\text{A. Penjumlahan dan Pengurangan}$.

$\begin{array}{rl}\text{Jika}& \text{diketahui sebagai misal}\\ \overline{u}& =a\overline{i}+b\overline{j}+c\overline{k}\text{dan}\\ \overline{v}& =p\overline{i}+q\overline{j}+r\overline{k}\\ \text{mak}& \text{a}\\ \mathbf{\text{Pen}}& \mathbf{\text{jumlahan dua vektor di atas adalah}}\\ \overline{u}+\overline{v}& =(a+p)\overline{i}+(b+q)\overline{j}+(c+r)\overline{k}\\ \mathbf{\text{dem}}& \mathbf{\text{ikian juga untuk pengurangan}}\\ \overline{u}-\overline{v}& =(a-p)\overline{i}+(b-q)\overline{j}+(c-r)\overline{k}\end{array}$.

$\text{CONTOH SOAL}$.

$\begin{array}{l}\\ 1.& \text{Jika diketahui}\overline{a}=\left(\begin{array}{c}1\\ 3\\ 7\end{array}\right)\text{dan}\overline{b}=\left(\begin{array}{c}8\\ -2\\ 0\end{array}\right)\\ & \text{Tentukanlah hasil dari}\\ & \text{a}.\overline{a}+\overline{b}\\ & \text{b}.\overline{a}-\overline{b}\\ \\ & \mathbf{\text{Jawab}}\\ & \begin{array}{rl}& \text{Diketahui bahwa}\\ & \overline{a}=\left(\begin{array}{c}1\\ 3\\ 7\end{array}\right)\text{dan}\overline{b}=\left(\begin{array}{c}8\\ -2\\ 0\end{array}\right),\\ & \text{maka}\\ & \overline{a}+\overline{b}=\left(\begin{array}{c}1\\ 3\\ 7\end{array}\right)+\left(\begin{array}{c}8\\ -2\\ 0\end{array}\right)\\ & =\left(\begin{array}{c}1+8\\ 3+(-2)\\ 7+0\end{array}\right)=\left(\begin{array}{c}9\\ 1\\ 7\end{array}\right)\\ & \text{Dan untuk}\overline{a}-\overline{b}\text{adalah}:\\ & \overline{a}-\overline{b}=\left(\begin{array}{c}1\\ 3\\ 7\end{array}\right)-\left(\begin{array}{c}8\\ -2\\ 0\end{array}\right)\\ & =\left(\begin{array}{c}1-8\\ 3-(-2)\\ 7-0\end{array}\right)=\left(\begin{array}{c}-7\\ 5\\ 7\end{array}\right)\end{array}\end{array}$.

$\text{B. 1. Perkalian Skalar dengan Vektor}$.

Misalkan suatu skalar   $m$  dan suatu vektor  $\overline{u}=a\overline{i}+b\overline{j}+c\overline{k}$, maka perkalian $m$  dengan vektor  $\overline{u}$ tersebut adalah  $\overline{u}=ma\overline{i}+mb\overline{j}+mc\overline{k}$.

$\text{CONTOH SOAL}$.

$\begin{array}{ll}2.& \text{Jika}\overline{a}=\left(\begin{array}{c}2022\\ 2021\\ 2020\end{array}\right),\text{tentukanlah nilai}\\ & \text{dari}2\overline{a}\text{dan}-3\overline{a}\\ \\ & \mathbf{\text{Jawab}}\\ & \begin{array}{rl}2\overline{a}& =2\left(\begin{array}{c}2022\\ 2021\\ 2020\end{array}\right)=\left(\begin{array}{c}4044\\ 4042\\ 4040\end{array}\right),\text{dan}\\ -3\overline{a}& =-3\left(\begin{array}{c}2022\\ 2021\\ 2020\end{array}\right)=\left(\begin{array}{c}-6066\\ -6063\\ -6060\end{array}\right)\end{array}\end{array}$

$\text{F. 2. Perkalian Skalar Dua Vektor}$.

Hasil dari perkalian skalar dua vektor $\overline{a}$  dan  $\overline{b}$ adalah :  $\overline{a}\bullet \overline{b}$.

Dengan

$\overline{a}\bullet \overline{b}=\left|\overline{a}\right|\left|\overline{b}\right|\cos \theta$.  sehingga

$\begin{array}{rl}& \text{Tanda dari hasil skalar ini adalah}\\ & \begin{array}{|lll|}\hline \mathbf{\text{Besar sudut}}\theta & \mathbf{\text{Tanda}}& \text{Bentuk}\\ 0^{\circ }\le \theta <{90}^{\circ }& \text{Positif}& \text{Lancip}\\ \theta ={90}^{\circ }& \text{Nol}& \text{Siku-siku}\\ {90}^{\circ }<\theta \le {180}^{\circ }& \text{Negatif}& \text{Tumpul}\\ \hline\end{array}\\ & \text{Untuk}\theta \text{berupa sudut istimewa}:\\ & \begin{array}{|ccccccc|}\hline \theta & 0^{\circ }& {30}^{\circ }& {45}^{\circ }& {60}^{\circ }& {90}^{\circ }& {180}^{\circ }\\ \cos \theta & 1& {\displaystyle \frac{1}{2}\sqrt{3}}& {\displaystyle \frac{1}{2}\sqrt{2}}& {\displaystyle \frac{1}{2}}& 0& -1\\ \hline\end{array}\end{array}$

Adapun secara rumus untuk menentukan besar sudutnya adalah:

$\cos \theta ={\displaystyle \frac{\overline{a}\bullet \overline{b}}{\left|\overline{a}\right|\left|\overline{b}\right|}}$.

Sebagai ilustrasinya perhatikanlah gambar berikut

Selain hasil di atas ada cara lain menyelesaikan perkalian skalar dua vektor, yaitu:

$\begin{array}{rl}\text{Jika}& \text{diketahui sebagai misal}\\ \overline{u}& =a\overline{i}+b\overline{j}+c\overline{k}\text{dan}\\ \overline{v}& =p\overline{i}+q\overline{j}+r\overline{k}\\ \text{mak}& \text{a}\\ \mathbf{\text{Per}}& \mathbf{\text{kalian skalar dua vektor adalah}}:\\ \overline{u}\bullet & \overline{v}=(a\overline{i}+b\overline{j}+c\overline{k})(p\overline{i}+q\overline{j}+r\overline{k})\\ & =ap.\overline{i}\bullet \overline{i}+aq.\overline{i}\bullet \overline{j}+ar.\overline{i}\bullet \overline{k}\\ & +bp.\overline{j}\bullet \overline{i}+bq.\overline{j}\bullet \overline{j}+br.\overline{j}\bullet \overline{k}\\ & +cp.\overline{k}\bullet \overline{i}+cq.\overline{k}\bullet \overline{j}+cr.\overline{k}\bullet \overline{k}\\ & =ap+0+0+0+bq+0+0+0+cr\\ & =ap+bq+cr\end{array}$

$\begin{array}{rl}& \text{Sebagai penjelasannya adalah}:\\ & ▹\overline{i}\bullet \overline{i}=\left|\overline{i}\right|\left|\overline{i}\right|\cos 0^{\circ }=1.1.1=1\\ & ▹\overline{i}\bullet \overline{j}=\left|\overline{i}\right|\left|\overline{j}\right|\cos {90}^{\circ }=1.1.0=0\\ & ▹\overline{i}\bullet \overline{k}=\left|\overline{i}\right|\left|\overline{k}\right|\cos {90}^{\circ }=1.1.0=0\\ & ▹\overline{j}\bullet \overline{i}=\overline{i}\bullet \overline{j}=0\\ & ▹\overline{j}\bullet \overline{j}=\left|\overline{j}\right|\left|\overline{j}\right|\cos 0^{\circ }=1.1.1=1\\ & ▹\overline{j}\bullet \overline{k}=\left|\overline{j}\right|\left|\overline{k}\right|\cos {90}^{\circ }=1.1.0=0\\ & ▹\overline{k}\bullet \overline{i}=\overline{i}\bullet \overline{k}=0\\ & ▹\overline{k}\bullet \overline{j}=\overline{j}\bullet \overline{k}=0\\ & ▹\overline{k}\bullet \overline{k}=\left|\overline{k}\right|\left|\overline{k}\right|\cos {90}^{\circ }=1.1.1=1\end{array}$

$\begin{array}{rl}& \text{Atau jika ditabelkan nilainya}\\ & \begin{array}{|cccc|}\hline \overline{u}\bullet \overline{v}& p\overline{i}& q\overline{j}& r\overline{k}\\ a\overline{k}& ap& 0& 0\\ b\overline{j}& 0& bq& 0\\ c\overline{k}& 0& 0& cr\\ \hline\end{array}\end{array}$

$\text{CONTOH SOAL}$.

$\begin{array}{ll}3.& \text{Jika}\overrightarrow{a}=\left(\begin{array}{c}1\\ 2\\ 4\end{array}\right),\text{dan}\overrightarrow{b}=\left(\begin{array}{c}5\\ 4\\ 0\end{array}\right)\\ & \text{tentukanlah nilai}\text{dari}\overrightarrow{a}\bullet \overrightarrow{b}\\ \\ & \mathbf{\text{Jawab}}\\ & \begin{array}{rl}\overrightarrow{a}\bullet \overrightarrow{b}& =1.5+2.4+4.0=5+8+0=13\end{array}\end{array}$

$\begin{array}{ll}4.& \text{Jika diketahui}\overrightarrow{a}=\overrightarrow{i}-2\overrightarrow{j}+3\overrightarrow{k},\\ & \text{dan}\overrightarrow{b}=3\overrightarrow{i}-4\overrightarrow{j}+m\overrightarrow{k}\text{serta}\\ & \text{nilai}\overrightarrow{a}\bullet \overrightarrow{b}=-4,\text{maka tentukan}\\ & \text{nilai}m\\ \\ & \mathbf{\text{Jawab}}\\ & \text{Diketahui bahwa}\\ & ▹\overrightarrow{a}=\overrightarrow{i}-2\overrightarrow{j}+3\overrightarrow{k}=\left(\begin{array}{c}1\\ -2\\ 3\end{array}\right),\text{dan}\\ & ▹\overrightarrow{b}=3\overrightarrow{i}-4\overrightarrow{j}+m\overrightarrow{k}=\left(\begin{array}{c}3\\ -4\\ m\end{array}\right)\\ & \begin{array}{rl}\overrightarrow{a}\bullet \overrightarrow{b}& =1.3+3.(-4)+3.m\\ -4& =3+8+3m\\ -3m& =11+4\\ m& =-{\displaystyle \frac{15}{3}}\\ & =-5\end{array}\end{array}$

$\begin{array}{ll}5.& \text{Diketahui}\left|\overrightarrow{a}\right|=10,\left|\overrightarrow{b}\right|=6.\\ & \text{Jika}\overrightarrow{a}\text{dan}\overrightarrow{b}\text{membentuk sudut}\\ & {60}^{\circ }.\text{Tentukanlah nilai}\overrightarrow{a}\bullet \overrightarrow{b}\\ \\ & \mathbf{\text{Jawab}}\\ & \begin{array}{rl}\overrightarrow{a}\bullet \overrightarrow{b}& =\left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right|\cos \theta \\ & =10.6.\cos {60}^{\circ }\\ & =60.\left({\displaystyle \frac{1}{2}}\right)\\ & =30\\ \text{Jadi}& \text{hasil kali skalarnya adalah 30}\end{array}\end{array}$.