Contoh 8 Vektor

Contoh soal sebelumnya di sini Contoh Soal 7

$\begin{array}{l}\\ 34.& \text{Vektor satuan untuk}\overrightarrow{a}=\left(\begin{array}{c}2\\ -1\\ 4\end{array}\right)\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.{\displaystyle \frac{1}{2}\sqrt{3}\left(\begin{array}{c}2\\ -1\\ 4\end{array}\right)}& \\ \text{b}.{\displaystyle \frac{1}{3}\sqrt{3}\left(\begin{array}{c}2\\ -1\\ 4\end{array}\right)}& \\ \text{c}.{\displaystyle \frac{1}{5}\sqrt{5}\left(\begin{array}{c}2\\ -1\\ 4\end{array}\right)}& \\ \text{d}.{\displaystyle \frac{1}{7}\sqrt{7}\left(\begin{array}{c}2\\ -1\\ 4\end{array}\right)}\\ \text{e}.{\displaystyle \frac{1}{21}\sqrt{21}\left(\begin{array}{c}2\\ -1\\ 4\end{array}\right)}\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}\text{Vektor satuan}\overrightarrow{a}& \text{adalah}{\overrightarrow{e}}_{\overrightarrow{a}},\text{yaitu}:\\ {\overrightarrow{e}}_{\overrightarrow{a}}& ={\displaystyle \frac{\overrightarrow{a}}{\left|\overrightarrow{a}\right|}}\\ & ={\displaystyle \frac{\left(\begin{array}{c}2\\ -1\\ 4\end{array}\right)}{\sqrt{2^2+(-1{)}^2+4^2}}}\\ & ={\displaystyle \frac{1}{\sqrt{21}}\left(\begin{array}{c}2\\ -1\\ 4\end{array}\right)}\\ & ={\displaystyle \frac{1}{21}\sqrt{21}\left(\begin{array}{c}2\\ -1\\ 4\end{array}\right)}\end{array}\end{array}$.

$\begin{array}{l}\\ 35.& \text{Posisi suatu titik dalam ruang saat }\\ & \text{waktu}t\text{ditunjukkan oleh vektor}\\ & \left(\begin{array}{c}t\\ t^2\\ -t\end{array}\right).\text{Jika pada saat}t=1\text{titik }\\ & \text{tersebut berada di titik P dan pada}\\ & \text{saat}t=2\text{titik tersebut berada }\\ & \text{di titik Q, maka jarak titik P dari Q}\\ & \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.\sqrt{24}-\sqrt{3}& & \text{d}.\sqrt{11}\\ \text{b}.2-\sqrt{2}& \text{c}.3& \text{e}.\sqrt{43}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{a}}\\ & \begin{array}{rl}\left|\overrightarrow{PQ}\right|& =\sqrt{(x_q-x_p{)}^2+(y_q-y_p{)}^2+(y_q-y_p{)}^2}\\ & =\sqrt{(2-1{)}^2+(2^2-1^2{)}^2+((-2)-(-1){)}^2}\\ & =\sqrt{1^2+3^2+(-1{)}^2}\\ & =\sqrt{1+9+1}\\ & =\sqrt{11}\end{array}\end{array}$

$\begin{array}{l}\\ 36.& \text{Jika diketahui}\left|\overrightarrow{a}\right|=4\sqrt{3},\left|\overrightarrow{b}\right|=5,\\ & \text{ dan}(\overrightarrow{a}+\overrightarrow{b}).(\overrightarrow{a}+\overrightarrow{b})=13,\\ & \text{maka}\mathrm{\angle }(\overrightarrow{a},\overrightarrow{b})=....\\ & \begin{array}{l}\\ \text{a}.{\displaystyle {30}^{\circ }}& & \text{d}.{\displaystyle {135}^{\circ }}\\ \text{b}.{\displaystyle {60}^{\circ }}& \text{c}.{\displaystyle {120}^{\circ }}& \text{e}.{\displaystyle {150}^{\circ }}\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}(\overrightarrow{a}+\overrightarrow{b}).(\overrightarrow{a}+\overrightarrow{b})& =13\\ \overrightarrow{a}.\overrightarrow{a}+\overrightarrow{a}.\overrightarrow{b}+\overrightarrow{b}.\overrightarrow{a}+\overrightarrow{b}.\overrightarrow{b}& =13\\ {\left|\overrightarrow{a}\right|}^2+2\overrightarrow{a}.\overrightarrow{b}+{\left|\overrightarrow{b}\right|}^2& =13,\\ \text{ingat bahwa}\overrightarrow{a}.\overrightarrow{b}=\overrightarrow{b}.\overrightarrow{a}& \\ {\left(4\sqrt{3}\right)}^2+2\left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right|\cos \mathrm{\angle }(\overrightarrow{a},\overrightarrow{b})+5^2& =13\\ 48+2.(4\sqrt{3}).5.\cos \mathrm{\angle }(\overrightarrow{a},\overrightarrow{b})+25& =13\\ 40\sqrt{3}\cos \mathrm{\angle }(\overrightarrow{a},\overrightarrow{b})& =13-25-48\\ \cos \mathrm{\angle }(\overrightarrow{a},\overrightarrow{b})& ={\displaystyle \frac{-60}{40\sqrt{3}}}\\ & =-{\displaystyle \frac{1}{2}\sqrt{3}}\\ & =-\cos 30\\ & =\cos ({180}^{\circ }-{30}^{\circ })\\ \cos \mathrm{\angle }(\overrightarrow{a},\overrightarrow{b})& =\cos {150}^{\circ }\\ \mathrm{\angle }(\overrightarrow{a},\overrightarrow{b})& ={150}^{\circ }\end{array}\end{array}$.

$\begin{array}{l}\\ 37.& \text{Jika diketahui titik}A(2,-1,4),B(4,1,3),\\ & \text{ dan}C(2,0,5),\text{maka}\sin \mathrm{\angle }(\overline{AB},\overline{AC})=....\\ & \begin{array}{l}\\ \text{a}.{\displaystyle \frac{1}{7}\sqrt{5}}& & \text{d}.{\displaystyle \frac{1}{6}\sqrt{3}}\\ \\ \text{b}.{\displaystyle \frac{1}{6}\sqrt{34}}& \text{c}.{\displaystyle \frac{2}{3}\sqrt{2}}& \text{e}.{\displaystyle \frac{1}{6}\sqrt{2}}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& \cos \mathrm{\angle }(\overline{AB},\overline{AC})={\displaystyle \frac{\overline{AB}.\overline{AC}}{\left|\overline{AB}\right|.\left|\overline{AC}\right|}}\\ & ={\displaystyle \frac{(\overrightarrow{b}-\overrightarrow{a}).(\overrightarrow{c}-\overrightarrow{a})}{\sqrt{x_{(\overrightarrow{b}-\overrightarrow{a})}^2+y_{(\overrightarrow{b}-\overrightarrow{a})}^2+z_{(\overrightarrow{b}-\overrightarrow{a})}^2}.\sqrt{x_{(\overrightarrow{c}-\overrightarrow{a})}^2+y_{(\overrightarrow{c}-\overrightarrow{a})}^2+z_{(\overrightarrow{c}-\overrightarrow{a})}^2}}}\\ & ={\displaystyle \frac{\left(\begin{array}{c}4-2\\ 1+1\\ 3-4\end{array}\right).\left(\begin{array}{c}2-2\\ 0+1\\ 5-4\end{array}\right)}{\sqrt{(4-2{)}^2+(1+1{)}^2+(3-4{)}^2}.\sqrt{(2-2{)}^2+(0+1{)}^2+(5-4{)}^2}}}\\ & ={\displaystyle \frac{2.0+2.1+-1.1}{\sqrt{4+4+1}.\sqrt{0+1+1}}}\\ & ={\displaystyle \frac{1}{3\sqrt{2}}}\\ & ={\displaystyle \frac{1}{6}\sqrt{2}}\end{array}\\ & \begin{array}{rl}& \text{Sehingga},\\ & \sin \mathrm{\angle }(\overline{AB},\overline{AC})\\ & =\sqrt{1-{\cos }^2\mathrm{\angle }(\overline{AB},\overline{AC})}\\ & =\sqrt{1-{\left({\displaystyle \frac{1}{6}\sqrt{2}}\right)}^2}\\ & =\sqrt{1-{\displaystyle \frac{2}{36}}}\\ & =\sqrt{{\displaystyle \frac{34}{36}}}\\ & ={\displaystyle \frac{1}{6}\sqrt{34}}\end{array}\end{array}$.

$\begin{array}{l}\\ 38.& \text{Diketahui segitiga ABC. Titik M }\\ & \text{di tengah AC, dan titik N pada BC}\\ & \text{Jika}\overrightarrow{AB}=\overrightarrow{c},\overrightarrow{AC}=\overrightarrow{b},\overrightarrow{BC}=\overrightarrow{a},\\ & \text{maka}\overrightarrow{MN}=....\\ & \begin{array}{l}\\ \text{a}.{\displaystyle \frac{1}{2}(\overrightarrow{b}-\overrightarrow{c})}& \\ \text{b}.{\displaystyle \frac{1}{2}(-\overrightarrow{b}+\overrightarrow{c})}& \\ \text{c}.{\displaystyle \frac{1}{2}(-\overrightarrow{a}+\overrightarrow{c})}\\ \text{d}.{\displaystyle \frac{1}{2}(\overrightarrow{a}-\overrightarrow{b})}\\ \text{e}.{\displaystyle \frac{1}{2}(-\overrightarrow{a}+\overrightarrow{b})}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}\overrightarrow{MN}& =\overrightarrow{MC}+\overrightarrow{CN}\\ & ={\displaystyle \frac{1}{2}\overrightarrow{AC}+{\displaystyle \frac{1}{2}(-\overrightarrow{BC})}}\\ & ={\displaystyle \frac{1}{2}(\overrightarrow{AC}-\overrightarrow{BC})}\\ & ={\displaystyle \frac{1}{2}(\overrightarrow{b}-\overrightarrow{a})\mathbf{\text{atau}}}\\ & ={\displaystyle \frac{1}{2}(-\overrightarrow{a}+\overrightarrow{b})}\end{array}\end{array}$.

$\begin{array}{l}\\ 39.& \text{Jika titik berat segitiga ABC adalah Z}\\ & \text{dengan}\text{A}(1,0,2),\text{B}(5,4,10),\text{C}(0,-1,6),\\ & \text{maka koordinat titik Z tersebut adalah}....\\ & \begin{array}{l}\\ \text{a}.(-2,-1,6)& & \text{d}.(3,2,6)\\ \text{b}.(2,1,6)& \text{c}.(3,-1,6)& \text{e}.(6,4,12)\end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{b}}\\ & \text{Coba perhatikanlah ilustrasi berikut}\end{array}$

$.\begin{array}{r}\\ & \begin{array}{rl}\text{koordinat}& {\text{titik A}}^{\prime }\\ & ={\displaystyle \frac{1}{2}(5+0,4-1,10+6)}\\ & =(\frac{5}{2},\frac{3}{2},8)\end{array}\\ & \begin{array}{rl}\text{Dalam se}& \text{gitiga ABC untuk titik berat Z }\\ \text{berlaku}& \text{ketentuan sebagai berikut}\\ AZ:Z{A}^{\prime }& =2:1\\ \overrightarrow{AZ}:\overrightarrow{Z{A}^{\prime }}& =2:1\\ \overrightarrow{OZ}& ={\displaystyle \frac{\overrightarrow{OA}+2\overrightarrow{O{A}^{\prime }}}{3}}\\ & ={\displaystyle \frac{\left(\begin{array}{c}1\\ 0\\ 2\end{array}\right)+2\left(\begin{array}{c}\frac{5}{2}\\ \frac{3}{2}\\ 8\end{array}\right)}{3}}\\ & =\left(\begin{array}{c}2\\ 1\\ 6\end{array}\right)\\ \text{Jadi},& \text{koordinat titik Z adalah}(2,1,6)\end{array}\end{array}$.

$\begin{array}{l}\\ 40.& \text{Diketahui titik}\text{A}(-4,-1,-2),\text{B}(-6,4,3),\\ & \text{C}(2,3,5).\text{Jika titik M membagi}\overrightarrow{AB}\\ & \text{sehingga}\overrightarrow{AM}:\overrightarrow{MB}=3:2\\ & \text{maka vektor yang diwakili oleh}\overrightarrow{MC}=....\\ & \begin{array}{l}\\ \text{a}.(-4,1,4)& & \text{d}.(6,4,1)\\ \text{b}.(-2,2,1)& \text{c}.(0,5,6)& \text{e}.(4,1,4)\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & \text{Coba perhatikanlah ilustrasi berikut}\end{array}$

$.\begin{array}{|cc|}\hline \text{Diketahui}& \text{Proses Penyelesaian}\\ \begin{array}{rl}\overrightarrow{AM}:\overrightarrow{MB}& =3:2\\ \overrightarrow{m}& ={\displaystyle \frac{2\overrightarrow{a}+3\overrightarrow{b}}{5}}\\ & \\ & \\ & \\ & \\ & \\ & \\ & \\ & \\ & \\ & \\ & \end{array}& \begin{array}{rl}\overrightarrow{MC}& =\overrightarrow{c}-\overrightarrow{m}\\ & =\overrightarrow{c}-{\displaystyle \frac{2\overrightarrow{a}+3\overrightarrow{b}}{5}}\\ & ={\displaystyle \frac{5\overrightarrow{c}-2\overrightarrow{a}-3\overrightarrow{b}}{5}}\\ & ={\displaystyle \frac{5\left(\begin{array}{c}2\\ 3\\ 5\end{array}\right)-2\left(\begin{array}{c}4\\ -1\\ -2\end{array}\right)-3\left(\begin{array}{c}-6\\ 4\\ 3\end{array}\right)}{5}}\\ & ={\displaystyle \frac{\left(\begin{array}{c}20\\ 5\\ 20\end{array}\right)}{5}}\\ & =\left(\begin{array}{c}4\\ 1\\ 4\end{array}\right)\end{array}\\ \hline\end{array}$

Proyeksi Ortoganal Suatu Vektor di Dimensi Tiga

 Materinya sama dengan proyeksi ortogonal pada dimensi dua klik di sini

Uraian berikut sebagai pengingat saja

$\begin{array}{rl}▹& \mathbf{\text{Proyeksi skalar vektor}}\\ & \left|\overrightarrow{c}\right|={\displaystyle \frac{\overrightarrow{a}\bullet \overrightarrow{b}}{\left|\overrightarrow{b}\right|}}\\ ▹& \mathbf{\text{Vektor proyeksi ortogonal}}\\ & \overrightarrow{c}={\displaystyle \frac{\overrightarrow{a}\bullet \overrightarrow{b}}{{\left|\overrightarrow{b}\right|}^2}.\overrightarrow{b}}\end{array}$

Sebagai penjelasannya adalah sebagai berikut:

Penjelasan pertama berkaitan dengan proyeksi skalar vektor di dimensi tiga, yaitu:

Diberikan sebuah ilustrasi berikut,

Perhatikan ilustrasi gambar di atas!

$\begin{array}{|cc|}\hline \mathrm{△}\text{OAC}& \mathrm{\angle }(\overrightarrow{a},\overrightarrow{b})\\ \begin{array}{rl}\cos \theta & ={\displaystyle \frac{\left|\overrightarrow{c}\right|}{\left|\overrightarrow{a}\right|}}\\ \iff \left|\overrightarrow{c}\right|& =\left|\overrightarrow{a}\right|\cos \theta ........(1)\end{array}& \begin{array}{rl}\cos \theta & ={\displaystyle \frac{\overrightarrow{a}\overrightarrow{b}}{\left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right|}........(2)}\end{array}\\ \hline\end{array}$. 

$\begin{array}{rl}\text{Dari}(1)& \text{dan}(2)\text{diperoleh}\\ \left|\overrightarrow{c}\right|& =\left|\overrightarrow{a}\right|\cos \theta \\ & =\left|\overrightarrow{a}\right|\left({\displaystyle \frac{\overrightarrow{a}\overrightarrow{b}}{\left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right|}}\right)\\ & =\left|{\displaystyle \frac{\overrightarrow{a}\overrightarrow{b}}{\left|\overrightarrow{b}\right|}}\right|\end{array}$

Dan penjelasan kedua berkaitan dengan vektor proyeksi ortogonalnya, yaitu:

$\begin{array}{rl}& \text{Perhatikan pula misal}\hat{c}\\ & \text{adalah vektor satuan dari}\overrightarrow{c}\text{dan}\overrightarrow{b},\\ & \text{maka}\\ & \begin{array}{rl}\overrightarrow{c}& =\left|\overrightarrow{c}\right|\hat{c}\end{array},\text{dan}\\ & \begin{array}{rl}\overrightarrow{b}& =\left|\overrightarrow{b}\right|\hat{b}=\left|\overrightarrow{b}\right|\hat{c}\end{array}\end{array}$.

$\begin{array}{rl}\text{Sehingga}& \mathbf{\text{proyeksi ortogonal vektor}}\overrightarrow{a}\text{pada}\overrightarrow{b}\text{adalah}:\\ \overrightarrow{c}& =\left|\overrightarrow{c}\right|\hat{b}\\ & =\left({\displaystyle \frac{\overrightarrow{a}\overrightarrow{b}}{\left|\overrightarrow{b}\right|}}\right)\left({\displaystyle \frac{\overrightarrow{b}}{\left|\overrightarrow{b}\right|}}\right)\\ & =\left({\displaystyle \frac{\overrightarrow{a}\overrightarrow{b}}{{\left|\overrightarrow{b}\right|}^2}}\right)\overrightarrow{b}\end{array}$

$\text{CONTOH SOAL}$.

$\begin{array}{l}\\ 1.& \text{Diketahui}\overrightarrow{a}=\left(\begin{array}{c}2\\ -3\\ 1\end{array}\right)\text{dan}\overrightarrow{b}=\left(\begin{array}{c}-4\\ 2\\ 2\end{array}\right).\\ & \text{Tentukanlah}\\ & \text{a}.\text{proyeksi skalar}\overrightarrow{a}\text{pada}\overrightarrow{b}\\ & \text{b}.\text{vektor proyeksi}\overrightarrow{a}\text{pada}\overrightarrow{b}\\ & \text{c}.\text{proyeksi skalar}\overrightarrow{b}\text{pada}\overrightarrow{a}\\ & \text{d}.\text{vektor proyeksi}\overrightarrow{b}\text{pada}\overrightarrow{a}\\ \\ & \mathbf{\text{Jawab}}:\\ & \text{Misalkan proyeksi skalar}\overrightarrow{a}\text{pada}\overrightarrow{b}\\ & \text{adalah}\left|\overrightarrow{c}\right|,\text{dan}\\ & \text{misalkan juga proyeksi skalar}\overrightarrow{b}\text{pada}\overrightarrow{a}\\ & \text{adalah}\left|\overrightarrow{d}\right|,\text{maka}\\ & \begin{array}{rl}\text{a}.\left|\overrightarrow{c}\right|& ={\displaystyle \frac{\overrightarrow{a}\bullet \overrightarrow{b}}{\left|\overrightarrow{b}\right|}}\\ & ={\displaystyle \frac{\left(\begin{array}{c}2\\ -3\\ 1\end{array}\right)\left(\begin{array}{c}-4\\ 2\\ 2\end{array}\right)}{\sqrt{(-4{)}^2+2^2+2^2}}}\\ & ={\displaystyle \frac{-8-6+2}{\sqrt{24}}=-\frac{12}{24}\sqrt{24}=-\sqrt{6}}\\ & \text{Karena hasilnya berupa panjang, maka}\\ & \text{diharga mutlak/positif}\\ & \left|\overrightarrow{c}\right|=|-\sqrt{6}|=\sqrt{6}\end{array}\\ & \begin{array}{rl}\text{b}.\overrightarrow{c}& ={\displaystyle \frac{\overrightarrow{a}\bullet \overrightarrow{b}}{{\left|\overrightarrow{b}\right|}^2}\times \overrightarrow{b}}\\ & ={\displaystyle \frac{-12}{(\sqrt{24}{)}^2}\times \left(\begin{array}{c}-4\\ 2\\ 2\end{array}\right)}\\ & =-{\displaystyle \frac{1}{2}\left(\begin{array}{c}-4\\ 2\\ 2\end{array}\right)=\left(\begin{array}{c}2\\ -1\\ -1\end{array}\right)}\end{array}\\ & \begin{array}{rl}\text{c}.\left|\overrightarrow{d}\right|& ={\displaystyle \frac{\overrightarrow{b}\bullet \overrightarrow{a}}{\left|\overrightarrow{a}\right|}}\\ & ={\displaystyle \frac{\left(\begin{array}{c}-4\\ 2\\ 2\end{array}\right)\left(\begin{array}{c}2\\ -3\\ 1\end{array}\right)}{\sqrt{2^2+(-3{)}^2+1^2}}}\\ & ={\displaystyle \frac{-8-6+2}{\sqrt{14}}=-\frac{12}{14}\sqrt{14}=-\frac{6}{7}\sqrt{14}}\\ & \text{Karena hasilnya berupa panjang, maka}\\ & \text{diharga mutlak/positif}\\ & \left|\overrightarrow{d}\right|=|-{\displaystyle \frac{6}{7}\sqrt{14}}|={\displaystyle \frac{6}{7}\sqrt{14}}\end{array}\\ & \begin{array}{rl}\text{d}.\overrightarrow{d}& ={\displaystyle \frac{\overrightarrow{b}\bullet \overrightarrow{a}}{{\left|\overrightarrow{a}\right|}^2}\times \overrightarrow{a}}\\ & ={\displaystyle \frac{-12}{(\sqrt{14}{)}^2}\times \left(\begin{array}{c}2\\ -3\\ 1\end{array}\right)}\\ & =-{\displaystyle \frac{12}{14}\left(\begin{array}{c}2\\ -3\\ 1\end{array}\right)=\left(\begin{array}{c}-\frac{12}{7}\\ \frac{18}{7}\\ -\frac{6}{7}\end{array}\right)}\end{array}\end{array}$

$\begin{array}{l}\\ 2.& \text{Diketahui}\overrightarrow{a}=\left(\begin{array}{c}-3\\ -2\\ m\end{array}\right)\text{dan}\overrightarrow{b}=\left(\begin{array}{c}2\\ -1\\ -2\end{array}\right).\\ & \text{Jika proyeksi skalar}\overrightarrow{a}\text{pada}\overrightarrow{b}\text{adalah}\\ & \text{bernilai}-{\displaystyle \frac{2}{3},\text{maka tentukan nilai}m}\\ \\ & \mathbf{\text{Jawab}}:\\ & \text{Misalkan proyeksi skalar}\overrightarrow{a}\text{pada}\overrightarrow{b}\text{adalah}\left|\overrightarrow{f}\right|,\\ & \text{maka}\\ & \begin{array}{rl}\left|\overrightarrow{f}\right|& ={\displaystyle \frac{\overrightarrow{a}\bullet \overrightarrow{b}}{\left|\overrightarrow{b}\right|}}\\ \iff -{\displaystyle \frac{2}{3}}& ={\displaystyle \frac{\left(\begin{array}{c}-3\\ -2\\ m\end{array}\right)\left(\begin{array}{c}2\\ -1\\ -2\end{array}\right)}{\sqrt{2^2+(-1{)}^2+(-2{)}^2}}}\\ \iff -{\displaystyle \frac{2}{3}}& ={\displaystyle \frac{-6+2-2m}{\sqrt{9}}=\frac{-4-2m}{3}}\\ \iff -2& =-4-2m\\ \iff 1& =2+m\\ \iff -m& =2-1=1\\ \iff m& =-1\end{array}\end{array}$.

$\begin{array}{l}\\ 4.& \text{Diketahui vektor}\overrightarrow{a}=3\overline{i}-2\overline{j}+2\overline{k}\text{dan}\\ & \overrightarrow{b}=2\overline{i}-2\overline{j}+\overline{k}.\text{Tentukanlah panjang }\\ & \text{vektor proyeksi ortogonal}\\ & \text{a}.\overrightarrow{a}\text{pada}\overrightarrow{b}\\ & \text{b}.\overrightarrow{a}\text{pada}(\overrightarrow{a}+\overrightarrow{b})\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}\text{a}.\text{Misal}& \text{kan}\overrightarrow{g}\text{adalah vektor proyeksi }\\ \text{yang}& \text{dimaksud, maka panjanynya}\\ (\text{lang}& \text{sung diharga mutlak})\\ \left|\overrightarrow{g}\right|& =\left|{\displaystyle \frac{\overrightarrow{a}\overrightarrow{b}}{\left|\overrightarrow{b}\right|}}\right|\\ & =\left|{\displaystyle \frac{\left(\begin{array}{c}3\\ -2\\ 2\end{array}\right).\left(\begin{array}{c}2\\ -2\\ 1\end{array}\right)}{\sqrt{2^2+(-2{)}^2+1^2}}}\right|\\ & =\left|\frac{3.2+(-2).(-2)+2.1}{\sqrt{4+4+1}}\right|\\ & =\left|{\displaystyle \frac{12}{3}}\right|=4\end{array}\\ & \begin{array}{rl}\text{b}.\text{Misal}& \text{kan}\overrightarrow{h}\text{adalah vektor proyeksi }\\ \text{yang}& \text{dimaksud, maka panjanynya}\\ (\text{lang}& \text{sung diharga mutlak})\\ \left|\overrightarrow{h}\right|& =\left|{\displaystyle \frac{\overrightarrow{a}(\overrightarrow{a}+\overrightarrow{b})}{|\overrightarrow{a}+\overrightarrow{b}|}}\right|\\ & =\left|{\displaystyle \frac{\left(\begin{array}{c}3\\ -2\\ 2\end{array}\right).\left(\begin{array}{c}3+2\\ -2+(-2)\\ 2+1\end{array}\right)}{\sqrt{(3+2{)}^2+(-2+(-2){)}^2+(2+1{)}^2}}}\right|\\ & =\left|\frac{3.5+(-2).(-4)+2.3}{\sqrt{25+16+9}}\right|\\ & =\left|\frac{29}{\sqrt{50}}\right|\\ & ={\displaystyle \frac{29}{5\sqrt{2}}=\frac{29}{10}\sqrt{5}}\end{array}\end{array}$

DAFTAR PUSTAKA

1. Johanes, Kastolan, Sulasim. 2006. Kompetensi Matematika 3A SMA Kelas XII Semester Pertama Program IPA. Jakarta: YUDHISTIRA.

 

Lanjutan 2 Materi Operasi Vektor Berdimensi Tiga

 $\text{G. Perbandingan Vektor }$.

Rumus perbandingan vektor yang berlaku pada dimensi dua juga berlaku untuk perbandingan vektor di dimensi tiga. Misalkan suatu  $\overline{AB}$  dan  titik  T pada  $\overline{AB}$ dengan  $\overline{AT}:\overline{TB}=m:n$ .

Karena titik T pada ruas garis AB, maka titik T membagi ruas AB dengan  $\overrightarrow{a}$  dan  $\overrightarrow{b}$ sebagai vektor posisi dari masing-masing titik A dan B dan vektor posisi titik T dapat ditentukan, yaitu:

$\overrightarrow{t}={\displaystyle \frac{n\overrightarrow{a}+m\overrightarrow{b}}{m+n}}$

Sebagai ilustrasinya adalah gambar berikut

Untuk bukti silahkan merujuk di sini 

(dengan menyesuaikan posisi titiknya)

Jika titik T pada perpanjangan garis  $\overline{AB}$ , maka vektor posisi titik T-nya adalah:

$\overrightarrow{t}={\displaystyle \frac{-n\overrightarrow{a}+m\overrightarrow{b}}{m-n}}$.

$\text{CONTOH SOAL}$.

$\begin{array}{l}\\ 1.& \text{Jika titik}A(12,12,0)\text{dan}B(6,6,12)\\ & \text{serta}P\text{membagi garis dengan}\\ & \overline{AP}:\overline{PB}=1:2.\text{Tentukanlah koordinat}\\ & \text{titik}P\text{jika},\\ & P\text{membagi di dalam}\\ & P\text{membagi di luar}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}\text{a}.& P\text{membagi di dalam}\\ & \overrightarrow{p}={\displaystyle \frac{n\overrightarrow{a}+m\overrightarrow{b}}{m+n}}\\ & ={\displaystyle \frac{2\overrightarrow{a}+1\overrightarrow{p}}{1+2}=\frac{2\left(\begin{array}{c}12\\ 12\\ 0\end{array}\right)+\left(\begin{array}{c}6\\ 6\\ 12\end{array}\right)}{3}}\\ & ={\displaystyle \frac{\left(\begin{array}{c}30\\ 30\\ 12\end{array}\right)}{3}=\left(\begin{array}{c}10\\ 10\\ 4\end{array}\right)}\\ \text{b}.& P\text{membagi di luar}\\ & \overrightarrow{p}={\displaystyle \frac{-n\overrightarrow{a}+m\overrightarrow{b}}{m-n}}\\ & ={\displaystyle \frac{-2\left(\begin{array}{c}12\\ 12\\ 0\end{array}\right)+\left(\begin{array}{c}6\\ 6\\ 12\end{array}\right)}{1-2}}\\ & =-\left(\begin{array}{c}-18\\ -18\\ 12\end{array}\right)=\left(\begin{array}{c}18\\ 18\\ -12\end{array}\right)\end{array}\end{array}$.

$\begin{array}{l}\\ 2.& \text{Jika titik}A(3,2,-1)\text{dan}B(1,-2,1)\\ & \text{dan}C(7,m-1,-5).\text{Tentukan nilai}m\\ & \text{agar ketiga titik itu segaris}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& \text{Diketahui}\\ & \overrightarrow{a}=\left(\begin{array}{c}3\\ 2\\ -1\end{array}\right),\overrightarrow{b}=\left(\begin{array}{c}1\\ -2\\ 1\end{array}\right),\left(\begin{array}{c}7\\ m-1\\ -5\end{array}\right)\\ & \text{Agar ketiga titik segaris, maka}\overline{AB}=k\overline{BC}\\ & \overrightarrow{b}-\overrightarrow{a}=k(\overrightarrow{c}-\overrightarrow{b})\\ & \left(\begin{array}{c}1\\ -2\\ 1\end{array}\right)-\left(\begin{array}{c}3\\ 2\\ -1\end{array}\right)=k(\begin{array}{c}7\\ m-1\\ -5\end{array}-\left(\begin{array}{c}1\\ -2\\ 1\end{array}\right))\\ & \iff \left(\begin{array}{c}-2\\ -4\\ 2\end{array}\right)=k\left(\begin{array}{c}6\\ m+1\\ -6\end{array}\right)\\ & \iff \{\begin{array}{ll}-2& =6k\\ -4& =k(m+1)\\ 2& =-6k\end{array}\iff -2=6k\\ & \iff k=-{\displaystyle \frac{1}{3}}\\ & \text{maka}\\ & \iff -4=k(m+1)\iff -4=-{\displaystyle \frac{1}{3}(m+1)}\\ & \iff 12=m+1\\ & \iff m=11\end{array}\end{array}$

Lanjutan 1 Materi Operasi Vektor Berdimensi Tiga (Hasil Kali Vektor)

 $\text{F. 4. Perkalian Silang Vektor (Pengayaan)}$.

Pada ruang dimensi tiga khususnya pada vektor akan berlaku perkalian silang (cross vektor) adalah perkalian antara dua vektor yang menghasilkan vektor tunggal. Misalkan diketahui  $\overrightarrow{u}$  dan  $\overrightarrow{v}$  adalah dua vektor sembarang dan keduanya membentuk sudut  $\theta$, maka hasil kali kedua vektor tersebut adalah sebuah vektor baru dengan dinotasiakan sebagai  $\overrightarrow{u}\times \overrightarrow{v}$. Tentunya sebagai syarat kedua vektor tersebut masing-masing tidak berupa vektor nol.

Jika  $\overrightarrow{u}\times \overrightarrow{v}=\overrightarrow{c}$ , maka

$\begin{array}{rl}\overrightarrow{u}& \times \overrightarrow{v}=\overrightarrow{c}=\left|\begin{array}{ccc}\overrightarrow{i}& \overrightarrow{j}& \overrightarrow{k}\\ x_1& y_1& z_1\\ x_2& y_2& z_2\end{array}\right|\\ & =(y_1{z}_2-z_1{y}_2)\overrightarrow{i}-(x_1{z}_2-z_1{x}_2)\overrightarrow{j}+(x_1{y}_2-y_1{x}_2)\overrightarrow{k}\end{array}$

Lalu kalau sudah demikian berapa besarnya? dan ke mana arahnya?

Besarnya adalah  $|\overrightarrow{u}\times \overrightarrow{v}|=\left|\overrightarrow{u}\right|\left|\overrightarrow{v}\right|\sin \theta$  dan arahnya tegak lurus terhadap  $\overrightarrow{u}$  dan  $\overrightarrow{v}$.

Sebagai ilustrasi perhatikanlah gambar berikut  untuk dua buah vektor sebagai misal  $\overrightarrow{a}$  dan  $\overrightarrow{b}$.

Jika putarannya dibalik, maka akan mendapatkan hasil sebagai mana ilustrasi berikut

Sehingga perlu diingat bahwa :  $\overrightarrow{a}\times \overrightarrow{b}=-\overrightarrow{b}\times \overrightarrow{a}$.

Pada hasil kali silang dua vektor berlaku

1. tidak bersifat komutatif , karena  $\overrightarrow{a}\times \overrightarrow{b}=-\overrightarrow{b}\times \overrightarrow{a}$.
2. distributif terhadap penjumlahan : $\overrightarrow{a}\times (\overrightarrow{b}+\overrightarrow{c})=\overrightarrow{a}\times \overrightarrow{b}+\overrightarrow{a}\times \overrightarrow{c}$.
3. pada perkalian dengan skalar : $k(\overrightarrow{a}\times \overrightarrow{b})=\left(k\overrightarrow{a}\right)\times \overrightarrow{b}=\overrightarrow{a}\times \left(k\overrightarrow{b}\right)$.
4. berlaku untuk sembarang vektor : $\overrightarrow{a}\times \overrightarrow{a}=0$.
5. jika kedua vektor sejajar, maka hasil kalinya adalah = 0.
6. Nilai dari perkalian kedua vektor terbut adalah sama dengan hasil luas jajar genjang.
7. Nilai dari poin 6 jika dibagi 2 akan berupa hasil luas sebuah segitiga yang dibentuk oleh kedua vektor tersebut.
8. berlaku identitas Lagrange : ${|\overrightarrow{a}\times \overrightarrow{b}|}^2={\left|\overrightarrow{a}\right|}^2.{\left|\overrightarrow{b}\right|}^2-{(\overrightarrow{a}\bullet \overrightarrow{b})}^2$.

$\text{CONTOH SOAL}$.

$\begin{array}{l}\\ 1.& \text{Diketahui}\overrightarrow{a}=4\overrightarrow{i}+3\overrightarrow{j}\text{dan}\overrightarrow{b}=4\overrightarrow{i}-3\overrightarrow{k}\\ & \text{Tentukanlah hasil}\overrightarrow{a}\times \overrightarrow{b}\text{dan}\overrightarrow{b}\times \overrightarrow{a}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}\overrightarrow{a}& \times \overrightarrow{b}=\left|\begin{array}{ccc}\overrightarrow{i}& \overrightarrow{j}& \overrightarrow{k}\\ x_1& y_1& z_1\\ x_2& y_2& z_2\end{array}\right|\\ & =(y_1{z}_2-z_1{y}_2)\overrightarrow{i}-(x_1{z}_2-z_1{x}_2)\overrightarrow{j}+(x_1{y}_2-y_1{x}_2)\overrightarrow{k}\\ & =\left|\begin{array}{ccc}\overrightarrow{i}& \overrightarrow{j}& \overrightarrow{k}\\ 4& 3& 0\\ 4& 0& -3\end{array}\right|\\ & =(-9-0)\overrightarrow{i}-(-12-0)\overrightarrow{j}+(0-12)\overrightarrow{k}\\ & =-9\overrightarrow{i}+12\overrightarrow{j}-12\overrightarrow{k}\end{array}\\ & \begin{array}{rl}\overrightarrow{b}& \times \overrightarrow{a}=\left|\begin{array}{ccc}\overrightarrow{i}& \overrightarrow{j}& \overrightarrow{k}\\ x_2& y_2& z_2\\ x_1& y_1& z_1\end{array}\right|\\ & =\left|\begin{array}{ccc}\overrightarrow{i}& \overrightarrow{j}& \overrightarrow{k}\\ 4& 0& -3\\ 4& 3& 0\end{array}\right|\\ & =(0-(-9))\overrightarrow{i}-(0-(-12))\overrightarrow{j}+(12-0)\overrightarrow{k}\\ & =9\overrightarrow{i}-12\overrightarrow{j}+12\overrightarrow{k}\end{array}\end{array}$.

$\begin{array}{l}\\ 2.& \text{Diketahui}\overrightarrow{a}=6\overrightarrow{i}+2\overrightarrow{j}+10\overrightarrow{k}\text{dan}\overrightarrow{b}=4\overrightarrow{i}+\overrightarrow{j}+9\overrightarrow{k}\\ & \text{Tentukanlah hasil}\overrightarrow{a}\times \overrightarrow{b}\text{dan}\overrightarrow{b}\times \overrightarrow{a}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}\overrightarrow{a}& \times \overrightarrow{b}=\left|\begin{array}{ccc}\overrightarrow{i}& \overrightarrow{j}& \overrightarrow{k}\\ x_1& y_1& z_1\\ x_2& y_2& z_2\end{array}\right|\\ & =(y_1{z}_2-z_1{y}_2)\overrightarrow{i}-(x_1{z}_2-z_1{x}_2)\overrightarrow{j}+(x_1{y}_2-y_1{x}_2)\overrightarrow{k}\\ & =\left|\begin{array}{ccc}\overrightarrow{i}& \overrightarrow{j}& \overrightarrow{k}\\ 6& 2& 10\\ 4& 1& 9\end{array}\right|\\ & =(18-10)\overrightarrow{i}-(54-40)\overrightarrow{j}+(6-8)\overrightarrow{k}\\ & =8\overrightarrow{i}-14\overrightarrow{j}-2\overrightarrow{k}\end{array}\\ & \begin{array}{rl}\overrightarrow{b}& \times \overrightarrow{a}=\left|\begin{array}{ccc}\overrightarrow{i}& \overrightarrow{j}& \overrightarrow{k}\\ x_2& y_2& z_2\\ x_1& y_1& z_1\end{array}\right|\\ & =\left|\begin{array}{ccc}\overrightarrow{i}& \overrightarrow{j}& \overrightarrow{k}\\ 4& 1& 9\\ 6& 2& 10\end{array}\right|\\ & =(10-18)\overrightarrow{i}-(40-54)\overrightarrow{j}+(8-6)\overrightarrow{k}\\ & =-8\overrightarrow{i}+14\overrightarrow{j}+2\overrightarrow{k}\end{array}\end{array}$.

$\begin{array}{l}\\ 3.& \text{Tentukanlah luas segitiga}ABC\text{jika}\\ & \text{diketahui}A(2,1,-2),B(0,-1,0),\text{dan}\\ & C(-1,2,-1)\\ \\ & \mathbf{\text{Jawab}}:\\ & \text{Misalkan luas segitiga}{\displaystyle \frac{1}{2}|\overrightarrow{p}\times \overrightarrow{q}|,\text{dengan}}\\ & \{\begin{array}{ll}\overrightarrow{p}& =\overline{AB}=\overline{OB}-\overline{OA}=\left(\begin{array}{c}0\\ -1\\ 0\end{array}\right)-\left(\begin{array}{c}2\\ 1\\ -2\end{array}\right)=\left(\begin{array}{c}-2\\ -2\\ 2\end{array}\right)\\ \overrightarrow{q}& =\overline{AC}=\overline{OC}-\overline{OA}=\left(\begin{array}{c}-1\\ 2\\ -1\end{array}\right)-\left(\begin{array}{c}2\\ 1\\ -2\end{array}\right)=\left(\begin{array}{c}-3\\ 1\\ 1\end{array}\right)\end{array}\\ & \begin{array}{rl}\overrightarrow{p}& \times \overrightarrow{q}=\left|\begin{array}{ccc}\overrightarrow{i}& \overrightarrow{j}& \overrightarrow{k}\\ x_1& y_1& z_1\\ x_2& y_2& z_2\end{array}\right|\\ & =(y_1{z}_2-z_1{y}_2)\overrightarrow{i}-(x_1{z}_2-z_1{x}_2)\overrightarrow{j}+(x_1{y}_2-y_1{x}_2)\overrightarrow{k}\\ & =\left|\begin{array}{ccc}\overrightarrow{i}& \overrightarrow{j}& \overrightarrow{k}\\ -2& -2& 2\\ -3& 1& 1\end{array}\right|\\ & =(-2-2)\overrightarrow{i}-(-2-(-6))\overrightarrow{j}+(-2-6)\overrightarrow{k}\\ & =-4\overrightarrow{i}-4\overrightarrow{j}-8\overrightarrow{k}\\ & \text{Sehingga}\\ & |\overrightarrow{p}\times \overrightarrow{q}|=\sqrt{(-4{)}^2+(-4{)}^2+(-8{)}^2}\\ & =\sqrt{16+16+64}=\sqrt{96}=4\sqrt{6}\\ & \text{Maka luas segi tiganya adalah}:\\ & \text{luas}\mathrm{△}ABC={\displaystyle \frac{1}{2}|\overrightarrow{p}\times \overrightarrow{q}|={\displaystyle \frac{1}{2}\left(4\sqrt{6}\right)=2\sqrt{6}}}\end{array}\end{array}$

DAFTAR PUSTAKA

1. Yuana, R.A., Indriyastuti. 2017. Perspektif Matematika untuk Kelas X SMA dan MA Kelompok Peminatan Matematika dan Ilmu Alam. Solo: PT. TIGA SERANGKAI PUSTAKA MANDIRI

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

1. Johanes, Kastolan, Sulasim. 2006. Kompetensi Matematika Program IPA 3A SMA Kelas XII Semester Pertama. Jakarta: YUDHISTIRA.
2. Kanginan, M., Nurdiansyah, H., Akhmad, G. 2016. Matematika untuk Siswa SMA/MA Kelas X Kelompok Peminatan Matematika dan Ilmu-Ilmu Alam. Bandung: YRAMA WIDYA.
3. Noormandiri, Sucipto, E. 2003. Buku Pelajaran Matematika SMU untuk Kelas 3 Program IPA. Jakarta: ERLANGGA.
4. Yuana, R.A., Indriyastuti. 2017. Perspektif Matematika untuk Kelas X SMA dan MA Kelompok Peminatan Matematika dan Ilmu Alam. Solo: PT. TIGA SERANGKAI PUSTAKA MANDIRI.

Lingkaran

$\text{A. Definisi Lingkaran}$.

Secara definisi lingkaran adalah tempat kedudukam titik-titik yang berjarak sama terhadap satu titik tertentu. Selanjutnya titik tertentu disebut sebagai pusat lingkaran sedangkan jarak yang salalu sama terhadapa titik tertentu tersebut disebut sebagai jari-jari atau radius (r).

Sebagai ilustrasi berikut diberikan gambar berkaitan kedudukan titik-titik tersebut

$\text{B. Persamaan Lingkaran Berpusat di O(0,0) }$.

Persamaan sebuah lingkaran dengan dengan jari-jari  $r$  dan berpusat di titik pusat koordinat dapat dilustrasikan sebagai berikut

Misalkan sebuah titik  $\text{P}(x,y)$  terletak pada sebuah lingkaran yang berpusat di O(0,0). Dan titik ${\text{P}}^{\prime }(x,0)$ adalah proyeksi titik  P  pada sumbu-X sehingga  $△{\text{OP}}^{\prime }\text{P}$   berupa sebuah segitiga siku-siku di ${\text{P}}^{\prime }$. Dengan rumus Pythagoras kita mendapatkan

$\begin{array}{rl}& O{P}^2=(O{P}^{\prime }{)}^2+(P{P}^{\prime }{)}^2\\ & \iff r^2=x^2+y^2\\ & \iff r=\sqrt{x^2+y^2}\end{array}$

Untuk lebih memudahkan pemahaman Anda, perhatikanlah ilustrasi berikut

Sehingga dapat disimpulkan persamaan lingkaran yang berpusat di O(0,0) adalah:

$\begin{array}{|ccc|}\hline & & \\ & x^2+y^2=r^2& \\ & & \\ \hline\end{array}$.

$\text{C. Persamaan Lingkaran Berpusat di (a,b)}$.

Perhatikanlah ilustrasi berikut

Pada ilustrasi gambar di atas ditunjukkan sebuah lingkaran berpusat di $N(a,b)$ dengan jari-jari  $r$, misalkan kita ambil sebuah titik $P(x,y)$ pada keliling lingkaran, maka $NP=r$.

$\begin{array}{rl}& \sqrt{(x-a{)}^2+(y-b{)}^2}=r^2\\ & (x-a{)}^2+(y-b{)}^2=r^2\\ & \text{persamaan di atas adalah}\mathbf{\text{Bentuk Umum}}\\ & \text{dari}\mathbf{\text{Persamaan Lingkaran}}\text{yang}\\ & \text{berpusat di}(a,b)\end{array}$

Selanjutnya perhatikanlah rangkuman berikut

$\begin{array}{|lcc|}\hline \text{Lingkaran}& x^2+y^2=r^2& (x-p{)}^2+(y-q{)}^2=r^2\\ \text{Pusat}& (0,0)& (p,q)\\ \text{Jari-jari}& r& r\\ \begin{array}{rl}& \text{Pesamaan garis}\\ & \text{singgung melalui}\\ & \text{titik}(x_1,y_1)\\ & \text{pada lingkaran}\end{array}& x_{1}x+y_{1}y=r^2& \begin{array}{rl}& (x_1-p)(x-p)\\ & +(y_1-q)(y-q)=r^2\end{array}\\ \begin{array}{rl}& \text{Persamaan garis}\\ & \text{singgung dengan}\\ & \text{gradien}m\end{array}& \begin{array}{rl}& y=mx\\ & \pm r\sqrt{m^2+1}\end{array}& \begin{array}{rl}& (y-q)=m(x-a)\\ & \pm r\sqrt{m^2+1}\end{array}\\ \hline\end{array}$.

Kusus untuk yang pusat  $(a,b)$ adalah:

$\begin{array}{|lc|}\hline \text{Lingkaran}& x^2+y^2+Ax+By+C=0\\ \text{Pusat}& (-\frac{1}{2}A,-\frac{1}{2}B)\\ \text{Jari-jari}& r=\sqrt{{\displaystyle \frac{1}{4}(A^2+B^2)-C}}\\ \begin{array}{rl}& \text{Pesamaan garis}\\ & \text{singgung melalui}\\ & \text{titik}(x_1,y_1)\\ & \text{pada lingkaran}\end{array}& \begin{array}{rl}& x_{1}x+y_{1}y\\ & +{\displaystyle \frac{A}{2}(x_1+x)}\\ & +{\displaystyle \frac{B}{2}(y_1+y)+C=0}\end{array}\\ \begin{array}{rl}& \text{Persamaan garis}\\ & \text{singgung dengan}\\ & \text{gradien}m\end{array}& \begin{array}{rl}& y+\frac{1}{2}B=m(x+\frac{1}{2}A)\\ & \pm \sqrt{{\displaystyle \frac{1}{4}(A^2+B^2)-C}}.\sqrt{m^2+1}\end{array}\\ \hline\end{array}$

$\text{D. Kedudukan Titik Terhadap Lingkaran }$.

Kedudukan sebuah titik terhadap sebuah lingkaran yang berpusat di O(0,0) memiliki 3 kemungkinan, yaitu:

• jika titik A(x,y) di dalam lingkaran, maka berlaku  $x^2+y^2<r^2$.
• jika titik A(x,y) pada lingkaran, maka berlaku  $x^2+y^2=r^2$, dan
• jika titik A(x,y) di luar lingkaran, maka berlaku  $x^2+y^2>r^2$.

Demikian juga kedudukan sebuah titik terhadap sebuah lingkaran yang berpusat di $(a,b)$ memiliki 3 kemungkinan, yaitu:

• jika titik A(x,y) di dalam lingkaran, maka berlaku $(x-a{)}^2+(y-b{)}^2<r^2$  atau  $x^2+y^2+Ax+By+C<0$.
• jika titik A(x,y) pada lingkaran, maka berlaku $(x-a{)}^2+(y-b{)}^2=r^2$  atau  $x^2+y^2+Ax+By+C=0$.
• jika titik A(x,y) di luar lingkaran, maka berlaku $(x-a{)}^2+(y-b{)}^2>r^2$  atau  $x^2+y^2+Ax+By+C>0$.

$\text{E. Kedudukan Garis Terhadap Lingkaran }$.

Posisi garis terhadap lingkaran tergantung nilai Diskriminan (D) hasil substitusi persamaan garis ke persamaan lingkaran.

$\{\begin{array}{ll}\bullet & \text{memotong lingkaran di dua titik}(D>0)\\ & \text{ada garis dan titik polar}\\ \bullet & \text{menyinggung lingkaran}(D=0)\\ \bullet & \text{tidak memotong ataupun menyinggung}(D<0)\end{array}$.

Berikut Ilustrasi gambarnya

$\text{F. Jarak Garis ke Pusat Lingkaran}$.

$\begin{array}{|ll|}\hline \text{Jarak titik}M(p,q)\text{terhadap pusat}& \\ \text{lingkaran}N(a,b)& \left|MN\right|=r\\ r=\left|{\displaystyle \frac{Ap+Bq+C}{\sqrt{A^2+B^2}}}\right|& \\ \hline\end{array}$.

$\text{CONTOH SOAL}$.

$\begin{array}{l}\\ 1.& \text{Sebuah lingkaran yang berpusat pada }\\ & \text{pangkal koordinat}\\ & \text{a}.\text{Tentukanlah persamaan lingkaran }\\ & \text{yang berjari-jari 5}\\ & \text{b}.\text{Gambarlah lingkaran (pada soal a.) }\\ & \text{pada kertas grafiks}\\ & \text{c}.\text{Lukislah titik-titik dari},\\ & A(2,3),B(4,3),\text{dan}C(3,6).\\ & \text{d}.\text{Nyatakan kedudukan titik-titik}\\ & A,B,\text{dan}C\text{terhadap lingkaran. }\\ & \text{Di dalam, pada, atau}\\ & \text{beradakah di luar lingkaran}\\ & \mathbf{\text{Jawab}}:\\ & \text{Perhatikanlah ilustrasi berikut}\end{array}$.

$\begin{array}{rl}\text{a}.& \text{Diketahui}r=5\\ & \begin{array}{rl}& x^2+y^2=5^2\\ & \updownarrow \\ & x^2+y^2=25\\ & \text{atau}\\ & L\equiv \{(x,y)|x^2+y^2=25\}\end{array}\\ \text{b}.& \text{Lihat gambar di atas}\\ \text{c}.& \text{Lihat juga gambar di atas}\\ \text{d}.& \text{Dari gambar jelas bahwa}:\\ & \begin{array}{c}\bullet \text{Titik}A(2,3)\text{berada di dalam lingkaran}\\ \bullet \text{Titik}A(4,3)\text{berada pada lingkaran}\\ \bullet \text{Titik}A(3,6)\text{berada di luar lingkaran}\end{array}\end{array}$

$\begin{array}{rl}\text{a}.& \text{Diketahui}r=5\\ & \begin{array}{rl}& x^2+y^2=5^2\\ & \updownarrow \\ & x^2+y^2=25\\ & \text{atau}\\ & L\equiv \{(x,y)|x^2+y^2=25\}\end{array}\\ \text{b}.& \text{Lihat gambar di atas}\\ \text{c}.& \text{Lihat juga gambar di atas}\\ \text{d}.& \text{Dari gambar jelas bahwa}:\\ & \begin{array}{c}\bullet \text{Titik}A(2,3)\text{berada di dalam lingkaran}\\ \text{atau}:(2{)}^2+(3{)}^2=4+9=13<25\\ \bullet \text{Titik}A(4,3)\text{berada pada lingkaran}\\ \text{atau}:(4{)}^2+(3{)}^2=16+9=25=25\\ \bullet \text{Titik}A(3,6)\text{berada di luar lingkaran}\\ \text{atau}:(3{)}^2+(6{)}^2=9+36=45>25\end{array}\end{array}$.

$\begin{array}{l}\\ 2.& \text{Tentukanlah persamaan lingkaran}\\ & \text{yang berpusat di pangkal koordinat}\\ & \text{dan melalui titik}P(5,-3)\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}\text{Diketahui}& \text{pusat lingkaran di pangkal }\\ \text{koordinat}& O(0,0)\text{serta lingkaran}\\ \text{yang mela}& \text{lui titik}P(5,-3),\text{maka}\\ r& =\sqrt{(x_p-0{)}^2+(y_p-0{)}^2}\\ & =\sqrt{5^2+(-3{)}^2}\\ & =\sqrt{25+9}\\ & =\sqrt{34}\\ \text{Sehingga }& ,\text{persamaan lingkarannya adalah}\\ L& \equiv x^2+y^2=r^2\iff x^2+y^2=34\end{array}\end{array}$.

$\begin{array}{l}\\ 3.& \text{Tentukanlah persamaan lingkaran yang }\\ & \text{berpusat di pangkal koordinat dan}\\ & \text{menyinggung}k\equiv 2x+y-5=0\\ \\ & \mathbf{\text{Jawab}}:\\ & \text{Perhatikan ilustrasi berikut}\end{array}$.

menjadi

$\begin{array}{rl}& \text{Diketahui}\text{bahwa titik}O\text{ke garis}k\text{adalah}\\ & r=OA={\displaystyle \left|\frac{a{x}_1+b{y}_1+c}{\sqrt{a^2+b^2}}\right|}\\ & ={\displaystyle \left|\frac{2(0)+(0)-5}{\sqrt{2^2+1^2}}\right|}\\ & ={\displaystyle \left|\frac{-5}{\sqrt{5}}\right|}\\ & =|-\sqrt{5}|\\ & =-(-\sqrt{5})=\sqrt{5}\\ & \text{(ingat, nilai mutlak bilangan negatif adalah bilngan positif)}\\ & \text{Sehingga persamaan lingkarannya adalah}:\\ & L\equiv x^2+y^2=r^2\iff x^2+y^2=5\end{array}$.

$\begin{array}{l}\\ 4.& \text{Tentukanlah pusat dan jari-jari lingkaran berikut?}\\ & \text{a}.L\equiv (x+1{)}^2+(y+2{)}^2=9\\ & \text{b}.L\equiv (x+1{)}^2+(y-2{)}^2=9\\ & \text{c}.L\equiv (x-1{)}^2+(y+2{)}^2=9\\ & \text{d}.L\equiv (x-1{)}^2+(y-2{)}^2=9\\ & \text{e}.L\equiv (x+3{)}^2+(y-3{)}^2=9\\ & \text{f}.L\equiv (x-1{)}^2+(y-2{)}^2=25\\ & \text{g}.L\equiv (x-1{)}^2+y^2=27\\ & \text{h}.L\equiv x^2+(y-1{)}^2=27\\ \\ & \text{Jawab}:\\ & L\equiv (x+1{)}^2+(y+2{)}^2=9,\text{pusat di}(-1,-2)\\ & \text{dan jari-jarinya adalah}\sqrt{9}=3\\ & \text{Soal yang belum dibahas silahkan }\\ & \text{diselesaikan sendiri sebagai latihan}\end{array}$.

$\begin{array}{l}\\ 5.& \text{Tentukanlah persamaan lingkaran yang }\\ & \text{berpusat di}A(2,-1)\text{dan menginggung}\\ & \text{garis}4y+3x-12=0\text{di titik}P\\ \\ & \mathbf{\text{Jawab}}:\\ & \text{Perhatikan ilustrasi berikut}\end{array}$.

$\begin{array}{rl}& \text{Sehingga}\\ & r=AP=\left|\frac{3(2)+4(1)-12}{\sqrt{3^2+4^2}}\right|\\ & =\left|\frac{-10}{5}\right|=|-2|=2\\ & \text{Sehingga persamaan lingkarannya adalah}\\ & L\equiv (x-2{)}^2+(y+1{)}^2=4\end{array}$.

$\begin{array}{l}\\ 6.& \text{Tentukanlah pusat dan jari-jari dari }\\ & \text{persamaan lingkaran}\\ & L\equiv 2x^2+2y^2-2x+6y-3=0\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& \text{Persamaan lingkaran}\\ & L\equiv 2x^2+2y^2-2x+6y-3=0\\ & \iff x^2+y^2-x+3y-{\displaystyle \frac{3}{2}=0\{\begin{array}{ll}A& =-1\\ B& =3\\ C& =-{\displaystyle \frac{3}{2}}\end{array}}\\ & \text{maka}\{\begin{array}{ll}\text{Pusat}& =(-{\displaystyle \frac{-1}{2},-\frac{3}{2}})=\left({\displaystyle \frac{1}{2},-\frac{3}{2}}\right)\\ \text{Jari-jari}& =r=\sqrt{{\displaystyle \frac{(-1{)}^2}{4}+\frac{3^2}{4}-(-\frac{3}{2})}}\\ & =\sqrt{{\displaystyle \frac{1}{4}+\frac{9}{4}+\frac{6}{4}}}=\sqrt{4}=2\end{array}\\ & \text{Jadi, lingkaran}2x^2+2y^2-2x+6y-3=0\\ & \text{berpusat di}\left({\displaystyle \frac{1}{2},-\frac{3}{2}}\right)\text{dan berjari-jari}2\end{array}\end{array}$.

$\begin{array}{l}\\ 7.& \text{Diketahui persamaan lingkaran}\\ & L\equiv 2x^2+2y^2-4x+3py-30=0\\ & \text{dan melalui titik}(-2,1).\text{Tentukanlah }\\ & \text{persamaan lingkaran baru yang}\\ & \text{kosentris(sepusat) dan panjang jari-jarinya}\\ & \text{dua kali panjang jari-jari lingkaran semula?}\\ \\ & \mathbf{\text{Jawab}}\\ & \begin{array}{rl}& \text{Diketahui persamaan lingkaran}\\ & 2x^2+2y^2-4x+3py-30=0,\text{melalui}\\ & (-2,1),\text{maka}\\ & \begin{array}{rl}& \text{kita tentukan harga}p\text{dulu, yaitu}:\\ & 2(-2{)}^2+2(1{)}^2-4(-2)+3p(1)-30=0\\ & \iff 8+2+8+3p-30=0\\ & \iff 3p=12\\ & \iff p=4\end{array}\\ & \text{Akibatnya persamaan lingkaran menjadi}\\ & \begin{array}{rl}& 2x^2+2y^2-4x+12y-30=0\\ & \iff x^2+y^2-2x+6y-15=0\end{array}\\ & \begin{array}{rl}& \{\begin{array}{l}\text{Pusat}:\\ (-{\displaystyle \frac{1}{2}A,-\frac{1}{2}B})\\ =(-{\displaystyle \frac{1}{2}.(-2),-\frac{1}{2},6})\\ =(1,-3)\\ \\ \text{Jari-jari }:\\ \begin{array}{rl}r& =\sqrt{{(-\frac{1}{2}A)}^2+(-\frac{1}{2}B)-C}\\ & =\sqrt{1^2+(-3{)}^2-(-15)}\\ & =\sqrt{1+9+15}=5\end{array}\end{array}\\ & \end{array}\\ & \text{Selanjutnya}\\ & \begin{array}{rl}& \text{Persamaan}\text{lingkaran baru }\\ & \text{dengan pusat}(1,-3)\text{dan jari-jari}\\ & r_{\text{baru}}=2r=2.5=10\\ & (x-1{)}^2+(y+3{)}^2=(10{)}^2\\ & \iff x^2-2x+1+y^2+6x+9=100\\ & \iff x^2+y^2-2x+6y-90=0\end{array}\end{array}\end{array}$

Berikut ilustrasi gambarnya

$\begin{array}{l}\\ 8.& \text{Tentukanlah nilai}p\text{supaya lingkaran}\\ & x^2+y^2-px-10y+4=0\\ & \text{a}.\text{menyinggung sumbu x}\\ & \text{b}.\text{memotong sumbu x di dua titik}\\ & \text{c}.\text{tidak memotong dan tidak menyinggung }\\ & \text{sumbu x}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{r}\\ & \text{Persamaan lingkaran}:\\ & x^2+y^2-px-10y+4=0\\ & \text{saat menyinggung}\text{sumbu x},\text{maka}y=0\\ & \text{adalah gar}\text{is yang sejajar sumbu x, maka}\\ & y=0\Rightarrow x^2+y^2-px-10y+4=0\\ & \iff x^2+0^2-px-0+4=0\\ & \iff x^2-px+4\end{array}\\ & \text{Selanjutnya}\\ & \begin{array}{|ccc|}\hline \text{Menyinggung}& \text{memotong}& \text{Tidak keduanya}\\ \begin{array}{rl}D& =b^2-4ac=0\\ & \iff p^2-4.1.4=0\\ & \iff p^2=16\\ & \iff p=\pm 4\\ & \end{array}& \begin{array}{rl}D& >0\\ & \iff b^2-4ac>0\\ & \iff p^2-16>0\\ & \iff (p+4)(p-4)>0\\ & \therefore p<-4\text{atau}p>4\end{array}& \begin{array}{rl}D& <0\\ & \iff b^2-4ac<0\\ & \iff p^2-16<0\\ & \iff (p+4)(p-4)<0\\ & \therefore -4<p<4\end{array}\\ \hline\end{array}\end{array}$.

$\begin{array}{l}\\ 9.& \text{Tentukanlah nilai}a\text{supaya lingkaran}\\ & x^2+y^2=1\text{dan garis}y=ax+2\\ & \text{a}.\text{bersinggungan}\\ & \text{b}.\text{berpotongan}\\ & \text{c}.\text{tidak berpotongan maupun bersinggungan}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& \text{Di sini yang kita bahas adalah yang poin b, }\\ & \text{yaitu untuk}y=ax+2,\text{maka}\\ & x^2+y^2=1\\ & x^2+(ax+2{)}^2=1\\ & x^2+a^2{x}^2+4ax+4=1\\ & (1+a^2)x^2+4ax+3=0\\ & \text{syarat berpotongan}D=b^2-4ac\ge 0\\ & (\text{artinya bersinggungan sekaligus berpotongan di 2 titik})\\ & (4a{)}^2-4(1+a^2)(3)\ge 0\\ & 16a^2-12a^2-12\ge 0\\ & 4a^2-12\ge 0\\ & a^2-3\ge 0\\ & (a+\sqrt{3})(a-\sqrt{3})\ge 0\\ & \therefore a\le -\sqrt{3}\text{atau}a\ge \sqrt{3}\end{array}\end{array}$.

$\begin{array}{l}\\ 10.& \text{Tentukanlah persamaan garis singgung lingkaran}\\ & x^2+y^2=12\text{dan melalui titik}P(0,4)\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& \text{Diketahui persamaan lingkaran}\\ & x^2+y^2=12\\ & \text{Persamaan garis singgung lingkaran}\\ & \text{melalui titik}(x_1,y_1)\text{adalah}:\\ & \begin{array}{rl}{x}^2+y^2& =12\\ xx+yy& =12\\ x_{1}x+y_{1}y& =12\\ \text{garis ini melalui}& \text{titik}\\ P(0,4)& ,\text{maka}\\ x_1.0+y_1.4& =12\\ y_1& =3......(1)\end{array}\\ & \text{Karena titik}(x_1,y_1)\text{pada lingkaran}\\ & \text{maka},\\ & x_1^2+y_1^2=12......(2)\\ & \text{Selanjutnya dari persamaan}(1)\mathrm{\&}(2)\\ & \text{akan diperoleh}\\ & \begin{array}{rl}& x_1^2+y_1^2=12\\ y_1=3\Rightarrow & x_1^2+(3{)}^2=12\\ \iff & x_1^2+9=12\\ \iff & x_1^2=3\\ \iff & x_1=\pm \sqrt{3}\end{array}\\ & \begin{array}{rl}& \text{Sehingga persa}\text{maan garis singgungnya}\\ & (x_{1}x+y_{1}y=12)\text{adalah}:\\ & \{\begin{array}{ll}\text{di titik}& (x_1,y_1)=(\sqrt{3},3)\Rightarrow \sqrt{3}x+3y=12\\ \text{di titik}& (x_1,y_1)=(-\sqrt{3},3)\Rightarrow -\sqrt{3}x+3y=12\end{array}\\ & \end{array}\end{array}\end{array}$

Berikut ilustrasi gambarnya

DAFTAR PUSTAKA

1. Kartini, Suprapto, Subandi, Setiadi, U. 2005. Matematika Program Studi Ilmu Alam Kelas XI untuk SMA dan MA. Klaten: INTAN PARIWARA.
2. Sobirin. 2006. Kompas Matematika Strategi Praktis Menguasai Tes Matematika SMA Kelas 2. Jakarta: KAWAN PUSTAKA.
3. Wirodikromo, S. 2007. Matematika Jilid 2 IPA untuk Kelas XI. Jakarta: ERLANGGA.

Vektor di Dimensi Tiga (Ruang)

 $\text{A. Letak Titik Dalam Ruang}$

Perhatikan titik P dalam ruang berikut

Kita ambil sistem ortogonal ruang yang terdiri atas tiga bidang yang saling berpotongan tegak lurus menurut tiga potong garis sebagaimana ilustrasi gambar di atas.Ketiga garis potong tersebut berturut-turut adalah sumbu X, sumbu Y, dan sumbu Z. 

Letak suatu titik P terhadap bidang OYZ, OXZ, dan OXY. Titik P(a,b,c) berarti P berjarak $\text{a}$  terhadap OYZ  dan berjarak  $\text{b}$ terhadap bidang OXZ, serta berjarak  $\text{c}$  terhadap bidang  OXY. Jadi, pasangan tiga buah bilangan riil berurutan menyatakan suatu titik dalam ruang dan demikian sebaliknya.

$\text{B. Vektor Dalam Ruang}$

Sebagaimana halnya vektor dalam bidang (dimensi dua), maka vektor dalam ruang atau ${\text{R}}^3$   juga dapat dinyatakan dengan 3 bilangan riil yang berbeda.

Sebagai misal

$\begin{array}{rl}P& (x,y,z),\text{maka vektor posisi}\overline{OP}\\ & \text{dalam ruang adalah}\\ & \overline{OP}=(x,y,z)\text{atau}\\ & \overline{OP}=\left(\begin{array}{c}x\\ y\\ z\end{array}\right)\text{ataupun juga dengan}\\ & \text{dengan vektor satuannya dalam}\\ & \text{hal ini adalah}:\overline{i},\overline{j},\mathrm{\&}\overline{k}\text{adalah}:\\ & \overline{OP}=x\overline{i}+y\overline{j}+z\overline{k}\\ \\ & \text{Selanjutnya}\\ & \text{vektor-vektor satuan di atas}\\ & \text{dalam ruang adalah}\\ & \overline{i}=\left(\begin{array}{c}1\\ 0\\ 0\end{array}\right)\text{dan}\\ & \overline{j}=\left(\begin{array}{c}0\\ 1\\ 0\end{array}\right),\text{serta}\\ & \overline{k}=\left(\begin{array}{c}0\\ 0\\ 1\end{array}\right)\end{array}$.

$\text{C. Vektor Basis Dalam Ruang}$

Perhatikan ilustrasi berikut 

Dengan penjelasan hampir kurang lebih sama pada poin B di atas, maka vektor basis suatu titik P adalah vektor yang dinyatakan dengan vektor satuan, yaitu: $\overline{OP}=x\overline{i}+y\overline{j}+z\overline{k}$.

Sebagai contoh titik P(3,5,8) jika ilustrasikan adalah sebagai berikut

dan jika dinyakatan dengan vektor basis adalah : $\overline{OP}=3\overline{i}+5\overline{j}+8\overline{k}$.

$\text{CONTOH SOAL}$.

Nyatakan vektor berikut dalam bentuk vektor baris, kolom dan basis

$\begin{array}{rl}& \mathbf{\text{Jawab}}:\\ & \text{Diktahui vektor adalah}\overrightarrow{w},\text{boleh juga}\\ & \text{disebut sebagai vektor posisi, sehingga}\\ & \text{dapat tuliskan juga dengan}:\overrightarrow{w}=\overline{OW}\\ & \text{Dan vektor posisi tersebut dapat dinyatakan}\\ & \text{dengan}:\\ & \text{a}.\text{vektor baris}=(8,-6,13)\\ & \text{b}.\text{vektor kolom}=\left(\begin{array}{c}8\\ -6\\ -13\end{array}\right)\\ & \text{c}.\text{vektor basis}=8\overrightarrow{i}-6\overrightarrow{j}-13\overrightarrow{k}\end{array}$

$\text{D. Vektor Posisi dan Vektor Bebas dalam Ruang}$

Vektor posisi suatu titik adalah vektor yang titik pangkalnya terletak dititik pusat koordinat dan titik ujungntya pada titik tersebut. Sedangkan vektor bebas di sini adalah sembarang vektor yang titik pangkalnya tidak berada pada pusat koordinat dan berujung pada suatu titik. Perhatikanlah ilustrasi berikut ini

Pada ilustrasi gambar di atas ada dua vektor posisi yaitu $\overrightarrow{OA}=\overline{a}$  dan  $\overrightarrow{OB}=\overline{b}$, sedangkan vektor bebas atau vektor sembarangnya adalah $\overrightarrow{AB}$.

$\begin{array}{rl}& \text{Perhatikan bahwa}\\ & \text{pada gambar di atas}\\ & \overrightarrow{OA}+\overrightarrow{AB}=\overrightarrow{OB}\\ & \overline{a}+\overrightarrow{AB}=\overline{b}\\ & \overrightarrow{AB}=\overline{b}-\overline{a}\end{array}$.

$\text{E. Modulus Vektor Dalam Ruang}$

Pengertian modulus vektor dalam ruang memiliki pengertian yang sama dalam bidang cuma yang membedakan adalah kondisinya saja. Karena baik di dalam ruang maupun bidang dalam menentukan modulus/bebsar/panjang suatu vektor adalah sama saja.

Jika suatu titik  $A(x_1,y_1,z_1)$  dan  $B(x_2,y_2,z_2)$, maka modulus dari dari kedua titik itu adalah jarak antara kedua titik tersebut, yaitu:

$\left|\overline{AB}\right|=\sqrt{(x_1-x_2{)}^2+(y_1-y_2{)}^2+(z_1-z_2{)}^2}$.

$\text{CONTOH SOAL}$

$\begin{array}{l}\\ 1.& \text{Jika di}{\text{R}}^3\text{diketahui titik}\\ & A(0,0,0),B(1,2,3),\text{dan}C(4,5,6)\\ & \text{Tentukanlah panjang}\\ & \text{a}.\overline{AB}\\ & \text{b}.\overline{AC}\\ & \text{c}.\overline{BC}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}\text{a}.\overline{AB}& =\left(\begin{array}{c}1-0\\ 2-0\\ 3-0\end{array}\right)=\left(\begin{array}{c}1\\ 2\\ 3\end{array}\right)\\ \left|\overline{AB}\right|& =\sqrt{1^2+2^2+3^2}\\ & =\sqrt{1+4+9}\\ & =\sqrt{14}\end{array}\\ & \begin{array}{rl}\text{b}.\overline{AC}& =\left(\begin{array}{c}4-0\\ 5-0\\ 6-0\end{array}\right)=\left(\begin{array}{c}4\\ 5\\ 6\end{array}\right)\\ \left|\overline{AC}\right|& =\sqrt{4^2+5^2+6^2}\\ & =\sqrt{16+25+36}\\ & =\sqrt{77}\end{array}\\ & \begin{array}{rl}\text{c}.\overline{BC}& =\left(\begin{array}{c}4-1\\ 5-2\\ 6-3\end{array}\right)=\left(\begin{array}{c}3\\ 3\\ 3\end{array}\right)\\ \left|\overline{BC}\right|& =\sqrt{3^2+3^2+3^2}\\ & =\sqrt{9+9+9}\\ & =\sqrt{27}\end{array}\end{array}$.

$\begin{array}{l}\\ 2.& \text{Tentukanlah panjang dari vektor}\\ & \overline{q}=3\overline{i}-\overline{j}-7\overline{k}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}\overline{q}& =3\overline{i}-\overline{j}-7\overline{k}=\left(\begin{array}{c}3\\ -1\\ -7\end{array}\right)\\ \left|\overline{q}\right|& =\sqrt{3^2+(-1{)}^2+(-7{)}^2}\\ & =\sqrt{9+1+49}\\ & =\sqrt{59}\end{array}\end{array}$.

DAFTAR PUSTAKA

1. Koesmartono, Rawuh (editor). 1973. Matematika Pendahuluan (Seri Matematika). Bandung: ITB

Contoh Soal 4 Fungsi Komposisi dan Fungsi Invers

$\begin{array}{l}\\ 16.& \text{Jika}f\text{dan}g\text{adalah fungsi yang }\\ & \text{mempunyai invers dan memenuhi}\\ & f(2x)=g(x-3),\text{maka}{f}^{-1}(x)\\ & \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& g^{-1}\left({\displaystyle \frac{x}{2}-\frac{2}{3}}\right)\\ \text{b}.& g^{-1}\left({\displaystyle \frac{x}{2}}\right)-{\displaystyle \frac{2}{3}}\\ \text{c}.& g^{-1}(2x+6)\\ \text{d}.& 2g^{-1}(x)-6\\ \text{e}.& 2g^{-1}(x)+6\end{array}\\ & (\mathbf{\text{SBMPTN 2016 Mat Das}})\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}& \text{Misalkan bahwa}\\ & f(2x)=g(x-3)=x,\\ & \text{maka}\\ & \{\begin{array}{ll}{f}^{-1}(x)& =2x\\ g^{-1}(x)& =x-3\end{array}\end{array}\\ & \begin{array}{|cc|}\hline \text{Sintak}& \text{Hasil}\\ \begin{array}{rl}{g}^{-1}(x)& =x-3\\ x& =g^{-1}(x)+3\\ & \end{array}& \begin{array}{rl}{f}^{-1}(x)& =2x\\ & =2(g^{-1}(x)+3)\\ & =2g^{-1}(x)+6\end{array}\\ \hline\end{array}\end{array}$.

$\begin{array}{l}\\ 17.& \text{Jika}{f}^{-1}(x)={\displaystyle \frac{x-1}{5}\text{dan}{g}^{-1}(x)={\displaystyle \frac{3-x}{2},}}\\ & \text{maka}{(f\circ g)}^{-1}(6)=....\\ & \begin{array}{l}\\ \text{a}.& -1& & & \text{d}.& 2\\ \text{b}.& 0& \text{c}.& 1& \text{e}.& 3\end{array}\\ & (\mathbf{\text{UMPTN 1995}})\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}\text{Diketahui b}& \text{ahwa}:\\ & \{\begin{array}{ll}{f}^{-1}(x)& ={\displaystyle \frac{x-1}{5}}\\ g^{-1}(x)& ={\displaystyle \frac{3-x}{2}}\end{array}\\ {(f\circ g)}^{-1}(x)& =(g^{-1}\circ f^{-1})(x)\\ & ={\displaystyle \frac{3-\left({\displaystyle \frac{x-1}{5}}\right)}{2}}\\ {(f\circ g)}^{-1}(6)& ={\displaystyle \frac{3-\left({\displaystyle \frac{6-1}{5}}\right)}{2}}\\ & ={\displaystyle \frac{3-1}{2}=\frac{2}{2}}\\ & =1\end{array}\end{array}$.

$\begin{array}{l}\\ 18.& \text{Invers dari}f(x)={125}^x\text{adalah}{f}^{-1}(x)\\ & \text{Nilai dari}{f}^{-1}\left(5\sqrt{5}\right)=....\\ & \begin{array}{l}\\ \text{a}.& 1& & & \text{d}.& {\displaystyle \frac{3}{5}}\\ \text{b}.& {\displaystyle \frac{1}{2}}& \text{c}.& {\displaystyle \frac{1}{6}}& \text{e}.& -{\displaystyle \frac{1}{2}}\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}& \text{Diketahui bahwa}f(x)={125}^x,\text{maka}\\ & f(x)=y={125}^x\\ & \text{Kedua ruas dilogkan masing-masing}\\ & \log y=\log {125}^x\\ & \iff \log y=x\log 125\\ & \iff x\log 125=\log y\\ & \iff x={\displaystyle \frac{\log y}{\log 125}}\\ & \iff x={}^{125}\log y\\ & \iff f^{-1}(x)={}^{125}\log x\\ & \text{Selanjutnya}\\ & f^{-1}(x)={}^{125}\log x\\ & \iff f^{-1}(5\sqrt{5})={}^{125}\log (5\sqrt{5})\\ & \iff f^{-1}(5\sqrt{5})={}^{5^3}\log 5^{{.}^{\frac{3}{2}}}\\ & \iff f^{-1}(5\sqrt{5})={\displaystyle \frac{{\displaystyle \frac{3}{2}}}{3}{}^5\log 5}\\ & \iff f^{-1}(5\sqrt{5})={\displaystyle \frac{1}{2}.1}\\ & \iff f^{-1}(5\sqrt{5})={\displaystyle \frac{1}{2}}\end{array}\end{array}$