Lanjutan Materi Vektor : Proyeksi Ortogonal Suatu Vektor

$\color{blue}\begin{aligned}\textrm{A}.\quad&\textrm{Panjang Proyeksi Ortogonal Suatu}\\ &\textrm{Vektor pada vektor lain} \end{aligned}$.

Perhatikanlah ilstrasi gambar yang dibentuk dari dua vektor berikut

Pada gambar di atas 

$\begin{array}{|c|c|}\hline \triangle \textrm{OAC}&\angle \left ( \overrightarrow{a},\overrightarrow{b} \right )\\\hline \begin{aligned}\cos \theta &=\displaystyle \frac{\left | \overrightarrow{c} \right |}{\left | \overrightarrow{a} \right |}\\ \Leftrightarrow \left | \overrightarrow{c} \right |&=\left | \overrightarrow{a} \right |\cos \theta \: \: ........(1) \end{aligned}&\begin{aligned}\cos \theta &=\displaystyle \frac{\overrightarrow{a}\overrightarrow{b}}{\left | \overrightarrow{a} \right |\left | \overrightarrow{b} \right |}\: \: ........(2) \end{aligned}\\\hline \end{array}$.

$\begin{aligned}\textrm{Dari}\: \: (1)\: \: &\textrm{dan} \: \: (2)\: \: \textrm{diperoleh}\\ \left | \overrightarrow{c} \right |&=\left | \overrightarrow{a} \right |\cos \theta \\ &=\left | \overrightarrow{a} \right |\left ( \displaystyle \frac{\overrightarrow{a}\overrightarrow{b}}{\left | \overrightarrow{a} \right |\left | \overrightarrow{b} \right |} \right )\\ &=\color{red}\left |\displaystyle \frac{\overrightarrow{a}\overrightarrow{b}}{\left | \overrightarrow{b} \right |} \right | \end{aligned}$

$\color{blue}\begin{aligned}\textrm{B}.\quad&\textrm{Proyeksi Ortogonal Suatu Vektor}\\ &\textrm{pada vektor lain} \end{aligned}$.

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

$\LARGE\colorbox{yellow}{CONTOH SOAL}$.

$\begin{array}{ll}\\ 1.&\textrm{Diketahui vektor}\: \: \overrightarrow{a}=\begin{pmatrix} 3\\ 2 \end{pmatrix}\: \: \textrm{dan}\: \: \overrightarrow{b}=\begin{pmatrix} -2\\ 1 \end{pmatrix}.\\ & \textrm{Tentukanlah proyeksi ortogonal vektor}\\ &\overrightarrow{a}\: \: \textrm{pada}\: \: \overrightarrow{b}\: \: \textrm{dan tentukanlah panjangnya} \\\\ &\textrm{Jawab}:\\ &\begin{aligned}\textrm{Misalkan}&\: \: \overrightarrow{c}\: \: \textrm{adalah vektor proyeksi yang dimaksud, }\\ &\textrm{maka}\\ \overrightarrow{c}&=\left ( \displaystyle \frac{\overrightarrow{a}\overrightarrow{b}}{\left | \overrightarrow{b} \right |^{2}} \right )\overrightarrow{b}\\ &=\displaystyle \frac{\begin{pmatrix} 3\\ 2 \end{pmatrix}.\begin{pmatrix} -2\\ 1 \end{pmatrix}}{(-2)^{2}+1^{2}}.\overrightarrow{b}=\frac{3.(-2)+2.1}{4+1}\begin{pmatrix} -2\\ 1 \end{pmatrix}\\ &=-\frac{4}{5}\begin{pmatrix} -2\\ 1 \end{pmatrix}=\begin{pmatrix} \frac{8}{5}\\ -\frac{4}{5} \end{pmatrix}\quad \textbf{atau}\\ &=\color{red}\frac{8}{5}\bar{i}-\frac{4}{5}\bar{j} \end{aligned}\\ &\begin{aligned}\textrm{Dan panjang}\: \: &\textrm{vektor proyeksi yang dimaksud adalah}:\\ \left |\overrightarrow{c} \right |&= \left |\displaystyle \frac{\overrightarrow{a}\overrightarrow{b}}{\left | \overrightarrow{b} \right |} \right |\\ &=\left |\displaystyle \frac{\begin{pmatrix} 3\\ 2 \end{pmatrix}.\begin{pmatrix} -2\\ 1 \end{pmatrix}}{\sqrt{(-2)^{2}+1^{2}}} \right |=\left |\frac{3.(-2)+2.1}{\sqrt{4+1}} \right |\\ &=\left |-\frac{4}{\sqrt{5}} \right |\\ &=\displaystyle \frac{4}{\sqrt{5}}=\color{red}\frac{4}{5}\sqrt{5} \end{aligned} \end{array}$.

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