Contoh 7 Vektor

$\begin{array}{ll}\\ 31.&\textrm{Jika}\: \: \vec{p}=\begin{pmatrix} 2\\ -5 \end{pmatrix},\: \vec{q}=\begin{pmatrix} 4\\ 3 \end{pmatrix},\: \textrm{maka }\\ &\textrm{proyeksi skalar ortogonal vektor}\: \vec{p}\\ &\textrm{pada}\: \: \vec{q}\: \: \textrm{adalah}....\\ &\textrm{a}.\quad \displaystyle \frac{3}{5}\\\\ &\textrm{b}.\quad \color{red}\displaystyle \frac{7}{5}\\\\ &\textrm{c}.\quad \displaystyle \frac{8}{5}\\\\ &\textrm{d}.\quad \displaystyle \frac{9}{5}\\\\ &\textrm{e}.\quad \displaystyle 2\\\\ &\textrm{Jawab}\\ &\begin{aligned}\left | \vec{r} \right |&=\displaystyle \frac{\vec{p}.\vec{q}}{\left | \vec{q} \right |}\\ &=\displaystyle \frac{\begin{pmatrix} 2\\ -5 \end{pmatrix}.\begin{pmatrix} 4\\ 3 \end{pmatrix}}{\sqrt{4^{2}+3^{2}}}\\ &=\displaystyle \frac{8+(-15)}{\sqrt{25}}\\ &=\left | \displaystyle \frac{-7}{5} \right |\\ &=\color{red}\displaystyle \frac{7}{5} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 32.&\textrm{Panjang Proyeksi vektor}\: \: \vec{a}=\begin{pmatrix} 5\\ 1 \end{pmatrix}\\ & \textrm{pada}\: \: \vec{b}=\begin{pmatrix} 0\\ 4 \end{pmatrix}\: \: \textrm{adalah}....\\ &\textrm{a}.\quad \displaystyle -1\\ &\textrm{b}.\quad -\displaystyle \frac{1}{2}\\ &\textrm{c}.\quad \color{red}1\\ &\textrm{d}.\quad \displaystyle 2\\ &\textrm{e}.\quad 4\\\\ &\textrm{Jawab}\\ &\begin{aligned}\left |\vec{c} \right |&=\left |\displaystyle \frac{\vec{a}.\vec{b}}{\left | \vec{b} \right |} \right |\\ &=\left |\displaystyle \frac{\begin{pmatrix} 5\\ 1 \end{pmatrix}\bullet \begin{pmatrix} 0\\ -4 \end{pmatrix}}{\left | \sqrt{0^{2}+(-4)^{2}} \right |} \right |\\ &=\left |\displaystyle \frac{0-4}{4} \right |=\left |-1 \right |=\color{red}1 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 33.&\textrm{Proyeksi vektor ortogonal}\: \: \vec{a}=\begin{pmatrix} 2\\ -4 \end{pmatrix}\\ & \textrm{pada}\: \: \vec{b}=\begin{pmatrix} -1\\ 2 \end{pmatrix}\: \: \textrm{adalah}....\\ &\textrm{a}.\quad \begin{pmatrix} 2\\ -1 \end{pmatrix}\\ &\textrm{b}.\quad \begin{pmatrix} 2\\ -2 \end{pmatrix}\\ &\textrm{c}.\quad \color{red}\begin{pmatrix} 2\\ -4 \end{pmatrix}\\ &\textrm{d}.\quad \begin{pmatrix} -1\\ 2 \end{pmatrix}\\ &\textrm{e}.\quad \begin{pmatrix} -2\\ 4 \end{pmatrix}\\\\ &\textrm{Jawab}\\ &\begin{aligned}\vec{c}&=\left ( \displaystyle \frac{\vec{a}\bullet \vec{b}}{\left |\vec{b} \right |^{2}} \right ).\vec{b}\\ &=\left (\displaystyle \frac{\begin{pmatrix} 2\\ -4 \end{pmatrix}\bullet \begin{pmatrix} -1\\ 2 \end{pmatrix}}{(-1)^{2}+2^{2}} \right ).\begin{pmatrix} -1\\ 2 \end{pmatrix}\\ &=\left (\displaystyle \frac{-2-8}{1+4} \right ).\begin{pmatrix} -1\\ 2 \end{pmatrix}\\ &=-2\begin{pmatrix} -1\\ 2 \end{pmatrix}\\ &=\color{red}\begin{pmatrix} 2\\ -4 \end{pmatrix} \end{aligned} \end{array}$

Contoh 6 Vektor

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

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

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

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

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

$\begin{array}{ll}\\ 30.&\textrm{Diketahui}\: \: \left | \vec{a} \right |=\sqrt{3}\: ,\left | \vec{b} \right |=1,\: \: \textrm{dan}\: \: \left | \vec{a}-\vec{b} \right |=1\\ &\textrm{maka panjang vektor}\: \: \vec{a}+\vec{b}\: \: \textrm{adalah}\: ....\\ &\begin{array}{llllllll}\\ \textrm{a}.&\sqrt{3}&&&\textrm{d}.&2\sqrt{2}\\ \textrm{b}.&\sqrt{5}&\textrm{c}.&\color{red}\sqrt{7}&\textrm{e}.&3 \end{array}\\\\ &\textbf{Jawab}\\ &\begin{aligned}\textrm{Diketahui}&\: \, \textrm{sebagaimana pada soal}\\ \left | \vec{a}-\vec{b} \right |^{2}&=\left | \vec{a} \right |^{2}+\left | \vec{b} \right |^{2}-2\left | \vec{a} \right |\left | \vec{b} \right |\cos \theta \\ 1^{2}&=\left ( \sqrt{3} \right )^{2}+1^{2}-2.\sqrt{3}.1.\cos \theta \\ 2\sqrt{3}\cos \theta &=3\\ \textrm{maka pan}&\textrm{jang vektor}\: \: \vec{a}+\vec{b}\: \: \textrm{adalah}\\ \left | \vec{a}+\vec{b} \right |&=\sqrt{\left | \vec{a} \right |^{2}+\left | \vec{b} \right |^{2}+2\left | \vec{a} \right |\left | \vec{b} \right |\cos \theta}\\ &=\sqrt{\left ( \sqrt{3} \right )^{2}+1^{2}+3}\\ &=\sqrt{3+1+3}\\ &=\color{red}\sqrt{7} \end{aligned} \end{array}$

Contoh 5 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 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 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 2 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 Vektor

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

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

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

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

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

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

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

Contoh Soal 2 Limit Fungsi Aljabar

$\begin{array}{ll}\\ 6.&\textrm{Diketahui bahwa}\: \: f(x)=x^{2}-2,\\ & \textrm{maka nilai}\: \: \: \underset{h\rightarrow 0 }{\textrm{Lim}}\: \: \displaystyle \frac{f(x+h)-f(x)}{h}=....\\ &\begin{array}{lll}\\ \textrm{a}.\quad \displaystyle x^{2}-2&&\textrm{d}.\quad \displaystyle x\\\\ \textrm{b}.\quad \displaystyle x^{2}\quad &\textrm{c}.\quad \displaystyle 2x\quad &\textrm{e}.\quad \displaystyle 2x-2 \end{array}\\\\ &\textrm{Jawab}:\\ &\begin{aligned}\textrm{Diketahui}&\: \textrm{bahwa}\: f(x)=x^{2}-2,\\ \textrm{maka nila}&\textrm{i untuk}\\ \underset{h\rightarrow 0 }{\textrm{Lim}}\: &\: \displaystyle \frac{f(x+h)-f(x)}{h}\, \, \, \, \\ &=\underset{h\rightarrow 0 }{\textrm{Lim}}\: \: \displaystyle \frac{\left ((x+h)^{2}-2 \right )-\left ( x^{2}-2 \right )}{h}\\ &=\underset{h\rightarrow 0 }{\textrm{Lim}}\: \: \displaystyle \displaystyle \frac{x^{2}+2xh+h^{2}-2-x^{2}+2}{h}\\ &=\underset{h\rightarrow 0 }{\textrm{Lim}}\: \: \displaystyle \displaystyle \frac{2xh+h^{2}}{h}\\ &=\underset{h\rightarrow 0 }{\textrm{Lim}}\: \: \displaystyle \left (2x+h \right )\\ &=\color{red}2x \end{aligned} \end{array}$

$\begin{array}{ll}\\ 7.&\textrm{Diketahui}\: \: f(x)=\sqrt{x-1},\\ & \textrm{maka nilai}\: \: \: \underset{h\rightarrow 0 }{\textrm{Lim}}\: \:\displaystyle \frac{f(2+h)-f(2)}{h}=....\\ &\begin{array}{lll}\\ \textrm{a}.\quad \color{red}\displaystyle \frac{1}{2}&&\textrm{d}.\quad \displaystyle 1\\\\ \textrm{b}.\quad -\displaystyle \frac{1}{2}\quad &\textrm{c}.\quad \displaystyle 0\quad &\textrm{e}.\quad \displaystyle -1\end{array}\\\\ &\textrm{Jawab}:\\ &\begin{aligned}\textrm{Diketa}&\textrm{hui bahwa}\: f(x)=\sqrt{x-1},\\ \textrm{maka}\: \: & \textrm{nilai untuk}\\ \underset{h\rightarrow 0 }{\textrm{Lim}}\: &\: \displaystyle \frac{f(2+h)-f(2)}{h}\\ &=\underset{h\rightarrow 0 }{\textrm{Lim}}\: \: \displaystyle \frac{\sqrt{(2+h)-1}-\sqrt{2-1}}{h}\\ &=\underset{h\rightarrow 0 }{\textrm{Lim}}\: \: \displaystyle \displaystyle \frac{\sqrt{h+1}-1}{h}\\ &=\underset{h\rightarrow 0 }{\textrm{Lim}}\: \: \displaystyle \displaystyle \frac{\sqrt{h+1}-1}{h}\times \displaystyle \frac{\sqrt{h+1}+1}{\sqrt{h+1}+1}\\ &=\underset{h\rightarrow 0 }{\textrm{Lim}}\: \: \displaystyle \frac{(h+1)-1}{h\times \left ( \sqrt{h+1}+1 \right )}\\ &=\underset{h\rightarrow 0 }{\textrm{Lim}}\: \: \displaystyle \frac{1}{\left ( \sqrt{h+1}+1 \right )}\\ &=\displaystyle \frac{1}{\sqrt{0+1}+1}\\ &=\color{red}\displaystyle \frac{1}{2} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 8.&\textbf{(Mat Das SIMAK UI 2013)}\\ &\textrm{Nilai}\: \: \underset{x\rightarrow 5}{\textrm{Lim}}\: \: \displaystyle \frac{\sqrt{x+2\sqrt{x+1}}}{\sqrt{x-2\sqrt{x+1}}}=....\\ &\begin{array}{lll}\\ \textrm{a}.\quad \sqrt{3}+\sqrt{2}&&\textrm{d}.\quad 5\\ \textrm{b}.\quad 5-2\sqrt{6}\quad &\textrm{c}.\quad 2\sqrt{6}\quad &\textrm{e}.\quad \color{red}5+2\sqrt{6}\end{array}\\\\ &\textrm{Jawab}:\\ & \begin{aligned}\underset{x\rightarrow 5}{\textrm{Lim}}\: &\: \displaystyle \frac{\sqrt{x+2\sqrt{x+1}}}{\sqrt{x-2\sqrt{x+1}}}\\ &=\displaystyle \frac{\sqrt{5+2\sqrt{5+1}}}{\sqrt{5-2\sqrt{5+1}}}\\ &=\displaystyle \frac{\sqrt{5+2\sqrt{6}}}{\sqrt{5-2\sqrt{6}}}\\ &=\displaystyle \frac{\sqrt{3+2+2\sqrt{3.2}}}{\sqrt{3+2-2\sqrt{3.2}}}\\ &=\displaystyle \frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}\\ &=\displaystyle \frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}\times \displaystyle \frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}+\sqrt{2}}\\ &=\displaystyle \frac{3+2+2\sqrt{6}}{3-2}\\ &=\color{red}5+2\sqrt{6} \end{aligned}\end{array}$

$\begin{array}{ll}\\ 9.&\textbf{(Mat IPA SBMPTN 2014)}\\\\ &\textrm{Jika}\: \: \underset{x\rightarrow a}{\textrm{Lim}}\: \: \displaystyle \left ( f(x)+\frac{1}{g(x)} \right )=4\\ & \textrm{dan}\: \: \underset{x\rightarrow a}{\textrm{Lim}}\: \: \displaystyle \left ( f(x)-\frac{1}{g(x)} \right )=-3,\\\\ &\textrm{maka nilai}\: \: \underset{x\rightarrow a}{\textrm{Lim}}\: \: \displaystyle f(x).g(x)=....\\ &\begin{array}{lll}\\ \textrm{a}.\quad \displaystyle \frac{1}{14}&&\textrm{d}.\quad \displaystyle \frac{4}{14}\\\\ \textrm{b}.\quad \color{red}\displaystyle \frac{2}{14}\quad &\textrm{c}.\quad \displaystyle \frac{3}{14}\quad &\textrm{e}.\quad \displaystyle \frac{5}{14}\end{array}\\\\ &\textrm{Jawab}:\\ &\begin{aligned}&\textrm{Perhatikan bahwa}\: ,\\ &\begin{array}{lll}\\ \underset{x\rightarrow a}{\textrm{Lim}}\: \: \displaystyle \left ( f(x)+\frac{1}{g(x)} \right )=4&&\\ &&\\ \underset{x\rightarrow a}{\textrm{Lim}}\: \: \displaystyle \left ( f(x)-\frac{1}{g(x)} \right )=-3&+&\\ &&\\\hline &&\\ 2\underset{x\rightarrow a}{\textrm{Lim}}\: \: \displaystyle f(x)\qquad\qquad\: \: =1&&\\ &&\\ \qquad\qquad\: \: \: \: \underset{x\rightarrow a}{\textrm{Lim}}\: \: f(x)=\displaystyle \frac{1}{2},&&\\\end{array}\\ &\textrm{sehingga}\: \: \underset{x\rightarrow a}{\textrm{Lim}}\: \: f(g)=\displaystyle \frac{2}{7}\\ &&\\ &\textrm{maka},\\ &\underset{x\rightarrow a}{\textrm{Lim}}\: \: f(x).g(x)=\displaystyle \frac{1}{2}\times \frac{2}{7}=\color{red}\displaystyle \frac{2}{14} \end{aligned} \end{array}$

Contoh Soal 1 Limit Fungsi Aljabar

$\begin{array}{ll}\\ 1&\textrm{Nilai}\: \: \underset{x\rightarrow 2}{\textrm{Lim}}\: \left ( \displaystyle \frac{6-x}{x^{2}-4}-\frac{1}{x-2} \right )=....\\ &\begin{array}{lll}\\ \textrm{a}.\quad \color{red}-\displaystyle \frac{1}{2}&&\textrm{d}.\quad \displaystyle \frac{1}{4}\\\\ \textrm{b}.\quad -\displaystyle \frac{1}{4}\quad &\textrm{c}.\quad 0\quad &\textrm{e}.\quad \displaystyle \frac{1}{2}\end{array}\\\\ &\textrm{Jawab}:\\ &\begin{aligned}\underset{x\rightarrow 2}{\textrm{Lim}}\: \left ( \displaystyle \frac{6-x}{x^{2}-4}-\frac{1}{x-2} \right )&= \left ( \displaystyle \frac{6-2}{2^{2}-4}-\frac{1}{2-2} \right )\\ &=\left ( \displaystyle \frac{4}{0}-\frac{1}{0} \right )=\color{blue}\infty -\infty \\ &\: \: \textrm{hal ini tidak diperkenankan}\\ \textrm{Sehingga},\, \qquad\qquad\qquad &\\ \underset{x\rightarrow 2}{\textrm{Lim}}\: \left ( \displaystyle \frac{6-x}{x^{2}-4}-\frac{1}{x-2} \right )&=\underset{x\rightarrow 2}{\textrm{Lim}}\: \left ( \displaystyle \frac{6-x}{x^{2}-4}-\frac{(x+2)}{(x-2)(x+2)} \right )\\ &=\underset{x\rightarrow 2}{\textrm{Lim}}\: \left ( \displaystyle \frac{6-x}{x^{2}-4}-\frac{x+2}{x^{2}-4} \right )\\ &=\underset{x\rightarrow 2}{\textrm{Lim}}\: \left ( \displaystyle \frac{4-2x}{x^{2}-4} \right )\\ &=\underset{x\rightarrow 2}{\textrm{Lim}}\: \left ( \displaystyle \frac{-2(x-2)}{(x+2)(x-2)} \right )\\ &=\underset{x\rightarrow 2}{\textrm{Lim}}\: \left ( \displaystyle \frac{-2}{x+2} \right )\\ &=\displaystyle -\frac{2}{(2+2)}\\ &=\color{red}-\displaystyle \frac{1}{2} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 2.&\textrm{Nilai}\: \: \underset{x\rightarrow 4}{\textrm{Lim}}\: \: \displaystyle \frac{x-4}{2\sqrt{x}-x}=....\\ &\begin{array}{lll}\\ \textrm{a}.\quad \color{red}-\displaystyle 2&&\textrm{d}.\quad \displaystyle \frac{1}{2}\\\\ \textrm{b}.\quad -\displaystyle \frac{1}{2}\quad &\textrm{c}.\quad 0\quad &\textrm{e}.\quad \displaystyle 2\end{array}\\\\ &\textrm{Jawab}:\\ &\begin{aligned}\underset{x\rightarrow 4}{\textrm{Lim}}\: \: \displaystyle \frac{x-4}{2\sqrt{x}-x}&= \left ( \displaystyle \frac{4-4}{2^{4}-4} \right )\\ &=\color{blue}\displaystyle \frac{0}{0} \\ &\: \: \textrm{hal ini juga tidak diperkenankan}\\ \textrm{Sehingga},\: \: \: \: \quad&\\ \underset{x\rightarrow 4}{\textrm{Lim}}\: \: \displaystyle \frac{x-4}{2\sqrt{x}-x}&=\underset{x\rightarrow 4}{\textrm{Lim}}\: \: \displaystyle \frac{\left ( \sqrt{x}+2 \right )\left ( \sqrt{x}-2 \right )}{\sqrt{x}\left ( 2-\sqrt{x} \right )} \\ &=\underset{x\rightarrow 4}{\textrm{Lim}}\: \: \displaystyle \frac{\left ( \sqrt{x}+2 \right )\left ( \sqrt{x}-2 \right )}{-\sqrt{x}\left ( \sqrt{x}-2 \right )}\\ &=\underset{x\rightarrow 4}{\textrm{Lim}}\: \: -\displaystyle \frac{\left ( \sqrt{x}+2 \right )}{\sqrt{x}}\\ &=-\displaystyle \frac{\sqrt{4}+2}{\sqrt{4}}\\ &=-\displaystyle \frac{2+2}{2}\\ &=\color{red}-2 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 3.&\textrm{Nilai}\: \: \underset{x\rightarrow 1}{\textrm{Lim}}\: \: \displaystyle \frac{\left ( 2x-3\sqrt{x}+1 \right )\left ( \sqrt{x}-1 \right )}{\left ( \sqrt{x}-1 \right )^{2}}=....\\ &\begin{array}{lll}\\ \textrm{a}.\quad \displaystyle \frac{1}{4}&&\textrm{d}.\quad \displaystyle 2\\\\ \textrm{b}.\quad \displaystyle \frac{1}{2}\quad &\textrm{c}.\quad \color{red}1\quad &\textrm{e}.\quad \displaystyle 4\end{array}\\\\ &\textrm{Jawab}:\\ &\begin{aligned}\underset{x\rightarrow 1}{\textrm{Lim}}\:& \: \displaystyle \frac{\left ( 2x-3\sqrt{x}+1 \right )\left ( \sqrt{x}-1 \right )}{\left ( \sqrt{x}-1 \right )^{2}}\\ &= \displaystyle \frac{\left ( 2-3+1 \right )\left ( 1-1 \right )}{\left ( 1-1 \right )^{2}} \\ &=\displaystyle \frac{0\times 0}{0^{2}}\\ &=\color{blue}\displaystyle \frac{0}{0} \\ &\: \: \textrm{hal ini juga tidak diperkenankan}\\ \textrm{Sehi}&\textrm{ngga},\\ \underset{x\rightarrow 1}{\textrm{Lim}}\: &\: \displaystyle \frac{\left ( 2x-3\sqrt{x}+1 \right )\left ( \sqrt{x}-1 \right )}{\left ( \sqrt{x}-1 \right )^{2}}\\ &=\underset{x\rightarrow 1}{\textrm{Lim}}\: \: \displaystyle \frac{\left ( \left ( 2\sqrt{x}-1 \right )\times \left ( \sqrt{x}-1 \right ) \right )\left ( \sqrt{x}-1 \right )}{\left ( \sqrt{x}-1 \right )\left ( \sqrt{x}-1 \right )}\\ &=\underset{x\rightarrow 1}{\textrm{Lim}}\: \: \left ( 2\sqrt{x}-1 \right )\\ &=2.1-1\\ &=2-1\\ &=\color{red}1 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 4.&\textrm{Nilai}\: \: \underset{x\rightarrow 3}{\textrm{Lim}}\: \: \displaystyle \frac{\sqrt{x+4}-\sqrt{2x+1}}{x-3}=....\\ &\begin{array}{lll}\\ \textrm{a}.\quad \color{red}-\displaystyle \frac{1}{14}\sqrt{7}&&\textrm{d}.\quad \displaystyle \frac{1}{7}\sqrt{7}\\\\ \textrm{b}.\quad -\displaystyle \frac{1}{7}\sqrt{7}\quad &\textrm{c}.\quad 0\quad &\textrm{e}.\quad \displaystyle \frac{1}{14}\sqrt{7}\end{array}\\\\ &\textrm{Jawab}:\\ &\begin{aligned}\underset{x\rightarrow 3}{\textrm{Lim}}\: &\: \displaystyle \frac{\sqrt{x+4}-\sqrt{2x+1}}{x-3}\\ &=\underset{x\rightarrow 3}{\textrm{Lim}}\: \: \displaystyle \frac{\sqrt{x+4}-\sqrt{2x+1}}{x-3}\times \displaystyle \frac{\sqrt{x+4}+\sqrt{2x+1}}{\sqrt{x+4}+\sqrt{2x+1}}\\ &=\underset{x\rightarrow 3}{\textrm{Lim}}\: \: \displaystyle \frac{\left ( x+4 \right )-\left ( 2x+1 \right )}{\left ( x-3 \right )\left ( \sqrt{x+4}+\sqrt{2x+1} \right )}\\ &=\underset{x\rightarrow 3}{\textrm{Lim}}\: \: -\displaystyle \frac{-x+3}{\left ( x-3 \right )\left ( \sqrt{x+4}+\sqrt{2x+1} \right )}\\ &=\underset{x\rightarrow 3}{\textrm{Lim}}\: \: \displaystyle \frac{-1}{\left ( \sqrt{x+4}+\sqrt{2x+1} \right )}\\ &=-\displaystyle \frac{1}{\left ( \sqrt{7}+\sqrt{7} \right )}\\ &=-\displaystyle \frac{1}{2\sqrt{7}}\\ &=-\displaystyle \frac{1}{2\sqrt{7}}\times \displaystyle \frac{\sqrt{7}}{\sqrt{7}}\\ &=\color{red}-\displaystyle \frac{1}{14}\sqrt{7} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 5.&\textrm{Jika}\: \: \underset{x\rightarrow 2}{\textrm{Lim}}\: \: \displaystyle \frac{ax-2a}{\sqrt{2x}-x}=6,\: \: \textrm{maka nilai}\: \: a\: \: \textrm{adalah}....\\ &\begin{array}{lll}\\ \textrm{a}.\quad \displaystyle 2&&\textrm{d}.\quad \displaystyle -2\\\\ \textrm{b}.\quad \displaystyle 1\quad &\textrm{c}.\quad -1\quad &\textrm{e}.\quad \displaystyle \color{red}-3\end{array}\\\\ &\textrm{Jawab}:\\ &\begin{aligned}\underset{x\rightarrow 2}{\textrm{Lim}}\: &\: \displaystyle \frac{ax-2a}{\sqrt{2x}-x}=6\\ &\textrm{dengan bantuan limit kanan }\\ &\textrm{yaitu}\: \: x=2+h\: \Rightarrow \: h\rightarrow 0\\ \underset{h\rightarrow 0}{\textrm{Lim}}\: &\: \displaystyle \frac{a\left ( 2+h \right )-2a}{\sqrt{2\left ( 2+h \right )}-\left ( 2+h \right )}=6\\ 6&=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{2a+ah-2a}{\sqrt{4+2h}-\left ( 2+h \right )}\\ 6&=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{ah}{\sqrt{4+2h}-\left ( 2+h \right )}\times \displaystyle \frac{\left ( \sqrt{4+2h}+\left ( 2+h \right ) \right )}{\left (\sqrt{4+2h} +\left ( 2+h \right ) \right )}\\ 6&=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{ah\times \left (\sqrt{4+2h} +\left ( 2+h \right ) \right )}{4+2h-\left ( 4+4h+h^{2} \right )}\\ 6&=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{ah\times \left (\sqrt{4+2h} +\left ( 2+h \right ) \right )}{-2h-h^{2}}\\ 6&=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{a\times \left (\sqrt{4+2h} +\left ( 2+h \right ) \right )}{-2-h}\\ 6&=\displaystyle \frac{a\times \left (\sqrt{4+0} +\left ( 2+0 \right ) \right )}{-2-0}\\ 6&=\displaystyle \frac{a\left ( \sqrt{4}+2 \right )}{-2}\\ \displaystyle \frac{a(4)}{-2}&=6\\ a(-2)&=6\\ a&=\color{red}-3 \end{aligned} \end{array}$

MATEMATIKA WAJIB MA/SMA untuk KONDISI KHUSUS TAHUN 2020-2021

 $\color{blue}\textrm{A. Kelas X (Sepuluh)}$

$\color{red}\begin{aligned}&\textrm{A.1 Persamaan dan pertidaksamaan}\\ &\qquad \textrm{nilai mutlak dari bentuk linear}\\ &\qquad \textrm{satu variabel} \end{aligned}$

• materi persamaan nilai mutlak
• contoh soal 1 , contoh 2 , contoh 3 , contoh 4 , contoh 5 
• materi pertidaksamaan nilai mutlak
• contoh soal 1 , contoh 2 , contoh 3 , 

$\color{red}\begin{aligned}&\textrm{A.2 Pertidaksamaan rasional}\\ &\qquad \textrm{dan irasional satu variabel} \end{aligned}$

• materi pertidaksamaan rasional dan irasional satu variabel
• contoh soal 1 , contoh 2 , contoh 3 , contoh 4 , 

$\color{red}\begin{aligned}&\textrm{A.3 Sistem persamaan linear tiga}\\ &\qquad \textrm{variabel} \end{aligned}$

• materi sistem persamaan linear tiga variabel (SPLTV)
• lanjutan materi SPLTV
• contoh soal 1 , contoh 2 , contoh 3 , contoh 4 

$\color{red}\begin{aligned}&\textrm{A.4 Sistem pertidaksamaan dua}\\ &\qquad \textrm{variabel-linear-linear} \end{aligned}$

• materi sistem pertidaksamaan dua variabel linear-linear
• contoh soal 1 , contoh 2 , contoh 3 , contoh 4 

$\color{black}\begin{aligned}&\textrm{A.5 Fungsi} \end{aligned}$

• Materi fungsi
• Lanjutan materi fungsi
• contoh soal 1 , contoh 2 , contoh 3

$\color{black}\begin{aligned}&\textrm{A.6 Fungsi Komposisi dan invers} \end{aligned}$

• Materi dan contoh soal
• contoh soal 1 , contoh 2 , contoh 3 , contoh 4 

$\color{black}\begin{aligned}&\textrm{A.7 Rasio trigonometri}\\ &\qquad \textrm{pada segitiga siku-siku} \end{aligned}$

$\color{black}\begin{aligned}&\textrm{A.8 Rasio trigonometri}\\ &\qquad \textrm{sudut-sudut diberbagai}\\ &\qquad \textrm{kuadran} \end{aligned}$

$\color{black}\begin{aligned}&\textrm{A.9 Aturan sinus dan cosinus} \end{aligned}$

$\color{blue}\textrm{B. Kelas XI (Sebelas)}$

$\color{red}\begin{aligned}&\textrm{B.1 Program linear} \end{aligned}$

• materi program linear
• contoh soal 1 , contoh 2 , contoh 3 , contoh 4 , contoh 5 

$\color{red}\begin{aligned}&\textrm{B.2 Matriks} \end{aligned}$

• materi matriks 
• lanjutan materi 1
• lanjutan materi 2
• contoh soal 1 , contoh 2 , contoh 3 

$\color{red}\begin{aligned}&\textrm{B.3 Determinan dan invers matriks}\\ &\qquad \textrm{ordo 2x2} \end{aligned}$

• materi determinan
• materi invers matriks ordo 2x2
• contoh soal 4 , contoh 5 , contoh 6 

$\color{red}\begin{aligned}&\textrm{B.4 Pola bilangan dan jumlah}\\ &\qquad \textrm{pada barisan aritmetika}\\ &\qquad \textrm{dan geometri} \end{aligned}$

• materi pola bilangan - induksi matematika
• contoh soal 1 , contoh 2 , contoh 3 , contoh 4 , contoh 5

$\color{black}\begin{aligned}&\textrm{B.5 Limit fungsi aljabar} \end{aligned}$

$\color{black}\begin{aligned}&\textrm{B.6 Turunan fungsi aljabar} \end{aligned}$

$\color{black}\begin{aligned}&\textrm{B.7 Keberkaitan Turunan fungsi} \end{aligned}$

$\color{black}\begin{aligned}&\textrm{B.8 Integral tak tentu}\\ &\qquad \textrm{fungsi aljabar} \end{aligned}$

$\color{blue}\textrm{C. Kelas XII (Duabelas)}$

$\color{red}\begin{aligned}&\textrm{C.1 Jarak dalam ruang} \end{aligned}$

• materi mengitung jarak di dimensi tiga
• contoh soal 1 , contoh 2 , contoh 3 , contoh 4 , contoh 5 

$\color{red}\begin{aligned}&\textrm{C.2 Statistika} \end{aligned}$

• materi statistika
• contoh soal 1 , contoh 2 , contoh 3 , contoh 4 , contoh 5 , contoh 6 , contoh 7 , contoh 8 , contoh 9 

$\color{black}\begin{aligned}&\textrm{C.3 Aturan pencacahan} \end{aligned}$

• materi 
• lanjutan materi
• lanjutan materi (faktorial, permutasi, kombinasi)
• contoh soal 1, contoh 2, contoh 3
• lanjutan materi (binomial newton)
• pengayaan : fungsi pembangkit untuk kombinasi

$\color{black}\begin{aligned}&\textrm{C.4 Peluang kejadian majumuk} \end{aligned}$

• materi
• lanjutan materi (peluang kejadian tunggal)
• lanjutan materi (peluang komplemen, peluang dua kejadian saling bebas)
• lanjutan materi (peluang kejadian saling lepas)
• contoh soal 1, contoh 2, contoh 3

MATEMATIKA PEMINATAN MA/SMA untuk KONDISI KHUSUS TAHUN 2020-2021

 $\color{blue}\textrm{A. Kelas X (Sepuluh)}$

$\color{red}\begin{aligned}&\textrm{A.1 Fungsi eksponensial}\\ &\qquad \textrm{dan fungsi logaritma} \end{aligned}$

• materi eksponen 1
• contoh soal eksponen 1 , contoh 2 , contoh 3 , contoh 4 , contoh 5 , contoh 6 , contoh 7 , contoh 8 , contoh 9 , contoh 10 
• materi logaritma 
• materi logaritma lanjutan
• contoh soal 1 , contoh 2 , contoh 3 , contoh 4 , contoh 5 , contoh 6
• materi logaritma lanjutan 2
• contoh soal 7 , contoh 8 , contoh 9 , contoh 10 

$\color{black}\begin{aligned}&\textrm{A.2 Vektor} \end{aligned}$

• materi
• operasi vektor pada bidang
• lanjutan materi pada bidang
• contoh soal 
• materi perkalian skalar dua vektor pada bidang
• contoh soal 2, contoh 3
• Proyeksi ortoganl suatu vektor
• Vektor dimensi tiga
• Contoh 1, contoh 2, contoh 3, contoh 4, contoh 5, contoh 6, contoh 7

$\color{blue}\textrm{B. Kelas XI (Sebelas)}$

$\color{red}\begin{aligned}&\textrm{B.1 Persamaan trigonometri} \end{aligned}$

• materi persamaan trigonometri
• lanjutan materi 1 , materi 2 , materi 3
• contoh soal 1 , contoh 2 , contoh 3 , contoh 4 , contoh 5 , contoh 6

$\color{black}\begin{aligned}&\textrm{B.2 Polinom} \end{aligned}$

• materi
• lanjutan materi (operasi polinom)
• lanjutan materi (teorema sisa dan teorema faktor)
• lanjutan materi (akar-akar persamaan polinom)
• contoh soal 1, contoh 2, contoh 3, contoh 4

$\color{blue}\textrm{C. Kelas XII (Duabelas)}$

$\color{red}\begin{aligned}&\textrm{C.1 Turunan fungsi trigonometri} \end{aligned}$

• materi turunan fungsi trigonometri
• lanjutan materi 
• contoh soal  (bagaian 1)
• lanjutan materi 2 
• lanjutan materi 3
• lanjutan materi 4 (sifat-sifat turunan)
• lanjutan materi 5
• contoh soal 1 (bagain 2) , contoh 2 , contoh 3 contoh 4 , contoh 5
• lanjutan materi 6 (persamaan garis singgung kurva)-turunan pertama
• lanjutan materi 7 (fungsi naik dan fungsi turun)-turunan pertama
• lanjutan materi 8 (nilai stasioner)-turunan pertama
• lanjutan materi 9 (aplikasi nilai stasioner)-turunan pertama
• contoh 6 , contoh 7 , contoh 8 , contoh 9
• turunan kedua fungsi trigonometri (Fungsi naik dan fungsi turun, titik belok serta selang kecekungan)
• contoh 10 , contoh 11 , contoh 12  

$\color{black}\begin{aligned}&\textrm{C.2 Distribusi peluang binomial} \end{aligned}$

• materi distribusi binomial
• lanjutan materi 1 (distribusi peluang diskrit)
• Contoh soal 1 , contoh 2 (variabel acak)
• lanjutan materi 2 (distribusi peluang kontinu)
• lanjutan materi (distribusi binomial-distribusi poisson)
• contoh soal 1 (distibusi binomial)
• contoh soal 2, contoh 3

Contoh Soal 3 Distribusi Binomial

$\begin{array}{ll}\\ 11.&\textrm{Suatu tes dengan pilihan jawaban }\\ &\textrm{benar-salah berjumlah 8 soal}\\ &\textrm{Supaya lulus tes, peserta diharuskan }\\ &\textrm{menjawab benar minimal 50}\%\\ &\textrm{Peluang seseorang dianggap lulus tes }\\ &\textrm{adalah}\: ....\\ &\textrm{a}.\quad \displaystyle 0,2188\qquad\qquad\quad\qquad \quad\textrm{d}.\quad 0,6367\\ &\textrm{b}.\quad \displaystyle \color{red}0,2734\quad \: \color{black}\textrm{c}.\quad 0,3633\quad\quad \textrm{e}.\quad 0,7266\\\\ &\textrm{Jawab}:\\ &\begin{aligned}&p=\textbf{Peluang benar}=\displaystyle \frac{1}{2},\qquad \textrm{dan}\: \: \\ &q=\textbf{Peluang Salah}=1-\displaystyle \frac{1}{2}=\frac{1}{2}\\ &f(x)=P(X=x)=\begin{pmatrix} n\\ x \end{pmatrix}p^{x}.q^{n-x}\\ &\textrm{maka}\\ &P\left ( X=50\%(8)=4 \right )=\begin{pmatrix} 8\\ 4 \end{pmatrix}\times \left ( \displaystyle \frac{1}{2} \right )^{4}\times \left ( \frac{1}{2} \right )^{8-4}\\ &\qquad =\displaystyle \frac{8!}{4!\times 4!}\left ( \displaystyle \frac{1}{2} \right )^{4+4}\\ &\qquad =70\times \displaystyle \frac{1}{256}\\ &\qquad =\color{red}0,2734 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 12.&\textrm{Sebuah kotak berisi 20 bola dengan }\\ &\textrm{rincian 12 boal berwarna kuning dan }\\ &\textrm{sisanya berwarna hijau. Dari kotak} \\ &\textrm{diambil 6 bola secara acak. Peluang}\\ &\textrm{terambil 4 bola hijau adalah}....\\ &\textrm{a}.\quad \displaystyle 0,1238\quad\quad\qquad\qquad \qquad\textrm{d}.\quad 0,8132\\ &\textrm{b}.\quad \color{red}\displaystyle 0,1382\: \quad \color{black}\textrm{c}.\quad 0,3110\quad\quad \textrm{e}.\quad 0,9590\\\\ &\textrm{Jawab}:\\ &\begin{aligned}&p=\textbf{Peluang bola kuning}\\ &\: \: =\displaystyle \frac{C_{1}^{12}}{C_{1}^{20}}=\displaystyle \frac{12}{20}=\frac{3}{5},\\ &q=\textbf{Peluang bola hijau}=1-\displaystyle \frac{3}{5}=\frac{2}{5}\\ &f(x)=P\left ( X=x \right )=\begin{pmatrix} n\\ x \end{pmatrix}p^{x}.q^{n-x}\\ &\textrm{maka}\\ &f(4)=\begin{pmatrix} 6\\ 4 \end{pmatrix}\times \left ( \displaystyle \frac{2}{5} \right )^{4}\times \left ( \frac{3}{5} \right )^{6-4}\\ &\qquad =\displaystyle \frac{6!}{2!\times 4!}\left ( \displaystyle \frac{16}{625} \right )\times \left ( \displaystyle \frac{9}{25} \right )\\ &\qquad =15\times \displaystyle \frac{144}{15625}=\frac{2160}{15625}\\ &\qquad =\color{red}0,1382 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 13.&\textrm{Dua dadu dilambungkan 5 kali}\\ &\textrm{Peluang muncul pasangan mata dadu}\\ &\textrm{berjumlah 4 sampai dengan 7 }\\ &\textrm{sebanyak 4 kali adalah}\: ....\\ &\textrm{a}.\quad \displaystyle 0,1503\: \: \: \: \qquad\qquad\quad\quad \quad\textrm{d}.\quad 0,1583\\ &\textrm{b}.\quad \displaystyle 0,1553\quad \textrm{c}.\quad \color{red}0,1563\quad\quad \color{black}\textrm{e}.\quad 0,1593\\\\ &\textrm{Jawab}:\\ &\begin{aligned}&p=\textbf{Peluang mata dadu berjumlah 4 sampai 7}\\ &\: \: =\displaystyle \frac{18}{36}=\frac{1}{2},\qquad \textrm{dan}\: \: \\ &q=\textbf{Peluang bola hijau}=1-\displaystyle \frac{1}{2}=\frac{1}{2}\\ &f(x)=P\left ( X=x \right )=\begin{pmatrix} n\\ x \end{pmatrix}p^{x}.q^{n-x}\\ &f(4)=P\left ( X=4 \right )=\begin{pmatrix} 5\\ 4 \end{pmatrix}\times \left ( \displaystyle \frac{1}{2} \right )^{4}\times \left ( \frac{1}{2} \right )^{5-4}\\ &\qquad =\displaystyle \frac{5!}{1!\times 4!}\left ( \displaystyle \frac{1}{16} \right )\times \left (\frac{1}{2} \right )\\ &\qquad =5\times \displaystyle \frac{1}{32}=\frac{5}{32}\\ &\qquad =\color{red}0,1563 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 14.&\textrm{Peluang seseorang sembih dari }\\ &\textrm{penyakit jantung adalah 0,6}\\ &\textrm{Jika 7 orang penderita ini menjalani }\\ &\textrm{operasi, maka peluang 3 sampai}\\ &\textrm{6 orang sembuh adalah}... .\\ &\textrm{a}.\quad \displaystyle 0,0629\qquad\qquad\quad\qquad \quad\textrm{d}.\quad \color{red}0,6822\\ &\textrm{b}.\quad \displaystyle 0,2613\quad \textrm{c}.\quad 0,2898\quad\quad \: \textrm{e}.\quad 0,9720\\\\ &\textrm{Jawab}:\\ &\begin{aligned}&p=\textbf{Peluang sembuh}=0,6,\qquad \textrm{maka}\: \: \\ &q=\textbf{Peluang tidak sembuh}=1-0,6=0,4\\ &f(x)=P\left ( X=x \right )=\begin{pmatrix} n\\ x \end{pmatrix}p^{x}.q^{n-x}\\ &\textrm{maka}\\ &P\left ( 3\leq X\leq 6 \right )=P\left ( X\leq 6 \right )-P\left ( X\leq 3 \right )\\ &=C_{4}^{7}(0,6)^{4}(0,4)^{3}+C_{5}^{7}(0,6)^{5}(0,4)^{2}+C_{6}^{7}(0,6)^{6}(0,4)^{1}\\ &=35\times 0,0082944+21\times 0,0124416+7\times 0,0186624\\ &=0,290304+0,2612736+0,1306368\\ &=\color{red}0,6822144 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 15.&\textrm{Peluang seseorang mendapatkan reaksi }\\ &\textrm{buruk setelah disuntik adalah 0,0005}\\ &\textrm{Dari 4000 orang yang disuntik, maka }\\ &\textrm{peluang seseorang mendapatkan reaksi}\\ & \textrm{ada 2 orang adalah}.....\\ &\textrm{a}.\quad \displaystyle \frac{1}{2}e^{-2}\\ &\textrm{b}.\quad e^{-2}\\ &\textrm{c}.\quad \color{red}2e^{-2}\\ &\textrm{d}.\quad \displaystyle \frac{1}{2}e^{2}\\ &\textrm{e}.\quad 2e^{2}\\\\ &\textrm{Jawab}:\\ &\begin{aligned}&\textrm{Di atas adalah contoh kasus }\\ &\textrm{permasalahan}\: \: \textbf{Distribusi Poisson}\\ &P\left ( X=x \right )=f(x)=\left\{\begin{matrix} \displaystyle \frac{e^{-\lambda }.\lambda ^{x}}{x!}\: \: ,\: \: x=0,1,2,3,\cdots \\\ 0,\quad \textrm{untuk}\: \: x\: \: \textrm{yang lain} \end{matrix}\right.\\ &P\left ( X=2 \right )=\displaystyle \frac{e^{-np}.(np)^{2}}{2!}\\ &\qquad =\displaystyle \frac{e^{-(4000.0,0005)}.(4000.0,0005)^{2}}{2!}\\ &\qquad =\displaystyle \frac{e^{-2}.2^{2}}{2}\\ &\qquad =\color{red}2e^{-2} \end{aligned} \end{array}$

Contoh Soal 2 Distribusi Binomial

 $\begin{array}{ll}\\ 6.&\textrm{Pengundian terhadap mata uang }\\ &\textrm{yang homogen sebanyak 10 kali}\\ &\textrm{Peluang untuk mendapatkan 6 }\\ &\textrm{muka angka adalah}\: ....\\ &\textrm{a}.\quad 0,1172\\ &\textrm{b}.\quad \color{red}0,2051\\ &\textrm{c}.\quad 0,2461\\ &\textrm{d}.\quad 0,2651\\ &\textrm{e}.\quad 0,2852\\\\ &\textrm{Jawab}:\\ &\begin{aligned}&p=\textbf{Peluang Angka}=\displaystyle \frac{1}{2},\quad \textrm{dan}\: \: \\ &q=\textbf{Bukan Angka}\\ &\: \: =\textbf{Peluang Gambar}=1-\displaystyle \frac{1}{2}=\frac{1}{2}\\ &f(x)=P(x;n;p)=P(X=x)=\begin{pmatrix} n\\ x \end{pmatrix}p^{x}q^{n-x}\\ &\textrm{maka}\\ &f(x)=P\left ( X=x \right )=\begin{pmatrix} n\\ x \end{pmatrix}p^{x}.q^{n-x}\\ &f(6)=P\left ( X=6 \right )=\begin{pmatrix} 10\\ 6 \end{pmatrix}\times \left ( \displaystyle \frac{1}{2} \right )^{6}\times \left ( \frac{1}{2} \right )^{10-6}\\ &\qquad =\displaystyle \frac{10!}{6!\times 4!}\left ( \displaystyle \frac{1}{2} \right )^{6+4}\\ &\qquad =210\times \displaystyle \frac{1}{1024}\\ &\qquad =\color{red}0,2051 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 7.&\textrm{Pada pengundian terhadap mata uang identik},\\ &\textrm{sebanyak 10 kali, peluang distribusi binomial} \\ &\textrm{untuk mendapatkan 7 muka gambar adalah}\: ....\\ &\begin{array}{lllllll}\\ \textrm{a}.\quad \displaystyle 0,2653&&\textrm{d}.\quad \displaystyle 0,7522\\ \textrm{b}.\quad \displaystyle \color{red}0,1172&\textrm{c}.\quad \displaystyle 0,2653&\textrm{e}.\quad 0,2422 \end{array}\\\\ &\textrm{Jawab}:\\ &\textrm{Uraian berikut sekaligus tambahan}\\ &\textrm{penjelasan pada uraian jawaban}\\ &\color{blue}\textrm{soal no. 6 di atas}\\ &\begin{aligned}f(x)&=P(x;n;p)=\begin{pmatrix} n\\ x \end{pmatrix}p^{x}q^{n-x}\\ &\textrm{Ingat sebuah koin ada 2 muka}\\ &\textrm{yaitu muka gambar (G) dan angka (A)}\\ &\color{red}\textrm{misalkan}\\ &A=\textrm{kejadian muncul muka gambar}\\ &\textrm{maka peluangnya adalah}\: \: \displaystyle \frac{1}{2}\\ &\textrm{Selanjutnya di sini disimbolkan dengan}\: \: \: \color{blue}p=\displaystyle \frac{1}{2}\\ &\color{red}\textrm{Demikian juga misalkan}\\ &B=\textrm{kejadian muncul muka angka}\\ &\textrm{maka peluang juga}\: \displaystyle \frac{1}{2}\\ &\textrm{Di sini dituliskan dengan}\: \: \: \color{blue}q=\displaystyle \frac{1}{2}\\ f(7)&=\begin{pmatrix} 10\\ 7 \end{pmatrix}\left ( \displaystyle \frac{1}{2} \right )^{7}\left ( \displaystyle \frac{1}{2} \right )^{10-7}\\ &=\begin{pmatrix} 10\\ 7 \end{pmatrix}\left ( \displaystyle \frac{1}{2} \right )^{7}\left ( \displaystyle \frac{1}{2} \right )^{3}\\ &=\displaystyle \frac{10!}{7!\times (10-7)!}\left ( \displaystyle \frac{1}{2} \right )^{7+3}\\ &=\displaystyle \frac{10.9.8.\not{7!}}{\not{7!}.3.2.1}\left ( \displaystyle \frac{1}{1024} \right ) \\ &=\color{red}0,1172 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 8.&\textrm{Sebuah uang logam dilempar sebanyak 8}\\ &\textrm{kali. Peluang muncul gambar sebanyak}\\ &\textrm{5 kali adalah}\: ....\\ &\begin{array}{llllllll}\\ \textrm{a}.&\displaystyle \frac{3}{32}&&&\textrm{d}.&\displaystyle \color{red}\frac{7}{32}\\\\ \textrm{b}.&\displaystyle \frac{4}{32}&\textrm{c}.&\displaystyle \frac{5}{32}&\textrm{e}.&\displaystyle \frac{9}{32} \end{array}\\\\ &\textrm{Jawab}:\\ &\begin{aligned}f(x)&=P(x;n;p)=\begin{pmatrix} n\\ x \end{pmatrix}p^{x}q^{n-x}\\ f(5)&=\begin{pmatrix} 8\\ 5 \end{pmatrix}\left ( \displaystyle \frac{1}{2} \right )^{5}\left ( \displaystyle \frac{1}{2} \right )^{8-5}\\ &=\begin{pmatrix} 8\\ 5 \end{pmatrix}\left ( \displaystyle \frac{1}{2} \right )^{5}\left ( \displaystyle \frac{1}{2} \right )^{3}\\ &=\displaystyle \frac{8!}{5!\times (8-5)!}\left ( \displaystyle \frac{1}{2} \right )^{5+3}\\ &=\displaystyle \frac{8.7.6.5!}{5!.3.2.1}\left ( \displaystyle \frac{1}{256} \right ) \\ &=\displaystyle \frac{8.7}{256}\\ &=\color{red}\displaystyle \frac{7}{32} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 9.&\textrm{Pada pelemparan sebuah koin sebanyak 4 kali}\\ &\textrm{Peluang didapatkannya dua angka pada} \\ &\textrm{pelemparan tersebut adalah}\: ....\\ &\begin{array}{lllllll}\\ \textrm{a}.\quad \displaystyle 0,123&&\textrm{d}.\quad \displaystyle 0,232\\ \textrm{b}.\quad \displaystyle 0,135&\textrm{c}.\quad \displaystyle 0,154&\textrm{e}.\quad \color{red}0,375 \end{array}\\\\ &\textrm{Jawab}:\\ &\begin{aligned}f(x)&=P(x;n;p)=\begin{pmatrix} n\\ x \end{pmatrix}p^{x}q^{n-x}\\ f(2)&=\begin{pmatrix} 4\\ 2 \end{pmatrix}\left ( \displaystyle \frac{1}{2} \right )^{2}\left ( \displaystyle \frac{1}{2} \right )^{4-2}\\ &=\begin{pmatrix} 4\\ 2 \end{pmatrix}\left ( \displaystyle \frac{1}{2} \right )^{2}\left ( \displaystyle \frac{1}{2} \right )^{2}\\ &=\displaystyle \frac{4!}{2!\times (4-2)!}\left ( \displaystyle \frac{1}{2} \right )^{2+2}\\ &=\displaystyle \frac{4.3.2!}{2!.2.1}\left ( \displaystyle \frac{1}{16} \right ) \\ &=\color{red}0,375 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 10.&\textrm{Dari data survei didapatkan bahwa}\\ &\textrm{satu dari lima orang telah berkunjung}\\ &\textrm{ke dokter dalam sembarang bulan yang}\\ &\textrm{ditanyakan. Jika 10 orang dipilih secara}\\ &\textrm{acak, peluang 3 orang telah berkunjung}\\ &\textrm{ke dokter bulan lalu adalah}\: ....\\ &\begin{array}{llllllll}\\ \textrm{a}.&\displaystyle 0,125&&&\textrm{d}.&\displaystyle \color{red}0,201\\\\ \textrm{b}.&\displaystyle 0,174&\textrm{c}.&\displaystyle 0,182&\textrm{e}.&\displaystyle 0,423 \end{array}\\\\ &\textrm{Jawab}:\\ &\begin{aligned}f(x)&=P(x;n;p)=\begin{pmatrix} n\\ x \end{pmatrix}p^{x}q^{n-x}\\ f(3)&=\begin{pmatrix} 10\\ 3 \end{pmatrix}\left ( \displaystyle \frac{1}{5} \right )^{3}\left ( \displaystyle \frac{4}{5} \right )^{10-3}\\ &=\begin{pmatrix} 10\\ 3 \end{pmatrix}\left ( \displaystyle \frac{1}{5} \right )^{3}\left ( \displaystyle \frac{4}{5} \right )^{7}\\ &=\displaystyle \frac{10!}{3!\times 7!}\left ( \displaystyle \frac{1}{125} \right )\left ( \displaystyle \frac{4^{7}}{5^{7}} \right )\\ &=\cdots \\ &=\color{red}\displaystyle 0,201 \end{aligned} \end{array}$