Contoh 8 Vektor

Contoh soal sebelumnya di sini Contoh Soal 7

$\begin{array}{l}\\ 34.& \text{Vektor satuan untuk}\overrightarrow{a}=\left(\begin{array}{c}2\\ -1\\ 4\end{array}\right)\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.{\displaystyle \frac{1}{2}\sqrt{3}\left(\begin{array}{c}2\\ -1\\ 4\end{array}\right)}& \\ \text{b}.{\displaystyle \frac{1}{3}\sqrt{3}\left(\begin{array}{c}2\\ -1\\ 4\end{array}\right)}& \\ \text{c}.{\displaystyle \frac{1}{5}\sqrt{5}\left(\begin{array}{c}2\\ -1\\ 4\end{array}\right)}& \\ \text{d}.{\displaystyle \frac{1}{7}\sqrt{7}\left(\begin{array}{c}2\\ -1\\ 4\end{array}\right)}\\ \text{e}.{\displaystyle \frac{1}{21}\sqrt{21}\left(\begin{array}{c}2\\ -1\\ 4\end{array}\right)}\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}\text{Vektor satuan}\overrightarrow{a}& \text{adalah}{\overrightarrow{e}}_{\overrightarrow{a}},\text{yaitu}:\\ {\overrightarrow{e}}_{\overrightarrow{a}}& ={\displaystyle \frac{\overrightarrow{a}}{\left|\overrightarrow{a}\right|}}\\ & ={\displaystyle \frac{\left(\begin{array}{c}2\\ -1\\ 4\end{array}\right)}{\sqrt{2^2+(-1{)}^2+4^2}}}\\ & ={\displaystyle \frac{1}{\sqrt{21}}\left(\begin{array}{c}2\\ -1\\ 4\end{array}\right)}\\ & ={\displaystyle \frac{1}{21}\sqrt{21}\left(\begin{array}{c}2\\ -1\\ 4\end{array}\right)}\end{array}\end{array}$.

$\begin{array}{l}\\ 35.& \text{Posisi suatu titik dalam ruang saat }\\ & \text{waktu}t\text{ditunjukkan oleh vektor}\\ & \left(\begin{array}{c}t\\ t^2\\ -t\end{array}\right).\text{Jika pada saat}t=1\text{titik }\\ & \text{tersebut berada di titik P dan pada}\\ & \text{saat}t=2\text{titik tersebut berada }\\ & \text{di titik Q, maka jarak titik P dari Q}\\ & \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.\sqrt{24}-\sqrt{3}& & \text{d}.\sqrt{11}\\ \text{b}.2-\sqrt{2}& \text{c}.3& \text{e}.\sqrt{43}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{a}}\\ & \begin{array}{rl}\left|\overrightarrow{PQ}\right|& =\sqrt{(x_q-x_p{)}^2+(y_q-y_p{)}^2+(y_q-y_p{)}^2}\\ & =\sqrt{(2-1{)}^2+(2^2-1^2{)}^2+((-2)-(-1){)}^2}\\ & =\sqrt{1^2+3^2+(-1{)}^2}\\ & =\sqrt{1+9+1}\\ & =\sqrt{11}\end{array}\end{array}$

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

$\begin{array}{l}\\ 37.& \text{Jika diketahui titik}A(2,-1,4),B(4,1,3),\\ & \text{ dan}C(2,0,5),\text{maka}\sin \mathrm{\angle }(\overline{AB},\overline{AC})=....\\ & \begin{array}{l}\\ \text{a}.{\displaystyle \frac{1}{7}\sqrt{5}}& & \text{d}.{\displaystyle \frac{1}{6}\sqrt{3}}\\ \\ \text{b}.{\displaystyle \frac{1}{6}\sqrt{34}}& \text{c}.{\displaystyle \frac{2}{3}\sqrt{2}}& \text{e}.{\displaystyle \frac{1}{6}\sqrt{2}}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& \cos \mathrm{\angle }(\overline{AB},\overline{AC})={\displaystyle \frac{\overline{AB}.\overline{AC}}{\left|\overline{AB}\right|.\left|\overline{AC}\right|}}\\ & ={\displaystyle \frac{(\overrightarrow{b}-\overrightarrow{a}).(\overrightarrow{c}-\overrightarrow{a})}{\sqrt{x_{(\overrightarrow{b}-\overrightarrow{a})}^2+y_{(\overrightarrow{b}-\overrightarrow{a})}^2+z_{(\overrightarrow{b}-\overrightarrow{a})}^2}.\sqrt{x_{(\overrightarrow{c}-\overrightarrow{a})}^2+y_{(\overrightarrow{c}-\overrightarrow{a})}^2+z_{(\overrightarrow{c}-\overrightarrow{a})}^2}}}\\ & ={\displaystyle \frac{\left(\begin{array}{c}4-2\\ 1+1\\ 3-4\end{array}\right).\left(\begin{array}{c}2-2\\ 0+1\\ 5-4\end{array}\right)}{\sqrt{(4-2{)}^2+(1+1{)}^2+(3-4{)}^2}.\sqrt{(2-2{)}^2+(0+1{)}^2+(5-4{)}^2}}}\\ & ={\displaystyle \frac{2.0+2.1+-1.1}{\sqrt{4+4+1}.\sqrt{0+1+1}}}\\ & ={\displaystyle \frac{1}{3\sqrt{2}}}\\ & ={\displaystyle \frac{1}{6}\sqrt{2}}\end{array}\\ & \begin{array}{rl}& \text{Sehingga},\\ & \sin \mathrm{\angle }(\overline{AB},\overline{AC})\\ & =\sqrt{1-{\cos }^2\mathrm{\angle }(\overline{AB},\overline{AC})}\\ & =\sqrt{1-{\left({\displaystyle \frac{1}{6}\sqrt{2}}\right)}^2}\\ & =\sqrt{1-{\displaystyle \frac{2}{36}}}\\ & =\sqrt{{\displaystyle \frac{34}{36}}}\\ & ={\displaystyle \frac{1}{6}\sqrt{34}}\end{array}\end{array}$.

$\begin{array}{l}\\ 38.& \text{Diketahui segitiga ABC. Titik M }\\ & \text{di tengah AC, dan titik N pada BC}\\ & \text{Jika}\overrightarrow{AB}=\overrightarrow{c},\overrightarrow{AC}=\overrightarrow{b},\overrightarrow{BC}=\overrightarrow{a},\\ & \text{maka}\overrightarrow{MN}=....\\ & \begin{array}{l}\\ \text{a}.{\displaystyle \frac{1}{2}(\overrightarrow{b}-\overrightarrow{c})}& \\ \text{b}.{\displaystyle \frac{1}{2}(-\overrightarrow{b}+\overrightarrow{c})}& \\ \text{c}.{\displaystyle \frac{1}{2}(-\overrightarrow{a}+\overrightarrow{c})}\\ \text{d}.{\displaystyle \frac{1}{2}(\overrightarrow{a}-\overrightarrow{b})}\\ \text{e}.{\displaystyle \frac{1}{2}(-\overrightarrow{a}+\overrightarrow{b})}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}\overrightarrow{MN}& =\overrightarrow{MC}+\overrightarrow{CN}\\ & ={\displaystyle \frac{1}{2}\overrightarrow{AC}+{\displaystyle \frac{1}{2}(-\overrightarrow{BC})}}\\ & ={\displaystyle \frac{1}{2}(\overrightarrow{AC}-\overrightarrow{BC})}\\ & ={\displaystyle \frac{1}{2}(\overrightarrow{b}-\overrightarrow{a})\mathbf{\text{atau}}}\\ & ={\displaystyle \frac{1}{2}(-\overrightarrow{a}+\overrightarrow{b})}\end{array}\end{array}$.

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

$.\begin{array}{r}\\ & \begin{array}{rl}\text{koordinat}& {\text{titik A}}^{\prime }\\ & ={\displaystyle \frac{1}{2}(5+0,4-1,10+6)}\\ & =(\frac{5}{2},\frac{3}{2},8)\end{array}\\ & \begin{array}{rl}\text{Dalam se}& \text{gitiga ABC untuk titik berat Z }\\ \text{berlaku}& \text{ketentuan sebagai berikut}\\ AZ:Z{A}^{\prime }& =2:1\\ \overrightarrow{AZ}:\overrightarrow{Z{A}^{\prime }}& =2:1\\ \overrightarrow{OZ}& ={\displaystyle \frac{\overrightarrow{OA}+2\overrightarrow{O{A}^{\prime }}}{3}}\\ & ={\displaystyle \frac{\left(\begin{array}{c}1\\ 0\\ 2\end{array}\right)+2\left(\begin{array}{c}\frac{5}{2}\\ \frac{3}{2}\\ 8\end{array}\right)}{3}}\\ & =\left(\begin{array}{c}2\\ 1\\ 6\end{array}\right)\\ \text{Jadi},& \text{koordinat titik Z adalah}(2,1,6)\end{array}\end{array}$.

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

$.\begin{array}{|cc|}\hline \text{Diketahui}& \text{Proses Penyelesaian}\\ \begin{array}{rl}\overrightarrow{AM}:\overrightarrow{MB}& =3:2\\ \overrightarrow{m}& ={\displaystyle \frac{2\overrightarrow{a}+3\overrightarrow{b}}{5}}\\ & \\ & \\ & \\ & \\ & \\ & \\ & \\ & \\ & \\ & \\ & \end{array}& \begin{array}{rl}\overrightarrow{MC}& =\overrightarrow{c}-\overrightarrow{m}\\ & =\overrightarrow{c}-{\displaystyle \frac{2\overrightarrow{a}+3\overrightarrow{b}}{5}}\\ & ={\displaystyle \frac{5\overrightarrow{c}-2\overrightarrow{a}-3\overrightarrow{b}}{5}}\\ & ={\displaystyle \frac{5\left(\begin{array}{c}2\\ 3\\ 5\end{array}\right)-2\left(\begin{array}{c}4\\ -1\\ -2\end{array}\right)-3\left(\begin{array}{c}-6\\ 4\\ 3\end{array}\right)}{5}}\\ & ={\displaystyle \frac{\left(\begin{array}{c}20\\ 5\\ 20\end{array}\right)}{5}}\\ & =\left(\begin{array}{c}4\\ 1\\ 4\end{array}\right)\end{array}\\ \hline\end{array}$