Contoh 8 Vektor

Contoh soal sebelumnya di sini Contoh Soal 7

$\begin{array}{ll}\\ 34.&\textrm{Vektor satuan untuk}\: \: \vec{a}=\begin{pmatrix} 2\\ -1\\ 4 \end{pmatrix}\: \textrm{adalah}....\\ &\begin{array}{lll}\\ \textrm{a}.\quad \displaystyle \frac{1}{2}\sqrt{3}\begin{pmatrix} 2\\ -1\\ 4 \end{pmatrix}&\\ \textrm{b}.\quad \displaystyle \frac{1}{3}\sqrt{3}\begin{pmatrix} 2\\ -1\\ 4 \end{pmatrix}&\\ \textrm{c}.\quad \displaystyle \frac{1}{5}\sqrt{5}\begin{pmatrix} 2\\ -1\\ 4 \end{pmatrix}&\\ \textrm{d}.\quad \displaystyle \frac{1}{7}\sqrt{7}\begin{pmatrix} 2\\ -1\\ 4 \end{pmatrix}\\ \color{red}\textrm{e}.\quad \displaystyle \frac{1}{21}\sqrt{21}\begin{pmatrix} 2\\ -1\\ 4 \end{pmatrix} \end{array}\\\\ &\textrm{Jawab}:\\ &\begin{aligned}\textrm{Vektor satuan}\: \: \vec{a}&\: \: \textrm{adalah}\: \: \vec{e}_{\vec{a}},\: \textrm{yaitu}:\\ \vec{e}_{\vec{a}}&=\displaystyle \frac{\vec{a}}{\left | \vec{a} \right |}\\ &=\displaystyle \frac{\begin{pmatrix} 2\\ -1\\ 4 \end{pmatrix}}{\sqrt{2^{2}+(-1)^{2}+4^{2}}}\\ &=\displaystyle \frac{1}{\sqrt{21}}\begin{pmatrix} 2\\ -1\\ 4 \end{pmatrix}\\ &=\displaystyle \frac{1}{21}\sqrt{21}\begin{pmatrix} 2\\ -1\\ 4 \end{pmatrix} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 35.&\textrm{Posisi suatu titik dalam ruang saat }\\ &\textrm{waktu}\: \: t\: \: \textrm{ditunjukkan oleh vektor}\\ &\begin{pmatrix} t\\ t^{2}\\ -t \end{pmatrix}.\: \textrm{Jika pada saat}\: \: t=1\: \: \textrm{titik }\\ &\textrm{tersebut berada di titik P dan pada}\\ &\textrm{saat}\: \: t=2\: \: \textrm{titik tersebut berada }\\ &\textrm{di titik Q, maka jarak titik P dari Q}\\ & \textrm{adalah}\: ...\: .\\ &\begin{array}{lll}\\ \textrm{a}.\quad \sqrt{24}-\sqrt{3}&&\color{red}\textrm{d}.\quad \sqrt{11}\\ \textrm{b}.\quad 2-\sqrt{2}&\textrm{c}.\quad 3&\textrm{e}.\quad \sqrt{43}\end{array}\\\\ &\textbf{Jawab}:\quad \textbf{a}\\ &\begin{aligned}\left | \overrightarrow{PQ} \right |&=\sqrt{(x_{q}-x_{p})^{2}+(y_{q}-y_{p})^{2}+(y_{q}-y_{p})^{2}}\\ &=\sqrt{(2-1)^{2}+(2^{2}-1^{2})^{2}+((-2)-(-1))^{2}}\\ &=\sqrt{1^{2}+3^{2}+(-1)^{2}}\\ &=\sqrt{1+9+1}\\ &=\sqrt{11} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 36.&\textrm{Jika diketahui}\: \: \left | \vec{a} \right |=4\sqrt{3},\: \left | \vec{b} \right |=5,\\ & \textrm{ dan}\: \left ( \vec{a}+\vec{b} \right ).\left ( \vec{a}+\vec{b} \right )=13,\\ & \textrm{maka}\: \: \angle \left ( \vec{a},\: \vec{b} \right )=....\\ &\begin{array}{lll}\\ \textrm{a}.\quad \displaystyle 30^{\circ}&&\textrm{d}.\quad \displaystyle 135^{\circ}\\ \textrm{b}.\quad \displaystyle 60^{\circ}&\textrm{c}.\quad \displaystyle 120^{\circ}&\color{red}\textrm{e}.\quad \displaystyle 150^{\circ} \end{array}\\\\ &\textrm{Jawab}:\\ &\begin{aligned}\left ( \vec{a}+\vec{b} \right ).\left ( \vec{a}+\vec{b} \right )&=13\\ \vec{a}.\vec{a}+\vec{a}.\vec{b}+\vec{b}.\vec{a}+\vec{b}.\vec{b}&=13\\ \left | \vec{a} \right |^{2}+2\vec{a}.\vec{b}+\left | \vec{b} \right |^{2}&=13,\\ \textrm{ingat bahwa}\: \: \vec{a}.\vec{b}=\vec{b}.\vec{a}&\\ \left ( 4\sqrt{3} \right )^{2}+2\left | \vec{a} \right |\left | \vec{b} \right |\cos \angle \left ( \vec{a},\: \vec{b} \right )+5^{2}&=13\\ 48+2.(4\sqrt{3}).5.\cos \angle \left ( \vec{a},\: \vec{b} \right )+25&=13\\ 40\sqrt{3}\cos \angle \left ( \vec{a},\: \vec{b} \right )&=13-25-48\\ \cos \angle \left ( \vec{a},\: \vec{b} \right )&=\displaystyle \frac{-60}{40\sqrt{3}}\\ &=-\displaystyle \frac{1}{2}\sqrt{3}\\ &=-\cos 30\\ &=\cos \left ( 180^{\circ}-30^{\circ} \right )\\ \cos \angle \left ( \vec{a},\: \vec{b} \right )&=\cos 150^{\circ}\\ \angle \left ( \vec{a},\: \vec{b} \right )&=150^{\circ} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 37.&\textrm{Jika diketahui titik}\: \: A(2,-1,4),\: B(4,1,3),\\ & \textrm{ dan}\: \: C(2,0,5),\: \: \textrm{maka}\: \: \sin \angle \left ( \overline{AB},\: \overline{AC} \right )=....\\ &\begin{array}{lll}\\ \textrm{a}.\quad \displaystyle \frac{1}{7}\sqrt{5}&&\textrm{d}.\quad \displaystyle \frac{1}{6}\sqrt{3}\\\\ \color{red}\textrm{b}.\quad \displaystyle \frac{1}{6}\sqrt{34}&\textrm{c}.\quad \displaystyle \frac{2}{3}\sqrt{2}&\textrm{e}.\quad \displaystyle \frac{1}{6}\sqrt{2} \end{array}\\\\ &\textbf{Jawab}:\\ &\begin{aligned}&\cos \angle \left ( \overline{AB},\: \overline{AC} \right )=\displaystyle \frac{\overline{AB}.\, \overline{AC}}{\left | \overline{AB} \right |.\left | \overline{AC} \right |} \\ &=\displaystyle \frac{(\vec{b}-\vec{a}).(\vec{c}-\vec{a})}{\sqrt{x^{2}_{\left (\vec{b}-\vec{a} \right )}+y^{2}_{\left ( \vec{b}-\vec{a} \right )}+z^{2}_{\left ( \vec{b}-\vec{a} \right )}}.\sqrt{x^{2}_{\left ( \vec{c}-\vec{a} \right )}+y^{2}_{\left ( \vec{c}-\vec{a} \right )}+z^{2}_{\left ( \vec{c}-\vec{a} \right )}}} \\ &=\displaystyle \frac{\begin{pmatrix} 4-2\\ 1+1\\ 3-4 \end{pmatrix}.\begin{pmatrix} 2-2\\ 0+1\\ 5-4 \end{pmatrix}}{\sqrt{(4-2)^{2}+(1+1)^{2}+(3-4)^{2}}.\sqrt{(2-2)^{2}+(0+1)^{2}+(5-4)^{2}}}\\ &=\displaystyle \frac{2.0+2.1+-1.1}{\sqrt{4+4+1}.\sqrt{0+1+1}}\\ &=\displaystyle \frac{1}{3\sqrt{2}}\\ &=\color{blue}\displaystyle \frac{1}{6}\sqrt{2} \end{aligned}\\ &\begin{aligned}&\textrm{Sehingga},\qquad \qquad \\ &\sin \angle \left ( \overline{AB},\: \overline{AC} \right )\\ &=\sqrt{1-\cos ^{2}\angle \left ( \overline{AB},\: \overline{AC} \right )}\\ &=\sqrt{1-\left ( \displaystyle \frac{1}{6}\sqrt{2} \right )^{2}}\\ &=\sqrt{1-\displaystyle \frac{2}{36}}\\ &=\sqrt{\displaystyle \frac{34}{36}}\\ &=\color{red}\displaystyle \frac{1}{6}\sqrt{34} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 38.&\textrm{Diketahui segitiga ABC. Titik M }\\ &\textrm{di tengah AC, dan titik N pada BC}\\ &\textrm{Jika}\: \: \overrightarrow{AB}=\vec{c}\: ,\: \overrightarrow{AC}=\vec{b}\: ,\: \overrightarrow{BC}=\vec{a}\: ,\\ &\textrm{maka}\: \: \overrightarrow{MN}=\, ....\\ &\begin{array}{lll}\\ \textrm{a}.\quad \displaystyle \frac{1}{2}\left ( \vec{b}-\vec{c} \right )&\\ \textrm{b}.\quad \displaystyle \frac{1}{2}\left ( -\vec{b}+\vec{c} \right )&\\ \textrm{c}.\quad \displaystyle \frac{1}{2}\left ( -\vec{a}+\vec{c} \right )\\ \textrm{d}.\quad \displaystyle \frac{1}{2}\left ( \vec{a}-\vec{b} \right )\\ \color{red}\textrm{e}.\quad \displaystyle \frac{1}{2}\left ( -\vec{a}+\vec{b} \right ) \end{array}\\\\ &\textbf{Jawab}:\\ &\begin{aligned}\overrightarrow{MN}&=\overrightarrow{MC}+\overrightarrow{CN}\\ &=\displaystyle \frac{1}{2}\overrightarrow{AC}+\displaystyle \frac{1}{2}\left ( -\overrightarrow{BC} \right )\\ &=\displaystyle \frac{1}{2}\left ( \overrightarrow{AC}-\overrightarrow{BC} \right )\\ &=\displaystyle \frac{1}{2}\left ( \vec{b}-\vec{a} \right )\qquad \textbf{atau}\\ &=\displaystyle \frac{1}{2}\left ( -\vec{a}+\vec{b} \right ) \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 39.&\textrm{Jika titik berat segitiga ABC adalah Z}\\ &\textrm{dengan}\: \: \textrm{A}(1,0,2),\: \textrm{B}(5,4,10),\: \textrm{C}(0,-1,6),\\ &\textrm{maka koordinat titik Z tersebut adalah}\, ....\\ &\begin{array}{lll}\\ \textrm{a}.\quad (-2,-1,6)&&\textrm{d}.\quad (3,2,6)\\ \color{red}\textrm{b}.\quad (2,1,6)&\textrm{c}.\quad (3,-1,6)&\textrm{e}.\quad (6,4,12) \end{array}\\\\ &\textbf{Jawab}:\quad \textbf{b}\\ &\textrm{Coba perhatikanlah ilustrasi berikut} \end{array}$

$.\: \qquad\begin{aligned}\\ &\begin{aligned}\textrm{koordinat}&\: \textrm{titik A}'\\ &=\displaystyle \frac{1}{2}\left ( 5+0,4-1,10+6 \right )\\ &=\color{blue}\left ( \frac{5}{2},\frac{3}{2},8 \right ) \end{aligned}\\ &\begin{aligned}\textrm{Dalam se}&\textrm{gitiga ABC untuk titik berat Z }\\ \textrm{berlaku}\: \: \: &\textrm{ketentuan sebagai berikut}\\ AZ:ZA'&=2:1\\ \overrightarrow{AZ}:\overrightarrow{ZA'}&=2:1\\ \overrightarrow{OZ}&=\displaystyle \frac{\overrightarrow{OA}+2\overrightarrow{OA'}}{3}\\ &=\displaystyle \frac{\begin{pmatrix} 1\\ 0\\ 2 \end{pmatrix}+2\begin{pmatrix} \frac{5}{2}\\ \frac{3}{2}\\ 8 \end{pmatrix}}{3}\\ &=\begin{pmatrix} 2\\ 1\\ 6 \end{pmatrix}\\ \textrm{Jadi},\: &\textrm{koordinat titik Z adalah}\: \color{red}(2,1,6) \end{aligned} \end{aligned}$.

$\begin{array}{ll}\\ 40.&\textrm{Diketahui titik}\: \: \textrm{A}(-4,-1,-2),\: \textrm{B}(-6,4,3),\\ &\textrm{C}(2,3,5).\: \: \textrm{Jika titik M membagi}\: \: \overrightarrow{AB}\\ &\textrm{sehingga}\: \: \overrightarrow{AM}:\overrightarrow{MB}=3:2\\ &\textrm{maka vektor yang diwakili oleh}\: \: \overrightarrow{MC}=....\\ &\begin{array}{lll}\\ \textrm{a}.\quad (-4,1,4)&&\textrm{d}.\quad (6,4,1)\\ \textrm{b}.\quad (-2,2,1)&\textrm{c}.\quad (0,5,6)&\color{red}\textrm{e}.\quad (4,1,4) \end{array}\\\\ &\textbf{Jawab}:\\ &\textrm{Coba perhatikanlah ilustrasi berikut} \end{array}$

$.\: \qquad\begin{array}{|c|c|}\hline \textrm{Diketahui}&\textrm{Proses Penyelesaian}\\\hline \begin{aligned}\overrightarrow{AM}:\overrightarrow{MB}&=3:2\\ \vec{m}&=\displaystyle \frac{2\vec{a}+3\vec{b}}{5}\\ &\\ &\\ &\\ &\\ &\\ &\\ &\\ &\\ &\\ &\\ & \end{aligned}&\begin{aligned}\overrightarrow{MC}&=\vec{c}-\vec{m}\\ &=\vec{c}-\displaystyle \frac{2\vec{a}+3\vec{b}}{5}\\ &=\displaystyle \frac{5\vec{c}-2\vec{a}-3\vec{b}}{5}\\ &=\displaystyle \frac{5\begin{pmatrix} 2\\ 3\\ 5 \end{pmatrix}-2\begin{pmatrix} 4\\ -1\\ -2 \end{pmatrix}-3\begin{pmatrix} -6\\ 4\\ 3 \end{pmatrix}}{5}\\ &=\displaystyle \frac{\begin{pmatrix} 20\\ 5\\ 20 \end{pmatrix}}{5}\\ &=\begin{pmatrix} 4\\ 1\\ 4 \end{pmatrix} \end{aligned}\\\hline \end{array}$