Materinya sama dengan proyeksi ortogonal pada dimensi dua klik di sini
Uraian berikut sebagai pengingat saja
$\begin{aligned}\triangleright \quad&\textbf{Proyeksi skalar vektor} \\ &\left | \vec{c} \right |=\displaystyle \frac{\vec{a}\bullet \vec{b}}{\left | \vec{b} \right |}\\ \triangleright \quad&\textbf{Vektor proyeksi ortogonal} \\ &\vec{c}=\displaystyle \frac{\vec{a}\bullet \vec{b}}{\left | \vec{b} \right |^{2}}.\vec{b} \end{aligned}$
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!
$\begin{array}{|c|c|}\hline \triangle \textrm{OAC}&\angle \left ( \overrightarrow{a},\overrightarrow{b} \right )\\\hline \begin{aligned}\cos \theta &=\displaystyle \frac{\left | \overrightarrow{c} \right |}{\left | \overrightarrow{a} \right |}\\ \Leftrightarrow \left | \overrightarrow{c} \right |&=\left | \overrightarrow{a} \right |\cos \theta \: \: ........(1) \end{aligned}&\begin{aligned}\cos \theta &=\displaystyle \frac{\overrightarrow{a}\overrightarrow{b}}{\left | \overrightarrow{a} \right |\left | \overrightarrow{b} \right |}\: \: ........(2) \end{aligned}\\\hline \end{array}$.
$\begin{aligned}\textrm{Dari}\: \: (1)\: \: &\textrm{dan} \: \: (2)\: \: \textrm{diperoleh}\\ \left | \overrightarrow{c} \right |&=\left | \overrightarrow{a} \right |\cos \theta \\ &=\left | \overrightarrow{a} \right |\left ( \displaystyle \frac{\overrightarrow{a}\overrightarrow{b}}{\left | \overrightarrow{a} \right |\left | \overrightarrow{b} \right |} \right )\\ &=\left |\displaystyle \frac{\overrightarrow{a}\overrightarrow{b}}{\left | \overrightarrow{b} \right |} \right | \end{aligned}$
Dan penjelasan kedua berkaitan dengan vektor proyeksi ortogonalnya, yaitu:
$\begin{aligned}&\color{red}\textrm{Perhatikan pula misal}\: \: \color{black}\hat{c}\\ & \textrm{adalah vektor satuan dari}\: \: \overrightarrow{c}\: \: \textrm{dan}\: \: \overrightarrow{b},\\ & \textrm{maka}\\ &\begin{aligned}\overrightarrow{c}&=\left | \overrightarrow{c} \right |\hat{c} \end{aligned},\: \: \textrm{dan}\\ &\begin{aligned}\overrightarrow{b}&=\left | \overrightarrow{b} \right |\hat{b}=\left | \overrightarrow{b} \right |\hat{c} \end{aligned} \end{aligned}$.
$\begin{aligned}\textrm{Sehingga}&\: \: \textbf{proyeksi ortogonal vektor}\: \: \overrightarrow{a}\: \: \textrm{pada}\: \: \overrightarrow{b}\: \: \textrm{adalah}:\\ \overrightarrow{c}&=\left | \overrightarrow{c} \right |\hat{b}\\ &=\left ( \displaystyle \frac{\overrightarrow{a}\overrightarrow{b}}{\left | \overrightarrow{b} \right |} \right )\left ( \displaystyle \frac{\overrightarrow{b}}{\left | \overrightarrow{b} \right |} \right )\\ &=\left (\displaystyle \frac{\overrightarrow{a}\overrightarrow{b}}{\left | \overrightarrow{b} \right |^{2}} \right )\overrightarrow{b} \end{aligned}$
$\LARGE\colorbox{yellow}{CONTOH SOAL}$.
$\begin{array}{ll}\\ 1.&\textrm{Diketahui}\: \: \vec{a}=\begin{pmatrix} 2\\ -3\\ 1 \end{pmatrix}\: \: \textrm{dan}\: \: \vec{b}=\begin{pmatrix} -4\\ 2\\ 2 \end{pmatrix}.\\ &\textrm{Tentukanlah}\\ &\textrm{a}.\quad \textrm{proyeksi skalar}\: \: \vec{a}\: \: \textrm{pada}\: \: \vec{b}\\ &\textrm{b}.\quad \textrm{vektor proyeksi}\: \: \vec{a}\: \: \textrm{pada}\: \: \vec{b}\\ &\textrm{c}.\quad \textrm{proyeksi skalar}\: \: \vec{b}\: \: \textrm{pada}\: \: \vec{a}\\ &\textrm{d}.\quad \textrm{vektor proyeksi}\: \: \vec{b}\: \: \textrm{pada}\: \: \vec{a}\\\\ &\textbf{Jawab}:\\ &\textrm{Misalkan proyeksi skalar}\: \: \vec{a}\: \: \textrm{pada}\: \: \vec{b}\\ &\textrm{adalah}\: \: \left | \vec{c} \right |,\: \: \textrm{dan}\\ &\textrm{misalkan juga proyeksi skalar}\: \: \vec{b}\: \: \textrm{pada}\: \: \vec{a}\\ &\textrm{adalah}\: \: \left | \vec{d} \right |,\: \: \textrm{maka}\\ &\begin{aligned}\textrm{a}.\quad\left | \vec{c} \right |&=\displaystyle \frac{\vec{a}\bullet \vec{b}}{\left | \vec{b} \right |}\\ &=\displaystyle \frac{\begin{pmatrix} 2\\ -3\\ 1 \end{pmatrix}\begin{pmatrix} -4\\ 2\\ 2 \end{pmatrix}}{\sqrt{(-4)^{2}+2^{2}+2^{2}}}\\ &=\displaystyle \frac{-8-6+2}{\sqrt{24}}=-\frac{12}{24}\sqrt{24}=-\sqrt{6}\\ &\textrm{Karena hasilnya berupa panjang, maka}\\ &\textrm{diharga mutlak/positif}\\ &\left | \vec{c} \right |=\left |-\sqrt{6} \right |=\color{red}\sqrt{6} \end{aligned}\\ &\begin{aligned}\textrm{b}.\quad \vec{c}&=\displaystyle \frac{\vec{a}\bullet \vec{b}}{\left | \vec{b} \right |^{2}}\times \vec{b}\\ &=\displaystyle \frac{-12}{(\sqrt{24})^{2}}\times \begin{pmatrix} -4\\ 2\\ 2 \end{pmatrix}\\ &=-\displaystyle \frac{1}{2}\begin{pmatrix} -4\\ 2\\ 2 \end{pmatrix}=\color{red}\begin{pmatrix} 2\\ -1\\ -1 \end{pmatrix} \end{aligned}\\ &\begin{aligned}\textrm{c}.\quad\left | \vec{d} \right |&=\displaystyle \frac{\vec{b}\bullet \vec{a}}{\left | \vec{a} \right |}\\ &=\displaystyle \frac{\begin{pmatrix} -4\\ 2\\ 2 \end{pmatrix}\begin{pmatrix} 2\\ -3\\ 1 \end{pmatrix}}{\sqrt{2^{2}+(-3)^{2}+1^{2}}}\\ &=\displaystyle \frac{-8-6+2}{\sqrt{14}}=-\frac{12}{14}\sqrt{14}=-\frac{6}{7}\sqrt{14}\\ &\textrm{Karena hasilnya berupa panjang, maka}\\ &\textrm{diharga mutlak/positif}\\ &\left | \vec{d} \right |=\left |-\displaystyle \frac{6}{7}\sqrt{14} \right |=\color{red}\displaystyle \frac{6}{7}\sqrt{14} \end{aligned}\\ &\begin{aligned}\textrm{d}.\quad \vec{d}&=\displaystyle \frac{\vec{b}\bullet \vec{a}}{\left | \vec{a} \right |^{2}}\times \vec{a}\\ &=\displaystyle \frac{-12}{(\sqrt{14})^{2}}\times \begin{pmatrix} 2\\ -3\\ 1 \end{pmatrix}\\ &=-\displaystyle \frac{12}{14}\begin{pmatrix} 2\\ -3\\ 1 \end{pmatrix}=\color{red}\begin{pmatrix} -\frac{12}{7}\\ \frac{18}{7}\\ -\frac{6}{7} \end{pmatrix} \end{aligned} \end{array}$
$\begin{array}{ll}\\ 2.&\textrm{Diketahui}\: \: \vec{a}=\begin{pmatrix} -3\\ -2\\ \color{blue}m \end{pmatrix}\: \: \textrm{dan}\: \: \vec{b}=\begin{pmatrix} 2\\ -1\\ -2 \end{pmatrix}.\\ & \textrm{Jika proyeksi skalar}\: \: \vec{a}\: \: \textrm{pada}\: \: \vec{b}\: \: \textrm{adalah}\\ &\textrm{bernilai}\: \: -\displaystyle \frac{2}{3},\: \: \textrm{maka tentukan nilai}\: \: \color{blue}m\\\\ &\textbf{Jawab}:\\ &\textrm{Misalkan proyeksi skalar}\: \: \vec{a}\: \: \textrm{pada}\: \: \vec{b}\: \: \textrm{adalah}\: \: \left | \vec{f} \right |,\\ &\textrm{maka}\\ &\begin{aligned}\left | \vec{f} \right |&=\displaystyle \frac{\vec{a}\bullet \vec{b}}{\left | \vec{b} \right |}\\ \Leftrightarrow \: -\displaystyle \frac{2}{3}&=\displaystyle \frac{\begin{pmatrix} -3\\ -2\\ \color{blue}m \end{pmatrix}\begin{pmatrix} 2\\ -1\\ -2 \end{pmatrix}}{\sqrt{2^{2}+(-1)^{2}+(-2)^{2}}}\\ \Leftrightarrow \: -\displaystyle \frac{2}{3}&=\displaystyle \frac{-6+2-2\color{blue}m}{\sqrt{9}}=\frac{-4-2\color{blue}m}{3}\\ \Leftrightarrow \: -2&=-4-2\color{blue}m\\ \Leftrightarrow \: 1&=2+\color{blue}m\\ \Leftrightarrow \: -\color{blue}m&=2-1=1\\ \Leftrightarrow \: \color{blue}m&=-1 \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 4.&\textrm{Diketahui vektor}\: \: \overrightarrow{a}=3\bar{i}-2\bar{j}+2\bar{k}\: \: \textrm{dan}\\ & \overrightarrow{b}=2\bar{i}-2\bar{j}+\bar{k}.\: \textrm{Tentukanlah panjang }\\ &\textrm{vektor proyeksi ortogonal}\\ &\textrm{a}.\quad \overrightarrow{a}\: \: \textrm{pada}\: \: \overrightarrow{b}\\ & \textrm{b}.\quad \overrightarrow{a}\: \: \textrm{pada}\: \: \left ( \overrightarrow{a}+\overrightarrow{b} \right ) \\\\ &\textbf{Jawab}:\\ &\begin{aligned}\textrm{a}.\quad\textrm{Misal}&\textrm{kan}\: \: \overrightarrow{g}\: \: \textrm{adalah vektor proyeksi }\\ \textrm{yang}&\: \textrm{dimaksud, maka panjanynya}\\ (\textrm{lang}&\textrm{sung diharga mutlak})\\ \left |\overrightarrow{g} \right |&= \left |\displaystyle \frac{\overrightarrow{a}\overrightarrow{b}}{\left | \overrightarrow{b} \right |} \right | \\ &=\left |\displaystyle \frac{\begin{pmatrix} 3\\ -2\\ 2 \end{pmatrix}.\begin{pmatrix} 2\\ -2\\ 1 \end{pmatrix}}{\sqrt{2^{2}+(-2)^{2}+1^{2}}} \right |\\ &=\left |\frac{3.2+(-2).(-2)+2.1}{\sqrt{4+4+1}} \right |\\ &=\left | \displaystyle \frac{12}{3} \right |=4 \end{aligned}\\ &\begin{aligned}\textrm{b}.\quad\textrm{Misal}&\textrm{kan}\: \: \overrightarrow{h}\: \: \textrm{adalah vektor proyeksi }\\ \textrm{yang}&\: \textrm{dimaksud, maka panjanynya}\\ (\textrm{lang}&\textrm{sung diharga mutlak})\\ \left |\overrightarrow{h} \right |&= \left |\displaystyle \frac{\overrightarrow{a}\left (\overrightarrow{a}+\overrightarrow{b} \right )}{\left |\overrightarrow{a}+ \overrightarrow{b} \right |} \right |\\ &=\left |\displaystyle \frac{\begin{pmatrix} 3\\ -2\\ 2 \end{pmatrix}.\begin{pmatrix} 3+2\\ -2+(-2)\\ 2+1 \end{pmatrix}}{\sqrt{(3+2)^{2}+(-2+(-2))^{2}+(2+1)^{2}}} \right |\\ &=\left |\frac{3.5+(-2).(-4)+2.3}{\sqrt{25+16+9}} \right |\\ &=\left |\frac{29}{\sqrt{50}} \right |\\ &=\displaystyle \frac{29}{5\sqrt{2}}=\frac{29}{10}\sqrt{5} \end{aligned} \end{array}$
DAFTAR PUSTAKA
- Johanes, Kastolan, Sulasim. 2006. Kompetensi Matematika 3A SMA Kelas XII Semester Pertama Program IPA. Jakarta: YUDHISTIRA.
Tidak ada komentar:
Posting Komentar
Informasi