Contoh 6 Vektor

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

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

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

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

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

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