Proyeksi Ortogonal Suatu Vektor

 

$\begin{array}{rl}\text{A}.& \text{Panjang Proyeksi Ortogonal Suatu}\\ & \text{Vektor pada vektor lain}\end{array}$.

Perhatikanlah ilstrasi gambar yang dibentuk dari dua vektor berikut

Pada gambar di atas 

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

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

$\begin{array}{rl}\text{B}.& \text{Proyeksi Ortogonal Suatu Vektor}\\ & \text{pada vektor lain}\end{array}$.

$\begin{array}{|c|}\hline \text{Perhatikan pula misal}\hat{c}\text{adalah vektor satuan dari}\overrightarrow{c}\text{dan}\overrightarrow{b},\\ \begin{array}{rl}\text{maka}\overrightarrow{c}& =\left|\overrightarrow{c}\right|\hat{c}\end{array}\text{dan}\begin{array}{rl}\overrightarrow{b}& =\left|\overrightarrow{b}\right|\hat{b}=\left|\overrightarrow{b}\right|\hat{c}\end{array}\\ \begin{array}{rl}\text{Sehingga}& \mathbf{\text{proyeksi ortogonal vektor}}\overrightarrow{a}\text{pada}\overrightarrow{b}\text{adalah}:\\ \overrightarrow{c}& =\left|\overrightarrow{c}\right|\hat{b}\\ & =\left({\displaystyle \frac{\overrightarrow{a}\overrightarrow{b}}{\left|\overrightarrow{b}\right|}}\right)\left({\displaystyle \frac{\overrightarrow{b}}{\left|\overrightarrow{b}\right|}}\right)\\ & =\left({\displaystyle \frac{\overrightarrow{a}\overrightarrow{b}}{{\left|\overrightarrow{b}\right|}^2}}\right)\overrightarrow{b}\end{array}\\ \hline\end{array}$.

$\text{CONTOH SOAL}$.

$\begin{array}{l}\\ 1.& \text{Diketahui vektor}\overrightarrow{a}=\left(\begin{array}{c}3\\ 2\end{array}\right)\text{dan}\overrightarrow{b}=\left(\begin{array}{c}-2\\ 1\end{array}\right).\\ & \text{Tentukanlah proyeksi ortogonal vektor}\\ & \overrightarrow{a}\text{pada}\overrightarrow{b}\text{dan tentukanlah panjangnya}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}\text{Misalkan}& \overrightarrow{c}\text{adalah vektor proyeksi yang dimaksud, }\\ & \text{maka}\\ \overrightarrow{c}& =\left({\displaystyle \frac{\overrightarrow{a}\overrightarrow{b}}{{\left|\overrightarrow{b}\right|}^2}}\right)\overrightarrow{b}\\ & ={\displaystyle \frac{\left(\begin{array}{c}3\\ 2\end{array}\right).\left(\begin{array}{c}-2\\ 1\end{array}\right)}{(-2{)}^2+1^2}.\overrightarrow{b}=\frac{3.(-2)+2.1}{4+1}\left(\begin{array}{c}-2\\ 1\end{array}\right)}\\ & =-\frac{4}{5}\left(\begin{array}{c}-2\\ 1\end{array}\right)=\left(\begin{array}{c}\frac{8}{5}\\ -\frac{4}{5}\end{array}\right)\mathbf{\text{atau}}\\ & =\frac{8}{5}\overline{i}-\frac{4}{5}\overline{j}\end{array}\\ & \begin{array}{rl}\text{Dan panjang}& \text{vektor proyeksi yang dimaksud adalah}:\\ \left|\overrightarrow{c}\right|& =\left|{\displaystyle \frac{\overrightarrow{a}\overrightarrow{b}}{\left|\overrightarrow{b}\right|}}\right|\\ & =\left|{\displaystyle \frac{\left(\begin{array}{c}3\\ 2\end{array}\right).\left(\begin{array}{c}-2\\ 1\end{array}\right)}{\sqrt{(-2{)}^2+1^2}}}\right|=\left|\frac{3.(-2)+2.1}{\sqrt{4+1}}\right|\\ & =|-\frac{4}{\sqrt{5}}|\\ & ={\displaystyle \frac{4}{\sqrt{5}}=\frac{4}{5}\sqrt{5}}\end{array}\end{array}$.

$\begin{array}{l}\\ 2.& \text{Diketahui vektor}\overrightarrow{a}=\left(\begin{array}{c}-5\\ 4\end{array}\right)\text{dan}\overrightarrow{b}=\left(\begin{array}{c}2\\ 6\end{array}\right).\\ & \text{Tentukanlah proyeksi ortogonal vektor}\\ & \overrightarrow{a}\text{pada}\overrightarrow{b}\text{dan tentukanlah panjangnya}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}\text{Misalkan}& \overrightarrow{c}\text{adalah vektor proyeksi yang dimaksud, }\\ & \text{maka}\\ \overrightarrow{c}& =\left({\displaystyle \frac{\overrightarrow{a}\overrightarrow{b}}{{\left|\overrightarrow{b}\right|}^2}}\right)\overrightarrow{b}\\ & ={\displaystyle \frac{\left(\begin{array}{c}-5\\ 4\end{array}\right).\left(\begin{array}{c}2\\ 6\end{array}\right)}{2^2+6^2}.\overrightarrow{b}=\frac{(-5).2+4.6}{4+36}\left(\begin{array}{c}2\\ 6\end{array}\right)}\\ & =\frac{7}{20}\left(\begin{array}{c}2\\ 6\end{array}\right)=\left(\begin{array}{c}\frac{7}{10}\\ \frac{21}{10}\end{array}\right)\mathbf{\text{atau}}\\ & =\frac{7}{10}\overline{i}+\frac{21}{10}\overline{j}\end{array}\\ & \begin{array}{rl}\text{Dan panjang}& \text{vektor proyeksi yang dimaksud adalah}:\\ \left|\overrightarrow{c}\right|& =\left|{\displaystyle \frac{\overrightarrow{a}\overrightarrow{b}}{\left|\overrightarrow{b}\right|}}\right|\\ & =\left|{\displaystyle \frac{\left(\begin{array}{c}-5\\ 4\end{array}\right).\left(\begin{array}{c}2\\ 6\end{array}\right)}{\sqrt{2^2+6^2}}}\right|=\left|\frac{(-5).2+4.6}{\sqrt{4+36}}\right|\\ & =\left|\frac{14}{\sqrt{40}}\right|\\ & ={\displaystyle \frac{14}{2\sqrt{10}}=\frac{7}{10}\sqrt{10}}\end{array}\end{array}$.

$.\text{Coba bandingkan dengan vektor di dimensi tiga berikut}$

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