PAS MATEMATIKA BLOG: Operasi Vektor Berdimensi Tiga

$\color{blue}\textrm{F. Operasi Vektor Dalam Ruang}$

Operasi vektor pada dimensi tiga kurang lebih sama dengan operasi pada vektor berdimensi dua.

$\color{blue}\textrm{F. 1. Penjumlahan dan Pengurangan}$.

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

$\begin{array}{ll}\\ 1.&\textrm{Jika diketahui}\: \: \bar{a}=\begin{pmatrix} 1\\ 3\\ 7 \end{pmatrix}\: \: \textrm{dan}\: \: \bar{b}=\begin{pmatrix} 8\\ -2\\ 0 \end{pmatrix}\\ &\textrm{Tentukanlah hasil dari}\\ &\textrm{a}.\quad \bar{a}+\bar{b}\\ &\textrm{b}.\quad \bar{a}-\bar{b}\\\\ &\textbf{Jawab}\\ &\begin{aligned}&\textrm{Diketahui bahwa}\\ &\bar{a}=\begin{pmatrix} 1\\ 3\\ 7 \end{pmatrix}\: \: \textrm{dan}\: \: \bar{b}=\begin{pmatrix} 8\\ -2\\ 0 \end{pmatrix},\\ &\textrm{maka}\\ &\bar{a}+\bar{b}=\begin{pmatrix} 1\\ 3\\ 7 \end{pmatrix}+\begin{pmatrix} 8\\ -2\\ 0 \end{pmatrix}\\ &\qquad =\begin{pmatrix} 1+8\\ 3+(-2)\\ 7+0 \end{pmatrix}=\begin{pmatrix} 9\\ 1\\ 7 \end{pmatrix}\\ &\textrm{Dan untuk}\: \: \bar{a}-\bar{b}\: \: \textrm{adalah}:\\ &\bar{a}-\bar{b}=\begin{pmatrix} 1\\ 3\\ 7 \end{pmatrix}-\begin{pmatrix} 8\\ -2\\ 0 \end{pmatrix}\\ &\qquad =\begin{pmatrix} 1-8\\ 3-(-2)\\ 7-0 \end{pmatrix}=\begin{pmatrix} -7\\ 5\\ 7 \end{pmatrix} \end{aligned} \end{array}$.

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

Misalkan suatu skalar $m$ dan suatu vektor $\bar{u}=a\bar{i}+b\bar{j}+c\bar{k}$, maka perkalian $m$ dengan vektor $\bar{u}$ tersebut adalah $\bar{u}=ma\bar{i}+mb\bar{j}+mc\bar{k}$.

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

$\begin{array}{ll} 2.&\textrm{Jika}\: \: \bar{a}=\begin{pmatrix} 2022\\ 2021\\ 2020 \end{pmatrix},\: \: \textrm{tentukanlah nilai}\\ &\textrm{dari}\: \: 2\bar{a}\: \: \: \textrm{dan}\: \: -3\bar{a}\\\\ &\textbf{Jawab}\\ &\begin{aligned}2\bar{a}&=2\begin{pmatrix} 2022\\ 2021\\ 2020 \end{pmatrix}=\begin{pmatrix} 4044\\ 4042\\ 4040 \end{pmatrix},\: \: \textrm{dan}\\ -3\bar{a}&=-3\begin{pmatrix} 2022\\ 2021\\ 2020 \end{pmatrix}=\begin{pmatrix} -6066\\ -6063\\ -6060 \end{pmatrix} \end{aligned} \end{array}$

$\color{blue}\textrm{F. 3. Perkalian Skalar Dua Vektor}$.

Hasil dari perkalian skalar dua vektor $\bar{a}$ dan $\bar{b}$ adalah : $\bar{a}\: \: \bullet\: \: \bar{b}$.

Dengan

$\bar{a}\: \: \bullet\: \: \bar{b}=\left | \bar{a} \right |\left | \bar{b} \right |\cos \theta$. sehingga

$\begin{aligned}&\textrm{Tanda dari hasil skalar ini adalah}\\ &\begin{array}{|l|l|l|}\hline \textbf{Besar sudut}\: \: \: \theta &\textbf{Tanda}&\textrm{Bentuk}\\\hline 0^{\circ}\leq \theta < 90^{\circ}&\textrm{Positif}&\color{red}\textrm{Lancip}\\\hline \theta =90^{\circ}&\textrm{Nol}&\textrm{Siku-siku}\\\hline 90^{\circ}< \theta \leq 180^{\circ}&\textrm{Negatif}&\color{blue}\textrm{Tumpul}\\\hline \end{array}\\ &\textrm{Untuk}\: \: \theta \: \: \textrm{berupa sudut istimewa}:\\ &\begin{array}{|c|c|c|c|c|c|c|}\hline \theta &0^{\circ}&30^{\circ}&45^{\circ}&60^{\circ}&90^{\circ}&180^{\circ}\\\hline \cos \theta &1&\displaystyle \frac{1}{2}\sqrt{3}&\displaystyle \frac{1}{2}\sqrt{2}&\displaystyle \frac{1}{2}&0&-1\\\hline \end{array} \end{aligned}$

Adapun secara rumus untuk menentukan besar sudutnya adalah:

$\cos \theta =\displaystyle \frac{\bar{a}\: \: \bullet\: \: \bar{b} }{\left | \bar{a} \right |\left | \bar{b} \right |}$.

Sebagai ilustrasinya perhatikanlah gambar berikut

Selain hasil di atas ada cara lain menyelesaikan perkalian skalar dua vektor, yaitu:

$\begin{aligned}\textrm{Jika}\: & \textrm{diketahui sebagai misal}\\ \bar{u}&=a\bar{i}+b\bar{j}+c\bar{k}\: \: \: \color{red}\textrm{dan}\\ \bar{v}&=p\bar{i}+q\bar{j}+r\bar{k}\\ \textrm{mak}&\textrm{a}\\ \textbf{Per}&\textbf{kalian skalar dua vektor adalah}:\\ \bar{u}\: \bullet \: &\bar{v}=\left ( a\bar{i}+b\bar{j}+c\bar{k} \right )\left ( p\bar{i}+q\bar{j}+r\bar{k} \right )\\ &\: \: =ap.\bar{i}\: \bullet \bar{i}+aq.\bar{i}\: \bullet \: \bar{j}+ar.\bar{i}\: \bullet \: \bar{k}\\ &\: \: \: \: \: \: \: +bp.\bar{j}\: \bullet \: \bar{i}+bq.\bar{j}\: \bullet \: \bar{j}+br.\bar{j}\: \bullet \: \bar{k}\\ &\: \: \: \: \: \: \: +cp.\bar{k}\: \bullet \: \bar{i}+cq.\bar{k}\: \bullet \: \bar{j}+cr.\bar{k}\: \bullet \: \bar{k}\\ &\: \: =ap+0+0+0+bq+0+0+0+cr\\ &\: \: =\color{red}a p+b q+c r \end{aligned}$

$\begin{aligned}&\textrm{Sebagai penjelasannya adalah}:\\ &\color{red}\triangleright \quad \color{black}\bar{i}\: \: \bullet \: \: \bar{i}=\left | \bar{i} \right |\left | \bar{i} \right |\cos 0^{\circ}=1.1.1=1\\ &\color{red}\triangleright \quad \color{black}\bar{i}\: \: \bullet \: \: \bar{j}=\left | \bar{i} \right |\left | \bar{j} \right |\cos 90^{\circ}=1.1.0=0\\ &\color{red}\triangleright \quad \color{black}\bar{i}\: \: \bullet \: \: \bar{k}=\left | \bar{i} \right |\left | \bar{k} \right |\cos 90^{\circ}=1.1.0=0\\ &\color{red}\triangleright \quad \color{black}\bar{j}\: \: \bullet \: \: \bar{i}=\bar{i}\: \: \bullet \: \: \bar{j}=0\\ &\color{red}\triangleright \quad \color{black}\bar{j}\: \: \bullet \: \: \bar{j}=\left | \bar{j} \right |\left | \bar{j} \right |\cos 0^{\circ}=1.1.1=1\\ &\color{red}\triangleright \quad \color{black}\bar{j}\: \: \bullet \: \: \bar{k}=\left | \bar{j} \right |\left | \bar{k} \right |\cos 90^{\circ}=1.1.0=0\\ &\color{red}\triangleright \quad \color{black}\bar{k}\: \: \bullet \: \: \bar{i}=\bar{i}\: \: \bullet \: \: \bar{k}=0\\ &\color{red}\triangleright \quad \color{black}\bar{k}\: \: \bullet \: \: \bar{j}=\bar{j}\: \: \bullet \: \: \bar{k}=0\\ &\color{red}\triangleright \quad \color{black}\bar{k}\: \: \bullet \: \: \bar{k}=\left | \bar{k} \right |\left | \bar{k} \right |\cos 90^{\circ}=1.1.1=1 \end{aligned}$

$\begin{aligned}&\textrm{Atau jika ditabelkan nilainya}\\ &\begin{array}{|c|c|c|c|}\hline \bar{u}\: \bullet \: \bar{v}&p\bar{i}&q\bar{j}&r\bar{k}\\\hline a\bar{k}&ap&0&0\\ b\bar{j}&0&bq&0\\ c\bar{k}&0&0&cr\\\hline \end{array} \end{aligned}$

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

$\begin{array}{ll} 3.&\textrm{Jika}\: \: \vec{a}=\begin{pmatrix} 1\\ 2\\ 4 \end{pmatrix},\: \: \textrm{dan}\: \: \vec{b}=\begin{pmatrix} 5\\ 4\\ 0 \end{pmatrix}\\\ & \textrm{tentukanlah nilai}\: \: \textrm{dari}\: \: \vec{a}\bullet \vec{b}\\\\ &\textbf{Jawab}\\ &\begin{aligned}\vec{a}\bullet \vec{b}&=1.5+2.4+4.0=5+8+0=13 \end{aligned} \end{array}$

$\begin{array}{ll} 4.&\textrm{Jika diketahui}\: \: \vec{a}=\vec{i}-2\vec{j}+3\vec{k},\\ & \textrm{dan}\: \: \vec{b}=3\vec{i}-4\vec{j}+m\vec{k}\: \: \textrm{serta}\\\ & \textrm{nilai}\: \: \vec{a}\bullet \vec{b}=-4,\: \: \textrm{maka tentukan}\\ &\textrm{nilai}\: \: m\\\\ &\textbf{Jawab}\\ &\textrm{Diketahui bahwa}\\ &\color{red}\triangleright \quad \color{black}\vec{a}=\vec{i}-2\vec{j}+3\vec{k}=\begin{pmatrix} 1\\ -2\\ 3 \end{pmatrix},\: \: \textrm{dan}\\ &\color{red}\triangleright \quad \color{black}\vec{b}=3\vec{i}-4\vec{j}+m\vec{k}=\begin{pmatrix} 3\\ -4\\ m \end{pmatrix}\\ &\begin{aligned}\vec{a}\bullet \vec{b}&=1.3+3.(-4)+3.m\\ -4&=3+8+3m\\ -3m&=11+4\\ m&=-\displaystyle \frac{15}{3}\\ &=\color{blue}-5 \end{aligned} \end{array}$

$\begin{array}{ll} 5.&\textrm{Diketahui}\: \: \left |\vec{a} \right |=10,\: \left | \vec{b} \right |=6.\\ & \textrm{Jika}\: \: \vec{a}\: \: \textrm{dan}\: \: \vec{b}\: \: \textrm{membentuk sudut}\\ &60^{\circ}.\: \textrm{Tentukanlah nilai}\: \: \vec{a}\bullet \vec{b}\\\\ &\textbf{Jawab}\\ &\begin{aligned}\vec{a}\bullet \vec{b}&=\left | \vec{a} \right |\left | \vec{b} \right |\cos \theta \\ &=10.6.\cos 60^{\circ}\\ &=60.\left ( \displaystyle \frac{1}{2} \right )\\ &=\color{blue}30\\ \textrm{Jadi}&\: \textrm{hasil kali skalarnya adalah 30} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 6.&\textrm{Diketahui}\: \: \overrightarrow{a}=\begin{pmatrix} -2\\ 1\\ 3 \end{pmatrix}\: \: \textrm{dan}\: \: \overrightarrow{b}=\begin{pmatrix} 4\\ -1\\ t \end{pmatrix},\\ & \textrm{jika}\: \: \overrightarrow{p}\: \: \textrm{tegak lurus}\: \: \overrightarrow{q},\: \: \textrm{maka tentukanlah}\\ &\textrm{nilai}\: \: t\: \: \textrm{adalah}\\\\ &\textbf{Jawab}:\\ &\begin{aligned}\textrm{Karena}&\: \textrm{kedua vektor tersebut saling }\\ \textrm{tegak l}& \textrm{urus maka}\\ \overrightarrow{a}.\overrightarrow{b}&=0\\ \begin{pmatrix} -2\\ 1\\ 3 \end{pmatrix}&\begin{pmatrix} 4\\ -1\\ t \end{pmatrix}=0\\ (-2).4&+1.(-1)+3.t=0\\ -8-1&+3t=0\\ 3t&=9\\ t&=\color{red}3 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 7.&\textrm{Tentukanlah nilai}\: \: \overrightarrow{a}.\overrightarrow{b}\: \: \textrm{jika}\\ &\textrm{a}.\quad \left | \overrightarrow{a} \right |=4,\: \left | \overrightarrow{b} \right |=6,\: \: \angle \left ( \overrightarrow{a},\overrightarrow{b} \right )=60^{\circ}\\ &\textrm{b}.\quad \overrightarrow{a}=2\vec{i}+\vec{j}-5\vec{k}\: \: \textrm{dan}\: \: \overrightarrow{b}=2\vec{i}-3\vec{k}\\ &\textrm{c}.\quad \overrightarrow{a}=\begin{pmatrix} 0\\ -1\\ 3 \end{pmatrix}\: \: \textrm{dan}\: \: \overrightarrow{b}=\begin{pmatrix} 4\\ -2\\ 1 \end{pmatrix}\\\\ &\textbf{Jawab}:\\ &\begin{aligned}\textrm{a}.\quad \overrightarrow{a}.\overrightarrow{b}&=\left | \overrightarrow{a} \right |\left | \overrightarrow{b} \right |\cos \angle \left ( \overrightarrow{a},\overrightarrow{b} \right )\\ &=4.6.\cos 60^{\circ}\\ &=24.\left ( \displaystyle \frac{1}{2} \right )\\ &=12 \end{aligned}\\ &\textrm{b}.\quad \overrightarrow{a}.\overrightarrow{b}=2.2+1.0+(-5).(-3)=4+15=19\\ &\textrm{c}.\quad \overrightarrow{a}.\overrightarrow{b}=0.4+(-1).(-2)+3.1=0+2+3=5 \end{array}$

$\begin{array}{ll} 8.&\textrm{Diketahui}\: \: \left |\vec{a} \right |=10,\: \left | \vec{b} \right |=3\\ & \textrm{dan}\: \: \vec{a}\bullet \vec{b}=15\sqrt{3}\: .\: \textrm{Tentukan sudut}\\ &\textrm{yang dibentuk oleh}\: \: \vec{a}\: \: \textrm{dan}\: \: \vec{b}\\\\ &\textbf{Jawab}\\ &\textrm{Dari bentuk}\\ &\begin{aligned}\vec{a}\bullet \vec{b}&=\left | \vec{a} \right |\left | \vec{b} \right |\cos \theta \\ \textrm{dipe}&\textrm{roleh bentuk}\\ \cos \theta &=\displaystyle \frac{\vec{a}\bullet \vec{b}}{\left | \vec{a} \right |\left | \vec{b} \right |}\\ \cos \theta &=\displaystyle \frac{15\sqrt{3}}{10.3}=\frac{15}{30}\sqrt{3}=\frac{1}{2}\sqrt{3}\\ \cos \theta&=\cos 30^{\circ}\\ \theta &=\color{red}30^{\circ}\\ \textrm{Jadi}&\: \textrm{sudut antara keduanya adalah}\: \: \color{red}30^{\circ} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 9.&\textrm{Tentukanlah besar sudut antara vektor}\\ &\overrightarrow{a}=\begin{pmatrix} -1\\ 1\\ 0\end{pmatrix}\: \: \textrm{dan}\: \: \overrightarrow{b}=\begin{pmatrix} 1\\ -2\\ 2 \end{pmatrix}\\\\ &\textbf{Jawab}:\\ &\begin{aligned}\cos \theta &=\displaystyle \frac{\overrightarrow{a}.\overrightarrow{b}}{\left | \overrightarrow{a} \right |\left | \overrightarrow{b} \right |}\\ &=\displaystyle \frac{\begin{pmatrix} -1\\ 1\\ 0 \end{pmatrix}\begin{pmatrix} 1\\ -2\\ 2 \end{pmatrix}}{\sqrt{(-1)^{2}+1^{2}}\sqrt{1^{2}+(-2)^{2}+2^{2}}}\\ &=\displaystyle \frac{-1-2+0}{\sqrt{2}\sqrt{9}}\\ &=-\displaystyle \frac{1}{\sqrt{2}}=-\displaystyle \frac{1}{2}\sqrt{2}\\ &=-\cos 45^{\circ}\\ &=\cos \left ( 180^{\circ}-45^{\circ} \right )\\ \cos \theta &=\cos 135^{\circ}\\ \therefore \: \theta &=\color{red}135^{\circ} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 10.&\textrm{Diketahui bahwa}\: \: \left |\overrightarrow{a} \right |=\sqrt{6} ,\: \: (\overrightarrow{a}-\overrightarrow{b})(\overrightarrow{a}+\overrightarrow{b})=0\\ & \textrm{dan}\: \: \overrightarrow{a}(\overrightarrow{a}-\overrightarrow{b})=3.\: \textrm{Tentukanlah besar}\\ &\textrm{sudut antara}\: \: \overrightarrow{a}\: \: \textrm{dan}\: \: \overrightarrow{b}\\\\ &\textbf{Jawab}:\\ &\begin{aligned}\textrm{Perhatikan}&\: \textrm{bahwa}\\ (\overrightarrow{a}-\overrightarrow{b})(\overrightarrow{a}+\overrightarrow{b})&=0\\ \left | \overrightarrow{a} \right |^{2}-\left | \overrightarrow{b} \right |^{2}&=0\\ \left | \overrightarrow{a} \right |^{2}&=\left | \overrightarrow{b} \right |^{2}\quad \Rightarrow \quad \left | \overrightarrow{a} \right |=\overrightarrow{b}=\sqrt{6}\\ \textrm{dan}\quad \overrightarrow{a}(\overrightarrow{a}-\overrightarrow{b})&=3\\ \left | \overrightarrow{a} \right |^{2}-\overrightarrow{a}\overrightarrow{b}&=3\\ 6-\overrightarrow{a}\overrightarrow{b}&=3\\ -\overrightarrow{a}\overrightarrow{b}&=3-6=-3\\ \overrightarrow{a}\overrightarrow{b}&=3\\ \left | \overrightarrow{a} \right |\left | \overrightarrow{b} \right |\cos \theta &=3\\ \cos \theta &=\displaystyle \frac{3}{\sqrt{6}\sqrt{6}}=\frac{3}{6}=\frac{1}{2}\\ \cos \theta &=\cos 60^{\circ}\\ \therefore \: \: \theta &=\color{red}60^{\circ} \end{aligned} \end{array}$

$\color{blue}\textrm{Berikut dua contoh untuk sudut tidak istimewa}$.

$\begin{array}{ll} 11.&\textrm{Diketahui}\: \: \vec{a} =\vec{i}+2\vec{j}+2\vec{k},\: \: \textrm{dan}\\ & \vec{b}=3\vec{i}+4\vec{j}\: .\: \textrm{Tentukan sudut}\\ &\textrm{yang dibentuk oleh}\: \: \vec{a}\: \: \textrm{dan}\: \: \vec{b}\\\\ &\textbf{Jawab}\\ &\begin{aligned}\textrm{Dik}&\textrm{etahui bahwa}\\ \triangleright \quad &\vec{a} =\vec{i}+2\vec{j}+2\vec{k}=\begin{pmatrix} 1\\ 2\\ 2 \end{pmatrix}\: \: \textrm{dan}\\ &\left | \vec{a} \right |=\sqrt{1^{2}+2^{2}+2^{2}}=\sqrt{9}=3\\ \triangleright \quad &\vec{b}=3\vec{i}+4\vec{j}=\begin{pmatrix} 3\\ 4\\ 0 \end{pmatrix}\: \: \textrm{dan}\\ &\left | \vec{b} \right |=\sqrt{3^{2}+4^{2}+0^{2}}=\sqrt{25}=5\\ \end{aligned}\\ &\textrm{Selanjutnya}\\ &\begin{aligned} \cos \theta &=\displaystyle \frac{\vec{a}\bullet \vec{b}}{\left | \vec{a} \right |\left | \vec{b} \right |}\\ \cos \theta &=\displaystyle \frac{\begin{pmatrix} 1\\ 2\\ 2 \end{pmatrix}\begin{pmatrix} 3\\ 4\\ 0 \end{pmatrix}}{3.5}=\frac{3+8+0}{15}=\frac{11}{15}\\ \cos \theta&=0,733\\ \theta &=\color{red}\arccos \left ( \displaystyle 0.733 \right )\\ &\quad \textrm{gunakan alat bantu tabel trigonometri}\\ &\quad \textrm{atau kalkulator scientific}\\ &=42,9^{\circ}\\ \textrm{Jadi}&\: \textrm{sudut antara keduanya adalah}\: \: 42,9^{\circ} \end{aligned} \end{array}$

$\begin{array}{ll} 12.&\textrm{Diketahui}\: \: \vec{p} =(1,2,2),\: \: \textrm{dan}\\ & \vec{q}=(3,-2,6)\: .\: \textrm{Tentukan sudut}\\ &\textrm{yang dibentuk oleh}\: \: \vec{p}\: \: \textrm{dan}\: \: \vec{q}\\\\ &\textbf{Jawab}\\ &\begin{aligned}\textrm{Dik}&\textrm{etahui bahwa}\\ \triangleright \quad &\vec{p} =(1,2,2)=\begin{pmatrix} 1\\ 2\\ 2 \end{pmatrix}\: \: \textrm{dan}\\ &\left | \vec{p} \right |=\sqrt{1^{2}+2^{2}+2^{2}}=\sqrt{9}=3\\ \triangleright \quad &\vec{q}=(3,-2,6)=\begin{pmatrix} 3\\ -2\\ 6 \end{pmatrix}\: \: \textrm{dan}\\ &\left | \vec{q} \right |=\sqrt{3^{2}+(-2)^{2}+6^{2}}=\sqrt{49}=7\\ \end{aligned}\\ &\textrm{Selanjutnya}\\ &\begin{aligned} \cos \theta &=\displaystyle \frac{\vec{p}\bullet \vec{q}}{\left | \vec{p} \right |\left | \vec{q} \right |}\\ \cos \theta &=\displaystyle \frac{\begin{pmatrix} 1\\ 2\\ 2 \end{pmatrix}\begin{pmatrix} 3\\ -2\\ 6 \end{pmatrix}}{3.7}=\frac{3-4+12}{21}=\frac{11}{21}\\ \cos \theta&=0,524\\ \theta &=\color{red}\arccos \left ( \displaystyle 0.524 \right )\\ &\quad \textrm{gunakan alat bantu tabel trigonometri}\\ &\quad \textrm{atau kalkulator scientific}\\ &=58,4^{\circ}\\ \textrm{Jadi}&\: \textrm{sudut antara keduanya adalah}\: \: 58,4^{\circ} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 13.&\textrm{Diketahui vektor}\: \: \overrightarrow{a}\: \: \textrm{dan}\: \: \overrightarrow{b}\: \: \textrm{memiliki }\\ &\textrm{panjang masing-masing adalah 2 dan 3}\\ &\textrm{serta}\: \: \angle \left ( \overrightarrow{a},\overrightarrow{b}\right )=60^{\circ}.\: \textrm{Carilah nilai}\\ &\textrm{a}.\quad \left | \overrightarrow{a}+\overrightarrow{b} \right |\\\\ &\textrm{b}.\quad \left | \overrightarrow{a}-\overrightarrow{b} \right |\\ &\textrm{b}\quad \textrm{besar sudut antara}\\ &\qquad \left ( \overrightarrow{a}+\overrightarrow{b} \right )\: \: \textrm{dan}\: \: \left ( \overrightarrow{a}-\overrightarrow{b} \right )\\\\ &\textbf{Jawab}:\\ &\begin{aligned}\textrm{a}.\quad&\left | \overrightarrow{a}+\overrightarrow{b} \right |^{2}\\ &=\left ( \overrightarrow{a}+\overrightarrow{b} \right )\left ( \overrightarrow{a}+\overrightarrow{b} \right )\\ &=\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\\ &\textrm{Jadi, nilainya adalah}\: \: \left | \overrightarrow{a}+\overrightarrow{b} \right |=\color{red}\sqrt{19} \end{aligned}\\ &\begin{aligned}\textrm{b}.\quad&\left | \overrightarrow{a}-\overrightarrow{b} \right |^{2}\\ &=\left ( \overrightarrow{a}-\overrightarrow{b} \right )\left ( \overrightarrow{a}-\overrightarrow{b} \right )\\ &=\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\\ &\textrm{Jadi, nilainya adalah}\: \: \left | \overrightarrow{a}-\overrightarrow{b} \right |=\color{red}\sqrt{7} \end{aligned}\\ &\begin{aligned}\textrm{c}.\quad \textrm{Untuk menentukan nilai}&\\ \cos \angle \left ( \overrightarrow{a}+\overrightarrow{b},\overrightarrow{a}-\overrightarrow{b} \right )&=\displaystyle \frac{\left (\overrightarrow{a}+\overrightarrow{b} \right ).\left (\overrightarrow{a}-\overrightarrow{b} \right )}{\left | \overrightarrow{a}+\overrightarrow{b} \right |.\left | \overrightarrow{a}-\overrightarrow{b} \right |}\\ &=\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}}\\ \angle \left ( \overrightarrow{a}+\overrightarrow{b},\overrightarrow{a}-\overrightarrow{b} \right )&=\color{red}\arccos \left ( -\frac{5}{\sqrt{133}} \right ) \end{aligned} \end{array}$

$\color{blue}\textrm{Berikut contoh untuk bentuk sudutnya}$.

$\begin{array}{ll} 14.&\textrm{Diketahui}\: \: \vec{p} =(x,3,2),\: \: \textrm{dan}\\ & \vec{q}=(2,-6,3)\: .\: \textrm{Tentukan nilai}\: \: x\\ &\textrm{agar kedua vektor}\\ &\textrm{a}\quad \textrm{membentuk sudut lancip}\\ &\textrm{b}\quad \textrm{membentuk sudut siku-siku}\\ &\textrm{c}\quad \textrm{membentuk sudut tumpul}\\ &\textrm{d}\quad \textrm{sama panjang}\\\\ &\textbf{Jawab}\\ &\begin{aligned}\textrm{Dik}&\textrm{etahui bahwa}\\ \triangleright \quad &\vec{p} =(x,3,2)=\begin{pmatrix} x\\ 3\\ 2 \end{pmatrix}\: \: \textrm{dan}\\ \triangleright \quad &\vec{q}=(2,-6,3)=\begin{pmatrix} 2\\ -6\\ 3 \end{pmatrix}\\ \end{aligned}\\ &\textrm{Selanjutnya}\\ &\vec{p}\bullet \vec{q}=\begin{pmatrix} x\\ 3\\ 2 \end{pmatrix}\begin{pmatrix} 2\\ -6\\ 3 \end{pmatrix}\\ &\quad =2x-18+6=2x-12\\ &\textrm{Selanjutnya}\\ &\begin{aligned} \textrm{a}\quad&\textbf{Syarat lancip},\: \textrm{yaitu}:\: \vec{p}\bullet \vec{q}>0\\ &2x-12>0\Leftrightarrow 2x>12\Leftrightarrow x>6\\ \textrm{b}\quad&\textbf{Syarat siku-siku},\: \textrm{yaitu}:\: \vec{p}\bullet \vec{q}=0\\ &2x-12=0\Leftrightarrow 2x=12\Leftrightarrow x=6\\ \textrm{c}\quad&\textbf{Syarat tumpul},\: \textrm{yaitu}:\: \vec{p}\bullet \vec{q}<0\\ &2x-12<0\Leftrightarrow 2x<12\Leftrightarrow x<6\\ \textrm{d}\quad&\textbf{Syarat panjang kedua vektor sama}\\ & \textrm{yaitu}:\: \left |\vec{p} \right |= \left |\vec{q} \right |,\: \textrm{maka}\\ &\begin{aligned}&\sqrt{x^{2}+3^{2}+2^{2}}=\sqrt{2^{2}+(-6)^{2}+3^{2}}\\ &x^{2}+9+4=4+36+9\\ &x^{2}=36\\ &x=\pm \sqrt{36}=\pm 6\\ &\textrm{Jadi},\: \color{red}x=-6\: \: \color{black}\textrm{atau}\: \: \color{red}x=6 \end{aligned} \end{aligned} \end{array}$

DAFTAR PUSTAKA

Johanes, Kastolan, Sulasim. 2006. Kompetensi Matematika Program IPA 3A SMA Kelas XII Semester Pertama. Jakarta: YUDHISTIRA.
Kanginan, M., Nurdiansyah, H., Akhmad, G. 2016. Matematika untuk Siswa SMA/MA Kelas X Kelompok Peminatan Matematika dan Ilmu-Ilmu Alam. Bandung: YRAMA WIDYA.
Noormandiri, Sucipto, E. 2003. Buku Pelajaran Matematika SMU untuk Kelas 3 Program IPA. Jakarta: ERLANGGA.
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.

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Operasi Vektor Berdimensi Tiga

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