Vektor di Dimensi Tiga

$\color{blue}\textrm{A. Vektor Di Ruang}$

Perhatikanlah ilustrasi gambar berikut

$\begin{array}{|c|c|}\hline \textrm{Nama}&\textbf{R}^{3}\\\hline \textrm{Vektor Satuan}&\textrm{Ruang (Bidang XYZ)}\\\hline \hat{e}_{\bar{a}}=\displaystyle \frac{\bar{a}}{\left | \bar{a} \right |}&\begin{cases} i= &\textrm{vektor satuan} \\ &\textrm{yang searah sumbu X}\\ j= &\textrm{Vektor satuan}\\ &\textrm{yang searah sumbu Y}\\ k=&\textrm{Vektor satuan}\\ &\textrm{searah sumbu Z} \end{cases} \\\hline \textrm{Vektor nol}&\overrightarrow{O}=\begin{pmatrix} 0\\ 0\\ 0 \end{pmatrix}\\\hline \textrm{Vektor posisi}&\overrightarrow{OP}=\vec{p}=\begin{pmatrix} p_{1}\\ p_{2}\\ p_{3} \end{pmatrix}=p_{1}\bar{i}+p_{2}\bar{j}+p_{3}\bar{j}\\\hline \textrm{Besar Vektor}&\overrightarrow{OP}=\sqrt{p_{1}^{2}+p_{2}^{2}+p_{3}^{2}}\\\hline \end{array}$

$\color{blue}\textrm{B. Operasi Vektor}$

$\color{blue}\textrm{1. Sifat-Sifat Aljabar Vektor}$

$\begin{array}{|l|l|l|}\hline 1.&\textrm{Komutatif penjumlahan}&\vec{a}+\vec{b}=\vec{b}+\vec{a}\\\hline 2.&\textrm{Asosiatif penjumlahan}&\left ( \vec{a}+\vec{b} \right )+\vec{c}=\vec{a}+\left ( \vec{b}+\vec{c} \right )\\\hline 3.&\textrm{Elemen Identitas}&\vec{a}+\vec{0}=\vec{0}+\vec{a}=\vec{a}\\\hline 4.&\textrm{Invers Penjumlahan}&\vec{a}+\left ( -\vec{a} \right )=\left ( -\vec{a} \right )+\vec{a}=\vec{0}\\\hline 5.&\textrm{Perkalian dengan skalar}&k\left ( l\vec{a} \right )=\left ( kl \right )\vec{a}\\ &&k\left ( \vec{a}+ \vec{b} \right )=k\vec{a}+k\vec{b}\\ &&k\left ( \vec{a}- \vec{b} \right )=k\vec{a}-k\vec{b}\\\hline 6.&\begin{aligned}&\textrm{Jika A, B, dan C segaris }\\ &\color{blue}\textrm{(Kolinear)} \end{aligned}&\begin{cases} \overrightarrow{AB}=k\overrightarrow{BC} \\ \overrightarrow{AC}=k\overrightarrow{AB} \\ dll \end{cases}\\\hline \end{array}$.

$\begin{array}{|c|c|}\hline \color{blue}\textrm{Vektor}&\color{blue}\textrm{Contoh}\\\hline \vec{z}=a\vec{i}+b\vec{j}+c\vec{k}&\begin{aligned}&\textrm{diketahui}\: \: \vec{p}=\vec{i}-2\vec{j}+2\vec{k}\\ &\textrm{maka pangjang vektor}\: \: \vec{p}\: \: \textrm{adalah}\\ &\left | \vec{p} \right |=\sqrt{1^{2}+(-2)^{2}+2^{2}}\\ &\quad\: \: =\sqrt{1+4+4}=\sqrt{9}=3 \end{aligned}\\\hline &\begin{aligned}&\textrm{Vektor satuan dari}\: \: \vec{p}\: \: \textrm{adalah}\\ &\vec{e}_{\vec{p}}=\frac{\vec{p}}{\left | \vec{p} \right |}=\displaystyle \frac{\begin{pmatrix} 1\\ -2\\ 2 \end{pmatrix}}{3}\\ &=\displaystyle \frac{1}{3}\begin{pmatrix} 1\\ -2\\ 2 \end{pmatrix}=\displaystyle \begin{pmatrix} \frac{1}{3}\\ -\frac{2}{3}\\ \frac{2}{3} \end{pmatrix} \end{aligned}\\\hline \end{array}$.

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

$\begin{array}{ll}\\ 1.&\textrm{Diketahui vektor-vektor}\: \overrightarrow{a}=\begin{pmatrix} 2\\ 1\\ -4 \end{pmatrix}\\ &\overrightarrow{b}=\begin{pmatrix} -3\\ -5\\ 2 \end{pmatrix},\: \: \textrm{dan}\: \: \overrightarrow{c}=\begin{pmatrix} 7\\ 0\\ 4 \end{pmatrix},\\ & \textrm{tentukanlah hasil dari}\\ &\textrm{a}.\quad \overrightarrow{a}+\overrightarrow{b}\\ &\textrm{b}.\quad 6\overrightarrow{a}+2\overrightarrow{b}\\ &\textrm{c}.\quad 2\overrightarrow{a}-\overrightarrow{b}+\overrightarrow{c}\\ &\textrm{d}.\quad \displaystyle \frac{1}{2}\overrightarrow{c}-\overrightarrow{a}+\displaystyle \frac{3}{4}\overrightarrow{b}\\\\ &\color{blue}\textrm{Jawab}:\\ &\begin{aligned}\textrm{a}\quad&\overrightarrow{a}+\overrightarrow{b}=\begin{pmatrix} 2\\ 1\\ -4 \end{pmatrix}+\begin{pmatrix} -3\\ -5\\ 2 \end{pmatrix}\\ &=\begin{pmatrix} 2+(-3)\\ 1+(-5)\\ (-4)+2 \end{pmatrix}\\ &=\begin{pmatrix} 2-3\\ 1-5\\ -4+2 \end{pmatrix}=\color{red}\begin{pmatrix} -1\\ -4\\ -2 \end{pmatrix}\\ \textrm{b}.\quad&6\overrightarrow{a}+2\overrightarrow{b}=6\begin{pmatrix} 2\\ 1\\ -4 \end{pmatrix}+2\begin{pmatrix} -3\\ -5\\ 2 \end{pmatrix}\\ &=\begin{pmatrix} 18-6\\ 6-10\\ -24+4 \end{pmatrix}=\color{red}\begin{pmatrix} 12\\ -4\\ -20 \end{pmatrix}\\ \textrm{c}.\quad&2\overrightarrow{a}-\overrightarrow{b}+\overrightarrow{c}\\ &2\begin{pmatrix} 2\\ 1\\ -4 \end{pmatrix}-\begin{pmatrix} -3\\ -5\\ 2 \end{pmatrix}+\begin{pmatrix} 7\\ 0\\ 4 \end{pmatrix}\\ &=\begin{pmatrix} 4+3+7\\ 2+5+0\\ -8-2+4 \end{pmatrix}=\color{red}\begin{pmatrix} 17\\ 7\\ -6 \end{pmatrix}\\ \textrm{d}.\quad&\displaystyle \frac{1}{2}\overrightarrow{c}-\overrightarrow{a}+\displaystyle \frac{3}{4}\overrightarrow{b}=\cdots \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 2.&\textrm{Diketahui vektor-vektor}\: \overrightarrow{a}=\begin{pmatrix} 2\\ 1\\ -4 \end{pmatrix}\\ &\textrm{tentukanlah}\: \: \left | \overrightarrow{a} \right |\\\\ &\color{blue}\textrm{Jawab}:\\ &\begin{aligned}\left | \overrightarrow{a} \right |&=\sqrt{2^{2}+1^{2}+(-4)^{2}}\\ &=\sqrt{4+1+16}\\ &=\color{red}\sqrt{21} \end{aligned} \end{array}$

$\color{blue}\textrm{2. Perkalian Skalar Dua Vektor}$

Konsep perkalian skalar dua buah vektor di ruang sama persis dengan konsep di bidang, yaitu:

$\color{red}\overrightarrow{a}\cdot \overrightarrow{b}=\left | \overrightarrow{a} \right |\left | \overrightarrow{b} \right |\cos \theta$.

Misalkan diketahui

$\color{red}\begin{aligned}&\overrightarrow{a}=\begin{pmatrix} a_{1}\\ a_{2}\\ a_{3} \end{pmatrix}, \overrightarrow{b}=\begin{pmatrix} b_{1}\\ b_{2}\\ b_{3} \end{pmatrix},\: \: \color{black}\textrm{maka}\\ & \overrightarrow{a} \cdot \overrightarrow{b} =\color{black}\begin{pmatrix} a_{1}\\ a_{2}\\ a_{3} \end{pmatrix}\cdot \begin{pmatrix} b_{1}\\ b_{2}\\ b_{3} \end{pmatrix}\\ &\qquad\quad =\color{black}a_{1}b_{1}+a_{2}b_{2}+a_{3}b_{3} \end{aligned}$

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

$\begin{array}{ll}\\ 1.&\textrm{Diketahui vektor-vektor}\: \overrightarrow{a}=\begin{pmatrix} 2\\ 1\\ -4 \end{pmatrix}\\ &\overrightarrow{b}=\begin{pmatrix} -3\\ -5\\ 2 \end{pmatrix},\: \: \textrm{dan}\: \: \overrightarrow{c}=\begin{pmatrix} 7\\ 0\\ 4 \end{pmatrix},\\ & \textrm{tentukanlah hasil dari}\\ &\textrm{a}.\quad \overrightarrow{a}\cdot \overrightarrow{b}\\ &\textrm{b}.\quad \overrightarrow{a}\cdot \overrightarrow{c}\\ &\textrm{c}.\quad \overrightarrow{b}\cdot \overrightarrow{c}\\\\ &\color{blue}\textrm{Jawab}:\\ &\begin{aligned}\textrm{a}\quad&\overrightarrow{a}\cdot \overrightarrow{b}=\begin{pmatrix} 2\\ 1\\ -4 \end{pmatrix}\cdot \begin{pmatrix} -3\\ -5\\ 2 \end{pmatrix}\\ &=(2)(-3)+(1)(-5)+(-4)(2)\\ &=-6-5-8=\color{red}-19\\ \textrm{b}\quad&\overrightarrow{a}\cdot \overrightarrow{c}=\begin{pmatrix} 2\\ 1\\ -4 \end{pmatrix}\cdot \begin{pmatrix} 7\\ 0\\ 4 \end{pmatrix}\\ &=(2)(7)+(1)(0)+(-4)(4)\\ &=14+0-16=\color{red}-2\\ \textrm{c}\quad&\overrightarrow{b}\cdot \overrightarrow{c}=\begin{pmatrix} -3\\ -5\\ 2 \end{pmatrix}\cdot \begin{pmatrix} 7\\ 0\\ 4 \end{pmatrix}\\ &=(-3)(7)+(-5)(0)+(2)(4)\\ &=-21+0+8=\color{red}-13 \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 2.&\textrm{Tentukanlah nilai}\: \: t\: \: \textrm{jika}\\ & \overrightarrow{p}=3\bar{i}+t\bar{j}+\bar{k}\: \: \textrm{dan}\: \: \overrightarrow{p}\cdot \overrightarrow{p}=13\\\\ &\color{blue}\textrm{Jawab}:\\ &\begin{aligned}&\overrightarrow{p}\cdot \overrightarrow{p}=13\\ &\overrightarrow{p}\cdot \overrightarrow{p}=\left | \overrightarrow{p} \right |\left | \overrightarrow{p} \right |\cos 0^{\circ}=13,\\ &\qquad\qquad \color{blue}\textrm{ingat bahwa}\: \: \angle \left ( \overrightarrow{p},\overrightarrow{p} \right )=0^{\circ}\\ &\qquad\qquad \color{blue}\textrm{dan nilai}\: \: \cos 0^{\circ}=1,\\ & \color{black}\textrm{maka}\\ &\overrightarrow{p}\cdot \overrightarrow{p}=\left | \overrightarrow{p} \right |^{2}.1=13\Leftrightarrow \left | \overrightarrow{p} \right |^{2}=13\\ &\Leftrightarrow \left (\sqrt{3^{2}+t^{2}+1^{2}} \right )^{2}=13\\ &\Leftrightarrow 3^{2}+t^{2}+1^{2}=13\\ &\Leftrightarrow 9+t^{2}+1=13\\ &\Leftrightarrow t^{2}=13-9-1=10\\ &\Leftrightarrow t^{2}=3\\ &\Leftrightarrow t=\color{red}\pm \sqrt{3} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 3.&\textrm{Diketahui}\: \: \overrightarrow{p}=\begin{pmatrix} -2\\ 1\\ 3 \end{pmatrix}\: \: \textrm{dan}\: \: \overrightarrow{q}=\begin{pmatrix} 4\\ -1\\ t \end{pmatrix}\\ &\textrm{Jika}\: \: \overrightarrow{p}\: \: \textrm{tegak lurus}\: \: \overrightarrow{q}\: \: \textrm{maka}\\ &\textrm{tentukanlah nilai}\: \: t\\\\ &\color{blue}\textrm{Jawab}:\\ &\begin{aligned}&\textrm{Diketahui bahwa}\\ &\overrightarrow{p}=\begin{pmatrix} -2\\ 1\\ 3 \end{pmatrix}\: \: \textrm{dan}\: \: \overrightarrow{q}=\begin{pmatrix} 4\\ -1\\ t \end{pmatrix}\\ &\textrm{dengan}\: \: \overrightarrow{p}\: \: \textrm{dan}\: \: \overrightarrow{q}\: \: \textrm{tegak lurus}\\ &\textrm{artinya}\: \: \color{blue}\angle \left ( \overrightarrow{p},\overrightarrow{q} \right )=90^{\circ}.\: \color{black}\textrm{Sehingga}\\ &\textrm{nilai}\: \: \color{blue}\cos 90^{\circ}=0\\ &\textrm{maka}\\ &\overrightarrow{p}\cdot \overrightarrow{q}=\left | \overrightarrow{p} \right |\left | \overrightarrow{q} \right |\cos \theta \\ &\overrightarrow{p}\cdot \overrightarrow{q}=\left | \overrightarrow{p} \right |\left | \overrightarrow{q} \right |\cdot 0=0\\ &\Leftrightarrow \: \begin{pmatrix} -2\\ 1\\ 3 \end{pmatrix}\cdot \begin{pmatrix} 4\\ -1\\ t \end{pmatrix}=0\\ &\Leftrightarrow \: (-2)(4)+(1)(-1)+(3)(t)=0\\ &\Leftrightarrow \: -8-1+3t=0\\ &\Leftrightarrow \: 3t=9\\ &\Leftrightarrow \: t=\color{red}3 \end{aligned} \end{array}$.

DAFTAR PUSTAKA

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