Contoh 6 Soal dan Pembahasan Materi Vektor

 $\begin{array}{l}\\ 26.& \text{Jika}\overrightarrow{p}=\left(\begin{array}{c}-2\\ 4\end{array}\right)\text{dan}\overrightarrow{q}=\left(\begin{array}{c}8\\ 4\end{array}\right),\text{maka}\\ & \text{sudut yang dibentuk vektor}\overrightarrow{p}\text{dan}\overrightarrow{q}\\ & \text{adalah}....\\ & \text{a}.0^{\circ }\\ & \text{b}.{60}^{\circ }\\ & \text{c}.{45}^{\circ }\\ & \text{d}.{60}^{\circ }\\ & \text{e}.{90}^{\circ }\\ \\ & \text{Jawab}\\ & \begin{array}{rl}\overrightarrow{p}.\overrightarrow{q}& ={\displaystyle \left|\begin{array}{c}\overrightarrow{p}\end{array}\right|.\left|\begin{array}{c}\overrightarrow{q}\end{array}\right|.\cos \mathrm{\angle }(\overrightarrow{p},\overrightarrow{q})}\\ \cos \mathrm{\angle }(\overrightarrow{p},\overrightarrow{q})& ={\displaystyle \frac{\overrightarrow{p}.\overrightarrow{q}}{\left|\overrightarrow{p}\right|.\left|\overrightarrow{q}\right|}}\\ & ={\displaystyle \frac{\left(\begin{array}{c}-2\\ 1\end{array}\right).\left(\begin{array}{c}8\\ 4\end{array}\right)}{\sqrt{(-2{)}^2+1^2}.\sqrt{8^2+4^2}}}\\ & ={\displaystyle \frac{-16+16}{\sqrt{20}.\sqrt{80}}}\\ & ={\displaystyle \frac{0}{40}}\\ & =0\\ \cos \mathrm{\angle }(\overrightarrow{p},\overrightarrow{q})& =\cos {90}^{\circ }\\ \mathrm{\angle }(\overrightarrow{p},\overrightarrow{q})& ={90}^{\circ }\end{array}\end{array}$

$\begin{array}{l}\\ 27.& \text{Jika}\overline{OA}=\left(\begin{array}{c}1\\ 2\end{array}\right),\overline{OB}=\left(\begin{array}{c}4\\ 2\end{array}\right),\text{dan}\\ & \theta =\mathrm{\angle }(\overline{OA},\overline{OB}),\text{maka}\tan \theta =....\\ \\ & \text{a}.{\displaystyle \frac{3}{5}}\\ \\ & \text{b}.{\displaystyle \frac{9}{16}}\\ \\ & \text{c}.{\displaystyle \frac{3}{4}}\\ \\ & \text{d}.{\displaystyle \frac{4}{3}}\\ \\ & \text{e}.{\displaystyle \frac{16}{9}}\\ \\ & \text{Jawab}\\ & \begin{array}{|cc|}\hline \begin{array}{rl}\cos \theta & ={\displaystyle \frac{\overrightarrow{a}.\overrightarrow{b}}{\left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right|}}\\ & ={\displaystyle \frac{\left(\begin{array}{c}1\\ 2\end{array}\right).\left(\begin{array}{c}4\\ 2\end{array}\right)}{\sqrt{1^2+2^2}\sqrt{4^2+2^2}}}\\ & ={\displaystyle \frac{4+4}{\sqrt{5}.\sqrt{20}}}\\ & ={\displaystyle \frac{8}{10}}\end{array}& \begin{array}{rl}\sin \theta & =\sqrt{1-{\cos }^2\theta }\\ & =\sqrt{1-{\left({\displaystyle \frac{8}{10}}\right)}^2}\\ & =\sqrt{{\displaystyle \frac{36}{100}}}\\ & ={\displaystyle \frac{6}{10}}\\ & \end{array}\\ \hline\end{array}\\ & \text{Selanjutnya}\\ & \begin{array}{rl}\tan \theta & ={\displaystyle \frac{\sin \theta }{\cos \theta }}\\ & ={\displaystyle \frac{{\displaystyle \frac{6}{10}}}{{\displaystyle \frac{8}{10}}}}\\ & ={\displaystyle \frac{3}{4}}\end{array}\end{array}$

$\begin{array}{l}\\ 28.& \text{Jika}\overrightarrow{a},\overrightarrow{b}\text{dan}\overrightarrow{c}\text{adalah vektor satuan dengan}\\ & \overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}=0.\text{Nilai dari}\overrightarrow{a}.\overrightarrow{b}+\overrightarrow{a}.\overrightarrow{c}+\overrightarrow{b}.\overrightarrow{c}\text{adalah}....\\ & \text{a}.{\displaystyle -3}\\ & \text{b}.{\displaystyle -\frac{3}{2}}\\ & \text{c}.{\displaystyle 0}\\ & \text{d}.{\displaystyle \frac{3}{2}}\\ & \text{e}.{\displaystyle 3}\\ \\ & \text{Jawab}\\ & \begin{array}{rl}\text{Karena}& \{\begin{array}{c}\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}\text{adalah vektor satuan, dan}\\ \overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}=0.\end{array}\\ \text{segitig}& \text{a ABC adalah segitiga sama sisi}\\ \overrightarrow{a}.\overrightarrow{b}=& \left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right|\cos {120}^0=1.1.(-\frac{1}{2})=-\frac{1}{2}\\ \overrightarrow{a}.\overrightarrow{c}=& \left|\overrightarrow{a}\right|\left|\overrightarrow{c}\right|\cos {120}^0=1.1.(-\frac{1}{2})=-\frac{1}{2}\\ \overrightarrow{b}.\overrightarrow{c}=& \left|\overrightarrow{b}\right|\left|\overrightarrow{c}\right|\cos {120}^0=1.1.(-\frac{1}{2})=-\frac{1}{2}\\ \text{Jadi, ni}& \text{lai dari}\overrightarrow{a}.\overrightarrow{b}+\overrightarrow{a}.\overrightarrow{c}+\overrightarrow{b}.\overrightarrow{c}\\ & =(-\frac{1}{2})+(-\frac{1}{2})+(-\frac{1}{2})=-\frac{3}{2}\end{array}\end{array}$

$.\text{berikut ilustrasinya}$

$\begin{array}{l}\\ 29.& \text{Jika}\mathrm{\angle }(\overrightarrow{a},\overrightarrow{b})={60}^{\circ },\left|\overrightarrow{a}\right|=4\text{dan}\\ & \left|\overrightarrow{b}\right|=3,\text{maka}\overrightarrow{a}(\overrightarrow{a}-\overrightarrow{b})\text{adalah}....\\ & \text{a}.2\\ & \text{b}.4\\ & \text{c}.6\\ & \text{d}.8\\ & \text{e}.10\\ \\ & \text{Jawab}\\ & \begin{array}{rl}\overrightarrow{a}(\overrightarrow{a}-\overrightarrow{b})& =\overrightarrow{a}.\overrightarrow{a}-\overrightarrow{a}.\overrightarrow{b}\\ & =\left|\overrightarrow{a}\right|\left|\overrightarrow{a}\right|\cos 0^{\circ }-\left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right|\cos {60}^{\circ }\\ & ={\left|\overrightarrow{a}\right|}^2-\left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right|.{\displaystyle \frac{1}{2}}\\ & =4^2-4.3.{\displaystyle \frac{1}{2}}\\ & =16-6\\ & =10\end{array}\end{array}$.

$\begin{array}{l}\\ 30.& \text{Tentukan}\bar{u}\bullet \bar{v},\text{jika diketahui}\\ & \text{a}.\left|\bar{u}\right|=10,\left|\bar{v}\right|=8\sqrt{3},\cos \mathrm{\angle }(\bar{v},\bar{u})={\displaystyle \frac{2}{5}\sqrt{3}}\\ & \text{b}.\left|\bar{u}\right|=6\sqrt{3},\left|\bar{v}\right|=4\sqrt{2},\cos (\bar{v},\bar{u})={30}^{\circ }\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}\text{a}.& \bar{u}\bullet \bar{v}=\left|\bar{u}\right|.\left|\bar{v}\right|.\cos \mathrm{\angle }(\bar{v},\bar{u})\\ & \bar{u}\bullet \bar{v}=10.(8\sqrt{3}).{\displaystyle \frac{2}{5}\sqrt{3}=2.8.2.3=96}\\ \text{b}.& \bar{u}\bullet \bar{v}=\left|\bar{u}\right|.\left|\bar{v}\right|.\cos \mathrm{\angle }(\bar{v},\bar{u})\\ & \bar{u}\bullet \bar{v}=(6\sqrt{3}).(4\sqrt{2}).{\displaystyle \cos {30}^{\circ }}\\ & =(6\sqrt{3}).(4\sqrt{2}).{\displaystyle \frac{1}{2}\sqrt{3}}\\ & ={\displaystyle \frac{6.4.3.\sqrt{2}}{2}=36\sqrt{2}}\end{array}\end{array}$.