PAS MATEMATIKA BLOG: Februari 2022

Contoh 6 Soal dan Pembahasan Materi Vektor

$\begin{array}{ll}\\ 26.&\textrm{Jika}\: \: \vec{p}=\begin{pmatrix} -2\\ 4 \end{pmatrix}\: \: \textrm{dan}\: \: \vec{q}=\begin{pmatrix} 8\\ 4 \end{pmatrix},\: \: \textrm{maka}\\ &\textrm{sudut yang dibentuk vektor}\: \: \vec{p}\: \: \textrm{dan}\: \: \vec{q}\\ &\textrm{adalah}....\\ &\textrm{a}.\quad 0^{\circ}\\ &\textrm{b}.\quad 60^{\circ}\\ &\textrm{c}.\quad 45^{\circ}\\ &\textrm{d}.\quad 60^{\circ}\\ &\textrm{e}.\quad \color{red}90^{\circ}\\\\ &\textrm{Jawab}\\ & \begin{aligned}\vec{p}.\vec{q}&=\displaystyle \begin{vmatrix} \vec{p} \end{vmatrix}.\begin{vmatrix} \vec{q} \end{vmatrix}.\cos \angle \left (\vec{p},\, \vec{q} \right )\\ \cos \angle \left (\vec{p},\, \vec{q} \right )&=\displaystyle \frac{\vec{p}.\vec{q}}{\left | \vec{p} \right |.\left | \vec{q} \right |}\\ &=\displaystyle \frac{\begin{pmatrix} -2\\ 1 \end{pmatrix}.\begin{pmatrix} 8\\ 4 \end{pmatrix}}{\sqrt{(-2)^{2}+1^{2}}.\sqrt{8^{2}+4^{2}}}\\ &=\displaystyle \frac{-16+16}{\sqrt{20}.\sqrt{80}}\\ &=\displaystyle \frac{0}{40}\\ &=0\\ \cos \angle \left (\vec{p},\, \vec{q} \right )&=\cos 90^{\circ}\\ \angle \left (\vec{p},\, \vec{q} \right )&=\color{red}90^{\circ} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 27.&\textrm{Jika}\: \: \overline{OA}=\begin{pmatrix} 1\\ 2 \end{pmatrix},\: \overline{OB}=\begin{pmatrix} 4\\ 2 \end{pmatrix},\: \textrm{dan}\\ &\theta =\angle \left ( \overline{OA},\: \overline{OB} \right ),\: \textrm{maka}\: \: \tan \theta =....\\\\ &\textrm{a}.\quad \displaystyle \frac{3}{5}\\\\ &\textrm{b}.\quad \displaystyle \frac{9}{16}\\\\ &\textrm{c}.\quad \color{red}\displaystyle \frac{3}{4}\\\\ &\textrm{d}.\quad \displaystyle \frac{4}{3}\\\\ &\textrm{e}.\quad \displaystyle \frac{16}{9}\\\\ &\textrm{Jawab}\\ &\begin{array}{|c|c|}\hline \begin{aligned}\cos \theta &=\displaystyle \frac{\vec{a}.\vec{b}}{\left | \vec{a} \right |\left | \vec{b} \right |}\\ &=\displaystyle \frac{\begin{pmatrix} 1\\ 2 \end{pmatrix}.\begin{pmatrix} 4\\ 2 \end{pmatrix}}{\sqrt{1^{2}+2^{2}}\sqrt{4^{2}+2^{2}}}\\ &=\displaystyle \frac{4+4}{\sqrt{5}.\sqrt{20}}\\ &=\displaystyle \frac{8}{10} \end{aligned}&\begin{aligned}\sin \theta &=\sqrt{1-\cos ^{2}\theta }\\ &=\sqrt{1-\left ( \displaystyle \frac{8}{10} \right )^{2}}\\ &=\sqrt{\displaystyle \frac{36}{100}}\\ &=\displaystyle \frac{6}{10}\\ & \end{aligned}\\\hline \end{array} \\ &\textrm{Selanjutnya}\\ &\begin{aligned}\tan \theta &=\displaystyle \frac{\sin \theta }{\cos \theta }\\ &=\displaystyle \frac{\displaystyle \frac{6}{10}}{\displaystyle \frac{8}{10}}\\ &=\color{red}\displaystyle \frac{3}{4} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 28.&\textrm{Jika}\: \: \vec{a},\: \vec{b}\: \: \textrm{dan}\: \: \vec{c}\: \: \textrm{adalah vektor satuan dengan}\\ & \vec{a}+\vec{b}+\vec{c}=0.\: \textrm{Nilai dari}\: \: \vec{a}.\vec{b}+\vec{a}.\vec{c}+\vec{b}.\vec{c}\: \: \textrm{adalah}....\\ &\textrm{a}.\quad \displaystyle -3\\ &\textrm{b}.\quad \color{red}\displaystyle -\frac{3}{2}\\ &\textrm{c}.\quad \displaystyle 0\\ &\textrm{d}.\quad \displaystyle \frac{3}{2}\\ &\textrm{e}.\quad \displaystyle 3\\\\ &\textrm{Jawab}\\ & \begin{aligned}\textrm{Karena}&\left\{\begin{matrix} \vec{a},\vec{b},\vec{c}\: \: \textrm{adalah vektor satuan, dan}\\ \vec{a}+\vec{b}+\vec{c}=0\qquad\qquad\qquad\qquad . \end{matrix}\right.\\ \textrm{segitig}&\textrm{a ABC adalah segitiga sama sisi}\\ \vec{a}.\vec{b}=&\left | \vec{a} \right |\left | \vec{b} \right |\cos 120^{0}=1.1.\left ( -\frac{1}{2} \right )=-\frac{1}{2}\\ \vec{a}.\vec{c}=&\left | \vec{a} \right |\left | \vec{c} \right |\cos 120^{0}=1.1.\left ( -\frac{1}{2} \right )=-\frac{1}{2}\\ \vec{b}.\vec{c}=&\left | \vec{b} \right |\left | \vec{c} \right |\cos 120^{0}=1.1.\left ( -\frac{1}{2} \right )=-\frac{1}{2}\\ \textrm{Jadi, ni}&\textrm{lai dari}\: \: \vec{a}.\vec{b}+\vec{a}.\vec{c}+\vec{b}.\vec{c}\\ &=\left ( -\frac{1}{2} \right )+\left ( -\frac{1}{2} \right )+\left ( -\frac{1}{2} \right )=\color{red}-\frac{3}{2} \end{aligned} \end{array}$

$.\: \qquad \color{blue}\textrm{berikut ilustrasinya}$

$\begin{array}{ll}\\ 29.&\textrm{Jika}\: \: \angle \left ( \vec{a},\vec{b} \right )=60^{\circ},\: \: \left |\vec{a} \right |=4\: \: \textrm{dan}\\ &\left |\vec{b} \right |=3,\: \: \textrm{maka}\: \: \vec{a}(\vec{a}-\vec{b})\: \: \textrm{adalah}....\\ &\textrm{a}.\quad 2\\ &\textrm{b}.\quad 4\\ &\textrm{c}.\quad 6\\ &\textrm{d}.\quad 8\\ &\textrm{e}.\quad \color{red}10\\\\ &\textrm{Jawab}\\ &\begin{aligned}\vec{a}(\vec{a}-\vec{b})&=\vec{a}.\vec{a}-\vec{a}.\vec{b}\\ &=\left | \vec{a} \right |\left | \vec{a} \right |\cos 0^{\circ}-\left | \vec{a} \right |\left | \vec{b} \right |\cos 60^{\circ}\\ &=\left | \vec{a} \right |^{2}-\left | \vec{a} \right |\left | \vec{b} \right |.\displaystyle \frac{1}{2}\\ &=4^{2}-4.3.\displaystyle \frac{1}{2}\\ &=16-6\\ &=\color{red}10 \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 30.&\textrm{Tentukan}\: \: \overline{u}\bullet \overline{v}\: ,\: \textrm{jika diketahui}\\ &\textrm{a}.\quad \left |\overline{u} \right |=10,\: \left |\overline{v} \right |=8\sqrt{3},\: \: \cos \angle (\overline{v},\, \overline{u})=\displaystyle \frac{2}{5}\sqrt{3}\\ &\textrm{b}.\quad \left |\overline{u} \right |=6\sqrt{3},\: \left |\overline{v} \right |=4\sqrt{2},\: \: \cos (\overline{v},\, \overline{u})=30^{\circ}\\\\ &\textbf{Jawab}:\\ &\begin{aligned}\textrm{a}.\quad &\overline{u}\bullet \overline{v}=\left | \overline{u} \right |.\left | \overline{v} \right |.\cos \angle (\overline{v},\, \overline{u})\\ &\overline{u}\bullet \overline{v}=10.(8\sqrt{3}).\displaystyle \frac{2}{5}\sqrt{3}=2.8.2.3=\color{red}96\\ \textrm{b}.\quad &\overline{u}\bullet \overline{v}=\left | \overline{u} \right |.\left | \overline{v} \right |.\cos \angle (\overline{v},\, \overline{u})\\ &\overline{u}\bullet \overline{v}=(6\sqrt{3}).(4\sqrt{2}).\displaystyle \cos 30^{\circ}\\ &\qquad =(6\sqrt{3}).(4\sqrt{2}).\displaystyle \frac{1}{2}\sqrt{3}\\ &\qquad=\displaystyle \frac{6.4.3.\sqrt{2}}{2} =\color{red}36\sqrt{2}\\ \end{aligned} \end{array}$.

Contoh 5 Soal dan Pembahasan Materi Vektor

$\begin{array}{ll}\\ 21.&\textrm{Jika}\: \: \vec{g}=\begin{pmatrix} 3^{x+y}\\ 5 \end{pmatrix}\: \: \textrm{dan}\: \: \vec{h}=\begin{pmatrix} 81\\ \displaystyle \frac{y+7}{2} \end{pmatrix}\\ & \textrm{sehingga}\: \: \vec{g}=\vec{h}\: \: \: \textrm{nilai dari}\: \: 4x-3y=.... \\ &\textrm{a}.\quad \color{red}-5\\ &\textrm{b}.\quad -1\\ &\textrm{c}.\quad 0\\ &\textrm{d}.\quad 5\\ &\textrm{e}.\quad 10\\\\ &\textrm{Jawab}\\ &\begin{aligned}\textrm{Diketahui}\: &\: \textrm{bahwa}:\quad \vec{g}=\vec{h}\\ \begin{pmatrix} 3^{x+y}\\ 5 \end{pmatrix}&=\begin{pmatrix} 81\\ \displaystyle \frac{y+7}{2} \end{pmatrix}\\ 3^{x+y}&=81=3^{4}\Leftrightarrow x+y=4\\ \displaystyle \frac{y+7}{2}&=5\Leftrightarrow y=10-7=3,\quad \textrm{sehingga}\\ x+y&=4\Leftrightarrow x+3=4\Leftrightarrow x=4-3=1,\\ &\textrm{maka}\\ 4x-3y&=4(1)-3(3)\\ &=4-9=\color{red}-5 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 22.&\textrm{Vektor}\: \: \vec{m}=\begin{pmatrix} -2\\ 5 \end{pmatrix}\: \: \textrm{searah }\\ &\textrm{dengan vektor}\: .... \\ &\textrm{a}.\quad \displaystyle \begin{pmatrix} 2\\ -5 \end{pmatrix}\\ &\textrm{b}.\quad \displaystyle \begin{pmatrix} 2\\ 5 \end{pmatrix}\\ &\textrm{c}.\quad \color{red}\displaystyle \begin{pmatrix} -6\\ 15 \end{pmatrix}\\ &\textrm{d}.\quad \displaystyle \begin{pmatrix} -4\\ 5 \end{pmatrix}\\ &\textrm{e}.\quad \displaystyle \begin{pmatrix} -3\\ 10 \end{pmatrix} \\\\ &\textrm{Jawab}\\ &\begin{aligned}&\textrm{Vektor}\quad \vec{m}\: \: \: \textrm{searah dengan vektor}\: \: k.\vec{m}\\ &\color{blue}k.\vec{m}=k\begin{pmatrix} -2\\ 5 \end{pmatrix}, \color{black}\textrm{dengan}\: \: k\: \: \textrm{skalar positif}\\ &\begin{array}{|c|c|c|}\hline \textrm{a}&\textrm{b}\\\hline \begin{pmatrix} 2\\ -5 \end{pmatrix}=-\begin{pmatrix} -2\\ 5 \end{pmatrix}&\begin{pmatrix} 2\\ 5 \end{pmatrix}=...\\\hline \textrm{c}&\textrm{d}\\\hline \color{red}\begin{pmatrix} -6\\ 15 \end{pmatrix}=3\begin{pmatrix} -2\\ 5 \end{pmatrix}&\begin{pmatrix} -4\\ 5 \end{pmatrix}=...\\\hline \textrm{e}&\\\hline \begin{pmatrix} -3\\ 10 \end{pmatrix}=...&\\\hline \end{array} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 23.&\textrm{Jika vektor}\: \: \overline{AC}=\vec{p},\: \overline{BC}=\vec{q}\: \: \textrm{dan}\\ &\overline{AD}:\overline{DC}=1:2,\: \: \textrm{maka vektor}\\ & \overline{BD}\: \: \textrm{bila dinyatakan}\\ &\textrm{dalam}\: \: \vec{p}\: \: \textrm{dan}\: \: \vec{q}\: \: \textrm{adalah}....\\ &\begin{array}{lll}\\ \textrm{a}.\quad \displaystyle \frac{1}{3}\left ( 3\vec{p}-2\vec{q} \right )\\ \textrm{b}.\quad \color{red}\displaystyle \frac{1}{3}\left ( 3\vec{q}-2\vec{p} \right )\\ \textrm{c}.\quad \displaystyle \frac{1}{3}\left ( \vec{p}-2\vec{q} \right )\\ \textrm{d}.\quad \displaystyle \left ( \vec{p}-\displaystyle \frac{1}{3}\vec{q} \right )\\ \textrm{e}.\quad \displaystyle \frac{1}{3}\left ( \vec{p}-\vec{q} \right ) \end{array}\\\\ &\textbf{Jawab}\\ &\begin{aligned}\overline{AC}:\overline{CD}&=3:-2\\ \overline{CD}&=-\displaystyle \frac{2}{3}\, \overline{AC}\\ &=\displaystyle -\frac{2}{3}\vec{p}\\ \textrm{maka}\, ,\qquad&\\ \overline{BD}&=\overline{BC}+\overline{CD}\\ &=\vec{q}+\left ( -\displaystyle \frac{2}{3}\vec{p} \right )\\ &=\color{red}\displaystyle \frac{1}{3}\left ( 3\vec{q}-2\vec{p} \right ) \end{aligned} \end{array}$

$.\: \quad \color{blue}\textrm{Gambar berikut untuk soal 24}$

$\begin{array}{ll}\\ 24.&\textrm{Jika vektor}\: \: \overline{AC}=\vec{p},\: \overline{BC}=\vec{q}\: \: \textrm{dan}\\ & \overline{AD}:\overline{DC}=1:2,\: \textrm{maka vektor}\: \: \overline{BD}\\ &\textrm{bila dinyatakan dalam}\: \: \vec{p}\: \: \textrm{dan}\: \: \vec{q}\: \: \textrm{adalah}.... \\ &\textrm{a}.\quad \displaystyle \frac{1}{3}\left ( 3\vec{p}-2\vec{q} \right )\\ &\textrm{b}.\quad \color{red}\displaystyle \frac{1}{3}\left ( 3\vec{q}-2\vec{p} \right )\\ &\textrm{c}.\quad \displaystyle \left ( \vec{p}-\displaystyle \frac{1}{3}\vec{q} \right )\\ &\textrm{d}.\quad \displaystyle \frac{1}{3}\left ( \vec{p}-2\vec{q} \right )\\ &\textrm{e}.\quad \displaystyle \frac{1}{3}\left ( \vec{p}-\vec{q} \right )\\\\ &\textrm{Jawab}\\ &\begin{aligned}\overline{AC}:\overline{CD}&=3:-2\\ \overline{CD}&=-\displaystyle \frac{2}{3}\, \overline{AC}\\ &=\displaystyle -\frac{2}{3}\vec{p}\\ \textrm{maka}\, ,\qquad&\\ \overline{BD}&=\overline{BC}+\overline{CD}\\ &=\vec{q}+\left ( -\displaystyle \frac{2}{3}\vec{p} \right )\\ &=\color{red}\displaystyle \frac{1}{3}\left ( 3\vec{q}-2\vec{p} \right ) \end{aligned} \end{array}$

$\begin{array}{ll}\\ 25.&\textrm{Jika titik}\: \: A(2,6)\: \: \textrm{dan}\: \: B(5,3)\: \: \textrm{demikian }\\ &\textrm{juga titik}\: \: P\: \: \textrm{terletak pada}\: \: \overline{AB}\\ &\textrm{dengan}\: \: \overline{AP}:\overline{PB}=2:1,\: \textrm{maka vektor }\\ &\textrm{posisi}\: \: \vec{p}\: \: \textrm{adalah}....\\ &\textrm{a}.\quad \color{red}\begin{pmatrix} 4\\ 4 \end{pmatrix}\\ &\textrm{b}.\quad \begin{pmatrix} 4\\ 5 \end{pmatrix}\\ &\textrm{c}.\quad \begin{pmatrix} -4\\ 4 \end{pmatrix}\\ &\textrm{d}.\quad \begin{pmatrix} 4\\ 2 \end{pmatrix}\\ &\textrm{e}.\quad \begin{pmatrix} -4\\ 6 \end{pmatrix}\\\\ &\textrm{Jawab}\\ &\begin{aligned}\overline{AP}:\overline{PB}&=2:1\\ \overline{AP}&=2\, \overline{PB}\\ \vec{p}-\vec{a}&=2\left ( \vec{b}-\vec{p} \right )\\ \vec{p}+2\vec{p}&=\vec{a}+2\vec{b}\\ 3\vec{p}&=\vec{a}+2\vec{b}\\ \vec{p}&=\displaystyle \frac{1}{3}\left ( \vec{a}+2\vec{b} \right )\\ &=\displaystyle \frac{1}{3}\begin{pmatrix} 2+2.5\\ 6+2.3 \end{pmatrix}\\ &=\displaystyle \frac{1}{3}\begin{pmatrix} 12\\ 12 \end{pmatrix}\\ &=\color{red}\begin{pmatrix} 4\\ 4 \end{pmatrix} \end{aligned} \end{array}$

Contoh 4 Soal dan Pembahasan Materi Vektor

$\begin{array}{ll}\\ 16.&\textrm{Diketahui titik A(-1,1,0) dan titik B(1,-2,2)}\\ &\textrm{maka panjang vektor}\: \: \overrightarrow{BA}\: \: \textrm{adalah}\: ....\\ &\begin{array}{llllllll}\\ \textrm{a}.&\sqrt{2}&&&\textrm{d}.&\color{red}\sqrt{17}\\ \textrm{b}.&\sqrt{5}&\textrm{c}.&\sqrt{9}&\textrm{e}.&\sqrt{21} \end{array}\\\\ &\textbf{Jawab}\\ &\begin{aligned}&\textrm{Diketahui}\: \, \textrm{sebagaimana pada soal}\\ &\textrm{maka pan}\textrm{jang vektor}\: \: \overrightarrow{BA}\: \: \textrm{adalah}\\ &\left | \overrightarrow{BA} \right |=\sqrt{(1-(-1))^{2}+(-2-1)^{2}+(2-0)^{2}}\\ &=\sqrt{2^{2}+(-3)^{2}+2^{2}}\\ &=\sqrt{4+9+4}\\ &=\color{red}\sqrt{17} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 17.&\textrm{Vektor satuan untuk vektor}\: \: \vec{a}=\begin{pmatrix} 2, & 1, &-2 \end{pmatrix}=\: ....\\ &\begin{array}{llllllll}\\ \textrm{a}.&\begin{pmatrix} -\displaystyle \frac{2}{3}, & -\displaystyle \frac{1}{3}, &\displaystyle \frac{2}{3} \end{pmatrix}&\\\\ \textrm{b}.&\color{red}\begin{pmatrix} \displaystyle \frac{2}{3}, & \displaystyle \frac{1}{3}, &-\displaystyle \frac{2}{3} \end{pmatrix}&\\\\ \textrm{c}.&\begin{pmatrix} \displaystyle \frac{2}{4}, & \displaystyle \frac{1}{4}, &-\displaystyle \frac{2}{4} \end{pmatrix}&\\\\ \textrm{d}.&\begin{pmatrix} -\displaystyle \frac{2}{4}, & -\displaystyle \frac{1}{4}, &\displaystyle \frac{2}{4} \end{pmatrix}\\\\ \textrm{e}.&\begin{pmatrix} -\displaystyle \frac{2}{9}, &-\displaystyle \frac{1}{9}, & \displaystyle \frac{2}{9} \end{pmatrix} \end{array}\\\\ &\textbf{Jawab}\\ &\begin{aligned}\textrm{Vektor }&\: \textrm{satuan dari vektor}\: \: \vec{a}\: \: \textrm{adalah}:\\ \hat{a}&=\displaystyle \frac{\vec{a}}{\left | \vec{a} \right |}\\ &=\displaystyle \frac{\begin{pmatrix} 2, & 1, & -2 \end{pmatrix}}{\sqrt{2^{2}+1^{2}+(-2)^{2}}}\\ &=\displaystyle \frac{\begin{pmatrix} 2, & 1, & -2 \end{pmatrix}}{\sqrt{9}}\\ &=\displaystyle \frac{\begin{pmatrix} 2, & 1, & -2 \end{pmatrix}}{3}\\ &=\color{red}\begin{pmatrix} \displaystyle \frac{2}{3}, & \displaystyle \frac{1}{3}, &-\displaystyle \frac{2}{3} \end{pmatrix} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 18.&\textrm{Jika titik A(-2,3,5) dan B(4,1,-3)},\\ & \textrm{maka vektor posisi AB adalah}\: ....\\ &\begin{array}{llllllll}\\ \textrm{a}.&\begin{pmatrix} -6, & 2, &8 \end{pmatrix}&\\ \textrm{b}.&\begin{pmatrix} 8, & 2, &-6 \end{pmatrix}&\\ \textrm{c}.&\color{red}\begin{pmatrix} 6, & -2, &-8 \end{pmatrix}&\\ \textrm{d}.&\begin{pmatrix} -8 & -2, &6 \end{pmatrix}\\ \textrm{e}.&\begin{pmatrix} 2, &4, & 2 \end{pmatrix} \end{array}\\\\ &\textbf{Jawab}\\ &\begin{aligned}\textrm{Vektor posisi}&\: \textrm{dari}\: \: \overrightarrow{AB}\: \: \textrm{adalah}:\\ \overrightarrow{AB}&=\overrightarrow{OB}-\overrightarrow{OA}\\ &=\begin{pmatrix} 4\\ 1\\ -3 \end{pmatrix}-\begin{pmatrix} -2\\ 3\\ 5 \end{pmatrix}\\ &=\begin{pmatrix} 4+2\\ 1-3\\ -3-5 \end{pmatrix}\\ &=\begin{pmatrix} 6\\ -2\\ -8 \end{pmatrix}\quad\: \textbf{atau}\\ &=\color{red}\begin{pmatrix} 6, & -2, & -8 \end{pmatrix} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 19.&\textrm{Jika}\: \: \vec{p}=\begin{pmatrix} ^{2}\log 8x\\ \left ( ^{2}\log x \right )^{y} \end{pmatrix}\: \: \textrm{dan}\: \: \vec{q}=\begin{pmatrix} 5\\ 8 \end{pmatrix}\\ &\textrm{sehingga}\: \: \vec{p}=\vec{q}\: \: \: \textrm{nilai dari}\: \: x.y=.... \\ &\textrm{a}.\quad 6\\ &\textrm{b}.\quad \color{red}12\\ &\textrm{c}.\quad 18\\ &\textrm{d}.\quad 24\\ &\textrm{e}.\quad 30\\\\ &\textrm{Jawab}\\ &\begin{aligned}\textrm{Diketahui}\: &\: \textrm{bahwa}:\quad \vec{p}=\vec{q}\\ \begin{pmatrix} ^{2}\log 8x\\ \left ( ^{2}\log x \right )^{y} \end{pmatrix}&=\begin{pmatrix} 5\\ 8 \end{pmatrix},\quad\: \: \textrm{maka}\\ 8x&=2^{5}=32\\ \Leftrightarrow x&=\displaystyle \frac{32}{8}=4\\ \left (^{2}\log 4 \right )^{y}&=8\\ \Leftrightarrow 2^{y}&=8=2^{3}\\ \Leftrightarrow y&=3\\ \textrm{Sehingga}&\\ x.y&=4\times 3=\color{red}12 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 20.&\textrm{Jika}\: \: \vec{p}=\begin{pmatrix} 3x\\ 4x+y \end{pmatrix}\: \: \textrm{dan}\: \: \vec{q}=\begin{pmatrix} \displaystyle \frac{2x-4}{2}\\ 6 \end{pmatrix}\\ & \textrm{sehingga}\: \: \vec{p}=\vec{q}\: \: \: \textrm{nilai dari}\: \: 2x+y=.... \\ &\textrm{a}.\quad -12\\ &\textrm{b}.\quad 0\\ &\textrm{c}.\quad 8\\ &\textrm{d}.\quad \color{red}9\\ &\textrm{e}.\quad 19\\\\ &\textrm{Jawab}\\ &\begin{aligned}\textrm{Diketahui}\: &\: \textrm{bahwa}:\quad \vec{p}=\vec{q}\\ \begin{pmatrix} 3x\\ 4x+y \end{pmatrix}&=\begin{pmatrix} \displaystyle \frac{2x-4}{2}\\ 6 \end{pmatrix}\\ 3x&=\displaystyle \frac{2x-4}{2}\Leftrightarrow 6x=2x-4\\ \Leftrightarrow x&=-1\\ 4(-1)+y&=6\Leftrightarrow -4+y=6\\ \Leftrightarrow y&=6+4\\ y&=10\\ x+y&=(-1)+10=\color{red}9 \end{aligned} \end{array}$

Contoh 3 Soal dan Pembahasan Materi Vektor

$\begin{array}{ll}\\ 11.&\textrm{Jika vektor}\: \: \vec{a}=\begin{pmatrix} 6\\ -4 \end{pmatrix}\: \: \textrm{dan}\: \: \vec{b}=\begin{pmatrix} 3\\ 2 \end{pmatrix},\\ &\textrm{maka}\: \: 3\vec{a}-2\vec{b}\: \: \textrm{adalah}....\\ &\begin{array}{lll}\\ \textrm{a}.\quad \color{red}\begin{pmatrix} 12\\ -16 \end{pmatrix}&&\textrm{d}.\quad \begin{pmatrix} 24\\ 16 \end{pmatrix}\\ \textrm{b}.\quad \begin{pmatrix} 24\\ -16 \end{pmatrix}&\textrm{c}.\quad \begin{pmatrix} 12\\ 16 \end{pmatrix}&\textrm{e}.\quad \begin{pmatrix} -12\\ -16 \end{pmatrix} \end{array}\\\\ &\textbf{Jawab}\\ &\begin{aligned}3\vec{a}-2\vec{b}&=3\begin{pmatrix} 6\\ -4 \end{pmatrix}-2\begin{pmatrix} 3\\ 2 \end{pmatrix}\\ &=\begin{pmatrix} 18-6\\ -12-4 \end{pmatrix}\\ &=\color{red}\begin{pmatrix} 12\\ -16 \end{pmatrix} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 12.&\textrm{Diketahui jajar genjang ABCD }\\ &\textrm{dengan titik E adalah perpotongan }\\ &\textrm{diagonal jajar genjang}. \end{array}$

$\begin{array}{ll}\\ .\, \quad&\textrm{Jika}\: \: \overline{AB}=\vec{b}\: \: \textrm{dan}\: \: \overline{AD}=\vec{a},\: \textrm{maka}\: \: \overline{CE}\\ & \textrm{bila dinyatakan dalam}\: \: \vec{a}\: \: \textrm{dan}\: \: \vec{b}\: \: \textrm{adalah}....\\ &\begin{array}{lll}\\ \textrm{a}.\quad \displaystyle \frac{1}{2}\left ( \vec{a}+\vec{b} \right )&\\ \textrm{b}.\quad \displaystyle \frac{1}{2}\left ( \vec{a}-\vec{b} \right )&\\ \textrm{c}.\quad \displaystyle \frac{1}{2}\left ( \vec{b}-\vec{a} \right )&\\ \textrm{d}.\quad \color{red}\displaystyle -\frac{1}{2}\left ( \vec{a}+\vec{b} \right )\\ \textrm{e}.\quad -\displaystyle \frac{1}{2}\left ( 2\vec{a}+\vec{b} \right ) \end{array}\\\\ &\textbf{Jawab}\\ &\begin{aligned} \overline{AC}&=\overline{AD}+\overline{DC}\\ \overline{CA}&=\overline{CD}+\overline{DA}\\ \overline{CE}&=\displaystyle \frac{1}{2}\, \overline{CA}\\ &=\displaystyle \frac{1}{2}\left ( -\vec{b}-\vec{a} \right )\\ &=\color{red}-\displaystyle \frac{1}{2}\left ( \vec{a}+\vec{b} \right ) \end{aligned} \end{array}$

$\begin{array}{ll}\\ 13.&\textrm{Pada segi enam beraturan ABCDEF},\\ & \textrm{jika}\: \: \overrightarrow{AB}=\vec{u}\: \: \textrm{dan}\: \: \overrightarrow{AF}=\vec{v}\: \: \textrm{maka vektor}\\ &\overrightarrow{AB}+\overrightarrow{AC}+\overrightarrow{AD}+\overrightarrow{AE}+\overrightarrow{AF}=....\\ &\begin{array}{llllllll}\\ \textrm{a}.&2\vec{u}+2\vec{v}&&&\textrm{d}.&\color{red}6\vec{u}+6\vec{v}\\ \textrm{b}.&4\vec{u}+4\vec{v}&\textrm{c}.&5\vec{u}+5\vec{v}&\textrm{e}.&8\vec{u}+8\vec{v} \end{array}\\\\ &\textbf{Jawab}\\ &\textrm{Perhatikanlah ilustrasi gambar berikut} \end{array}$

$\begin{aligned}.\: \, \qquad &\overrightarrow{AB}+\overrightarrow{AC}+\overrightarrow{AD}+\overrightarrow{AE}+\overrightarrow{AF}\\ &=\overrightarrow{AB}+\left (\overrightarrow{AO}+\overrightarrow{OC} \right )+2\overrightarrow{AO}+\left (\overrightarrow{AO}+\overrightarrow{OE} \right )+\overrightarrow{AF}\\ &=\vec{u}+\left (2\vec{u}+\vec{v} \right )+2\left ( \vec{v}+\vec{u} \right )+\left ( 2\vec{v}+\vec{u} \right )+\vec{v}\\ &=\color{red}6\vec{u}+6\vec{v} \end{aligned}$

$\begin{array}{ll}\\ 14.&\textrm{Perhatikanlah juga ilustrasi gambar berikut} \end{array}$

$\begin{array}{ll}\\ .\, \quad&\textrm{maka vektor}\: \: \vec{w}\: \: \textrm{adalah}\: ....\\ &\begin{array}{llllllll}\\ \textrm{a}.&\color{red}8\vec{i}-6\vec{j}-13\vec{k}&\\ \textrm{b}.&8\vec{i}-13\vec{j}-6\vec{k}\\ \textrm{c}.&6\vec{i}-8\vec{j}-13\vec{k}&\\ \textrm{d}.&-6\vec{i}+8\vec{j}-13\vec{k}\\ \textrm{e}.&-6\vec{i}-13\vec{j}+8\vec{k} \end{array}\\\\ &\textbf{Jawab}\\ &\textrm{Kita perhatikan juga ilustrasi }\\ &\textrm{gambarnya semisal dengan soal No.1}\\ &\textrm{Misalkan titiknya adalah titik }\\ &\textrm{W dengan koordinat (8,-6,-13)},\\ &\textrm{maka vektor posisi titik }\\ &\textrm{W tersebut adalah}\: \: \overrightarrow{OW}=\vec{w}\\ & \textrm{di mana}\\ &\begin{aligned}&\textrm{Vektor}\: \: \vec{w}\: \textrm{jika dinyatakan }\\ &\textrm{dalam kombinasi linear adalah}\\ &\vec{w}=\color{red}8\vec{i}-6\vec{j}-13\vec{k} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 15.&\textrm{Jika titik Z(4,-5,2)},\: \textrm{maka panjang }\\ &\textrm{vektor posisi titik Z adalah}\: ....\\ &\begin{array}{llllllll}\\ \textrm{a}.&1&&&\textrm{d}.&5\sqrt{2}\\ \textrm{b}.&2\sqrt{5}&\textrm{c}.&\color{red}3\sqrt{5}&\textrm{e}.&5\sqrt{3} \end{array}\\\\ &\textbf{Jawab}\\ &\begin{aligned}\textrm{Vektor posisi}&\: \, \textrm{titik Z tersebut adalah}\\ \overrightarrow{OZ}=\vec{z}&=\begin{pmatrix} 4, & -5, & 2 \end{pmatrix},\\ \textrm{Dan panjang}&\: \, \textrm{vektor}\: \: \vec{z}\: \: \textrm{ini adalah}\\ \left | \vec{z} \right |&=\sqrt{4^{2}+(-5)^{2}+2^{2}}\\ &=\sqrt{16+25+4}\\ &=\sqrt{45}=\sqrt{9\times 5}\\ &=\color{red}3\sqrt{5} \end{aligned} \end{array}$.

Contoh Soal 16 (Segitiga dan Ketaksamaan)

$\begin{array}{ll}\\ 76.&\textbf{(IMO 1983)}\\ &\textrm{Jika}\: \: a,b,c\: \: \textrm{adalah panjang sisi-sisi segitiga}\\ &\textrm{tunjukkan bahwa}\\ &\quad a^{2}b(a-b)+b^{2}c(b-c)+c^{2}a(c-a)\geq 0\\\\ &\textbf{Bukti}:\\ &\textrm{Pada sebuah segitiga dengan sisi}\: \: a,b,c\\ &\textrm{berlaku}\: \: \begin{cases} a+b>c & \Rightarrow a>c-b\\ a+c>b & \Rightarrow c>b-c\\ b+c>a & \Rightarrow b>a-c \end{cases}\\ &\textrm{Sehingga untuk ketaksamaan pada soal}\\ &\color{red}a^{2}b(a-b)+b^{2}c(b-c)+c^{2}a(x-a)\\ &\geq a^{2}(a-c)(a-b)+b^{2}(b-c)(b-c)+c^{2}(c-b)(c-a)\color{red}\geq 0\\ &\textrm{Bentuk terakhir memenuhi bentuk dari}\\ &\textbf{Ketaksamaan Schur}\: \: \textrm{saat}\: \: \color{red}r=2.\\ &\textrm{Jadi},\\ &\quad a^{2}b(a-b)+b^{2}c(b-c)+c^{2}a(c-a)\geq 0\quad \blacksquare \end{array}$.

$\begin{array}{ll}\\ 77.&\textrm{Misalkan}\: \: a,b,c\: \: \textrm{bilangan real positif dengan}\\ &a+b+c=2\: ,\: \textrm{tunjukkan bahwa}\\ &\quad a^{4}+b^{4}+c^{4}+abc\geq a^{3}+b^{3}+c^{3}\\\\ &\textbf{Bukti}\\ &\textrm{Dengan}\: \: \textbf{Ketaksamaan Schur}\: \textrm{saat}\: \: \color{red}r=2\\ &\textrm{kita memiliki}\\ &a^{2}(a-b)(a-c)+b^{2}(b-a)(b-c)+c^{2}(c-a)(c-b)\geq 0\\ &\Leftrightarrow a^{4}+b^{4}+c^{4}+abc(a+b+c)-a^{3}(b+c)-b^{3}(a+c)-c^{3}(a+b)\geq 0\\ &\Leftrightarrow a^{4}+b^{4}+c^{4}+abc(a+b+c)\geq a^{3}(b+c)+b^{3}(a+c)+c^{3}(a+b)\\ &\Leftrightarrow a^{4}+b^{4}+c^{4}+abc(a+b+c)\geq (a^{3}+b^{3}+c^{3})(a+b+c)-(a^{4}+b^{4}+c^{4})\\ &\Leftrightarrow 2(a^{4}+b^{4}+c^{4})+abc(a+b+c)\geq (a^{3}+b^{3}+c^{3})(a+b+c)\\ &\Leftrightarrow 2(a^{4}+b^{4}+c^{4})+abc(2)\geq (a^{3}+b^{3}+c^{3})(2)\\ &\Leftrightarrow a^{4}+b^{4}+c^{4}+abc\geq a^{3}+b^{3}+c^{3}\qquad \blacksquare \\\\ &\color{blue}\textrm{Bentuk di atas kadang dituliskan dengan bentuk}\\ &\color{blue}\textrm{berikut}:\\ &\begin{aligned}&\textrm{Dengan}\: \: \textbf{Ketaksamaan Schur}\: \textrm{saat}\: \: \color{red}r=2\\ &\textrm{kita memiliki}\\ &\displaystyle \sum_{\textrm{siklik}}^{.}a^{2}(a-b)(a-c)\geq 0\\ &\Leftrightarrow \displaystyle \sum_{\textrm{siklik}}^{.}a^{4}+abc\displaystyle \sum_{\textrm{siklik}}^{.}a-\displaystyle \sum_{\textrm{siklik}}^{.}a^{3}(b+c)\geq 0\\ &\Leftrightarrow \displaystyle \sum_{\textrm{siklik}}^{.}a^{4}+abc\displaystyle \sum_{\textrm{siklik}}^{.}a\geq \displaystyle \sum_{\textrm{siklik}}^{.}a^{3}(b+c)\\ &\Leftrightarrow \displaystyle \sum_{\textrm{siklik}}^{.}a^{4}+abc\displaystyle \sum_{\textrm{siklik}}^{.}a\geq \left ( \displaystyle \sum_{\textrm{siklik}}^{.}a^{3} \right )\left ( \displaystyle \sum_{\textrm{siklik}}^{.}a \right )-\left ( \displaystyle \sum_{\textrm{siklik}}^{.}a^{4} \right )\\ &\Leftrightarrow 2\displaystyle \sum_{\textrm{siklik}}^{.}a^{4}+abc\displaystyle \sum_{\textrm{siklik}}^{.}a\geq \left ( \displaystyle \sum_{\textrm{siklik}}^{.}a^{3} \right )\left ( \displaystyle \sum_{\textrm{siklik}}^{.}a \right )\\ &\Leftrightarrow \displaystyle \sum_{\textrm{siklik}}^{.}a^{4}+abc\geq \left ( \displaystyle \sum_{\textrm{siklik}}^{.}a^{3} \right )\qquad \blacksquare \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 78.&\textrm{Misalkan}\: \: x,y,x\: \: \textrm{bilangan real positif dengan}\\ &\textrm{tunjukkan bahwa}\\ &\quad a^{2}+b^{2}+c^{2}+2abc+1\geq 2(ab+acb+c)\\&\qquad\qquad\qquad\qquad\qquad (\textbf{Darij Grinberg})\\\\ &\textbf{Bukti}\\ &\textrm{Dengan}\: \: \textbf{ketaksamaan AM-GM}\: \: \textrm{dan}\\ &\textrm{dilanjutkan dengan}\: \: \textbf{ketaksamaan Schur}\\ &\textrm{serta menggesernya ke ruas kiri, maka}\\ &a^{2}+b^{2}+c^{2}+2abc+1- 2(ab+ac+bc)\\ &\geq a^{2}+b^{2}+c^{2}+3(abc)^{^{\frac{2}{3}}}+1\geq 2(ab+ac+bc)\\ &\geq \left ((a)^{^{\frac{2}{3}}} \right )^{3}+\left ((b)^{^{\frac{2}{3}}} \right )^{3}+\left ((c)^{^{\frac{2}{3}}} \right )^{3}+3(abc)^{^{\frac{2}{3}}}-2(ab+ac+bc)\\ &\geq a^{.^{\frac{2}{3}}}b^{.^{\frac{2}{3}}}\left (a^{.^{\frac{2}{3}}} +b^{.^{\frac{2}{3}}} \right )+a^{.^{\frac{2}{3}}}c^{.^{\frac{2}{3}}}\left (a^{.^{\frac{2}{3}}} +c^{.^{\frac{2}{3}}} \right )\\ &\quad +b^{.^{\frac{2}{3}}}c^{.^{\frac{2}{3}}}\left (b^{.^{\frac{2}{3}}} +c^{.^{\frac{2}{3}}} \right )-2(ab+ac+bc)\\ &\geq a^{.^{\frac{2}{3}}}b^{.^{\frac{2}{3}}}\left (a^{.^{\frac{2}{3}}} +b^{.^{\frac{2}{3}}} \right )+a^{.^{\frac{2}{3}}}c^{.^{\frac{2}{3}}}\left (a^{.^{\frac{2}{3}}} +c^{.^{\frac{2}{3}}} \right )\\ &\quad +b^{.^{\frac{2}{3}}}c^{.^{\frac{2}{3}}}\left (b^{.^{\frac{2}{3}}} +c^{.^{\frac{2}{3}}} \right )-2(ab+ac+bc)\\ &= a^{.^{\frac{2}{3}}}b^{.^{\frac{2}{3}}}\left (a^{.^{\frac{2}{3}}} -b^{.^{\frac{2}{3}}} \right )^{2}+a^{.^{\frac{2}{3}}}c^{.^{\frac{2}{3}}}\left (a^{.^{\frac{2}{3}}} -c^{.^{\frac{2}{3}}} \right )^{2}\\ &\quad +b^{.^{\frac{2}{3}}}c^{.^{\frac{2}{3}}}\left (b^{.^{\frac{2}{3}}} -c^{.^{\frac{2}{3}}} \right )^{2}\geq 0\qquad \blacksquare \end{array}$.

$\begin{array}{ll}\\ 79.&(\textbf{APMO 2004})\\ &\textrm{Misalkan}\: \: x,y,x\: \: \textrm{bilangan real positif dengan}\\ &\textrm{tunjukkan bahwa}\\ &\quad (x^{2}+2)(y^{2}+2)(z^{2}+2)\geq 9(xy+yz+zx)\\\\ &\textbf{Bukti}\\ &\color{blue}\textbf{Alternatif 1}\\ &\textrm{Dengan menjabarkan akan didapatkan}\\ &x^{2}y^{2}z^{2}+2\displaystyle \sum_{\textrm{siklik}}^{.}x^{2}y^{2}+4\displaystyle \sum_{\textrm{siklik}}^{.}x^{2}+8\geq 9\displaystyle \sum_{\textrm{siklik}}^{.}xy\\ &\textrm{Perhatikan bahwa}\\ &\bullet \quad\color{red}2\displaystyle \sum_{\textrm{siklik}}^{.}x^{2}y^{2}-4\displaystyle \sum_{\textrm{siklik}}^{.}xy+6=2\displaystyle \sum_{\textrm{siklik}}^{.}(xy-1)^{2}\geq 0\\ &\bullet \quad \color{red}\displaystyle \sum_{\textrm{siklik}}^{.}x^{2}\geq \displaystyle \sum_{\textrm{siklik}}^{.}xy\: \: \color{black}\textrm{atau}\: \: \color{red}x^{2}+y^{2}+z^{2}\geq xy+xz+yz\\ &\textrm{Kita cukup membuktikan bahwa}\\ & x^{2}y^{2}z^{2}+\displaystyle \sum_{\textrm{siklik}}^{.}x^{2}+2\geq 2\displaystyle \sum_{\textrm{siklik}}^{.}xy\\ &\Leftrightarrow x^{2}y^{2}z^{2}+2\geq \displaystyle \sum_{\textrm{siklik}}^{.}xy\\ &\begin{aligned}&\textrm{Untuk}\: \: a,b,c\: \: \textrm{bilangan real positif},\\ &\textbf{Ketaksamaan Schur}\: \textrm{saat}\: \: \color{red}r=1\\ &\textrm{memberikan}\\ &\displaystyle \sum_{\textrm{siklik}}^{.}a^{3}+3abc\geq \displaystyle \sum_{\textrm{siklik}}^{.}a^{2}b+\displaystyle \sum_{\textrm{siklik}}^{.}ab^{2}\\ &\qquad\qquad\qquad =ab(a+b)+bc(b+c)+ca(c+a)\\ &\textrm{Dengan}\: \: \textbf{Ketaksamaan AM-GM}\: \: \textrm{didapakan}\\ &\displaystyle \sum_{\textrm{siklik}}^{.}a^{3}+3abc\geq 2\displaystyle \sum_{\textrm{siklik}}^{.}(ab)^{\frac{3}{2}}\\ &\textrm{Pilih}\: \: a=x^{\frac{2}{3}},\: b=y^{\frac{2}{3}},\: c=z^{\frac{2}{3}},\: \textrm{maka didapatkan}\\ &(x^{\frac{2}{3}})^{3}+(y^{\frac{2}{3}})^{3}+(z^{\frac{2}{3}})^{3}+3(xyz)^{\frac{2}{3}}\geq 2(xy+yz+zx)\\ &\textrm{Selanjutnya kita selesaikan ini}\: ,\: x^{2}y^{2}z^{2}+2\geq \displaystyle \sum_{\textrm{siklik}}^{.}xy\\ &\Leftrightarrow x^{2}y^{2}z^{2}+2 \geq 3(xyz)^{\frac{2}{3}}\\ &\textrm{Misalkan}\: \: (xyz)^{\frac{2}{3}}=t,\: \textrm{maka}\\ &t^{3}+2\geq 3t\Leftrightarrow t^{3}-3t+2\geq 0\\ &(t-1)^{2}(t+2)\geq 0\: \: \textrm{adalah hal benar} \end{aligned} \end{array}$.

$.\: \quad\begin{aligned}&\color{blue}\textbf{Alternatif 2}\\ &x^{2}y^{2}z^{2}+2\displaystyle \sum_{\textrm{siklik}}^{.}x^{2}y^{2}+4\displaystyle \sum_{\textrm{siklik}}^{.}x^{2}+8\geq 9\displaystyle \sum_{\textrm{siklik}}^{.}xy\\ &\textrm{atau dalam bentuk utuhnya, yaitu}\\ &x^{2}y^{2}z^{2}+2(x^{2}y^{2}+x^{2}z^{2}+y^{2}z^{2})+4(x^{2}+y^{2}+z^{2})+8\geq 9(xy+xz+yz)\\ &\textrm{Sekarang kira uraikan satu persatu bagian}\\ &\bullet\quad x^{2}y^{2}z^{2}+1+1\geq 3\sqrt[3]{(xyz)^{2}}\geq \displaystyle \frac{9abc}{a+b+c}=\frac{9r}{p}\\ &\qquad \textrm{ingat bahwa}\: \: \textbf{jika ada}\: \: \displaystyle \frac{9r}{p}\geq 4q-p^{2}\\ &\qquad =4(xy+xz+yz)-(x+y+z)^{2}\\ &\qquad \textrm{adalah}\: \: \textbf{ketaksamaan Schur saat}\: \: \color{red}r=1\\ &\bullet \quad x^{2}y^{2}+1+x^{2}z^{2}+1+y^{2}z^{2}+1\geq 2(xy+xz+yz)\\ &\bullet\quad x^{2}+y^{2}+z^{2}\geq xy+xz+yz\\ &\qquad \textrm{keduanya didapat dengan ketaksamaan}\: \: \textbf{AM-GM}\\&x^{2}y^{2}z^{2}+2(x^{2}y^{2}+x^{2}z^{2}+y^{2}z^{2})+4(x^{2}+y^{2}+z^{2})+8\\ &=x^{2}y^{2}z^{2}+2+2(x^{2}y^{2}+x^{2}z^{2}+y^{2}z^{2}+3)+4(x^{2}+y^{2}+z^{2})\\ &\geq 4(xy+xz+yz)-(x+y+z)^{2}+4(xy+xz+yz)+4(xy+xz+yz)\\ &\geq 12(xy+xz+yz)-(x+y+z)^{2}\\ &\geq 12(xy+xz+yz)-3(xy+xz+yz)\\ &=9(xy+xz+yz)\qquad \blacksquare \end{aligned}$.

$.\: \quad\begin{aligned}&\color{blue}\textbf{Alternatif 3}\\ &(x^{2}+2)(y^{2}+2)(z^{2}+2)- 9(xy+yz+zx)\\ &\textrm{Dengan}\: \: \textbf{ketaksamaan AM-GM}\\ &=x^{2}y^{2}z^{2}+2(x^{2}y^{2}+x^{2}z^{2}+y^{2}z^{2})\\ &\quad +4(x^{2}+y^{2}+z^{2})+8- 9(xy+xz+yz)\\ &=4(x^{2}+y^{2}+z^{2})+2\left ((x^{2}y^{2}+1) +(x^{2}z^{2}+1)+(y^{2}z^{2}+1) \right )\\ &\quad +(x^{2}y^{2}z^{2}+1)+1-9(xy+xz+yz)\\ &\geq 4(x^{2}+y^{2}+z^{2})+4(xy+xz+yz)\\ &\quad +2xyz+1-9(xy+xz+yz)\\ &=(x^{2}+y^{2}+z^{2})+3(x^{2}+y^{2}+z^{2})\\ &\quad +2xyz+1-5(x^{2}+y^{2}+z^{2})\\ &\geq x^{2}+y^{2}+z^{2}+3(xy+xz+yz)\\ &+2xyz-5(xy+xz+yz)\\ &=x^{2}+y^{2}+z^{2}+2xyz+1-2(xy+xz+yz)\geq 0\\ &\textrm{adalah benar dengan bukti ada pada}\\ &\textrm{nomor soal sebelumnya} \end{aligned}$.

$\begin{array}{ll}\\ 80.&\textrm{Misalkan}\: \: x,y,x\: \: \textrm{bilangan real positif dengan}\\ &\textrm{tunjukkan bahwa}\\ &\quad 2(a^{2}+b^{2}+c^{2})+abc+8\geq 5(a+b+c)\\&\qquad\qquad\qquad\qquad\qquad (\textbf{Tran Nam Dung})\\\\ &\textbf{Bukti}\\ &\textrm{Dengan}\: \: \textbf{ketaksamaan AM-GM}\: \: \textrm{dan}\\ &\textrm{menggeser ke ruas kiri dan masing-masing}\\ &\textrm{serta mengalikan semunya dengan 6, maka}\\ &12(a^{2}+b^{2}+c^{2})+6abc+48- 30(a+b+c)\\ &= 12(a^{2}+b^{2}+c^{2})+3(2abc+1)+45- 5.2.3(a+b+c)\\ &\geq 2(a^{2}+b^{2}+c^{2})+9\sqrt[3]{(abc)^{2}}+45- 5\left ((a+b+c)^{2}+9 \right )\\ &=12(a^{2}+b^{2}+c^{2})+\displaystyle \frac{9abc}{\sqrt[3]{abc}}-5\left ((a^{2}+b^{2}+c^{2})+2(ab+ac+bc) \right )\\ &=7(a^{2}+b^{2}+c^{2})+ \displaystyle \frac{9abc}{\sqrt[3]{abc}}-10(ab+ac+bc)\\ &\geq 7(a^{2}+b^{2}+c^{2})+\displaystyle \frac{27abc}{a+b+c}-10(ab+ac+bc)\\ &\begin{aligned}&\textrm{dengan}\: \: \textbf{ketaksamaan Schur},\: \: \textrm{yaitu}:\\ &p^{3}+9r\geq 4pq\Leftrightarrow \displaystyle \color{red}\frac{9r}{p}\geq 4q-p^{2}\\ &\textrm{maka ketaksamaan akan menjadi}\\ &\geq 7(a^{2}+b^{2}+c^{2})+\color{blue}3(4q-p^{2})-10q\\ &\geq 7(a^{2}+b^{2}+c^{2})+2q-3p^{2}\\ &=7(a^{2}+b^{2}+c^{2})+2(ab+ac+bc)-3(a+b+c)^{2}\\ &=7(a^{2}+b^{2}+c^{2})+2q-3\left ((a^{2}+b^{2}+c^{2})+2q \right )\\ &=4(a^{2}+b^{2}+c^{2})+2q-6q\\ &=4(a^{2}+b^{2}+c^{2})-4q\\ &=4(a^{2}+b^{2}+c^{2})-4(ab+ac+bc)\\ &=4(a^{2}+b^{2}+c^{2}-ab-ac-bc)\geq 0\quad \blacksquare \end{aligned} \end{array}$.

DAFTAR PUSTAKA

Tung, K.Y. 2013. Ayo Raih Medali Emas Olimpiade Matematika SMA. Yogyakarta: ANDI.
Venkatachala, B.J. 2009. Inequalities An Approach Through Problems (2nd). India: SPRINGER.
Vo Tranh Van..... Bat Dang Thuc Schur Va Phuong Phap Doi Bien P, Q, R.
Vo Quoc Ba Can. 2007. Bai Viet Ve Bat Dang Thuc Schur Va Vornicu Schur.
Young, B. 2009. Seri Buku Olimpiade Matematika Strategi Menyelesaikan Soal-Soal Olimpiade Matematika: Ketaksamaan (Inequality). Bandung: PAKAR RAYA.

Soal dan Pembahasan Seleksi Mandiri Madrasah (KSN-S) 2022

$\begin{array}{ll}\\ 1.&\textrm{Jika}\: \: a,b\: \: \textrm{bilangan real positif},\\ &\textrm{tunjukkan bahwa}\: \: a^{a}.b^{b}\geq a^{b}.b^{a}\\\\ &\textbf{Bukti}\\ &\begin{aligned}&\textrm{Diketahui}\: \: a^{a}.b^{b}\geq a^{b}.b^{a}\Leftrightarrow \displaystyle \frac{a^{a}.b^{b}}{a^{b}.b^{a}}\geq 1\\ &\Leftrightarrow \displaystyle \frac{a^{a-b}}{b^{a-b}}\geq 1\Leftrightarrow \left ( \displaystyle \frac{a}{b} \right )^{a-b}\geq 1\\ &\textrm{Selanjutnya akan ada dua kemungkinan}\\ &\textrm{yaitu}:\: a\geq b> 0\: \: \textrm{dan}\: \: b\geq a> 0 \end{aligned}\\ &\begin{array}{|c|l|l|}\hline \textrm{No}&\textrm{Kondisi}&\textrm{Akibat}\\\hline 1.&a\geq b> 0&\displaystyle \frac{a}{b}\geq 1\: \: \textrm{atau}\: \: a-b\geq 0\\ &&\textrm{maka}\: \: \begin{aligned}&\left (\displaystyle \frac{a}{b} \right )^{a-b}\geq 1 \end{aligned}\\\hline 2.&b\geq a> 0&\displaystyle \frac{a}{b}\leq 1\: \: \textrm{atau}\: \: a-b\leq 0\\ &&\textrm{maka}\: \: \begin{aligned}&\left (\displaystyle \frac{a}{b} \right )^{a-b}\geq 1 \end{aligned}\\\hline \end{array}\\ &\textrm{Karena keduanya memiliki hasil yang sama}\\ &\textrm{maka}\: \: a^{a}.b^{b}\geq a^{b}.b^{a}\qquad \blacksquare \end{array}$.

$\begin{array}{ll}\\ 2.&\textrm{Jika}\: \: a=\sqrt{\displaystyle \frac{b}{1-b}}\: \: \textrm{nyatakanlah}\: \: b\: \: \textrm{dalam}\: \: a\\\\ &\textbf{Jawab}:\\ &\begin{aligned}&\textrm{Diketahui}\\ &a=\sqrt{\displaystyle \frac{b}{1-b}}\Leftrightarrow a^{2}=\displaystyle \frac{b}{1-b}\Leftrightarrow a^{2}(1-b)=b\\ &\Leftrightarrow a^{2}-a^{2}b=b\Leftrightarrow b+a^{2}b=a^{2}\\ &\Leftrightarrow b(1-a^{2})=a^{2}\Leftrightarrow b=\color{blue}\displaystyle \frac{a^{2}}{1-a^{2}} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 3.&\textrm{Sederhanakanlah bentuk}\: \: 3.4^{4}+3.4^{4}+3.4^{4}\\\\ &\textbf{Jawab}:\\ &\begin{aligned} &3.4^{4}+3.4^{4}+3.4^{4}=3(3.4^{4})\\ &=9\times 4^{4} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 4.&\textrm{Diketahui}\: \: a^{2}+b^{2}=1\: \: \textrm{dan}\: \: x^{2}+y^{2}=1\\ &\textrm{Silahkan lanjutkan proses berikut}\\ &(a^{2}+b^{2})(x^{2}+y^{2})-(ax+by)^{2}=....\\ &\textrm{a}.\quad \textrm{Bagaimana hubungan}\: \: ax+by\: \: \textrm{dengan}\: \: 1\\ &\textrm{b}.\quad \textrm{Mengapa}\\\\ &\textbf{Jawab}:\\ &\begin{aligned}&\color{blue}\textrm{Perhatikan bahwa}\\ &(a^{2}+b^{2})(x^{2}+y^{2})-(ax+by)^{2}\\ &=a^{2}x^{2}+a^{2}y^{2}+b^{2}x^{2} +b^{2}y^{2}-a^{2}x^{2}-b^{2}y^{2}-2abxy\\ &=a^{2}y^{2}+b^{2}x^{2}-2abxy\\ &=(ay-bx)^{2}\\ &\textrm{a. Nilai}\: \: \color{red}ax+by\: \: \color{black}\textrm{selalu lebih kecil}\\ &\quad\: \textrm{atau sama dengan 1}\\ &\textrm{b. Kita memiliki}\: \: (ay-bx)^{2}\geq 0,\: \: \textrm{sehingga}\\ &\quad\Leftrightarrow \: (a^{2}+b^{2})(x^{2}+y^{2})-(ax+by)^{2}\geq 0\\ &\quad\Leftrightarrow \: \quad 1\times 1-(ax+by)^{2}\geq 0\\ &\quad\Leftrightarrow \: \quad\qquad 1-(ax+by)^{2}\geq 0\\ &\quad\Leftrightarrow \: \quad\qquad (ax+by)^{2}-1\leq 0\\ &\quad\Leftrightarrow \: \: \: \: \: \qquad\qquad (ax+by)^{2}\leq 1\qquad \blacksquare \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 5.&\textrm{Pada segi enam beraturan, berapa banyak}\\ &\textrm{segitiga yang dapat Anda temukan}\\\\ &\textbf{Jawab}:\\ &\begin{aligned}&\textrm{Setiap segitiga dapat terbentuk dari 3 buah}\\ &\textrm{sebarang, sehingga banyak segitiga bila}\\ &\textrm{diketahui}\: \: n=6,\: \: r=3\: \: \textrm{adalah}:\\ &\color{red}\textrm{kombinasi 3 titik dari 6 titik yang ada}\\ &\textrm{Adapun untuk rumus kombinasinya}:\\ &C_{r}^{n}=\begin{pmatrix} n\\ r \end{pmatrix}=\displaystyle \frac{n!}{r!(n-r)!}\\ &\textrm{dengan}\: \: n!=1\times 2\times 3\times 4\times ...\times n\\ &\textrm{maka}\\ &C_{3}^{6}=\begin{pmatrix} 6\\ 3 \end{pmatrix}=\displaystyle \frac{6!}{3!.3!}=\frac{6.5.4.3!}{1.2.3.3!}=\color{blue}20 \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 6.&\textrm{Tentukan nilai dari bentuk}\\ &\displaystyle \frac{1^{-3}+2^{-3}+3^{-3}+4^{-3}+5^{-3}+...}{1^{-3}+3^{-3}+5^{-3}+7^{-3}+9^{-3}+...}\\\\ &\textbf{Jawab}:\\ &\begin{aligned}&\textrm{Misalkan}\\ &x=1^{-3}+2^{-3}+3^{-3}+4^{-3}+5^{-3}+...\: \: \textrm{dan}\\ &y=1^{-3}+3^{-3}+5^{-3}+7^{-3}+9^{-3}+...\\ &\textrm{Sekarang perhatikan bahwa}\\ &x=1^{-3}+2^{-3}+3^{-3}+4^{-3}+5^{-3}+...\\ &\: \: \: =(1^{-3}+3^{-3}+5^{-3}+...)\: +\: (2^{-3}+4^{-3}+6^{-3}+...)\\ &\: \: \: =(1^{-3}+3^{-3}+5^{-3}+...)\: +\: 2^{-3}(1^{-3}+2^{-3}+3^{-3}+...)\\ &\: \: \: =y+\displaystyle \frac{1}{8}x\\ &x-\displaystyle \frac{1}{8}x=y\Leftrightarrow \displaystyle \frac{7}{8}x=y\Leftrightarrow \displaystyle \frac{x}{y}=\color{blue}\frac{8}{7}\\ \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 7.&\textrm{Jika}\: \: 60^{a}=3\: \: \textrm{dan}\: \: 60^{b}=5\\ &\textrm{maka hasil dari}\: \: 12^{^{\left ( \displaystyle \frac{1-x-y}{2(1-b)} \right )}}\\\\ &\textbf{Jawab}:\\ &\color{purple}\textrm{Perhatikanlah bahwa}\\ &\color{blue}\begin{aligned}&\begin{cases} 60^{a}=3 & \Rightarrow \: ^{60}\log 3=a \\ 60^{b}=5 & \Rightarrow \: ^{60}\log 5=b \end{cases} \end{aligned} \\ &\color{purple}\textrm{Selanjutnya}\\ &\color{purple}\textrm{Untuk}\: \: (1-a-b),\: \textrm{maka}\\ &\color{purple}\begin{aligned}1-a-b&=1-\: ^{60}\log 3-\: ^{60}\log 5\\ &=\: ^{60}\log 60-\: ^{60}\log 3-\: ^{60}\log 5\\ &=\: ^{60}\log \displaystyle \frac{60}{3\times 5}\\ &=\: ^{60}\log 4\\ &=\: ^{60}\log 2^{2} \end{aligned} \\ &\color{purple}\textrm{Untuk}\: \: 2(1-b),\: \textrm{maka}\\ &\color{purple}\begin{aligned}2(1-b)&=2\left ( 1-\: ^{60}\log 5 \right )\\ &=2\left ( ^{60}\log 60-\: ^{60}\log 5 \right )\\ &=2\left ( ^{60}\log \displaystyle \frac{60}{5} \right )\\ &=2\left ( ^{60}\log 12 \right )\\ &=\: ^{60}\log 12^{2} \end{aligned} \\ &\color{blue}\textrm{Untuk}\: \: \left ( \displaystyle \frac{1-x-y}{2(1-b)} \right ),\: \textrm{maka}\\ &\color{purple}\begin{aligned}\left ( \displaystyle \frac{1-x-y}{2(1-b)} \right )&=\displaystyle \frac{\: ^{60}\log 2^{2}}{\: ^{60}\log 12^{2}}\\ &=\displaystyle \frac{2\: ^{60}\log 2}{2\: ^{60}\log 12}\\ &=\displaystyle \frac{\: ^{60}\log 2}{\: ^{60}\log 12}\\ &=\: ^{12}\log 2 \end{aligned}\\ &\color{purple}\textrm{Jadi},\: \: 12^{^{\left ( \displaystyle \frac{1-x-y}{2(1-b)} \right )}}=12^{^{\: ^{12}\log 2}}=2 \end{array}$.

$\begin{array}{ll}\\ 8.&\textrm{Tentukan hasil dari bentuk}\\ &\sin ^{2}1^{\circ}+\sin ^{2}2^{\circ}+\sin ^{2}3^{\circ}+...+\sin ^{2}89^{\circ}+\sin ^{2}90^{\circ}\\\\ &\textbf{Jawab}:\\ &\textrm{Ingat sudut-sudut yang berelasi pada trigonometri}:\\ &\color{red}\sin \left ( 90^{0}-\alpha \right )=\cos \alpha \\ &\textrm{Ingat pula identitas trigonometri}:\\ & \color{red}\sin ^{2}\alpha +\cos ^{2}\alpha =1\\ &\underline{\textrm{maka bentuk}}\\ &\sin ^{2}\color{blue}1^{\circ}\color{black}=\sin^{2} \left (\color{blue}90^{\circ}-89^{\circ}\color{black} \right ) =\cos ^{2}89^{\circ},\\ & \textrm{dengan cara yang sama akan diperoleh}\\ &\sin ^{2}\color{blue}2^{\circ}\color{black}=\cos ^{2}88^{\circ}\\ &\sin ^{2}\color{blue}3^{\circ}\color{black}=\cos ^{2}87^{\circ}\\ &\sin ^{2}\color{blue}4^{\circ}\color{black}=\cos ^{2}86^{\circ}\\ &\qquad\qquad \vdots \\ &\sin ^{2}\color{blue}44^{\circ}\color{black}=\cos ^{2}46^{\circ}\\ &\textrm{Sehingga}\\ &\begin{aligned}&\sin ^{2}1^{\circ}+\sin ^{2}2^{\circ}+\sin ^{2}3^{\circ}+...+\sin ^{2}89^{\circ}+\sin ^{2}90^{\circ}\\ &=\sin ^{2}1^{\circ}+...+\sin ^{2}44^{\circ}+\sin ^{2}46^{\circ}+...+\sin ^{2}89^{\circ}\color{red}+\sin ^{2}45^{\circ}+\sin ^{2}90^{\circ}\\ &=\underset{44}{\underbrace{1+1+1+...+1}}\color{red}+\sin ^{2}45^{\circ}+\sin ^{2}90^{\circ}\\ &=44+\displaystyle \frac{1}{2}+1\\ &=\color{blue}45\displaystyle \frac{1}{2} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 9.&\textrm{Tentukan sisa pembagian}\: \: x^{3}-5x^{2}+3x-4\\ &\textrm{oleh}\: \: 2x+1\\\\ &\textbf{Jawab}:\\ &\color{blue}\textbf{Alternatif 1}\\ &\textrm{Misalkan}\: \: f(x)=x^{3}-5x^{2}+3x-4\\ &\textrm{Sisa pembagian}\: \: f(x)\: \: \textrm{oleh}\: \: q(x)=2x+1\: \: \textrm{adalah}:\\ &\textrm{ambil}\: \: q(x)=2x+1=0\Rightarrow x=-\displaystyle \frac{1}{2},\: \: \textrm{maka}\\ &\textrm{penentuan sisa pembagian cukup dengan}\\ &f\left ( -\displaystyle \frac{1}{2} \right )=\left ( -\displaystyle \frac{1}{2} \right )^{3}-5\left ( -\displaystyle \frac{1}{2} \right )^{2}+3\left ( -\displaystyle \frac{1}{2} \right )-4\\ &\qquad\qquad =-\displaystyle \frac{1}{8}-\displaystyle \frac{5}{4}-\frac{3}{2}-4\\ &\qquad\qquad =\displaystyle \frac{-1-10-12-32}{8}=\color{red}-\displaystyle \displaystyle \frac{55}{8}\\ &\color{blue}\textbf{Alternatif 2}\\ &\textrm{Dengan metode Horner, yaitu}:\\ &\begin{array}{c|cccccccccc}\\ x=-\displaystyle \frac{1}{2}&1&-5&3&-4&&&\\ &&&&&&&\\ &&-\displaystyle \frac{1}{2}&\displaystyle \frac{11}{4}&-\displaystyle \frac{23}{8}&&&+\\\hline &1&-5\displaystyle \frac{1}{2}&\displaystyle \frac{23}{4}&\color{red}-\displaystyle \frac{55}{8}&&& \end{array} \end{array}$.

$\begin{array}{ll}\\ 10.&\textrm{Fungsi}\: \: f\: \: \textrm{memiliki sifat untuk tiap bilangan}\\ &\textrm{bilangan real}\: \: x\: \: \textrm{berlaku}\: \: f(x)+f(x-1)=x^{2}\\ &\textrm{Jika}\: \: f(19)=94,\: \textrm{tentukan sisa pembagian}\\ & f(94)\: \: \textrm{oleh}\: \: 100\\\\ &\textbf{Jawab}:\\ &\textrm{Perhatikan bahwa}\: \: f(x)=x^{2}-f(x-1),\\ &\textrm{maka}\\ &\begin{aligned}f(94)&=94^{2}-f(93)=94^{2}-\left ( 93^{2}-f(92) \right )\\ &=94^{2}-93^{2}+f(92)=94^{2}-93^{2}+(92^{2}-f(91))\\ &=94^{2}-93^{2}+92^{2}-91^{2}+f(90)\\ &=94^{2}-93^{2}+92^{2}-91^{2}+90^{2}-\cdots +20^{2}-f(19)\\ &\quad \textrm{ingat bentuk}\: \: \color{red}a^{2}-b^{2}=(a+b)(a-b),\: \color{black}\textrm{maka}\\ &=(94+93)+(92+91)+\cdots +400-94\\ &=\underset{4255}{\underbrace{94+93+...+22+21}}+400-94\\ &=4255+306\\ &=\color{blue}4561 \end{aligned}\\ &\textrm{Sehingga sisa pembagian}\: \: f(94)\: \: \textrm{oleh 100 adalah}\: \: \color{blue}61 \end{array}$.

DAFTAR PUSTAKA

Idris, M., Rusdi, I. 2015. Langkah Awal Meraih Medali Emas Olimpiade Matematika SMA. Bandung: YRAMA WIDYA.
Susyanto, N. 2012. Tutor Senior Olimpiade Matematika Lima Benua Tingkat SMP. Yogyakarta: KENDI MAS MEDIA
..........Η Στήλη των Μαθηματικών, έτος 2007, τεύχη 46-94

Kumpulkan Materi Ketaksamaan

Kumpulan Materi Ketaksamaan

Konsep Ketaksamaan QM-AM-GM-HM
Materi Ketaksamaan QM-AM-GM-HM
Materi Ketaksamaan Cauchy-Schwarz-Engel (CS-E)
Materi Ketaksamaan Renata (Rearrangement)
Materi Ketaksamaan Chebyshev
Materi Ketaksamaan Holder
Materi Ketaksamaan Schur
Materi lanjutan Ketaksamaan Schur
Contoh Soal 1 Ketaksamaan
Contoh Soal 2 Ketaksamaan
Contoh Soal 3 Ketaksamaan
Contoh Soal 4 Ketaksamaan
Contoh Soal 5 Ketaksamaan
Contoh Soal 6 Ketaksamaan
Contoh Soal 7 Ketaksamaan
Contoh Soal 8 Ketaksamaan
Contoh Soal 9 Ketaksamaan
Contoh Soal 10 Ketaksamaan
Contoh Soal 11 Ketaksamaan
Contoh Soal 12 Ketaksamaan
Contoh Soal 13 Ketaksamaan
Contoh Soal 14 Ketaksamaan
Contoh Soal 15 Ketaksamaan

Contoh 2 Soal dan Pembahasan Materi Vektor

$\begin{array}{ll}\\ 6.&\textrm{Diketahui titik}\: \: P(n,2),\: Q(1,-2),\: n>0\\ &\textrm{dan panjang}\: \: \overline{PQ}=5,\: \: \textrm{maka nilai}\: \: n\\ & \textrm{adalah}....\\ &\textrm{a}.\quad 1\\ &\textrm{b}.\quad 2\\ &\textrm{c}.\quad 3\\ &\textrm{d}.\quad \color{red}4\\ &\textrm{e}.\quad 5\\\\ &\textrm{Jawab}\\ &\begin{aligned}\overline{PQ}&=5\\ \sqrt{(x_{Q}-x_{P})^{2}+(y_{Q}-y_{P})^{2}}&=5\\ (x_{Q}-x_{P})^{2}+(y_{Q}-y_{P})^{2}&=25\\ (1-n)^{2}+(-2-2)^{2}&=25\\ (1-n)^{2}+16&=25\\ (1-n)^{2}-3^{2}&=0\\ (1+3-n)(1-3-n)&=0\\ \color{red}n=4\: \: \color{black}\textrm{atau}\: \: \color{red}n=-2& \end{aligned} \end{array}$

$\begin{array}{ll}\\ 7.&\textrm{Diketahui vektor}\: \: \vec{u}=\begin{pmatrix} 3\\ 4 \end{pmatrix}\: \: \textrm{dan}\: \: \vec{v}=\begin{pmatrix} 2\\ -1 \end{pmatrix}.\\ & \textrm{Nilai}\: \: \left | \vec{u}+\vec{v} \right |\: \: \textrm{adalah}....\\ &\textrm{a}.\quad \sqrt{28}\\ &\textrm{b}.\quad \sqrt{30}\\ &\textrm{c}.\quad \color{red}\sqrt{34}\\ &\textrm{d}.\quad \sqrt{44}\\ &\textrm{e}.\quad \sqrt{50}\\\\ &\textrm{Jawab}\\ &\begin{aligned}\vec{u}+\vec{v}&=\begin{pmatrix} 3\\ 4 \end{pmatrix} + \begin{pmatrix} 2\\ -1 \end{pmatrix}\\ &=\begin{pmatrix} 3+2\\ 4+(-1) \end{pmatrix}\\ &=\begin{pmatrix} 5\\ 3 \end{pmatrix}\\ \left | \vec{u}+\vec{v} \right |&=\sqrt{5^{2}+3^{2}}\\ &=\sqrt{25+9}\\ &=\color{red}\sqrt{34} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 8.&\textrm{Vektor satuan}\: \: \vec{u}=\begin{pmatrix} -5\\ -12 \end{pmatrix}\: \: \textrm{adalah}.... \\ &\textrm{a}.\quad \color{red}\displaystyle -\frac{1}{13}\begin{pmatrix} 5\\ 12 \end{pmatrix}\\ &\textrm{b}.\quad \displaystyle \frac{1}{15}\begin{pmatrix} -5\\ -12 \end{pmatrix}\\ &\textrm{c}.\quad \displaystyle -\frac{1}{17}\begin{pmatrix} 5\\ 12 \end{pmatrix}\\ &\textrm{d}.\quad \displaystyle -\frac{1}{17}\begin{pmatrix} 5\\ 12 \end{pmatrix}\\ &\textrm{e}.\quad \displaystyle \frac{1}{2} \begin{pmatrix} 5\\ 12 \end{pmatrix}\\\\ &\textrm{Jawab}\\ &\begin{aligned}\vec{e}_{\vec{u}}&=\displaystyle \frac{\vec{u}}{\left | \vec{u} \right |},\quad \textrm{maka}\\ \vec{e}_{\begin{pmatrix} -5\\ -12 \end{pmatrix}}&=\displaystyle \frac{\begin{pmatrix} -5\\ -12 \end{pmatrix}}{\left | \begin{pmatrix} -5\\ -12 \end{pmatrix} \right |}\\ &=\displaystyle \frac{\begin{pmatrix} -5\\ -12 \end{pmatrix}}{\sqrt{(-5)^{2}+(-12)^{2}}}\\ &=\displaystyle \frac{-\begin{pmatrix} 5\\ 12 \end{pmatrix}}{\sqrt{169}}\\ &=\color{red}-\displaystyle \frac{1}{13}\begin{pmatrix} 5\\ 12 \end{pmatrix} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 9.&\textrm{Jika vektor}\: \: \vec{p}=\begin{pmatrix} 8\\ 7 \end{pmatrix}\: \: \textrm{dan}\: \: \vec{q}=\begin{pmatrix} -3\\ 9 \end{pmatrix}\\ & \textrm{Hasil dari}\: \: \vec{p}+\vec{q}\: \: \textrm{adalah}....\\ &\textrm{a}.\quad \begin{pmatrix} 6\\ 14 \end{pmatrix}\\ &\textrm{b}.\quad \begin{pmatrix} 6\\ 13 \end{pmatrix}\\ &\textrm{c}.\quad \begin{pmatrix} 6\\ 15 \end{pmatrix}\\ &\textrm{d}.\quad \color{red}\begin{pmatrix} 5\\ 16 \end{pmatrix}\\ &\textrm{e}.\quad \begin{pmatrix} 5\\ 48 \end{pmatrix}\\\\ &\textrm{Jawab}\\ &\begin{aligned}\vec{p}+\vec{q}&=\begin{pmatrix} 8\\ 7 \end{pmatrix}+\begin{pmatrix} -3\\ 9 \end{pmatrix}\\ &=\begin{pmatrix} 8-3\\ 7+9 \end{pmatrix}\\ &=\color{red}\begin{pmatrix} 5\\ 16 \end{pmatrix} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 10.&\textrm{Jika vektor}\: \: \vec{p}=\begin{pmatrix} ^{2}\log 32\\ -\, ^{3}\log \displaystyle \frac{1}{81} \end{pmatrix}\: \: \textrm{dan}\: \: \vec{q}=\begin{pmatrix} -9\\ -21 \end{pmatrix}\\ & \textrm{Hasil dari}\: \: \vec{p}+\vec{q}\: \: \textrm{adalah}....\\ &\textrm{a}.\quad \begin{pmatrix} 6\\ 17 \end{pmatrix}\\ &\textrm{b}.\quad \begin{pmatrix} -6\\ -17 \end{pmatrix}\\ &\textrm{c}.\quad \begin{pmatrix} 4\\ 17 \end{pmatrix}\\ &\textrm{d}.\quad \color{red}\begin{pmatrix} -4\\ -17 \end{pmatrix}\\ &\textrm{e}.\quad \begin{pmatrix} 5\\ -16 \end{pmatrix}\\\\ &\textrm{Jawab}\\ &\begin{aligned}\vec{p}+\vec{q}&=\begin{pmatrix} ^{2}\log 32\\ -\, ^{3}\log \displaystyle \frac{1}{81} \end{pmatrix}+\begin{pmatrix} -9\\ -21 \end{pmatrix}\\ &=\begin{pmatrix} 5-9\\ 4-21 \end{pmatrix}\\ &=\color{red}\begin{pmatrix} -4\\ -17 \end{pmatrix} \end{aligned} \end{array}$

Contoh 1 Soal dan Pembahasan Materi Vektor

$\begin{array}{ll}\\ 1.&\textrm{Perhatikanlah ilustrasi gambar berikut} \end{array}$

$\begin{array}{ll}\\ .\, \quad&\textrm{maka vektor}\: \: \vec{u}\: \: \textrm{adalah}\: ....\\ &\begin{array}{llllllll}\\ \textrm{a}.&3\vec{i}+5\vec{j}&&&\textrm{d}.&-3\vec{i}-5\vec{j}\\ \textrm{b}.&5\vec{i}+3\vec{j}&\textrm{c}.&\color{red}-3\vec{i}+5\vec{j}&\textrm{e}.&-5\vec{i}+3\vec{j} \end{array}\\\\ &\textbf{Jawab}\\ &\textrm{Kita perhatikan lagi gambarnya}\\ &\begin{aligned}&\textrm{Vektor}\: \: \vec{u}\: \: \textrm{jika dinyatakan sebagai }\\ &\textrm{kombinasi linear adalah}\: \: \vec{u}=\color{red}-3\vec{i}+5\vec{j} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 2.&\textrm{Panjang vektor}\: \: \vec{p}=\begin{pmatrix} 4\\ -8 \end{pmatrix}\: \: \textrm{adalah}....\\ &\textrm{a}.\quad \displaystyle \sqrt{4}\\ &\textrm{b}.\quad \sqrt{12}\\ &\textrm{b}.\quad \sqrt{20}\\ &\textrm{d}.\quad \color{red}\displaystyle \sqrt{80}\\ &\textrm{e}.\quad \sqrt{100}\\\\ &\textrm{Jawab}\\ &\begin{aligned}\vec{p}&=\begin{pmatrix} 4\\ -8 \end{pmatrix},\: \: \textrm{maka besar dari vektor}\\ \vec{p}&\: \: \textrm{adalah}=\left | \vec{p} \right |=\sqrt{x^{2}+y^{2}}\\ &\textrm{Yaitu}\\ \left | \vec{p} \right |&=\sqrt{4^{2}+(-8)^{2}}\\ &=\sqrt{16+64}=\color{red}\sqrt{80} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 3.&\textrm{Perhatikanlah gambar berikut}\\ & \end{array}$.

$\begin{array}{ll}\\ .\quad &\textrm{Panjang vektor}\: \: \vec{h}\: \: \textrm{tersebut di atas adalah}....\\ &\textrm{a}.\quad 5\\ &\textrm{b}.\quad 7\\ &\textrm{c}.\quad \color{red}10\\ &\textrm{d}.\quad 12\\ &\textrm{e}.\quad 15\\\\ &\textrm{Jawab}\\ &\begin{aligned}\textrm{Diketahui}&\quad \vec{h}=\overline{OH}=8\vec{i}+6\vec{j}\\ \vec{h}&=\sqrt{x_{H}^{2}+y_{H}^{2}}\\ &=\sqrt{8^{2}+6^{2}}\qquad \textbf{(ingat tripel Pythagoras)}\\ &=\sqrt{10^{2}}=\color{red}10 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 4.&\textrm{Vektor satuan dari}\: \: \vec{q}=3\vec{i}-4\vec{j}\: \: \textrm{adalah}....\\ &\textrm{a}.\quad \displaystyle \frac{4}{5}\vec{i}-\frac{3}{5}\vec{j}\\ &\textrm{b}.\quad \color{red}\displaystyle \frac{3}{5}\vec{i}-\frac{4}{5}\vec{j}\\ &\textrm{c}.\quad 3\vec{i}-4\vec{j}\\ &\textrm{d}.\quad 4\vec{i}-3\vec{j}\\ &\textrm{e}.\quad 15\vec{i}-20\vec{j}\\\\ &\textrm{Jawab}\\ &\begin{aligned}\vec{q}&=3\vec{i}-4\vec{j}\\ &\textrm{Vektor satuan dari vektor}\: \: \vec{q}\: \: \textrm{adalah}:\\ &\: \hat{e}_{_{\vec{q}}}=\displaystyle \frac{1}{\left | \vec{q} \right |}. \vec{q}\\ &\textrm{Sehingga}\\ \hat{e}_{_{\vec{q}}}&=\displaystyle \frac{1}{\sqrt{3^{2}+(-4)^{2}}}.\begin{pmatrix} 3\\ -4 \end{pmatrix}\\ &=\displaystyle \frac{1}{5}\begin{pmatrix} 3\\ -4 \end{pmatrix}\\ &\textrm{atau dalam vektor basis}\\ &=\color{red}\displaystyle \frac{3}{5}\vec{i}-\frac{4}{5}\vec{j} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 5.&\textrm{Vektor berikut yang memiliki panjang}\\ & 29\: \: \textrm{satuan adalah}....\\ &\textrm{a}.\quad \displaystyle 18\vec{i}-19\vec{j}\\ &\textrm{b}.\quad \displaystyle 19\vec{i}-20\vec{j}\\ &\textrm{c}.\quad 20\vec{i}-21\vec{j}\\ &\textrm{d}.\quad \color{red}21\vec{i}-22\vec{j}\\ &\textrm{e}.\quad 22\vec{i}-23\vec{j}\\\\ &\textrm{Jawab}\\ &\begin{aligned}\textrm{Ingat}&\textrm{lah akan tigaan Pythagoras}\\ &\begin{cases} (3,4,5) &\Rightarrow 3^{2}+4^{2}=5^{2} \\ (5,12,13) & \Rightarrow 5^{2}+12^{2}=13^{2} \\ (8,15,17) &\Rightarrow \cdots \\ (20,21,29)&\Rightarrow \cdots \\ \vdots & dll \end{cases}\\ \textrm{Sehin}&\textrm{gga}\\ \textrm{yang}&\: \: \textrm{paling mungkin adalah}\: : \: \\ &=\sqrt{20^{2}+21^{2}}=\sqrt{400+441}\\ &=\sqrt{841}\\ &=\color{red}29 \end{aligned} \end{array}$