Lanjutan Contoh Soal Vektor Dimensi Dua (Bagian 3)

$\begin{array}{ll}\\ 7&\textrm{Jika}\: \: \bar{a}=\begin{pmatrix} 4\\ -3 \end{pmatrix}\: \: \textrm{dan}\: \: \bar{b}=\begin{pmatrix} 1\\ 2 \end{pmatrix}\\ &\textrm{tentukanlah}\\ &\textrm{a}.\quad \bar{a}+\bar{b}\qquad\qquad\qquad \textrm{d}.\quad \left ( 3\bar{a}+3\bar{b} \right )\\ &\textrm{b}.\quad \bar{b}+\bar{a}\, \qquad\qquad\qquad \textrm{e}.\quad 2(\bar{a}-\bar{b})\\ &\textrm{c}.\quad 3(\bar{a}+ \bar{b})\: \: \, \, \: \qquad \qquad \textrm{f}.\quad 2 \bar{a}-2\bar{b}\\\\ &\textrm{Jawab}:\\ &\begin{aligned}\textrm{a}\quad &\bar{a}+\bar{b}=\begin{pmatrix} 4\\ -3 \end{pmatrix}+\begin{pmatrix} 1\\ 2 \end{pmatrix}=\begin{pmatrix} 4+1\\ -3+2 \end{pmatrix}=\begin{pmatrix} 5\\ -1 \end{pmatrix}\\ \textrm{b}\quad &\bar{b}+\bar{a}=\begin{pmatrix} 1\\ 2 \end{pmatrix}+\begin{pmatrix} 4\\ -3 \end{pmatrix}=\begin{pmatrix} 1+4\\ 2+(-3) \end{pmatrix}=\begin{pmatrix} 5\\ -1 \end{pmatrix}\\ \textrm{c}\quad &\cdots \\ \textrm{d}\quad &3\bar{a}+3\bar{b}=3\begin{pmatrix} 4\\ -3 \end{pmatrix}+3\begin{pmatrix} 1\\ 2 \end{pmatrix}\\ &=\begin{pmatrix} 12\\ -9 \end{pmatrix}+\begin{pmatrix} 3\\ 6 \end{pmatrix}=\begin{pmatrix} 12+3\\ -9+6 \end{pmatrix}=\begin{pmatrix} 15\\ -3 \end{pmatrix}\\ \textrm{e}\quad &\cdots \\ \textrm{f}\quad &\cdots \end{aligned} \end{array}$

$\begin{array}{ll}\\ 8.&\textrm{Perhatikanlah gambar berikut} \end{array}$.

$.\qquad\begin{array}{ll}\\ &\textrm{Pada Balok ABCD.EFGH diatas diketahui }\\ &\textrm{DA = 4 cm, DC = 5 cm, dan DH 3 cm}.\\ &\textrm{Misalkan}\: \: \vec{i}\: \: \textrm{adalah vektor satuan dengan arah }\\ &\textrm{sama dengan}\: \: \overrightarrow{DA},\: \vec{j}\: \: \textrm{adalah vektor satuan }\\ &\textrm{dengan arah sama}\: \: \overrightarrow{DC}\: ,\: \textrm{dan}\: \: \vec{k}\\ &\textrm{adalah vektor satuan dengan arah sama dengan}\: \: \overrightarrow{DH}.\: \\ &\textrm{Nyatakanlah vektor-vektor berikut dalam vektor}\: \: \vec{i},\: \vec{j}\: \textrm{dan}\: \: \vec{k}.\\ &\begin{array}{ll}\\ \textrm{a}.&\overrightarrow{DA},\: \: \overrightarrow{DC}\: \: \textrm{dan} \: \: \overrightarrow{DH}\\ \textrm{b}.&\overrightarrow{DB}\: \: \: \: \textrm{dan}\: \: \overrightarrow{DF}\\ \textrm{c}.&\overrightarrow{DP}\: \: \textrm{jika}\: \: P\: \: \textrm{titik tengan}\: \: EF\\ \textrm{d}.&\overrightarrow{DQ}\: \: \textrm{jika}\: \: Q\: \: \textrm{titik pada perpanjangan}\\ & FG\: \: \textrm{dengan}\: \: FG=GQ\end{array} \end{array}$.

$.\qquad\begin{aligned}&\textrm{Jawab}:\\ &\begin{array}{|l|l|}\hline \begin{aligned}\textrm{a}.&\begin{cases} \overrightarrow{DA}=4\vec{i}\\ \overrightarrow{DC}=5\vec{j} \\ \overrightarrow{DH}=3\vec{k} \end{cases}\\ &\\ &\\ \end{aligned}&\begin{aligned}\textrm{b}.\: \overrightarrow{DB}&=\overrightarrow{DC}+\overrightarrow{CB}\\ &=\overrightarrow{DC}+\overrightarrow{DA}\\ &=5\vec{j}+4\vec{i}\\ \overrightarrow{DF}&=\overrightarrow{DB}+\overrightarrow{BF}\\ &=4\vec{i}+5\vec{j}+3\vec{k} \end{aligned}\\\hline \end{array}\\ &\begin{array}{|l|l|}\hline \begin{aligned}\textrm{c}.\: \overrightarrow{DP}&=\overrightarrow{DE}+\overrightarrow{EP}\\ &=\overrightarrow{DA}+\overrightarrow{AE}+\overrightarrow{EP}\\ &=4\vec{i}+3\vec{k}+\frac{1}{2}\overrightarrow{EF}\\ &=4\vec{i}+3\vec{k}+\frac{1}{2}\overrightarrow{DC}\\ &= 4\vec{i}+3\vec{k}+\frac{1}{2}\left ( 5\vec{j} \right )\\ &=4\vec{i}+\frac{5}{2}\vec{j}+3\vec{k}\end{aligned}&\begin{aligned}\textrm{d}.\: \overrightarrow{DQ}&=\overrightarrow{DG}+\overrightarrow{GQ}\\ &=\overrightarrow{DC}+\overrightarrow{CG}+\overrightarrow{GQ}\\ &=\overrightarrow{DC}+\overrightarrow{DH}+\overrightarrow{AD}\\ &=\overrightarrow{DC}+\overrightarrow{DH}-\overrightarrow{DA}\\ &=5\vec{j}+3\vec{k}-4\vec{i}\\ &\\ &\\ & \end{aligned}\\\hline \end{array} \end{aligned}$

$\begin{array}{ll}\\ 9.&\textrm{Diketahui}\: \: \vec{a}=\begin{pmatrix} -5\\ 3 \end{pmatrix}\: \: \textrm{dan}\: \: \vec{b}=\begin{pmatrix} 6\\ -2 \end{pmatrix}.\\ & \textrm{Tentukanlah}\\ &\textrm{a}.\quad \vec{a}-\vec{b}\qquad\qquad \textrm{b}.\quad \vec{b}-\vec{a}\\ &\textrm{c}.\quad 4\left ( \vec{a}-\vec{b} \right )\qquad\qquad \textrm{d}.\quad 4\vec{a}-4\vec{b}\\\\ &\textrm{Jawab}:\\ &\begin{array}{llll}\\ \begin{aligned}a.\quad \vec{a}-\vec{b}&=\vec{a}+\left ( -\vec{b} \right )\\ &=\begin{pmatrix} -5\\ 3 \end{pmatrix}+\begin{pmatrix} -6\\ 2 \end{pmatrix}\\ &=\begin{pmatrix} -11\\ 5 \end{pmatrix} \end{aligned}&\\ \begin{aligned}b.\quad \vec{b}-\vec{a}&=\vec{b}+\left ( -\vec{a} \right )\\ &=\begin{pmatrix} 6\\ -2 \end{pmatrix}+\begin{pmatrix} 5\\ -3 \end{pmatrix}\\ &=\begin{pmatrix} 11\\ -5 \end{pmatrix} \end{aligned}\\ \begin{aligned}c.\quad 4\left ( \vec{a}-\vec{b} \right )&=4\begin{pmatrix} -11\\ 5 \end{pmatrix}\\ &=\begin{pmatrix} -44\\ 20 \end{pmatrix}\\ & \end{aligned}&\\ \end{array} \end{array}$

$\begin{array}{ll}\\ 10.&\textrm{Diketahui}\: \: \vec{p}=\begin{pmatrix} 4\\ -3 \end{pmatrix}\: \: \textrm{dan}\: \: \vec{q}=\begin{pmatrix} 1\\ 2 \end{pmatrix}.\\ & \textrm{Tentukanlah}\\ &\textrm{a}.\quad \vec{p}+\vec{q}\qquad\qquad \textrm{b}.\quad \vec{q}+\vec{p}\\ &\textrm{c}.\quad 4\left ( \vec{p}+\vec{q} \right )\qquad\qquad \textrm{d}.\quad 4\vec{p}+4\vec{q}\\ &\textrm{e}.\quad 4\left ( \vec{p}-\vec{q} \right )\: \: \quad\quad \textrm{f}.\quad 4\vec{p}-4\vec{q}\\\\ &\textrm{Jawab}:\\ &\begin{array}{llll}\\ \begin{aligned}a.\quad \vec{p}+\vec{q}&=\begin{pmatrix} 4\\ -3 \end{pmatrix}+\begin{pmatrix} 1\\ 2 \end{pmatrix}\\ &=\begin{pmatrix} 5\\ -1 \end{pmatrix} \end{aligned}&\begin{aligned}b.\quad \vec{q}+\vec{p}&=\begin{pmatrix} 1\\ 2 \end{pmatrix}+\begin{pmatrix} 4\\ -3 \end{pmatrix}\\ &=\begin{pmatrix} 5\\ -1 \end{pmatrix} \end{aligned}&&\\ \end{array} \end{array}$

$\begin{array}{ll}\\ 11.&\textrm{Pada contoh soal No. 8 tentukanlah }\\ &\textrm{panjang vektor}\: \: \overrightarrow{DA},\: \overrightarrow{DP}\: \textrm{dan}\: \: \overrightarrow{DQ}\\\\ &\textrm{Jawab}:\\ &\begin{aligned}&\left |\overrightarrow{DA} \right |=4\: cm,\\ &\left |\overrightarrow{DP} \right |^{2}=\left | \overrightarrow{DE} \right |^{2}+\left | \overrightarrow{EP} \right |^{2}\\ &=\left | \overrightarrow{DA} \right |^{2}+\left | \overrightarrow{AE} \right |^{2}+\left | \overrightarrow{EP} \right |^{2}\\ &=4^{2}+3^{2}+\left ( \frac{5}{2} \right )^{2}\\ &=16+9+\frac{25}{4}\\ &\left | \overrightarrow{DP} \right |=\sqrt{\frac{125}{4}}\\ &=\displaystyle \frac{5}{2}\sqrt{5}\: cm,\: \: \textrm{dan}\\ &\left | \overrightarrow{DQ} \right |^{2}=\left | \overrightarrow{DG} \right |^{2}+\left | \overrightarrow{GQ} \right |^{2}\\ &=\left | \overrightarrow{DC} \right |^{2}+\left | \overrightarrow{CG} \right |^{2}+\left | \overrightarrow{GQ} \right |^{2}\\ &=5^{2}+3^{2}+4^{2}\\ &=25+9+16\\ &=50\\ &\left | \overrightarrow{DQ} \right |&=\sqrt{50}\\ &=5\sqrt{2}\: cm \end{aligned} \end{array}$

$\begin{array}{ll}\\ 12.&\textrm{Perhatikanlah gambar di bawah ini}\\ &\textrm{dan nyatakan titik-titik pada gambar }\\ &\textrm{tersebut dalam vektor posisi}\\ \end{array}$

$.\qquad\begin{aligned}&\textrm{Jawab}:\\ &\textrm{Perhatikanlah gambar berikut} \end{aligned}$

$.\qquad\begin{array}{|l|l|}\hline \textrm{Titik}&\textrm{Vektor Posisi}\\\hline P(5,3)&\vec{p}=\begin{pmatrix} 5\\ 3 \end{pmatrix}\\\hline Q(2,-3)&\vec{q}=\begin{pmatrix} 2\\ -3 \end{pmatrix}\\\hline R(-5,-1)&\vec{r}=\begin{pmatrix} -5\\ -1 \end{pmatrix}\\\hline S(-3,7)&\vec{s}=\begin{pmatrix} -3\\ 7 \end{pmatrix}\\\hline \end{array}$

$\begin{array}{ll}\\ 13.&\textrm{Perhatikanlah gambar pada soal No. 12 di atas. }\\ &\textrm{Tentukanlah vektor-vektor berikut}\\ &\textrm{a}.\quad \overrightarrow{PQ}\\ &\textrm{b}.\quad \overrightarrow{PS}\\ &\textrm{c}.\quad \overrightarrow{QS}\\ &\textrm{d}.\quad \overrightarrow{QP}+\overrightarrow{PS}\\\\ &\textrm{Jawab}:\\ &\begin{array}{|l|l|}\hline \begin{aligned}\textrm{a}.\quad \overrightarrow{PQ}&=\overrightarrow{PO}+\overrightarrow{OQ}\\ &=-\overrightarrow{OP}+\overrightarrow{OQ}\\ &=\overrightarrow{OQ}-\overrightarrow{OP}\\ &=\vec{q}-\vec{p}\\ &=\begin{pmatrix} 2\\ -3 \end{pmatrix}-\begin{pmatrix} 5\\ 3 \end{pmatrix}\\ &=\begin{pmatrix} -3\\ -6 \end{pmatrix} \end{aligned}&\begin{aligned}\textrm{b}.\quad \overrightarrow{PS}&=\vec{s}-\vec{p}\\ \textrm{dengan }&\textrm{cara semisal poin a}\\ \overrightarrow{PS}&=\begin{pmatrix} -3\\ 7 \end{pmatrix}-\begin{pmatrix} 5\\ 3 \end{pmatrix}\\ &=\begin{pmatrix} -8\\ 4 \end{pmatrix}\\ &\\ & \end{aligned} \\\hline \end{array}\\ &\begin{array}{|l|l|}\hline \begin{aligned}\textrm{c}.\quad \overrightarrow{QS}&=\vec{s}-\vec{q}\\ &=\begin{pmatrix} -3\\ 7 \end{pmatrix}-\begin{pmatrix} 2\\ -3 \end{pmatrix}\\ &=\begin{pmatrix} -5\\ 10 \end{pmatrix} \end{aligned}&\begin{aligned}\textrm{d}.\quad \overrightarrow{QP}+\overrightarrow{PS}&=\overrightarrow{QS}\\ &=\begin{pmatrix} -5\\ 10 \end{pmatrix}\\ \textrm{lihat jawaban}&\textrm{ poin c} \end{aligned}\\\hline \end{array} \end{array}$