$\begin{array}{ll}\\ 11.&\textrm{Jika vektor}\: \: \vec{a}=\begin{pmatrix} 6\\ -4 \end{pmatrix}\: \: \textrm{dan}\: \: \vec{b}=\begin{pmatrix} 3\\ 2 \end{pmatrix},\\ &\textrm{maka}\: \: 3\vec{a}-2\vec{b}\: \: \textrm{adalah}....\\ &\begin{array}{lll}\\ \textrm{a}.\quad \color{red}\begin{pmatrix} 12\\ -16 \end{pmatrix}&&\textrm{d}.\quad \begin{pmatrix} 24\\ 16 \end{pmatrix}\\ \textrm{b}.\quad \begin{pmatrix} 24\\ -16 \end{pmatrix}&\textrm{c}.\quad \begin{pmatrix} 12\\ 16 \end{pmatrix}&\textrm{e}.\quad \begin{pmatrix} -12\\ -16 \end{pmatrix} \end{array}\\\\ &\textbf{Jawab}\\ &\begin{aligned}3\vec{a}-2\vec{b}&=3\begin{pmatrix} 6\\ -4 \end{pmatrix}-2\begin{pmatrix} 3\\ 2 \end{pmatrix}\\ &=\begin{pmatrix} 18-6\\ -12-4 \end{pmatrix}\\ &=\color{red}\begin{pmatrix} 12\\ -16 \end{pmatrix} \end{aligned} \end{array}$
$\begin{array}{ll}\\ 12.&\textrm{Diketahui jajar genjang ABCD }\\ &\textrm{dengan titik E adalah perpotongan }\\ &\textrm{diagonal jajar genjang}. \end{array}$
$\begin{array}{ll}\\ .\, \quad&\textrm{Jika}\: \: \overline{AB}=\vec{b}\: \: \textrm{dan}\: \: \overline{AD}=\vec{a},\: \textrm{maka}\: \: \overline{CE}\\ & \textrm{bila dinyatakan dalam}\: \: \vec{a}\: \: \textrm{dan}\: \: \vec{b}\: \: \textrm{adalah}....\\ &\begin{array}{lll}\\ \textrm{a}.\quad \displaystyle \frac{1}{2}\left ( \vec{a}+\vec{b} \right )&\\ \textrm{b}.\quad \displaystyle \frac{1}{2}\left ( \vec{a}-\vec{b} \right )&\\ \textrm{c}.\quad \displaystyle \frac{1}{2}\left ( \vec{b}-\vec{a} \right )&\\ \textrm{d}.\quad \color{red}\displaystyle -\frac{1}{2}\left ( \vec{a}+\vec{b} \right )\\ \textrm{e}.\quad -\displaystyle \frac{1}{2}\left ( 2\vec{a}+\vec{b} \right ) \end{array}\\\\ &\textbf{Jawab}\\ &\begin{aligned} \overline{AC}&=\overline{AD}+\overline{DC}\\ \overline{CA}&=\overline{CD}+\overline{DA}\\ \overline{CE}&=\displaystyle \frac{1}{2}\, \overline{CA}\\ &=\displaystyle \frac{1}{2}\left ( -\vec{b}-\vec{a} \right )\\ &=\color{red}-\displaystyle \frac{1}{2}\left ( \vec{a}+\vec{b} \right ) \end{aligned} \end{array}$
$\begin{array}{ll}\\ 13.&\textrm{Pada segi enam beraturan ABCDEF},\\ & \textrm{jika}\: \: \overrightarrow{AB}=\vec{u}\: \: \textrm{dan}\: \: \overrightarrow{AF}=\vec{v}\: \: \textrm{maka vektor}\\ &\overrightarrow{AB}+\overrightarrow{AC}+\overrightarrow{AD}+\overrightarrow{AE}+\overrightarrow{AF}=....\\ &\begin{array}{llllllll}\\ \textrm{a}.&2\vec{u}+2\vec{v}&&&\textrm{d}.&\color{red}6\vec{u}+6\vec{v}\\ \textrm{b}.&4\vec{u}+4\vec{v}&\textrm{c}.&5\vec{u}+5\vec{v}&\textrm{e}.&8\vec{u}+8\vec{v} \end{array}\\\\ &\textbf{Jawab}\\ &\textrm{Perhatikanlah ilustrasi gambar berikut} \end{array}$
$\begin{aligned}.\: \, \qquad &\overrightarrow{AB}+\overrightarrow{AC}+\overrightarrow{AD}+\overrightarrow{AE}+\overrightarrow{AF}\\ &=\overrightarrow{AB}+\left (\overrightarrow{AO}+\overrightarrow{OC} \right )+2\overrightarrow{AO}+\left (\overrightarrow{AO}+\overrightarrow{OE} \right )+\overrightarrow{AF}\\ &=\vec{u}+\left (2\vec{u}+\vec{v} \right )+2\left ( \vec{v}+\vec{u} \right )+\left ( 2\vec{v}+\vec{u} \right )+\vec{v}\\ &=\color{red}6\vec{u}+6\vec{v} \end{aligned}$
$\begin{array}{ll}\\ 14.&\textrm{Perhatikanlah juga ilustrasi gambar berikut} \end{array}$
$\begin{array}{ll}\\ .\, \quad&\textrm{maka vektor}\: \: \vec{w}\: \: \textrm{adalah}\: ....\\ &\begin{array}{llllllll}\\ \textrm{a}.&\color{red}8\vec{i}-6\vec{j}-13\vec{k}&\\ \textrm{b}.&8\vec{i}-13\vec{j}-6\vec{k}\\ \textrm{c}.&6\vec{i}-8\vec{j}-13\vec{k}&\\ \textrm{d}.&-6\vec{i}+8\vec{j}-13\vec{k}\\ \textrm{e}.&-6\vec{i}-13\vec{j}+8\vec{k} \end{array}\\\\ &\textbf{Jawab}\\ &\textrm{Kita perhatikan juga ilustrasi }\\ &\textrm{gambarnya semisal dengan soal No.1}\\ &\textrm{Misalkan titiknya adalah titik }\\ &\textrm{W dengan koordinat (8,-6,-13)},\\ &\textrm{maka vektor posisi titik }\\ &\textrm{W tersebut adalah}\: \: \overrightarrow{OW}=\vec{w}\\ & \textrm{di mana}\\ &\begin{aligned}&\textrm{Vektor}\: \: \vec{w}\: \textrm{jika dinyatakan }\\ &\textrm{dalam kombinasi linear adalah}\\ &\vec{w}=\color{red}8\vec{i}-6\vec{j}-13\vec{k} \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 15.&\textrm{Jika titik Z(4,-5,2)},\: \textrm{maka panjang }\\ &\textrm{vektor posisi titik Z adalah}\: ....\\ &\begin{array}{llllllll}\\ \textrm{a}.&1&&&\textrm{d}.&5\sqrt{2}\\ \textrm{b}.&2\sqrt{5}&\textrm{c}.&\color{red}3\sqrt{5}&\textrm{e}.&5\sqrt{3} \end{array}\\\\ &\textbf{Jawab}\\ &\begin{aligned}\textrm{Vektor posisi}&\: \, \textrm{titik Z tersebut adalah}\\ \overrightarrow{OZ}=\vec{z}&=\begin{pmatrix} 4, & -5, & 2 \end{pmatrix},\\ \textrm{Dan panjang}&\: \, \textrm{vektor}\: \: \vec{z}\: \: \textrm{ini adalah}\\ \left | \vec{z} \right |&=\sqrt{4^{2}+(-5)^{2}+2^{2}}\\ &=\sqrt{16+25+4}\\ &=\sqrt{45}=\sqrt{9\times 5}\\ &=\color{red}3\sqrt{5} \end{aligned} \end{array}$.
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