Contoh 7 Soal dan Pembahasan Materi Vektor

$\begin{array}{ll}\\ 31.&\textrm{Diketahui}\: \: \left | \vec{a} \right |=\sqrt{3}\: ,\left | \vec{b} \right |=1,\: \: \textrm{dan}\: \: \left | \vec{a}-\vec{b} \right |=1\\ &\textrm{maka panjang vektor}\: \: \vec{a}+\vec{b}\: \: \textrm{adalah}\: ....\\ &\begin{array}{llllllll}\\ \textrm{a}.&\sqrt{3}&&&\textrm{d}.&2\sqrt{2}\\ \textrm{b}.&\sqrt{5}&\textrm{c}.&\color{red}\sqrt{7}&\textrm{e}.&3 \end{array}\\\\ &\textbf{Jawab}\\ &\begin{aligned}\textrm{Diketahui}&\: \, \textrm{sebagaimana pada soal}\\ \left | \vec{a}-\vec{b} \right |^{2}&=\left | \vec{a} \right |^{2}+\left | \vec{b} \right |^{2}-2\left | \vec{a} \right |\left | \vec{b} \right |\cos \theta \\ 1^{2}&=\left ( \sqrt{3} \right )^{2}+1^{2}-2.\sqrt{3}.1.\cos \theta \\ 2\sqrt{3}\cos \theta &=3\\ \textrm{maka pan}&\textrm{jang vektor}\: \: \vec{a}+\vec{b}\: \: \textrm{adalah}\\ \left | \vec{a}+\vec{b} \right |&=\sqrt{\left | \vec{a} \right |^{2}+\left | \vec{b} \right |^{2}+2\left | \vec{a} \right |\left | \vec{b} \right |\cos \theta}\\ &=\sqrt{\left ( \sqrt{3} \right )^{2}+1^{2}+3}\\ &=\sqrt{3+1+3}\\ &=\color{red}\sqrt{7} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 32.&\textrm{Jika}\: \: \left | \overline{u} \right |=6\: ,\: \left | \overline{v} \right |=4\sqrt{3},\: \: \textrm{dan}\: \: \left | \overline{u}-\overline{v} \right |=8\\ &\textrm{tentukanlah nilai dari}\\ &\textrm{a}.\quad \overline{u}\bullet \overline{v}\\ &\textrm{b}.\quad \left | \overline{u}+\overline{v} \right |\\ &\textrm{c}.\quad \textbf{cosinus}\:  \: \textrm{sudut antara}\: \: \overline{u}\: \: \textrm{dan}\: \: \overline{v}\\\\ &\textbf{Jawab}:\\ &\begin{aligned}\textrm{a}.\quad &\overline{u}\bullet \overline{v}=\: \cdots \\ &\: 2.\overline{u}\bullet \overline{v}=\left | \overline{u} \right |^{2}+\left | \overline{v} \right |^{2}-\left | \overline{u}-\overline{v} \right |^{2}\\ &\: 2.\overline{u}\bullet \overline{v}=6^{2}+(4\sqrt{3})^{2}-8^{2}\\ &\: 2. \overline{u}\bullet \overline{v}=36+48-64=84-64=20\\ &\quad \overline{u}\bullet \overline{v}=\displaystyle \frac{20}{2}=\color{red}10\\ \textrm{b}.\quad &\left | \overline{u}+\overline{v} \right |^{2}=\left | \overline{u} \right |^{2}+\left | \overline{v} \right |^{2}+2.\overline{u}\bullet \overline{v}\\ &\left | \overline{u}+\overline{v} \right |^{2}=6^{2}+(4\sqrt{3})^{2}+20\\ &\: \: \quad\qquad =84+20=104\\ &\left | \overline{u}+\overline{v} \right |=\color{red}\sqrt{104}\\ \textrm{c}.\quad &\cos \angle (\overline{u},\, \overline{v})=\displaystyle \frac{\overline{u}\bullet \overline{v}}{\left |\overline{u}  \right |.\left | \overline{v} \right |}=\frac{10}{6.(4\sqrt{3})}\times \frac{\sqrt{3}}{\sqrt{3}}\\ &\quad\qquad\qquad =\displaystyle \frac{10\sqrt{3}}{72}=\color{red}\frac{5}{36}\sqrt{3}\\ &\color{blue}\textbf{Berikut ilustrasi gambarnya}  \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 33.&\textrm{Jika}\: \: \vec{p}=\begin{pmatrix} 2\\ -5 \end{pmatrix},\: \vec{q}=\begin{pmatrix} 4\\ 3 \end{pmatrix},\: \textrm{maka }\\ &\textrm{proyeksi skalar ortogonal vektor}\: \vec{p}\\ &\textrm{pada}\: \: \vec{q}\: \: \textrm{adalah}....\\ &\textrm{a}.\quad \displaystyle \frac{3}{5}\\\\ &\textrm{b}.\quad \color{red}\displaystyle \frac{7}{5}\\\\ &\textrm{c}.\quad \displaystyle \frac{8}{5}\\\\ &\textrm{d}.\quad \displaystyle \frac{9}{5}\\\\ &\textrm{e}.\quad \displaystyle 2\\\\ &\textrm{Jawab}\\ &\begin{aligned}\left | \vec{r} \right |&=\displaystyle \frac{\vec{p}.\vec{q}}{\left | \vec{q} \right |}\\ &=\displaystyle \frac{\begin{pmatrix} 2\\ -5 \end{pmatrix}.\begin{pmatrix} 4\\ 3 \end{pmatrix}}{\sqrt{4^{2}+3^{2}}}\\ &=\displaystyle \frac{8+(-15)}{\sqrt{25}}\\ &=\left | \displaystyle \frac{-7}{5} \right |\\ &=\color{red}\displaystyle \frac{7}{5} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 34.&\textrm{Panjang Proyeksi vektor}\: \: \vec{a}=\begin{pmatrix} 5\\ 1 \end{pmatrix}\\ & \textrm{pada}\: \: \vec{b}=\begin{pmatrix} 0\\ 4 \end{pmatrix}\: \: \textrm{adalah}....\\ &\textrm{a}.\quad \displaystyle -1\\ &\textrm{b}.\quad -\displaystyle \frac{1}{2}\\ &\textrm{c}.\quad \color{red}1\\ &\textrm{d}.\quad \displaystyle 2\\ &\textrm{e}.\quad 4\\\\ &\textrm{Jawab}\\ &\begin{aligned}\left |\vec{c} \right |&=\left |\displaystyle \frac{\vec{a}.\vec{b}}{\left | \vec{b} \right |} \right |\\ &=\left |\displaystyle \frac{\begin{pmatrix} 5\\ 1 \end{pmatrix}\bullet \begin{pmatrix} 0\\ -4 \end{pmatrix}}{\left | \sqrt{0^{2}+(-4)^{2}} \right |} \right |\\ &=\left |\displaystyle \frac{0-4}{4} \right |=\left |-1 \right |=\color{red}1 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 35.&\textrm{Proyeksi vektor ortogonal}\: \: \vec{a}=\begin{pmatrix} 2\\ -4 \end{pmatrix}\\ & \textrm{pada}\: \: \vec{b}=\begin{pmatrix} -1\\ 2 \end{pmatrix}\: \: \textrm{adalah}....\\ &\textrm{a}.\quad \begin{pmatrix} 2\\ -1 \end{pmatrix}\\ &\textrm{b}.\quad \begin{pmatrix} 2\\ -2 \end{pmatrix}\\ &\textrm{c}.\quad \color{red}\begin{pmatrix} 2\\ -4 \end{pmatrix}\\ &\textrm{d}.\quad \begin{pmatrix} -1\\ 2 \end{pmatrix}\\ &\textrm{e}.\quad \begin{pmatrix} -2\\ 4 \end{pmatrix}\\\\ &\textrm{Jawab}\\ &\begin{aligned}\vec{c}&=\left ( \displaystyle \frac{\vec{a}\bullet \vec{b}}{\left |\vec{b} \right |^{2}} \right ).\vec{b}\\ &=\left (\displaystyle \frac{\begin{pmatrix} 2\\ -4 \end{pmatrix}\bullet \begin{pmatrix} -1\\ 2 \end{pmatrix}}{(-1)^{2}+2^{2}} \right ).\begin{pmatrix} -1\\ 2 \end{pmatrix}\\ &=\left (\displaystyle \frac{-2-8}{1+4} \right ).\begin{pmatrix} -1\\ 2 \end{pmatrix}\\ &=-2\begin{pmatrix} -1\\ 2 \end{pmatrix}\\ &=\color{red}\begin{pmatrix} 2\\ -4 \end{pmatrix} \end{aligned} \end{array}$

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