Operasi Vektor Berdimensi Tiga

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

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

$\text{F. 1. Penjumlahan dan Pengurangan}$.

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

$\text{CONTOH SOAL}$.

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

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

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  $\overline{u}=ma\overline{i}+mb\overline{j}+mc\overline{k}$.

$\text{CONTOH SOAL}$.

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

$\text{F. 3. Perkalian Skalar Dua Vektor}$.

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:

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

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

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

$\text{CONTOH SOAL}$.

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

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

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

$\begin{array}{l}\\ 6.& \text{Diketahui}\overrightarrow{a}=\left(\begin{array}{c}-2\\ 1\\ 3\end{array}\right)\text{dan}\overrightarrow{b}=\left(\begin{array}{c}4\\ -1\\ t\end{array}\right),\\ & \text{jika}\overrightarrow{p}\text{tegak lurus}\overrightarrow{q},\text{maka tentukanlah}\\ & \text{nilai}t\text{adalah}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}\text{Karena}& \text{kedua vektor tersebut saling }\\ \text{tegak l}& \text{urus maka}\\ \overrightarrow{a}.\overrightarrow{b}& =0\\ \left(\begin{array}{c}-2\\ 1\\ 3\end{array}\right)& \left(\begin{array}{c}4\\ -1\\ t\end{array}\right)=0\\ (-2).4& +1.(-1)+3.t=0\\ -8-1& +3t=0\\ 3t& =9\\ t& =3\end{array}\end{array}$

$\begin{array}{l}\\ 7.& \text{Tentukanlah nilai}\overrightarrow{a}.\overrightarrow{b}\text{jika}\\ & \text{a}.\left|\overrightarrow{a}\right|=4,\left|\overrightarrow{b}\right|=6,\mathrm{\angle }(\overrightarrow{a},\overrightarrow{b})={60}^{\circ }\\ & \text{b}.\overrightarrow{a}=2\overrightarrow{i}+\overrightarrow{j}-5\overrightarrow{k}\text{dan}\overrightarrow{b}=2\overrightarrow{i}-3\overrightarrow{k}\\ & \text{c}.\overrightarrow{a}=\left(\begin{array}{c}0\\ -1\\ 3\end{array}\right)\text{dan}\overrightarrow{b}=\left(\begin{array}{c}4\\ -2\\ 1\end{array}\right)\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}\text{a}.\overrightarrow{a}.\overrightarrow{b}& =\left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right|\cos \mathrm{\angle }(\overrightarrow{a},\overrightarrow{b})\\ & =4.6.\cos {60}^{\circ }\\ & =24.\left({\displaystyle \frac{1}{2}}\right)\\ & =12\end{array}\\ & \text{b}.\overrightarrow{a}.\overrightarrow{b}=2.2+1.0+(-5).(-3)=4+15=19\\ & \text{c}.\overrightarrow{a}.\overrightarrow{b}=0.4+(-1).(-2)+3.1=0+2+3=5\end{array}$

$\begin{array}{ll}8.& \text{Diketahui}\left|\overrightarrow{a}\right|=10,\left|\overrightarrow{b}\right|=3\\ & \text{dan}\overrightarrow{a}\bullet \overrightarrow{b}=15\sqrt{3}.\text{Tentukan sudut}\\ & \text{yang dibentuk oleh}\overrightarrow{a}\text{dan}\overrightarrow{b}\\ \\ & \mathbf{\text{Jawab}}\\ & \text{Dari bentuk}\\ & \begin{array}{rl}\overrightarrow{a}\bullet \overrightarrow{b}& =\left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right|\cos \theta \\ \text{dipe}& \text{roleh bentuk}\\ \cos \theta & ={\displaystyle \frac{\overrightarrow{a}\bullet \overrightarrow{b}}{\left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right|}}\\ \cos \theta & ={\displaystyle \frac{15\sqrt{3}}{10.3}=\frac{15}{30}\sqrt{3}=\frac{1}{2}\sqrt{3}}\\ \cos \theta & =\cos {30}^{\circ }\\ \theta & ={30}^{\circ }\\ \text{Jadi}& \text{sudut antara keduanya adalah}{30}^{\circ }\end{array}\end{array}$.

$\begin{array}{l}\\ 9.& \text{Tentukanlah besar sudut antara vektor}\\ & \overrightarrow{a}=\left(\begin{array}{c}-1\\ 1\\ 0\end{array}\right)\text{dan}\overrightarrow{b}=\left(\begin{array}{c}1\\ -2\\ 2\end{array}\right)\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}\cos \theta & ={\displaystyle \frac{\overrightarrow{a}.\overrightarrow{b}}{\left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right|}}\\ & ={\displaystyle \frac{\left(\begin{array}{c}-1\\ 1\\ 0\end{array}\right)\left(\begin{array}{c}1\\ -2\\ 2\end{array}\right)}{\sqrt{(-1{)}^2+1^2}\sqrt{1^2+(-2{)}^2+2^2}}}\\ & ={\displaystyle \frac{-1-2+0}{\sqrt{2}\sqrt{9}}}\\ & =-{\displaystyle \frac{1}{\sqrt{2}}=-{\displaystyle \frac{1}{2}\sqrt{2}}}\\ & =-\cos {45}^{\circ }\\ & =\cos ({180}^{\circ }-{45}^{\circ })\\ \cos \theta & =\cos {135}^{\circ }\\ \therefore \theta & ={135}^{\circ }\end{array}\end{array}$.

$\begin{array}{l}\\ 10.& \text{Diketahui bahwa}\left|\overrightarrow{a}\right|=\sqrt{6},(\overrightarrow{a}-\overrightarrow{b})(\overrightarrow{a}+\overrightarrow{b})=0\\ & \text{dan}\overrightarrow{a}(\overrightarrow{a}-\overrightarrow{b})=3.\text{Tentukanlah besar}\\ & \text{sudut antara}\overrightarrow{a}\text{dan}\overrightarrow{b}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}\text{Perhatikan}& \text{bahwa}\\ (\overrightarrow{a}-\overrightarrow{b})(\overrightarrow{a}+\overrightarrow{b})& =0\\ {\left|\overrightarrow{a}\right|}^2-{\left|\overrightarrow{b}\right|}^2& =0\\ {\left|\overrightarrow{a}\right|}^2& ={\left|\overrightarrow{b}\right|}^2\Rightarrow \left|\overrightarrow{a}\right|=\overrightarrow{b}=\sqrt{6}\\ \text{dan}\overrightarrow{a}(\overrightarrow{a}-\overrightarrow{b})& =3\\ {\left|\overrightarrow{a}\right|}^2-\overrightarrow{a}\overrightarrow{b}& =3\\ 6-\overrightarrow{a}\overrightarrow{b}& =3\\ -\overrightarrow{a}\overrightarrow{b}& =3-6=-3\\ \overrightarrow{a}\overrightarrow{b}& =3\\ \left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right|\cos \theta & =3\\ \cos \theta & ={\displaystyle \frac{3}{\sqrt{6}\sqrt{6}}=\frac{3}{6}=\frac{1}{2}}\\ \cos \theta & =\cos {60}^{\circ }\\ \therefore \theta & ={60}^{\circ }\end{array}\end{array}$

$\text{Berikut dua contoh untuk sudut tidak istimewa}$.

$\begin{array}{ll}11.& \text{Diketahui}\overrightarrow{a}=\overrightarrow{i}+2\overrightarrow{j}+2\overrightarrow{k},\text{dan}\\ & \overrightarrow{b}=3\overrightarrow{i}+4\overrightarrow{j}.\text{Tentukan sudut}\\ & \text{yang dibentuk oleh}\overrightarrow{a}\text{dan}\overrightarrow{b}\\ \\ & \mathbf{\text{Jawab}}\\ & \begin{array}{rl}\text{Dik}& \text{etahui bahwa}\\ ▹& \overrightarrow{a}=\overrightarrow{i}+2\overrightarrow{j}+2\overrightarrow{k}=\left(\begin{array}{c}1\\ 2\\ 2\end{array}\right)\text{dan}\\ & \left|\overrightarrow{a}\right|=\sqrt{1^2+2^2+2^2}=\sqrt{9}=3\\ ▹& \overrightarrow{b}=3\overrightarrow{i}+4\overrightarrow{j}=\left(\begin{array}{c}3\\ 4\\ 0\end{array}\right)\text{dan}\\ & \left|\overrightarrow{b}\right|=\sqrt{3^2+4^2+0^2}=\sqrt{25}=5\end{array}\\ & \text{Selanjutnya}\\ & \begin{array}{rl}\cos \theta & ={\displaystyle \frac{\overrightarrow{a}\bullet \overrightarrow{b}}{\left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right|}}\\ \cos \theta & ={\displaystyle \frac{\left(\begin{array}{c}1\\ 2\\ 2\end{array}\right)\left(\begin{array}{c}3\\ 4\\ 0\end{array}\right)}{3.5}=\frac{3+8+0}{15}=\frac{11}{15}}\\ \cos \theta & =0,733\\ \theta & =\arccos \left({\displaystyle 0.733}\right)\\ & \text{gunakan alat bantu tabel trigonometri}\\ & \text{atau kalkulator scientific}\\ & =42,9^{\circ }\\ \text{Jadi}& \text{sudut antara keduanya adalah}42,9^{\circ }\end{array}\end{array}$

$\begin{array}{ll}12.& \text{Diketahui}\overrightarrow{p}=(1,2,2),\text{dan}\\ & \overrightarrow{q}=(3,-2,6).\text{Tentukan sudut}\\ & \text{yang dibentuk oleh}\overrightarrow{p}\text{dan}\overrightarrow{q}\\ \\ & \mathbf{\text{Jawab}}\\ & \begin{array}{rl}\text{Dik}& \text{etahui bahwa}\\ ▹& \overrightarrow{p}=(1,2,2)=\left(\begin{array}{c}1\\ 2\\ 2\end{array}\right)\text{dan}\\ & \left|\overrightarrow{p}\right|=\sqrt{1^2+2^2+2^2}=\sqrt{9}=3\\ ▹& \overrightarrow{q}=(3,-2,6)=\left(\begin{array}{c}3\\ -2\\ 6\end{array}\right)\text{dan}\\ & \left|\overrightarrow{q}\right|=\sqrt{3^2+(-2{)}^2+6^2}=\sqrt{49}=7\end{array}\\ & \text{Selanjutnya}\\ & \begin{array}{rl}\cos \theta & ={\displaystyle \frac{\overrightarrow{p}\bullet \overrightarrow{q}}{\left|\overrightarrow{p}\right|\left|\overrightarrow{q}\right|}}\\ \cos \theta & ={\displaystyle \frac{\left(\begin{array}{c}1\\ 2\\ 2\end{array}\right)\left(\begin{array}{c}3\\ -2\\ 6\end{array}\right)}{3.7}=\frac{3-4+12}{21}=\frac{11}{21}}\\ \cos \theta & =0,524\\ \theta & =\arccos \left({\displaystyle 0.524}\right)\\ & \text{gunakan alat bantu tabel trigonometri}\\ & \text{atau kalkulator scientific}\\ & =58,4^{\circ }\\ \text{Jadi}& \text{sudut antara keduanya adalah}58,4^{\circ }\end{array}\end{array}$.

$\begin{array}{l}\\ 13.& \text{Diketahui vektor}\overrightarrow{a}\text{dan}\overrightarrow{b}\text{memiliki }\\ & \text{panjang masing-masing adalah 2 dan 3}\\ & \text{serta}\mathrm{\angle }(\overrightarrow{a},\overrightarrow{b})={60}^{\circ }.\text{Carilah nilai}\\ & \text{a}.|\overrightarrow{a}+\overrightarrow{b}|\\ \\ & \text{b}.|\overrightarrow{a}-\overrightarrow{b}|\\ & \text{b}\text{besar sudut antara}\\ & (\overrightarrow{a}+\overrightarrow{b})\text{dan}(\overrightarrow{a}-\overrightarrow{b})\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}\text{a}.& {|\overrightarrow{a}+\overrightarrow{b}|}^2\\ & =(\overrightarrow{a}+\overrightarrow{b})(\overrightarrow{a}+\overrightarrow{b})\\ & =\overrightarrow{a}\overrightarrow{a}+2\overrightarrow{a}\overrightarrow{b}+\overrightarrow{b}\overrightarrow{b}\\ & ={\left|\overrightarrow{a}\right|}^2\cos 0^{\circ }+2\left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right|\cos {60}^{\circ }+{\left|\overrightarrow{b}\right|}^2\cos 0^{\circ }\\ & =2^2.1+2.2.3.{\displaystyle \frac{1}{2}+3^2.1}\\ & =4+6+9=19\\ & \text{Jadi, nilainya adalah}|\overrightarrow{a}+\overrightarrow{b}|=\sqrt{19}\end{array}\\ & \begin{array}{rl}\text{b}.& {|\overrightarrow{a}-\overrightarrow{b}|}^2\\ & =(\overrightarrow{a}-\overrightarrow{b})(\overrightarrow{a}-\overrightarrow{b})\\ & =\overrightarrow{a}\overrightarrow{a}-2\overrightarrow{a}\overrightarrow{b}+\overrightarrow{b}\overrightarrow{b}\\ & ={\left|\overrightarrow{a}\right|}^2\cos 0^{\circ }-2\left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right|\cos {60}^{\circ }+{\left|\overrightarrow{b}\right|}^2\cos 0^{\circ }\\ & =2^2.1-2.2.3.{\displaystyle \frac{1}{2}+3^2.1}\\ & =4-6+9=7\\ & \text{Jadi, nilainya adalah}|\overrightarrow{a}-\overrightarrow{b}|=\sqrt{7}\end{array}\\ & \begin{array}{rl}\text{c}.\text{Untuk menentukan nilai}& \\ \cos \mathrm{\angle }(\overrightarrow{a}+\overrightarrow{b},\overrightarrow{a}-\overrightarrow{b})& ={\displaystyle \frac{(\overrightarrow{a}+\overrightarrow{b}).(\overrightarrow{a}-\overrightarrow{b})}{|\overrightarrow{a}+\overrightarrow{b}|.|\overrightarrow{a}-\overrightarrow{b}|}}\\ & ={\displaystyle \frac{\overrightarrow{a}\overrightarrow{a}-\overrightarrow{a}\overrightarrow{b}+\overrightarrow{b}\overrightarrow{a}-\overrightarrow{b}\overrightarrow{b}}{\sqrt{19}.\sqrt{7}}}\\ & ={\displaystyle \frac{2^2-3^2}{\sqrt{133}}=-\frac{5}{\sqrt{133}}}\\ \mathrm{\angle }(\overrightarrow{a}+\overrightarrow{b},\overrightarrow{a}-\overrightarrow{b})& =\arccos (-\frac{5}{\sqrt{133}})\end{array}\end{array}$

$\text{Berikut contoh untuk bentuk sudutnya}$.

$\begin{array}{ll}14.& \text{Diketahui}\overrightarrow{p}=(x,3,2),\text{dan}\\ & \overrightarrow{q}=(2,-6,3).\text{Tentukan nilai}x\\ & \text{agar kedua vektor}\\ & \text{a}\text{membentuk sudut lancip}\\ & \text{b}\text{membentuk sudut siku-siku}\\ & \text{c}\text{membentuk sudut tumpul}\\ & \text{d}\text{sama panjang}\\ \\ & \mathbf{\text{Jawab}}\\ & \begin{array}{rl}\text{Dik}& \text{etahui bahwa}\\ ▹& \overrightarrow{p}=(x,3,2)=\left(\begin{array}{c}x\\ 3\\ 2\end{array}\right)\text{dan}\\ ▹& \overrightarrow{q}=(2,-6,3)=\left(\begin{array}{c}2\\ -6\\ 3\end{array}\right)\end{array}\\ & \text{Selanjutnya}\\ & \overrightarrow{p}\bullet \overrightarrow{q}=\left(\begin{array}{c}x\\ 3\\ 2\end{array}\right)\left(\begin{array}{c}2\\ -6\\ 3\end{array}\right)\\ & =2x-18+6=2x-12\\ & \text{Selanjutnya}\\ & \begin{array}{rl}\text{a}& \mathbf{\text{Syarat lancip}},\text{yaitu}:\overrightarrow{p}\bullet \overrightarrow{q}>0\\ & 2x-12>0\iff 2x>12\iff x>6\\ \text{b}& \mathbf{\text{Syarat siku-siku}},\text{yaitu}:\overrightarrow{p}\bullet \overrightarrow{q}=0\\ & 2x-12=0\iff 2x=12\iff x=6\\ \text{c}& \mathbf{\text{Syarat tumpul}},\text{yaitu}:\overrightarrow{p}\bullet \overrightarrow{q}<0\\ & 2x-12<0\iff 2x<12\iff x<6\\ \text{d}& \mathbf{\text{Syarat panjang kedua vektor sama}}\\ & \text{yaitu}:\left|\overrightarrow{p}\right|=\left|\overrightarrow{q}\right|,\text{maka}\\ & \begin{array}{rl}& \sqrt{x^2+3^2+2^2}=\sqrt{2^2+(-6{)}^2+3^2}\\ & x^2+9+4=4+36+9\\ & x^2=36\\ & x=\pm \sqrt{36}=\pm 6\\ & \text{Jadi},x=-6\text{atau}x=6\end{array}\end{array}\end{array}$

DAFTAR PUSTAKA

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