Contoh 9 Vektor

$\begin{array}{ll}\\ 41.&\textrm{Diketahui segi empat ABCD dengan}\: \: \overrightarrow{DA}=\vec{a},\\ & \overrightarrow{DB}=\vec{b},\: \: \textrm{dan}\: \: \overrightarrow{DC}=\vec{c}.\: \: \textrm{Jika titik H pada AB}\\ &\textrm{dengan}\: \: \overrightarrow{AH}:\overrightarrow{HB}=1:2,\: \textrm{dan titik J pada BC}\\ &\textrm{dengan}\: \: \overrightarrow{BJ}:\overrightarrow{JC}=1:2\: \: \textrm{maka}\: \: \overrightarrow{HJ}=....\\ &\begin{array}{lll}\\ \textrm{a}.\quad \displaystyle \frac{-\vec{a}+\vec{b}+\vec{c}}{3}\\ \textrm{b}.\quad \displaystyle \frac{\vec{a}+\vec{b}+\vec{c}}{3}\\ \color{red}\textrm{c}.\quad \displaystyle \frac{-2\vec{a}+\vec{b}+\vec{c}}{3}&\\ \textrm{d}.\quad \displaystyle \frac{-2\vec{a}-\vec{b}+\vec{c}}{3}\\ \textrm{e}.\quad \displaystyle \frac{-2\vec{a}+\vec{b}-2\vec{c}}{3} \end{array}\\\\ &\textbf{Jawab}:\\ &\textrm{Coba perhatikanlah ilustrasi berikut} \end{array}$

$.\: \qquad\begin{aligned} &\begin{array}{|c|c|c|}\hline \textrm{Diketahui 1}&\textrm{Diketahui 2}\\\hline \begin{aligned}\overrightarrow{AH}:\overrightarrow{HB}&=1:2\\ \vec{h}&=\displaystyle \frac{2\vec{a}+\vec{b}}{3}\\ & \end{aligned}&\begin{aligned}\overrightarrow{BJ}:\overrightarrow{JC}&=1:2\\ \vec{j}&=\displaystyle \frac{2\vec{b}+\vec{c}}{3}\\ & \end{aligned}\\\hline \end{array}\\ &\textrm{Proses Penyelesaian}\\ &\begin{aligned}\overrightarrow{HJ}&=\vec{j}-\vec{h}\\ &=\left ( \displaystyle \frac{2\vec{b}+\vec{c}}{3} \right )-\left ( \displaystyle \frac{2\vec{a}+\vec{b}}{3} \right )\\ &=\color{red}\displaystyle \frac{-2\vec{a}+\vec{b}+\vec{c}}{3} \end{aligned} \end{aligned}$.

$\begin{array}{ll}\\ 42.&\textrm{Supaya vektor}\: \: \vec{a}=\begin{pmatrix} x\\ 4\\ 7 \end{pmatrix},\: \: \textrm{dan}\: \: \vec{b}=\begin{pmatrix} 6\\ y\\ 14 \end{pmatrix},\\ & \textrm{segaris, harga}\: \: x-y=....\\ &\begin{array}{lll}\\ \color{red}\textrm{a}.\quad -5&&\textrm{d}.\quad 4\\ \textrm{b}.\quad -2&\textrm{c}.\quad 3&\textrm{e}.\quad 6 \end{array}\\\\ &\textbf{Jawab}:\\ &\begin{aligned}\textrm{Dikethui}&\: \textrm{bahwa}:\\ \textrm{vektor}\: \: & \vec{a}\: \: \textrm{dan}\: \: \vec{b}\: \: \textrm{segaris},\: \textrm{maka}\\ m\vec{a}&=\vec{b}\\ \textrm{dengan}\: \: &m\: \: \textrm{adalah skalar/faktor pengali}\\ m\begin{pmatrix} x\\ 4\\ 7 \end{pmatrix}&=\begin{pmatrix} 6\\ y\\ 14 \end{pmatrix}\\ \textrm{di}\textrm{dapat}&\textrm{kan}\: \: \begin{cases} mx &=6\: ...................(1) \\ 4m &=y\: ....................(2) \\ 7m &=14 \: ...................(3) \end{cases}\\ \textrm{Dari per}& \textrm{samaan}\: \: (3)\: \: \textrm{akan}\\ \textrm{didapat}& \textrm{kan nilai}\: \: \color{blue}m=2\\ \textrm{maka}&\: \textrm{akan didapatkan juga}\: \: \begin{cases} x & =3 \\ y & =8 \end{cases}\\ \textrm{sehin}&\textrm{gga nilai dari}\\ x-y&=3-8\\ &=\color{red}-5 \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 43.&\textrm{Diketahui}\: \: O\: \: \textrm{titik pangkal}\: \: A(0,1,2)\\ &\textrm{dan}\: \: B(3,4,5),\: \textrm{maka luas segitiga}\\ & OAB\: \: \textrm{sama dengan}\: ...\: .\\ &\begin{array}{lll}\\ \textrm{a}.\quad 3\sqrt{6}\\ \textrm{b}.\quad \displaystyle \frac{2}{3}\sqrt{6}\\ \textrm{c}.\quad \displaystyle \frac{4}{3}\sqrt{6}&\\ \color{red}\textrm{d}.\quad \displaystyle \frac{3}{2}\sqrt{6}\\ \textrm{e}.\quad 2\sqrt{6} \end{array}\\\\ &\textbf{Jawab}:\\ &\textrm{Coba perhatikanlah ilustrasi berikut}\\ &\begin{aligned}&\textrm{Misalkan luas segitiga}\: \: \displaystyle \frac{1}{2}\left | \vec{p}\times \vec{q} \right |,\\ & \textrm{dengan}\\ &\begin{cases} \vec{p} & =\overline{OA}=\begin{pmatrix} 0\\ 1\\ 2 \end{pmatrix}\\ \vec{q} & =\overline{OB}=\begin{pmatrix} 3\\ 4\\ 5 \end{pmatrix} \end{cases} \end{aligned}\\ &\begin{aligned}\vec{p}&\times \vec{q}=\begin{vmatrix} \vec{i} & \vec{j} & \vec{k}\\ x_{1} & y_{1} &z_{1} \\ x_{2} & y_{2} &z_{2} \end{vmatrix}\\ &=(y_{1}z_{2}-z_{1}y_{2})\vec{i}-(x_{1}z_{2}-z_{1}x_{2})\vec{j}+(x_{1}y_{2}-y_{1}x_{2})\vec{k}\\ &=\begin{vmatrix} \vec{i} & \vec{j} & \vec{k}\\ 0 & 1 &2 \\ 3 & 4 &5 \end{vmatrix}\\ &=(5-8)\vec{i}-(0-6)\vec{j}+(0-3)\vec{k}\\ &=-3\vec{i}+6\vec{j}-3\vec{k}\\ &\textrm{Sehingga}\\ &\left | \vec{p}\times \vec{q} \right |=\sqrt{(-3)^{2}+6^{2}+(-3)^{2}}\\ &\quad\qquad =\sqrt{9+36+9}=\sqrt{54}=3\sqrt{6}\\ &\textrm{Maka luas segi tiganya adalah}:\\ &\textrm{luas}\: \triangle ABC=\displaystyle \frac{1}{2}\left | \vec{p}\times \vec{q} \right |=\displaystyle \frac{1}{2}\left ( 3\sqrt{6} \right )=\color{red}\displaystyle \frac{3}{2}\sqrt{6} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 44.&\textrm{Proyeksi skalar ortogonal}\\ &\overrightarrow{a}=2\vec{i}-3\vec{j}+6\vec{k},\: \textrm{pada}\\ &\overrightarrow{b}=\vec{i}+2\vec{j}+2\vec{k}\: \: \textrm{adalah}\: ...\: .\\ &\begin{array}{lllllll}\\ \textrm{a}.\quad \displaystyle \frac{4}{3}&&&&&\textrm{d}.&\displaystyle \frac{16}{3}\\\\ \color{red}\textrm{b}.\quad \displaystyle \frac{8}{3}&&\textrm{c}&\displaystyle \frac{10}{3}&&\textrm{e}.&\displaystyle \frac{20}{3}\\ \end{array}\\\\ &\textbf{Jawab}:\\ &\textrm{Diketahui bahwa}\\ &\begin{cases} \overrightarrow{a} & =(2,-3,6) \\ \overrightarrow{b} & =(1,2,2) \end{cases}\\ &\textrm{Selanjutnya}\\ &\begin{aligned}\\ \left | \overrightarrow{c} \right |&=\left |\displaystyle \frac{\overrightarrow{a}\overrightarrow{b}}{\left | \overrightarrow{b} \right |} \right |\\ &=\displaystyle \frac{\begin{pmatrix} 2\\ -3\\ 6 \end{pmatrix}\begin{pmatrix} 1\\ 2\\ 2 \end{pmatrix}}{\sqrt{1^{2}+2^{2}+2^{2}}}\\ &=\displaystyle \frac{2-6+12}{\sqrt{9}}=\color{red}\frac{8}{3} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 45.&\textrm{Diketahui vektor}\: \: \overrightarrow{a}=3\vec{i}-4\vec{j}+p\vec{k}\\ &\textrm{dan}\: \: \overrightarrow{b}=2\vec{i}+2\vec{j}-3\vec{k}.\: \textrm{Jika panjang}\\ &\textrm{proyeksi vektor}\: \: \overrightarrow{a}\: \: \textrm{pada}\: \: \overrightarrow{b}\: \: \textrm{adalah}\\ &\displaystyle \frac{4}{\sqrt{17}},\: \textrm{maka nilai}\: \: p\: \: \textrm{adalah}\: ...\: .\\ &\begin{array}{lllllll}\\ \color{red}\textrm{a}.\quad \displaystyle -2&&&&&\textrm{d}.&\displaystyle 2\\\\ \textrm{b}.\quad \displaystyle -1&&\textrm{c}&\displaystyle 1&&\textrm{e}.&\displaystyle 3\\ \end{array}\\\\ &\textbf{Jawab}:\\ &\textrm{Panjang proyeksi skalar vektor}\: \: \overrightarrow{a}\: \: \textrm{pada}\: \: \overrightarrow{b}\\ &\begin{aligned}\\ \left | \overrightarrow{c} \right |&=\displaystyle \frac{\overrightarrow{a}\overrightarrow{b}}{\left | \overrightarrow{b} \right |} \\ \displaystyle \frac{4}{\sqrt{17}}&=\displaystyle \frac{\begin{pmatrix} 3\\ -4\\ p \end{pmatrix}\begin{pmatrix} 2\\ 2\\ -3 \end{pmatrix}}{\sqrt{2^{2}+2^{2}+(-3)^{2}}}\\ \displaystyle \frac{4}{\sqrt{17}}&=\displaystyle \frac{6-8-3p}{\sqrt{17}}\\ 4&=-2-3p\\ 6&=-3p\\ -3p&=6\\ p&=\color{red}-2 \end{aligned} \end{array}$