PAS MATEMATIKA BLOG: April 2021

Persamaan Garis Singgung Lingkaran (PGSL)

$\color{blue}\textrm{A. PGS melalui titik pada lingkaran pusat (0,0)}$

Misalkan titik $P(x_{1},y_{1})$ yang terletak pada lingkaran $x^{2}+y^{2}=r^{2}$. Gradien dari garis OP adalah $\displaystyle \frac{y_{1}}{x_{1}}$

Perhatikanlah ilustrasi gambar berikut

$\color{blue}\textrm{B. PGS melalui titik pada lingkaran pusat (p,q)}$

Kurang lebih dengan penjelasan yang sama dan persamaan garisnya di rumuskan

$\begin{cases} (p,q) & \equiv (x_{1}-p)(x-p)+(y_{1}-q)(y-q)=r^{2} \\ (p,q) & \equiv px+qy+\displaystyle \frac{1}{2}A(p+x)+\displaystyle \frac{1}{2}B(q+y)+C=0 \end{cases}$.

$\color{blue}\textrm{C. PGSL dengan gradien m}$

$\color{blue}\textrm{D. PGS melalui titik di luar lingkaran pusat (0,0)}$

Contoh 9 Vektor

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

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

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

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

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

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

Contoh 8 Vektor

Contoh soal sebelumnya di sini Contoh Soal 7

$\begin{array}{ll}\\ 34.&\textrm{Vektor satuan untuk}\: \: \vec{a}=\begin{pmatrix} 2\\ -1\\ 4 \end{pmatrix}\: \textrm{adalah}....\\ &\begin{array}{lll}\\ \textrm{a}.\quad \displaystyle \frac{1}{2}\sqrt{3}\begin{pmatrix} 2\\ -1\\ 4 \end{pmatrix}&\\ \textrm{b}.\quad \displaystyle \frac{1}{3}\sqrt{3}\begin{pmatrix} 2\\ -1\\ 4 \end{pmatrix}&\\ \textrm{c}.\quad \displaystyle \frac{1}{5}\sqrt{5}\begin{pmatrix} 2\\ -1\\ 4 \end{pmatrix}&\\ \textrm{d}.\quad \displaystyle \frac{1}{7}\sqrt{7}\begin{pmatrix} 2\\ -1\\ 4 \end{pmatrix}\\ \color{red}\textrm{e}.\quad \displaystyle \frac{1}{21}\sqrt{21}\begin{pmatrix} 2\\ -1\\ 4 \end{pmatrix} \end{array}\\\\ &\textrm{Jawab}:\\ &\begin{aligned}\textrm{Vektor satuan}\: \: \vec{a}&\: \: \textrm{adalah}\: \: \vec{e}_{\vec{a}},\: \textrm{yaitu}:\\ \vec{e}_{\vec{a}}&=\displaystyle \frac{\vec{a}}{\left | \vec{a} \right |}\\ &=\displaystyle \frac{\begin{pmatrix} 2\\ -1\\ 4 \end{pmatrix}}{\sqrt{2^{2}+(-1)^{2}+4^{2}}}\\ &=\displaystyle \frac{1}{\sqrt{21}}\begin{pmatrix} 2\\ -1\\ 4 \end{pmatrix}\\ &=\displaystyle \frac{1}{21}\sqrt{21}\begin{pmatrix} 2\\ -1\\ 4 \end{pmatrix} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 35.&\textrm{Posisi suatu titik dalam ruang saat }\\ &\textrm{waktu}\: \: t\: \: \textrm{ditunjukkan oleh vektor}\\ &\begin{pmatrix} t\\ t^{2}\\ -t \end{pmatrix}.\: \textrm{Jika pada saat}\: \: t=1\: \: \textrm{titik }\\ &\textrm{tersebut berada di titik P dan pada}\\ &\textrm{saat}\: \: t=2\: \: \textrm{titik tersebut berada }\\ &\textrm{di titik Q, maka jarak titik P dari Q}\\ & \textrm{adalah}\: ...\: .\\ &\begin{array}{lll}\\ \textrm{a}.\quad \sqrt{24}-\sqrt{3}&&\color{red}\textrm{d}.\quad \sqrt{11}\\ \textrm{b}.\quad 2-\sqrt{2}&\textrm{c}.\quad 3&\textrm{e}.\quad \sqrt{43}\end{array}\\\\ &\textbf{Jawab}:\quad \textbf{a}\\ &\begin{aligned}\left | \overrightarrow{PQ} \right |&=\sqrt{(x_{q}-x_{p})^{2}+(y_{q}-y_{p})^{2}+(y_{q}-y_{p})^{2}}\\ &=\sqrt{(2-1)^{2}+(2^{2}-1^{2})^{2}+((-2)-(-1))^{2}}\\ &=\sqrt{1^{2}+3^{2}+(-1)^{2}}\\ &=\sqrt{1+9+1}\\ &=\sqrt{11} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 36.&\textrm{Jika diketahui}\: \: \left | \vec{a} \right |=4\sqrt{3},\: \left | \vec{b} \right |=5,\\ & \textrm{ dan}\: \left ( \vec{a}+\vec{b} \right ).\left ( \vec{a}+\vec{b} \right )=13,\\ & \textrm{maka}\: \: \angle \left ( \vec{a},\: \vec{b} \right )=....\\ &\begin{array}{lll}\\ \textrm{a}.\quad \displaystyle 30^{\circ}&&\textrm{d}.\quad \displaystyle 135^{\circ}\\ \textrm{b}.\quad \displaystyle 60^{\circ}&\textrm{c}.\quad \displaystyle 120^{\circ}&\color{red}\textrm{e}.\quad \displaystyle 150^{\circ} \end{array}\\\\ &\textrm{Jawab}:\\ &\begin{aligned}\left ( \vec{a}+\vec{b} \right ).\left ( \vec{a}+\vec{b} \right )&=13\\ \vec{a}.\vec{a}+\vec{a}.\vec{b}+\vec{b}.\vec{a}+\vec{b}.\vec{b}&=13\\ \left | \vec{a} \right |^{2}+2\vec{a}.\vec{b}+\left | \vec{b} \right |^{2}&=13,\\ \textrm{ingat bahwa}\: \: \vec{a}.\vec{b}=\vec{b}.\vec{a}&\\ \left ( 4\sqrt{3} \right )^{2}+2\left | \vec{a} \right |\left | \vec{b} \right |\cos \angle \left ( \vec{a},\: \vec{b} \right )+5^{2}&=13\\ 48+2.(4\sqrt{3}).5.\cos \angle \left ( \vec{a},\: \vec{b} \right )+25&=13\\ 40\sqrt{3}\cos \angle \left ( \vec{a},\: \vec{b} \right )&=13-25-48\\ \cos \angle \left ( \vec{a},\: \vec{b} \right )&=\displaystyle \frac{-60}{40\sqrt{3}}\\ &=-\displaystyle \frac{1}{2}\sqrt{3}\\ &=-\cos 30\\ &=\cos \left ( 180^{\circ}-30^{\circ} \right )\\ \cos \angle \left ( \vec{a},\: \vec{b} \right )&=\cos 150^{\circ}\\ \angle \left ( \vec{a},\: \vec{b} \right )&=150^{\circ} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 37.&\textrm{Jika diketahui titik}\: \: A(2,-1,4),\: B(4,1,3),\\ & \textrm{ dan}\: \: C(2,0,5),\: \: \textrm{maka}\: \: \sin \angle \left ( \overline{AB},\: \overline{AC} \right )=....\\ &\begin{array}{lll}\\ \textrm{a}.\quad \displaystyle \frac{1}{7}\sqrt{5}&&\textrm{d}.\quad \displaystyle \frac{1}{6}\sqrt{3}\\\\ \color{red}\textrm{b}.\quad \displaystyle \frac{1}{6}\sqrt{34}&\textrm{c}.\quad \displaystyle \frac{2}{3}\sqrt{2}&\textrm{e}.\quad \displaystyle \frac{1}{6}\sqrt{2} \end{array}\\\\ &\textbf{Jawab}:\\ &\begin{aligned}&\cos \angle \left ( \overline{AB},\: \overline{AC} \right )=\displaystyle \frac{\overline{AB}.\, \overline{AC}}{\left | \overline{AB} \right |.\left | \overline{AC} \right |} \\ &=\displaystyle \frac{(\vec{b}-\vec{a}).(\vec{c}-\vec{a})}{\sqrt{x^{2}_{\left (\vec{b}-\vec{a} \right )}+y^{2}_{\left ( \vec{b}-\vec{a} \right )}+z^{2}_{\left ( \vec{b}-\vec{a} \right )}}.\sqrt{x^{2}_{\left ( \vec{c}-\vec{a} \right )}+y^{2}_{\left ( \vec{c}-\vec{a} \right )}+z^{2}_{\left ( \vec{c}-\vec{a} \right )}}} \\ &=\displaystyle \frac{\begin{pmatrix} 4-2\\ 1+1\\ 3-4 \end{pmatrix}.\begin{pmatrix} 2-2\\ 0+1\\ 5-4 \end{pmatrix}}{\sqrt{(4-2)^{2}+(1+1)^{2}+(3-4)^{2}}.\sqrt{(2-2)^{2}+(0+1)^{2}+(5-4)^{2}}}\\ &=\displaystyle \frac{2.0+2.1+-1.1}{\sqrt{4+4+1}.\sqrt{0+1+1}}\\ &=\displaystyle \frac{1}{3\sqrt{2}}\\ &=\color{blue}\displaystyle \frac{1}{6}\sqrt{2} \end{aligned}\\ &\begin{aligned}&\textrm{Sehingga},\qquad \qquad \\ &\sin \angle \left ( \overline{AB},\: \overline{AC} \right )\\ &=\sqrt{1-\cos ^{2}\angle \left ( \overline{AB},\: \overline{AC} \right )}\\ &=\sqrt{1-\left ( \displaystyle \frac{1}{6}\sqrt{2} \right )^{2}}\\ &=\sqrt{1-\displaystyle \frac{2}{36}}\\ &=\sqrt{\displaystyle \frac{34}{36}}\\ &=\color{red}\displaystyle \frac{1}{6}\sqrt{34} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 38.&\textrm{Diketahui segitiga ABC. Titik M }\\ &\textrm{di tengah AC, dan titik N pada BC}\\ &\textrm{Jika}\: \: \overrightarrow{AB}=\vec{c}\: ,\: \overrightarrow{AC}=\vec{b}\: ,\: \overrightarrow{BC}=\vec{a}\: ,\\ &\textrm{maka}\: \: \overrightarrow{MN}=\, ....\\ &\begin{array}{lll}\\ \textrm{a}.\quad \displaystyle \frac{1}{2}\left ( \vec{b}-\vec{c} \right )&\\ \textrm{b}.\quad \displaystyle \frac{1}{2}\left ( -\vec{b}+\vec{c} \right )&\\ \textrm{c}.\quad \displaystyle \frac{1}{2}\left ( -\vec{a}+\vec{c} \right )\\ \textrm{d}.\quad \displaystyle \frac{1}{2}\left ( \vec{a}-\vec{b} \right )\\ \color{red}\textrm{e}.\quad \displaystyle \frac{1}{2}\left ( -\vec{a}+\vec{b} \right ) \end{array}\\\\ &\textbf{Jawab}:\\ &\begin{aligned}\overrightarrow{MN}&=\overrightarrow{MC}+\overrightarrow{CN}\\ &=\displaystyle \frac{1}{2}\overrightarrow{AC}+\displaystyle \frac{1}{2}\left ( -\overrightarrow{BC} \right )\\ &=\displaystyle \frac{1}{2}\left ( \overrightarrow{AC}-\overrightarrow{BC} \right )\\ &=\displaystyle \frac{1}{2}\left ( \vec{b}-\vec{a} \right )\qquad \textbf{atau}\\ &=\displaystyle \frac{1}{2}\left ( -\vec{a}+\vec{b} \right ) \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 39.&\textrm{Jika titik berat segitiga ABC adalah Z}\\ &\textrm{dengan}\: \: \textrm{A}(1,0,2),\: \textrm{B}(5,4,10),\: \textrm{C}(0,-1,6),\\ &\textrm{maka koordinat titik Z tersebut adalah}\, ....\\ &\begin{array}{lll}\\ \textrm{a}.\quad (-2,-1,6)&&\textrm{d}.\quad (3,2,6)\\ \color{red}\textrm{b}.\quad (2,1,6)&\textrm{c}.\quad (3,-1,6)&\textrm{e}.\quad (6,4,12) \end{array}\\\\ &\textbf{Jawab}:\quad \textbf{b}\\ &\textrm{Coba perhatikanlah ilustrasi berikut} \end{array}$

$.\: \qquad\begin{aligned}\\ &\begin{aligned}\textrm{koordinat}&\: \textrm{titik A}'\\ &=\displaystyle \frac{1}{2}\left ( 5+0,4-1,10+6 \right )\\ &=\color{blue}\left ( \frac{5}{2},\frac{3}{2},8 \right ) \end{aligned}\\ &\begin{aligned}\textrm{Dalam se}&\textrm{gitiga ABC untuk titik berat Z }\\ \textrm{berlaku}\: \: \: &\textrm{ketentuan sebagai berikut}\\ AZ:ZA'&=2:1\\ \overrightarrow{AZ}:\overrightarrow{ZA'}&=2:1\\ \overrightarrow{OZ}&=\displaystyle \frac{\overrightarrow{OA}+2\overrightarrow{OA'}}{3}\\ &=\displaystyle \frac{\begin{pmatrix} 1\\ 0\\ 2 \end{pmatrix}+2\begin{pmatrix} \frac{5}{2}\\ \frac{3}{2}\\ 8 \end{pmatrix}}{3}\\ &=\begin{pmatrix} 2\\ 1\\ 6 \end{pmatrix}\\ \textrm{Jadi},\: &\textrm{koordinat titik Z adalah}\: \color{red}(2,1,6) \end{aligned} \end{aligned}$.

$\begin{array}{ll}\\ 40.&\textrm{Diketahui titik}\: \: \textrm{A}(-4,-1,-2),\: \textrm{B}(-6,4,3),\\ &\textrm{C}(2,3,5).\: \: \textrm{Jika titik M membagi}\: \: \overrightarrow{AB}\\ &\textrm{sehingga}\: \: \overrightarrow{AM}:\overrightarrow{MB}=3:2\\ &\textrm{maka vektor yang diwakili oleh}\: \: \overrightarrow{MC}=....\\ &\begin{array}{lll}\\ \textrm{a}.\quad (-4,1,4)&&\textrm{d}.\quad (6,4,1)\\ \textrm{b}.\quad (-2,2,1)&\textrm{c}.\quad (0,5,6)&\color{red}\textrm{e}.\quad (4,1,4) \end{array}\\\\ &\textbf{Jawab}:\\ &\textrm{Coba perhatikanlah ilustrasi berikut} \end{array}$

$.\: \qquad\begin{array}{|c|c|}\hline \textrm{Diketahui}&\textrm{Proses Penyelesaian}\\\hline \begin{aligned}\overrightarrow{AM}:\overrightarrow{MB}&=3:2\\ \vec{m}&=\displaystyle \frac{2\vec{a}+3\vec{b}}{5}\\ &\\ &\\ &\\ &\\ &\\ &\\ &\\ &\\ &\\ &\\ & \end{aligned}&\begin{aligned}\overrightarrow{MC}&=\vec{c}-\vec{m}\\ &=\vec{c}-\displaystyle \frac{2\vec{a}+3\vec{b}}{5}\\ &=\displaystyle \frac{5\vec{c}-2\vec{a}-3\vec{b}}{5}\\ &=\displaystyle \frac{5\begin{pmatrix} 2\\ 3\\ 5 \end{pmatrix}-2\begin{pmatrix} 4\\ -1\\ -2 \end{pmatrix}-3\begin{pmatrix} -6\\ 4\\ 3 \end{pmatrix}}{5}\\ &=\displaystyle \frac{\begin{pmatrix} 20\\ 5\\ 20 \end{pmatrix}}{5}\\ &=\begin{pmatrix} 4\\ 1\\ 4 \end{pmatrix} \end{aligned}\\\hline \end{array}$

Proyeksi Ortoganal Suatu Vektor di Dimensi Tiga

Materinya sama dengan proyeksi ortogonal pada dimensi dua klik di sini

Uraian berikut sebagai pengingat saja

$\begin{aligned}\triangleright \quad&\textbf{Proyeksi skalar vektor} \\ &\left | \vec{c} \right |=\displaystyle \frac{\vec{a}\bullet \vec{b}}{\left | \vec{b} \right |}\\ \triangleright \quad&\textbf{Vektor proyeksi ortogonal} \\ &\vec{c}=\displaystyle \frac{\vec{a}\bullet \vec{b}}{\left | \vec{b} \right |^{2}}.\vec{b} \end{aligned}$

Sebagai penjelasannya adalah sebagai berikut:

Penjelasan pertama berkaitan dengan proyeksi skalar vektor di dimensi tiga, yaitu:

Diberikan sebuah ilustrasi berikut,

Perhatikan ilustrasi gambar di atas!

$\begin{array}{|c|c|}\hline \triangle \textrm{OAC}&\angle \left ( \overrightarrow{a},\overrightarrow{b} \right )\\\hline \begin{aligned}\cos \theta &=\displaystyle \frac{\left | \overrightarrow{c} \right |}{\left | \overrightarrow{a} \right |}\\ \Leftrightarrow \left | \overrightarrow{c} \right |&=\left | \overrightarrow{a} \right |\cos \theta \: \: ........(1) \end{aligned}&\begin{aligned}\cos \theta &=\displaystyle \frac{\overrightarrow{a}\overrightarrow{b}}{\left | \overrightarrow{a} \right |\left | \overrightarrow{b} \right |}\: \: ........(2) \end{aligned}\\\hline \end{array}$.

$\begin{aligned}\textrm{Dari}\: \: (1)\: \: &\textrm{dan} \: \: (2)\: \: \textrm{diperoleh}\\ \left | \overrightarrow{c} \right |&=\left | \overrightarrow{a} \right |\cos \theta \\ &=\left | \overrightarrow{a} \right |\left ( \displaystyle \frac{\overrightarrow{a}\overrightarrow{b}}{\left | \overrightarrow{a} \right |\left | \overrightarrow{b} \right |} \right )\\ &=\left |\displaystyle \frac{\overrightarrow{a}\overrightarrow{b}}{\left | \overrightarrow{b} \right |} \right | \end{aligned}$

Dan penjelasan kedua berkaitan dengan vektor proyeksi ortogonalnya, yaitu:

$\begin{aligned}&\color{red}\textrm{Perhatikan pula misal}\: \: \color{black}\hat{c}\\ & \textrm{adalah vektor satuan dari}\: \: \overrightarrow{c}\: \: \textrm{dan}\: \: \overrightarrow{b},\\ & \textrm{maka}\\ &\begin{aligned}\overrightarrow{c}&=\left | \overrightarrow{c} \right |\hat{c} \end{aligned},\: \: \textrm{dan}\\ &\begin{aligned}\overrightarrow{b}&=\left | \overrightarrow{b} \right |\hat{b}=\left | \overrightarrow{b} \right |\hat{c} \end{aligned} \end{aligned}$.

$\begin{aligned}\textrm{Sehingga}&\: \: \textbf{proyeksi ortogonal vektor}\: \: \overrightarrow{a}\: \: \textrm{pada}\: \: \overrightarrow{b}\: \: \textrm{adalah}:\\ \overrightarrow{c}&=\left | \overrightarrow{c} \right |\hat{b}\\ &=\left ( \displaystyle \frac{\overrightarrow{a}\overrightarrow{b}}{\left | \overrightarrow{b} \right |} \right )\left ( \displaystyle \frac{\overrightarrow{b}}{\left | \overrightarrow{b} \right |} \right )\\ &=\left (\displaystyle \frac{\overrightarrow{a}\overrightarrow{b}}{\left | \overrightarrow{b} \right |^{2}} \right )\overrightarrow{b} \end{aligned}$

$\LARGE\colorbox{yellow}{CONTOH SOAL}$.

$\begin{array}{ll}\\ 1.&\textrm{Diketahui}\: \: \vec{a}=\begin{pmatrix} 2\\ -3\\ 1 \end{pmatrix}\: \: \textrm{dan}\: \: \vec{b}=\begin{pmatrix} -4\\ 2\\ 2 \end{pmatrix}.\\ &\textrm{Tentukanlah}\\ &\textrm{a}.\quad \textrm{proyeksi skalar}\: \: \vec{a}\: \: \textrm{pada}\: \: \vec{b}\\ &\textrm{b}.\quad \textrm{vektor proyeksi}\: \: \vec{a}\: \: \textrm{pada}\: \: \vec{b}\\ &\textrm{c}.\quad \textrm{proyeksi skalar}\: \: \vec{b}\: \: \textrm{pada}\: \: \vec{a}\\ &\textrm{d}.\quad \textrm{vektor proyeksi}\: \: \vec{b}\: \: \textrm{pada}\: \: \vec{a}\\\\ &\textbf{Jawab}:\\ &\textrm{Misalkan proyeksi skalar}\: \: \vec{a}\: \: \textrm{pada}\: \: \vec{b}\\ &\textrm{adalah}\: \: \left | \vec{c} \right |,\: \: \textrm{dan}\\ &\textrm{misalkan juga proyeksi skalar}\: \: \vec{b}\: \: \textrm{pada}\: \: \vec{a}\\ &\textrm{adalah}\: \: \left | \vec{d} \right |,\: \: \textrm{maka}\\ &\begin{aligned}\textrm{a}.\quad\left | \vec{c} \right |&=\displaystyle \frac{\vec{a}\bullet \vec{b}}{\left | \vec{b} \right |}\\ &=\displaystyle \frac{\begin{pmatrix} 2\\ -3\\ 1 \end{pmatrix}\begin{pmatrix} -4\\ 2\\ 2 \end{pmatrix}}{\sqrt{(-4)^{2}+2^{2}+2^{2}}}\\ &=\displaystyle \frac{-8-6+2}{\sqrt{24}}=-\frac{12}{24}\sqrt{24}=-\sqrt{6}\\ &\textrm{Karena hasilnya berupa panjang, maka}\\ &\textrm{diharga mutlak/positif}\\ &\left | \vec{c} \right |=\left |-\sqrt{6} \right |=\color{red}\sqrt{6} \end{aligned}\\ &\begin{aligned}\textrm{b}.\quad \vec{c}&=\displaystyle \frac{\vec{a}\bullet \vec{b}}{\left | \vec{b} \right |^{2}}\times \vec{b}\\ &=\displaystyle \frac{-12}{(\sqrt{24})^{2}}\times \begin{pmatrix} -4\\ 2\\ 2 \end{pmatrix}\\ &=-\displaystyle \frac{1}{2}\begin{pmatrix} -4\\ 2\\ 2 \end{pmatrix}=\color{red}\begin{pmatrix} 2\\ -1\\ -1 \end{pmatrix} \end{aligned}\\ &\begin{aligned}\textrm{c}.\quad\left | \vec{d} \right |&=\displaystyle \frac{\vec{b}\bullet \vec{a}}{\left | \vec{a} \right |}\\ &=\displaystyle \frac{\begin{pmatrix} -4\\ 2\\ 2 \end{pmatrix}\begin{pmatrix} 2\\ -3\\ 1 \end{pmatrix}}{\sqrt{2^{2}+(-3)^{2}+1^{2}}}\\ &=\displaystyle \frac{-8-6+2}{\sqrt{14}}=-\frac{12}{14}\sqrt{14}=-\frac{6}{7}\sqrt{14}\\ &\textrm{Karena hasilnya berupa panjang, maka}\\ &\textrm{diharga mutlak/positif}\\ &\left | \vec{d} \right |=\left |-\displaystyle \frac{6}{7}\sqrt{14} \right |=\color{red}\displaystyle \frac{6}{7}\sqrt{14} \end{aligned}\\ &\begin{aligned}\textrm{d}.\quad \vec{d}&=\displaystyle \frac{\vec{b}\bullet \vec{a}}{\left | \vec{a} \right |^{2}}\times \vec{a}\\ &=\displaystyle \frac{-12}{(\sqrt{14})^{2}}\times \begin{pmatrix} 2\\ -3\\ 1 \end{pmatrix}\\ &=-\displaystyle \frac{12}{14}\begin{pmatrix} 2\\ -3\\ 1 \end{pmatrix}=\color{red}\begin{pmatrix} -\frac{12}{7}\\ \frac{18}{7}\\ -\frac{6}{7} \end{pmatrix} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 2.&\textrm{Diketahui}\: \: \vec{a}=\begin{pmatrix} -3\\ -2\\ \color{blue}m \end{pmatrix}\: \: \textrm{dan}\: \: \vec{b}=\begin{pmatrix} 2\\ -1\\ -2 \end{pmatrix}.\\ & \textrm{Jika proyeksi skalar}\: \: \vec{a}\: \: \textrm{pada}\: \: \vec{b}\: \: \textrm{adalah}\\ &\textrm{bernilai}\: \: -\displaystyle \frac{2}{3},\: \: \textrm{maka tentukan nilai}\: \: \color{blue}m\\\\ &\textbf{Jawab}:\\ &\textrm{Misalkan proyeksi skalar}\: \: \vec{a}\: \: \textrm{pada}\: \: \vec{b}\: \: \textrm{adalah}\: \: \left | \vec{f} \right |,\\ &\textrm{maka}\\ &\begin{aligned}\left | \vec{f} \right |&=\displaystyle \frac{\vec{a}\bullet \vec{b}}{\left | \vec{b} \right |}\\ \Leftrightarrow \: -\displaystyle \frac{2}{3}&=\displaystyle \frac{\begin{pmatrix} -3\\ -2\\ \color{blue}m \end{pmatrix}\begin{pmatrix} 2\\ -1\\ -2 \end{pmatrix}}{\sqrt{2^{2}+(-1)^{2}+(-2)^{2}}}\\ \Leftrightarrow \: -\displaystyle \frac{2}{3}&=\displaystyle \frac{-6+2-2\color{blue}m}{\sqrt{9}}=\frac{-4-2\color{blue}m}{3}\\ \Leftrightarrow \: -2&=-4-2\color{blue}m\\ \Leftrightarrow \: 1&=2+\color{blue}m\\ \Leftrightarrow \: -\color{blue}m&=2-1=1\\ \Leftrightarrow \: \color{blue}m&=-1 \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 4.&\textrm{Diketahui vektor}\: \: \overrightarrow{a}=3\bar{i}-2\bar{j}+2\bar{k}\: \: \textrm{dan}\\ & \overrightarrow{b}=2\bar{i}-2\bar{j}+\bar{k}.\: \textrm{Tentukanlah panjang }\\ &\textrm{vektor proyeksi ortogonal}\\ &\textrm{a}.\quad \overrightarrow{a}\: \: \textrm{pada}\: \: \overrightarrow{b}\\ & \textrm{b}.\quad \overrightarrow{a}\: \: \textrm{pada}\: \: \left ( \overrightarrow{a}+\overrightarrow{b} \right ) \\\\ &\textbf{Jawab}:\\ &\begin{aligned}\textrm{a}.\quad\textrm{Misal}&\textrm{kan}\: \: \overrightarrow{g}\: \: \textrm{adalah vektor proyeksi }\\ \textrm{yang}&\: \textrm{dimaksud, maka panjanynya}\\ (\textrm{lang}&\textrm{sung diharga mutlak})\\ \left |\overrightarrow{g} \right |&= \left |\displaystyle \frac{\overrightarrow{a}\overrightarrow{b}}{\left | \overrightarrow{b} \right |} \right | \\ &=\left |\displaystyle \frac{\begin{pmatrix} 3\\ -2\\ 2 \end{pmatrix}.\begin{pmatrix} 2\\ -2\\ 1 \end{pmatrix}}{\sqrt{2^{2}+(-2)^{2}+1^{2}}} \right |\\ &=\left |\frac{3.2+(-2).(-2)+2.1}{\sqrt{4+4+1}} \right |\\ &=\left | \displaystyle \frac{12}{3} \right |=4 \end{aligned}\\ &\begin{aligned}\textrm{b}.\quad\textrm{Misal}&\textrm{kan}\: \: \overrightarrow{h}\: \: \textrm{adalah vektor proyeksi }\\ \textrm{yang}&\: \textrm{dimaksud, maka panjanynya}\\ (\textrm{lang}&\textrm{sung diharga mutlak})\\ \left |\overrightarrow{h} \right |&= \left |\displaystyle \frac{\overrightarrow{a}\left (\overrightarrow{a}+\overrightarrow{b} \right )}{\left |\overrightarrow{a}+ \overrightarrow{b} \right |} \right |\\ &=\left |\displaystyle \frac{\begin{pmatrix} 3\\ -2\\ 2 \end{pmatrix}.\begin{pmatrix} 3+2\\ -2+(-2)\\ 2+1 \end{pmatrix}}{\sqrt{(3+2)^{2}+(-2+(-2))^{2}+(2+1)^{2}}} \right |\\ &=\left |\frac{3.5+(-2).(-4)+2.3}{\sqrt{25+16+9}} \right |\\ &=\left |\frac{29}{\sqrt{50}} \right |\\ &=\displaystyle \frac{29}{5\sqrt{2}}=\frac{29}{10}\sqrt{5} \end{aligned} \end{array}$

DAFTAR PUSTAKA

Johanes, Kastolan, Sulasim. 2006. Kompetensi Matematika 3A SMA Kelas XII Semester Pertama Program IPA. Jakarta: YUDHISTIRA.

Lanjutan 2 Materi Operasi Vektor Berdimensi Tiga

$\color{blue}\textrm{G. Perbandingan Vektor }$.

Rumus perbandingan vektor yang berlaku pada dimensi dua juga berlaku untuk perbandingan vektor di dimensi tiga. Misalkan suatu $\overline{AB}$ dan titik T pada $\overline{AB}$ dengan $\overline{AT}:\overline{TB}=m:n$ .

Karena titik T pada ruas garis AB, maka titik T membagi ruas AB dengan $\vec{a}$ dan $\vec{b}$ sebagai vektor posisi dari masing-masing titik A dan B dan vektor posisi titik T dapat ditentukan, yaitu:

$\vec{t}=\displaystyle \frac{n\vec{a}+m\vec{b}}{m+n}$

Sebagai ilustrasinya adalah gambar berikut

Untuk bukti silahkan merujuk di sini

(dengan menyesuaikan posisi titiknya)

Jika titik T pada perpanjangan garis $\overline{AB}$ , maka vektor posisi titik T-nya adalah:

$\vec{t}=\displaystyle \frac{-n\vec{a}+m\vec{b}}{m-n}$.

$\LARGE\colorbox{yellow}{CONTOH SOAL}$.

$\begin{array}{ll}\\ 1.&\textrm{Jika titik}\: \: A(12,12,0)\: \: \textrm{dan}\: \: B(6,6,12)\\ &\textrm{serta}\: \: P\: \: \textrm{membagi garis dengan}\\ & \overline{AP}:\overline{PB}=1:2.\: \textrm{Tentukanlah koordinat}\\ &\textrm{titik}\: \: P\: \: \textrm{jika},\\ &P\: \: \textrm{membagi di dalam}\\ &P\: \: \textrm{membagi di luar}\\\\ &\textbf{Jawab}:\\ &\begin{aligned}\textrm{a}.\quad&P\: \: \textrm{membagi di dalam}\\ &\vec{p}=\displaystyle \frac{n\vec{a}+m\vec{b}}{m+n}\\ &\: \: =\displaystyle \frac{2\vec{a}+1\vec{p}}{1+2}=\frac{2\begin{pmatrix} 12\\ 12\\ 0 \end{pmatrix}+\begin{pmatrix} 6\\ 6\\ 12 \end{pmatrix}}{3}\\ &=\displaystyle \frac{\begin{pmatrix} 30\\ 30\\ 12 \end{pmatrix}}{3}=\begin{pmatrix} 10\\ 10\\ 4 \end{pmatrix}\\ \textrm{b}.\quad&P\: \: \textrm{membagi di luar}\\ &\vec{p}=\displaystyle \frac{-n\vec{a}+m\vec{b}}{m-n}\\ &\: \: =\displaystyle \frac{-2\begin{pmatrix} 12\\ 12\\ 0 \end{pmatrix}+\begin{pmatrix} 6\\ 6\\ 12 \end{pmatrix}}{1-2}\\ &\: \: =-\begin{pmatrix} -18\\ -18\\ 12 \end{pmatrix}=\begin{pmatrix} 18\\ 18\\ -12 \end{pmatrix} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 2.&\textrm{Jika titik}\: \: A(3,2,-1)\: \: \textrm{dan}\: \: B(1,-2,1)\\ &\textrm{dan}\: \: C(7,m-1,-5)\: .\: \textrm{Tentukan nilai}\: \: m\\ &\textrm{agar ketiga titik itu segaris}\\\\ &\textbf{Jawab}:\\ &\begin{aligned}&\textrm{Diketahui}\\ &\vec{a}=\begin{pmatrix} 3\\ 2\\ -1 \end{pmatrix},\: \vec{b}=\begin{pmatrix} 1\\ -2\\ 1 \end{pmatrix},\: \begin{pmatrix} 7\\ m-1\\ -5 \end{pmatrix}\\ &\textrm{Agar ketiga titik segaris, maka}\: \: \overline{AB}=k\overline{BC}\\ &\vec{b}-\vec{a}=k\left ( \vec{c}-\vec{b} \right )\\ &\begin{pmatrix} 1\\ -2\\ 1 \end{pmatrix}-\begin{pmatrix} 3\\ 2\\ -1 \end{pmatrix}=k\left (\begin{matrix} 7\\ m-1\\ -5 \end{matrix} -\begin{pmatrix} 1\\ -2\\ 1 \end{pmatrix} \right )\\ &\Leftrightarrow \: \begin{pmatrix} -2\\ -4\\ 2 \end{pmatrix}=k\begin{pmatrix} 6\\ m+1\\ -6 \end{pmatrix}\\ &\Leftrightarrow \: \begin{cases} -2 & =6k \\ -4 & =k(m+1) \\ 2 & =-6k \end{cases}\Leftrightarrow -2=6k\\ &\Leftrightarrow k=-\displaystyle \frac{1}{3}\\ &\textrm{maka}\\ &\Leftrightarrow -4=k(m+1)\Leftrightarrow -4=-\displaystyle \frac{1}{3}(m+1)\\ &\Leftrightarrow 12=m+1\\ &\Leftrightarrow m=\color{red}11 \end{aligned} \end{array}$

Lanjutan 1 Materi Operasi Vektor Berdimensi Tiga (Hasil Kali Vektor)

$\color{blue}\textrm{F. 4. Perkalian Silang Vektor (Pengayaan)}$.

Pada ruang dimensi tiga khususnya pada vektor akan berlaku perkalian silang (cross vektor) adalah perkalian antara dua vektor yang menghasilkan vektor tunggal. Misalkan diketahui $\vec{u}$ dan $\vec{v}$ adalah dua vektor sembarang dan keduanya membentuk sudut $\theta$, maka hasil kali kedua vektor tersebut adalah sebuah vektor baru dengan dinotasiakan sebagai $\vec{u}\times \vec{v}$. Tentunya sebagai syarat kedua vektor tersebut masing-masing tidak berupa vektor nol.

Jika $\vec{u}\times \vec{v}=\vec{c}$ , maka

$\begin{aligned}\vec{u}&\times \vec{v}=\vec{c}=\begin{vmatrix} \vec{i} & \vec{j} & \vec{k}\\ x_{1} & y_{1} &z_{1} \\ x_{2} & y_{2} &z_{2} \end{vmatrix}\\ &=(y_{1}z_{2}-z_{1}y_{2})\vec{i}-(x_{1}z_{2}-z_{1}x_{2})\vec{j}+(x_{1}y_{2}-y_{1}x_{2})\vec{k} \end{aligned}$

Lalu kalau sudah demikian berapa besarnya? dan ke mana arahnya?

Besarnya adalah $\left | \vec{u}\times \vec{v} \right |=\left | \vec{u} \right |\left | \vec{v} \right |\sin \theta$ dan arahnya tegak lurus terhadap $\vec{u}$ dan $\vec{v}$.

Sebagai ilustrasi perhatikanlah gambar berikut untuk dua buah vektor sebagai misal $\vec{a}$ dan $\vec{b}$.

Jika putarannya dibalik, maka akan mendapatkan hasil sebagai mana ilustrasi berikut

Sehingga perlu diingat bahwa : $\vec{a}\times \vec{b}=-\vec{b}\times \vec{a}$.

Pada hasil kali silang dua vektor berlaku

tidak bersifat komutatif , karena $\vec{a}\times \vec{b}=-\vec{b}\times \vec{a}$.
distributif terhadap penjumlahan : $\vec{a}\times \left (\vec{b}+\vec{c} \right )=\vec{a}\times \vec{b}+\vec{a}\times \vec{c}$.
pada perkalian dengan skalar : $k\left (\vec{a}\times \vec{b} \right )=\left (k\vec{a} \right )\times \vec{b}=\vec{a}\times \left ( k\vec{b} \right )$.
berlaku untuk sembarang vektor : $\vec{a}\times \vec{a} =0$.
jika kedua vektor sejajar, maka hasil kalinya adalah = 0.
Nilai dari perkalian kedua vektor terbut adalah sama dengan hasil luas jajar genjang.
Nilai dari poin 6 jika dibagi 2 akan berupa hasil luas sebuah segitiga yang dibentuk oleh kedua vektor tersebut.
berlaku identitas Lagrange : $\left | \vec{a}\times \vec{b} \right |^{2}=\left | \vec{a} \right |^{2}.\left | \vec{b} \right |^{2}-\left ( \vec{a}\: \bullet \: \vec{b} \right )^{2}$.

$\LARGE\colorbox{yellow}{CONTOH SOAL}$.

$\begin{array}{ll}\\ 1.&\textrm{Diketahui}\: \: \vec{a}=4\vec{i}+3\vec{j}\: \: \textrm{dan}\: \: \vec{b}=4\vec{i}-3\vec{k}\\ &\textrm{Tentukanlah hasil}\: \: \vec{a}\times \vec{b}\: \: \textrm{dan}\: \: \vec{b}\times \vec{a}\\\\ &\textbf{Jawab}:\\ &\begin{aligned}\vec{a}&\times \vec{b}=\begin{vmatrix} \vec{i} & \vec{j} & \vec{k}\\ x_{1} & y_{1} &z_{1} \\ x_{2} & y_{2} &z_{2} \end{vmatrix}\\ &=(y_{1}z_{2}-z_{1}y_{2})\vec{i}-(x_{1}z_{2}-z_{1}x_{2})\vec{j}+(x_{1}y_{2}-y_{1}x_{2})\vec{k}\\ &=\begin{vmatrix} \vec{i} & \vec{j} & \vec{k}\\ 4 & 3 &0 \\ 4 & 0 &-3 \end{vmatrix}\\ &=(-9-0)\vec{i}-(-12-0)\vec{j}+(0-12)\vec{k}\\ &=-9\vec{i}+12\vec{j}-12\vec{k} \end{aligned}\\ &\begin{aligned}\vec{b}&\times \vec{a}=\begin{vmatrix} \vec{i} & \vec{j} & \vec{k}\\ x_{2} & y_{2} &z_{2} \\ x_{1} & y_{1} &z_{1} \end{vmatrix}\\ &=\begin{vmatrix} \vec{i} & \vec{j} & \vec{k}\\ 4 & 0 &-3 \\ 4 & 3 &0 \end{vmatrix}\\ &=(0-(-9))\vec{i}-(0-(-12))\vec{j}+(12-0)\vec{k}\\ &=9\vec{i}-12\vec{j}+12\vec{k} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 2.&\textrm{Diketahui}\: \: \vec{a}=6\vec{i}+2\vec{j}+10\vec{k}\: \: \textrm{dan}\: \: \vec{b}=4\vec{i}+\vec{j}+9\vec{k}\\ &\textrm{Tentukanlah hasil}\: \: \vec{a}\times \vec{b}\: \: \textrm{dan}\: \: \vec{b}\times \vec{a}\\\\ &\textbf{Jawab}:\\ &\begin{aligned}\vec{a}&\times \vec{b}=\begin{vmatrix} \vec{i} & \vec{j} & \vec{k}\\ x_{1} & y_{1} &z_{1} \\ x_{2} & y_{2} &z_{2} \end{vmatrix}\\ &=(y_{1}z_{2}-z_{1}y_{2})\vec{i}-(x_{1}z_{2}-z_{1}x_{2})\vec{j}+(x_{1}y_{2}-y_{1}x_{2})\vec{k}\\ &=\begin{vmatrix} \vec{i} & \vec{j} & \vec{k}\\ 6 & 2 &10 \\ 4 & 1 &9 \end{vmatrix}\\ &=(18-10)\vec{i}-(54-40)\vec{j}+(6-8)\vec{k}\\ &=8\vec{i}-14\vec{j}-2\vec{k} \end{aligned}\\ &\begin{aligned}\vec{b}&\times \vec{a}=\begin{vmatrix} \vec{i} & \vec{j} & \vec{k}\\ x_{2} & y_{2} &z_{2} \\ x_{1} & y_{1} &z_{1} \end{vmatrix}\\ &=\begin{vmatrix} \vec{i} & \vec{j} & \vec{k}\\ 4 & 1 &9 \\ 6 & 2 &10 \end{vmatrix}\\ &=(10-18)\vec{i}-(40-54)\vec{j}+(8-6)\vec{k}\\ &=-8\vec{i}+14\vec{j}+2\vec{k} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 3.&\textrm{Tentukanlah luas segitiga}\: \: ABC\: \: \textrm{jika}\\ &\textrm{diketahui}\: \: A(2,1,-2),\: B(0,-1,0),\: \: \textrm{dan}\\ &C(-1,2,-1)\\\\ &\textbf{Jawab}:\\ &\textrm{Misalkan luas segitiga}\: \: \displaystyle \frac{1}{2}\left | \vec{p}\times \vec{q} \right |,\: \: \textrm{dengan}\\ &\begin{cases} \vec{p} & =\overline{AB}=\overline{OB}-\overline{OA}=\begin{pmatrix} 0\\ -1\\ 0 \end{pmatrix}-\begin{pmatrix} 2\\ 1\\ -2 \end{pmatrix}=\begin{pmatrix} -2\\ -2\\ 2 \end{pmatrix} \\ \vec{q} & =\overline{AC}=\overline{OC}-\overline{OA}=\begin{pmatrix} -1\\ 2\\ -1 \end{pmatrix}-\begin{pmatrix} 2\\ 1\\ -2 \end{pmatrix}=\begin{pmatrix} -3\\ 1\\ 1 \end{pmatrix} \end{cases}\\ &\begin{aligned}\vec{p}&\times \vec{q}=\begin{vmatrix} \vec{i} & \vec{j} & \vec{k}\\ x_{1} & y_{1} &z_{1} \\ x_{2} & y_{2} &z_{2} \end{vmatrix}\\ &=(y_{1}z_{2}-z_{1}y_{2})\vec{i}-(x_{1}z_{2}-z_{1}x_{2})\vec{j}+(x_{1}y_{2}-y_{1}x_{2})\vec{k}\\ &=\begin{vmatrix} \vec{i} & \vec{j} & \vec{k}\\ -2 & -2 &2 \\ -3 & 1 &1 \end{vmatrix}\\ &=(-2-2)\vec{i}-(-2-(-6))\vec{j}+(-2-6)\vec{k}\\ &=-4\vec{i}-4\vec{j}-8\vec{k}\\ &\textrm{Sehingga}\\ &\left | \vec{p}\times \vec{q} \right |=\sqrt{(-4)^{2}+(-4)^{2}+(-8)^{2}}\\ &\quad\qquad =\sqrt{16+16+64}=\sqrt{96}=4\sqrt{6}\\ &\textrm{Maka luas segi tiganya adalah}:\\ &\textrm{luas}\: \triangle ABC=\displaystyle \frac{1}{2}\left | \vec{p}\times \vec{q} \right |=\displaystyle \frac{1}{2}\left ( 4\sqrt{6} \right )=\color{red}2\sqrt{6} \end{aligned} \end{array}$

DAFTAR PUSTAKA

Yuana, R.A., Indriyastuti. 2017. Perspektif Matematika untuk Kelas X SMA dan MA Kelompok Peminatan Matematika dan Ilmu Alam. Solo: PT. TIGA SERANGKAI PUSTAKA MANDIRI

Operasi Vektor Berdimensi Tiga

$\color{blue}\textrm{F. Operasi Vektor Dalam Ruang}$

Operasi vektor pada dimensi tiga kurang lebih sama dengan operasi pada vektor berdimensi dua.

$\color{blue}\textrm{F. 1. Penjumlahan dan Pengurangan}$.

$\LARGE\colorbox{yellow}{CONTOH SOAL}$.

$\begin{array}{ll}\\ 1.&\textrm{Jika diketahui}\: \: \bar{a}=\begin{pmatrix} 1\\ 3\\ 7 \end{pmatrix}\: \: \textrm{dan}\: \: \bar{b}=\begin{pmatrix} 8\\ -2\\ 0 \end{pmatrix}\\ &\textrm{Tentukanlah hasil dari}\\ &\textrm{a}.\quad \bar{a}+\bar{b}\\ &\textrm{b}.\quad \bar{a}-\bar{b}\\\\ &\textbf{Jawab}\\ &\begin{aligned}&\textrm{Diketahui bahwa}\\ &\bar{a}=\begin{pmatrix} 1\\ 3\\ 7 \end{pmatrix}\: \: \textrm{dan}\: \: \bar{b}=\begin{pmatrix} 8\\ -2\\ 0 \end{pmatrix},\\ &\textrm{maka}\\ &\bar{a}+\bar{b}=\begin{pmatrix} 1\\ 3\\ 7 \end{pmatrix}+\begin{pmatrix} 8\\ -2\\ 0 \end{pmatrix}\\ &\qquad =\begin{pmatrix} 1+8\\ 3+(-2)\\ 7+0 \end{pmatrix}=\begin{pmatrix} 9\\ 1\\ 7 \end{pmatrix}\\ &\textrm{Dan untuk}\: \: \bar{a}-\bar{b}\: \: \textrm{adalah}:\\ &\bar{a}-\bar{b}=\begin{pmatrix} 1\\ 3\\ 7 \end{pmatrix}-\begin{pmatrix} 8\\ -2\\ 0 \end{pmatrix}\\ &\qquad =\begin{pmatrix} 1-8\\ 3-(-2)\\ 7-0 \end{pmatrix}=\begin{pmatrix} -7\\ 5\\ 7 \end{pmatrix} \end{aligned} \end{array}$.

$\color{blue}\textrm{F. 2. Perkalian Skalar dengan Vektor}$.

Misalkan suatu skalar $m$ dan suatu vektor $\bar{u}=a\bar{i}+b\bar{j}+c\bar{k}$, maka perkalian $m$ dengan vektor $\bar{u}$ tersebut adalah $\bar{u}=ma\bar{i}+mb\bar{j}+mc\bar{k}$.

$\LARGE\colorbox{yellow}{CONTOH SOAL}$.

$\begin{array}{ll} 2.&\textrm{Jika}\: \: \bar{a}=\begin{pmatrix} 2022\\ 2021\\ 2020 \end{pmatrix},\: \: \textrm{tentukanlah nilai}\\ &\textrm{dari}\: \: 2\bar{a}\: \: \: \textrm{dan}\: \: -3\bar{a}\\\\ &\textbf{Jawab}\\ &\begin{aligned}2\bar{a}&=2\begin{pmatrix} 2022\\ 2021\\ 2020 \end{pmatrix}=\begin{pmatrix} 4044\\ 4042\\ 4040 \end{pmatrix},\: \: \textrm{dan}\\ -3\bar{a}&=-3\begin{pmatrix} 2022\\ 2021\\ 2020 \end{pmatrix}=\begin{pmatrix} -6066\\ -6063\\ -6060 \end{pmatrix} \end{aligned} \end{array}$

$\color{blue}\textrm{F. 3. Perkalian Skalar Dua Vektor}$.

Hasil dari perkalian skalar dua vektor $\bar{a}$ dan $\bar{b}$ adalah : $\bar{a}\: \: \bullet\: \: \bar{b}$.

Dengan

$\bar{a}\: \: \bullet\: \: \bar{b}=\left | \bar{a} \right |\left | \bar{b} \right |\cos \theta$. sehingga

$\begin{aligned}&\textrm{Tanda dari hasil skalar ini adalah}\\ &\begin{array}{|l|l|l|}\hline \textbf{Besar sudut}\: \: \: \theta &\textbf{Tanda}&\textrm{Bentuk}\\\hline 0^{\circ}\leq \theta < 90^{\circ}&\textrm{Positif}&\color{red}\textrm{Lancip}\\\hline \theta =90^{\circ}&\textrm{Nol}&\textrm{Siku-siku}\\\hline 90^{\circ}< \theta \leq 180^{\circ}&\textrm{Negatif}&\color{blue}\textrm{Tumpul}\\\hline \end{array}\\ &\textrm{Untuk}\: \: \theta \: \: \textrm{berupa sudut istimewa}:\\ &\begin{array}{|c|c|c|c|c|c|c|}\hline \theta &0^{\circ}&30^{\circ}&45^{\circ}&60^{\circ}&90^{\circ}&180^{\circ}\\\hline \cos \theta &1&\displaystyle \frac{1}{2}\sqrt{3}&\displaystyle \frac{1}{2}\sqrt{2}&\displaystyle \frac{1}{2}&0&-1\\\hline \end{array} \end{aligned}$

Adapun secara rumus untuk menentukan besar sudutnya adalah:

$\cos \theta =\displaystyle \frac{\bar{a}\: \: \bullet\: \: \bar{b} }{\left | \bar{a} \right |\left | \bar{b} \right |}$.

Sebagai ilustrasinya perhatikanlah gambar berikut

Selain hasil di atas ada cara lain menyelesaikan perkalian skalar dua vektor, yaitu:

$\begin{aligned}\textrm{Jika}\: & \textrm{diketahui sebagai misal}\\ \bar{u}&=a\bar{i}+b\bar{j}+c\bar{k}\: \: \: \color{red}\textrm{dan}\\ \bar{v}&=p\bar{i}+q\bar{j}+r\bar{k}\\ \textrm{mak}&\textrm{a}\\ \textbf{Per}&\textbf{kalian skalar dua vektor adalah}:\\ \bar{u}\: \bullet \: &\bar{v}=\left ( a\bar{i}+b\bar{j}+c\bar{k} \right )\left ( p\bar{i}+q\bar{j}+r\bar{k} \right )\\ &\: \: =ap.\bar{i}\: \bullet \bar{i}+aq.\bar{i}\: \bullet \: \bar{j}+ar.\bar{i}\: \bullet \: \bar{k}\\ &\: \: \: \: \: \: \: +bp.\bar{j}\: \bullet \: \bar{i}+bq.\bar{j}\: \bullet \: \bar{j}+br.\bar{j}\: \bullet \: \bar{k}\\ &\: \: \: \: \: \: \: +cp.\bar{k}\: \bullet \: \bar{i}+cq.\bar{k}\: \bullet \: \bar{j}+cr.\bar{k}\: \bullet \: \bar{k}\\ &\: \: =ap+0+0+0+bq+0+0+0+cr\\ &\: \: =\color{red}a p+b q+c r \end{aligned}$

$\begin{aligned}&\textrm{Sebagai penjelasannya adalah}:\\ &\color{red}\triangleright \quad \color{black}\bar{i}\: \: \bullet \: \: \bar{i}=\left | \bar{i} \right |\left | \bar{i} \right |\cos 0^{\circ}=1.1.1=1\\ &\color{red}\triangleright \quad \color{black}\bar{i}\: \: \bullet \: \: \bar{j}=\left | \bar{i} \right |\left | \bar{j} \right |\cos 90^{\circ}=1.1.0=0\\ &\color{red}\triangleright \quad \color{black}\bar{i}\: \: \bullet \: \: \bar{k}=\left | \bar{i} \right |\left | \bar{k} \right |\cos 90^{\circ}=1.1.0=0\\ &\color{red}\triangleright \quad \color{black}\bar{j}\: \: \bullet \: \: \bar{i}=\bar{i}\: \: \bullet \: \: \bar{j}=0\\ &\color{red}\triangleright \quad \color{black}\bar{j}\: \: \bullet \: \: \bar{j}=\left | \bar{j} \right |\left | \bar{j} \right |\cos 0^{\circ}=1.1.1=1\\ &\color{red}\triangleright \quad \color{black}\bar{j}\: \: \bullet \: \: \bar{k}=\left | \bar{j} \right |\left | \bar{k} \right |\cos 90^{\circ}=1.1.0=0\\ &\color{red}\triangleright \quad \color{black}\bar{k}\: \: \bullet \: \: \bar{i}=\bar{i}\: \: \bullet \: \: \bar{k}=0\\ &\color{red}\triangleright \quad \color{black}\bar{k}\: \: \bullet \: \: \bar{j}=\bar{j}\: \: \bullet \: \: \bar{k}=0\\ &\color{red}\triangleright \quad \color{black}\bar{k}\: \: \bullet \: \: \bar{k}=\left | \bar{k} \right |\left | \bar{k} \right |\cos 90^{\circ}=1.1.1=1 \end{aligned}$

$\begin{aligned}&\textrm{Atau jika ditabelkan nilainya}\\ &\begin{array}{|c|c|c|c|}\hline \bar{u}\: \bullet \: \bar{v}&p\bar{i}&q\bar{j}&r\bar{k}\\\hline a\bar{k}&ap&0&0\\ b\bar{j}&0&bq&0\\ c\bar{k}&0&0&cr\\\hline \end{array} \end{aligned}$

$\LARGE\colorbox{yellow}{CONTOH SOAL}$.

$\begin{array}{ll} 3.&\textrm{Jika}\: \: \vec{a}=\begin{pmatrix} 1\\ 2\\ 4 \end{pmatrix},\: \: \textrm{dan}\: \: \vec{b}=\begin{pmatrix} 5\\ 4\\ 0 \end{pmatrix}\\\ & \textrm{tentukanlah nilai}\: \: \textrm{dari}\: \: \vec{a}\bullet \vec{b}\\\\ &\textbf{Jawab}\\ &\begin{aligned}\vec{a}\bullet \vec{b}&=1.5+2.4+4.0=5+8+0=13 \end{aligned} \end{array}$

$\begin{array}{ll} 4.&\textrm{Jika diketahui}\: \: \vec{a}=\vec{i}-2\vec{j}+3\vec{k},\\ & \textrm{dan}\: \: \vec{b}=3\vec{i}-4\vec{j}+m\vec{k}\: \: \textrm{serta}\\\ & \textrm{nilai}\: \: \vec{a}\bullet \vec{b}=-4,\: \: \textrm{maka tentukan}\\ &\textrm{nilai}\: \: m\\\\ &\textbf{Jawab}\\ &\textrm{Diketahui bahwa}\\ &\color{red}\triangleright \quad \color{black}\vec{a}=\vec{i}-2\vec{j}+3\vec{k}=\begin{pmatrix} 1\\ -2\\ 3 \end{pmatrix},\: \: \textrm{dan}\\ &\color{red}\triangleright \quad \color{black}\vec{b}=3\vec{i}-4\vec{j}+m\vec{k}=\begin{pmatrix} 3\\ -4\\ m \end{pmatrix}\\ &\begin{aligned}\vec{a}\bullet \vec{b}&=1.3+3.(-4)+3.m\\ -4&=3+8+3m\\ -3m&=11+4\\ m&=-\displaystyle \frac{15}{3}\\ &=\color{blue}-5 \end{aligned} \end{array}$

$\begin{array}{ll} 5.&\textrm{Diketahui}\: \: \left |\vec{a} \right |=10,\: \left | \vec{b} \right |=6.\\ & \textrm{Jika}\: \: \vec{a}\: \: \textrm{dan}\: \: \vec{b}\: \: \textrm{membentuk sudut}\\ &60^{\circ}.\: \textrm{Tentukanlah nilai}\: \: \vec{a}\bullet \vec{b}\\\\ &\textbf{Jawab}\\ &\begin{aligned}\vec{a}\bullet \vec{b}&=\left | \vec{a} \right |\left | \vec{b} \right |\cos \theta \\ &=10.6.\cos 60^{\circ}\\ &=60.\left ( \displaystyle \frac{1}{2} \right )\\ &=\color{blue}30\\ \textrm{Jadi}&\: \textrm{hasil kali skalarnya adalah 30} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 6.&\textrm{Diketahui}\: \: \overrightarrow{a}=\begin{pmatrix} -2\\ 1\\ 3 \end{pmatrix}\: \: \textrm{dan}\: \: \overrightarrow{b}=\begin{pmatrix} 4\\ -1\\ t \end{pmatrix},\\ & \textrm{jika}\: \: \overrightarrow{p}\: \: \textrm{tegak lurus}\: \: \overrightarrow{q},\: \: \textrm{maka tentukanlah}\\ &\textrm{nilai}\: \: t\: \: \textrm{adalah}\\\\ &\textbf{Jawab}:\\ &\begin{aligned}\textrm{Karena}&\: \textrm{kedua vektor tersebut saling }\\ \textrm{tegak l}& \textrm{urus maka}\\ \overrightarrow{a}.\overrightarrow{b}&=0\\ \begin{pmatrix} -2\\ 1\\ 3 \end{pmatrix}&\begin{pmatrix} 4\\ -1\\ t \end{pmatrix}=0\\ (-2).4&+1.(-1)+3.t=0\\ -8-1&+3t=0\\ 3t&=9\\ t&=\color{red}3 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 7.&\textrm{Tentukanlah nilai}\: \: \overrightarrow{a}.\overrightarrow{b}\: \: \textrm{jika}\\ &\textrm{a}.\quad \left | \overrightarrow{a} \right |=4,\: \left | \overrightarrow{b} \right |=6,\: \: \angle \left ( \overrightarrow{a},\overrightarrow{b} \right )=60^{\circ}\\ &\textrm{b}.\quad \overrightarrow{a}=2\vec{i}+\vec{j}-5\vec{k}\: \: \textrm{dan}\: \: \overrightarrow{b}=2\vec{i}-3\vec{k}\\ &\textrm{c}.\quad \overrightarrow{a}=\begin{pmatrix} 0\\ -1\\ 3 \end{pmatrix}\: \: \textrm{dan}\: \: \overrightarrow{b}=\begin{pmatrix} 4\\ -2\\ 1 \end{pmatrix}\\\\ &\textbf{Jawab}:\\ &\begin{aligned}\textrm{a}.\quad \overrightarrow{a}.\overrightarrow{b}&=\left | \overrightarrow{a} \right |\left | \overrightarrow{b} \right |\cos \angle \left ( \overrightarrow{a},\overrightarrow{b} \right )\\ &=4.6.\cos 60^{\circ}\\ &=24.\left ( \displaystyle \frac{1}{2} \right )\\ &=12 \end{aligned}\\ &\textrm{b}.\quad \overrightarrow{a}.\overrightarrow{b}=2.2+1.0+(-5).(-3)=4+15=19\\ &\textrm{c}.\quad \overrightarrow{a}.\overrightarrow{b}=0.4+(-1).(-2)+3.1=0+2+3=5 \end{array}$

$\begin{array}{ll} 8.&\textrm{Diketahui}\: \: \left |\vec{a} \right |=10,\: \left | \vec{b} \right |=3\\ & \textrm{dan}\: \: \vec{a}\bullet \vec{b}=15\sqrt{3}\: .\: \textrm{Tentukan sudut}\\ &\textrm{yang dibentuk oleh}\: \: \vec{a}\: \: \textrm{dan}\: \: \vec{b}\\\\ &\textbf{Jawab}\\ &\textrm{Dari bentuk}\\ &\begin{aligned}\vec{a}\bullet \vec{b}&=\left | \vec{a} \right |\left | \vec{b} \right |\cos \theta \\ \textrm{dipe}&\textrm{roleh bentuk}\\ \cos \theta &=\displaystyle \frac{\vec{a}\bullet \vec{b}}{\left | \vec{a} \right |\left | \vec{b} \right |}\\ \cos \theta &=\displaystyle \frac{15\sqrt{3}}{10.3}=\frac{15}{30}\sqrt{3}=\frac{1}{2}\sqrt{3}\\ \cos \theta&=\cos 30^{\circ}\\ \theta &=\color{red}30^{\circ}\\ \textrm{Jadi}&\: \textrm{sudut antara keduanya adalah}\: \: \color{red}30^{\circ} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 9.&\textrm{Tentukanlah besar sudut antara vektor}\\ &\overrightarrow{a}=\begin{pmatrix} -1\\ 1\\ 0\end{pmatrix}\: \: \textrm{dan}\: \: \overrightarrow{b}=\begin{pmatrix} 1\\ -2\\ 2 \end{pmatrix}\\\\ &\textbf{Jawab}:\\ &\begin{aligned}\cos \theta &=\displaystyle \frac{\overrightarrow{a}.\overrightarrow{b}}{\left | \overrightarrow{a} \right |\left | \overrightarrow{b} \right |}\\ &=\displaystyle \frac{\begin{pmatrix} -1\\ 1\\ 0 \end{pmatrix}\begin{pmatrix} 1\\ -2\\ 2 \end{pmatrix}}{\sqrt{(-1)^{2}+1^{2}}\sqrt{1^{2}+(-2)^{2}+2^{2}}}\\ &=\displaystyle \frac{-1-2+0}{\sqrt{2}\sqrt{9}}\\ &=-\displaystyle \frac{1}{\sqrt{2}}=-\displaystyle \frac{1}{2}\sqrt{2}\\ &=-\cos 45^{\circ}\\ &=\cos \left ( 180^{\circ}-45^{\circ} \right )\\ \cos \theta &=\cos 135^{\circ}\\ \therefore \: \theta &=\color{red}135^{\circ} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 10.&\textrm{Diketahui bahwa}\: \: \left |\overrightarrow{a} \right |=\sqrt{6} ,\: \: (\overrightarrow{a}-\overrightarrow{b})(\overrightarrow{a}+\overrightarrow{b})=0\\ & \textrm{dan}\: \: \overrightarrow{a}(\overrightarrow{a}-\overrightarrow{b})=3.\: \textrm{Tentukanlah besar}\\ &\textrm{sudut antara}\: \: \overrightarrow{a}\: \: \textrm{dan}\: \: \overrightarrow{b}\\\\ &\textbf{Jawab}:\\ &\begin{aligned}\textrm{Perhatikan}&\: \textrm{bahwa}\\ (\overrightarrow{a}-\overrightarrow{b})(\overrightarrow{a}+\overrightarrow{b})&=0\\ \left | \overrightarrow{a} \right |^{2}-\left | \overrightarrow{b} \right |^{2}&=0\\ \left | \overrightarrow{a} \right |^{2}&=\left | \overrightarrow{b} \right |^{2}\quad \Rightarrow \quad \left | \overrightarrow{a} \right |=\overrightarrow{b}=\sqrt{6}\\ \textrm{dan}\quad \overrightarrow{a}(\overrightarrow{a}-\overrightarrow{b})&=3\\ \left | \overrightarrow{a} \right |^{2}-\overrightarrow{a}\overrightarrow{b}&=3\\ 6-\overrightarrow{a}\overrightarrow{b}&=3\\ -\overrightarrow{a}\overrightarrow{b}&=3-6=-3\\ \overrightarrow{a}\overrightarrow{b}&=3\\ \left | \overrightarrow{a} \right |\left | \overrightarrow{b} \right |\cos \theta &=3\\ \cos \theta &=\displaystyle \frac{3}{\sqrt{6}\sqrt{6}}=\frac{3}{6}=\frac{1}{2}\\ \cos \theta &=\cos 60^{\circ}\\ \therefore \: \: \theta &=\color{red}60^{\circ} \end{aligned} \end{array}$

$\color{blue}\textrm{Berikut dua contoh untuk sudut tidak istimewa}$.

$\begin{array}{ll} 11.&\textrm{Diketahui}\: \: \vec{a} =\vec{i}+2\vec{j}+2\vec{k},\: \: \textrm{dan}\\ & \vec{b}=3\vec{i}+4\vec{j}\: .\: \textrm{Tentukan sudut}\\ &\textrm{yang dibentuk oleh}\: \: \vec{a}\: \: \textrm{dan}\: \: \vec{b}\\\\ &\textbf{Jawab}\\ &\begin{aligned}\textrm{Dik}&\textrm{etahui bahwa}\\ \triangleright \quad &\vec{a} =\vec{i}+2\vec{j}+2\vec{k}=\begin{pmatrix} 1\\ 2\\ 2 \end{pmatrix}\: \: \textrm{dan}\\ &\left | \vec{a} \right |=\sqrt{1^{2}+2^{2}+2^{2}}=\sqrt{9}=3\\ \triangleright \quad &\vec{b}=3\vec{i}+4\vec{j}=\begin{pmatrix} 3\\ 4\\ 0 \end{pmatrix}\: \: \textrm{dan}\\ &\left | \vec{b} \right |=\sqrt{3^{2}+4^{2}+0^{2}}=\sqrt{25}=5\\ \end{aligned}\\ &\textrm{Selanjutnya}\\ &\begin{aligned} \cos \theta &=\displaystyle \frac{\vec{a}\bullet \vec{b}}{\left | \vec{a} \right |\left | \vec{b} \right |}\\ \cos \theta &=\displaystyle \frac{\begin{pmatrix} 1\\ 2\\ 2 \end{pmatrix}\begin{pmatrix} 3\\ 4\\ 0 \end{pmatrix}}{3.5}=\frac{3+8+0}{15}=\frac{11}{15}\\ \cos \theta&=0,733\\ \theta &=\color{red}\arccos \left ( \displaystyle 0.733 \right )\\ &\quad \textrm{gunakan alat bantu tabel trigonometri}\\ &\quad \textrm{atau kalkulator scientific}\\ &=42,9^{\circ}\\ \textrm{Jadi}&\: \textrm{sudut antara keduanya adalah}\: \: 42,9^{\circ} \end{aligned} \end{array}$

$\begin{array}{ll} 12.&\textrm{Diketahui}\: \: \vec{p} =(1,2,2),\: \: \textrm{dan}\\ & \vec{q}=(3,-2,6)\: .\: \textrm{Tentukan sudut}\\ &\textrm{yang dibentuk oleh}\: \: \vec{p}\: \: \textrm{dan}\: \: \vec{q}\\\\ &\textbf{Jawab}\\ &\begin{aligned}\textrm{Dik}&\textrm{etahui bahwa}\\ \triangleright \quad &\vec{p} =(1,2,2)=\begin{pmatrix} 1\\ 2\\ 2 \end{pmatrix}\: \: \textrm{dan}\\ &\left | \vec{p} \right |=\sqrt{1^{2}+2^{2}+2^{2}}=\sqrt{9}=3\\ \triangleright \quad &\vec{q}=(3,-2,6)=\begin{pmatrix} 3\\ -2\\ 6 \end{pmatrix}\: \: \textrm{dan}\\ &\left | \vec{q} \right |=\sqrt{3^{2}+(-2)^{2}+6^{2}}=\sqrt{49}=7\\ \end{aligned}\\ &\textrm{Selanjutnya}\\ &\begin{aligned} \cos \theta &=\displaystyle \frac{\vec{p}\bullet \vec{q}}{\left | \vec{p} \right |\left | \vec{q} \right |}\\ \cos \theta &=\displaystyle \frac{\begin{pmatrix} 1\\ 2\\ 2 \end{pmatrix}\begin{pmatrix} 3\\ -2\\ 6 \end{pmatrix}}{3.7}=\frac{3-4+12}{21}=\frac{11}{21}\\ \cos \theta&=0,524\\ \theta &=\color{red}\arccos \left ( \displaystyle 0.524 \right )\\ &\quad \textrm{gunakan alat bantu tabel trigonometri}\\ &\quad \textrm{atau kalkulator scientific}\\ &=58,4^{\circ}\\ \textrm{Jadi}&\: \textrm{sudut antara keduanya adalah}\: \: 58,4^{\circ} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 13.&\textrm{Diketahui vektor}\: \: \overrightarrow{a}\: \: \textrm{dan}\: \: \overrightarrow{b}\: \: \textrm{memiliki }\\ &\textrm{panjang masing-masing adalah 2 dan 3}\\ &\textrm{serta}\: \: \angle \left ( \overrightarrow{a},\overrightarrow{b}\right )=60^{\circ}.\: \textrm{Carilah nilai}\\ &\textrm{a}.\quad \left | \overrightarrow{a}+\overrightarrow{b} \right |\\\\ &\textrm{b}.\quad \left | \overrightarrow{a}-\overrightarrow{b} \right |\\ &\textrm{b}\quad \textrm{besar sudut antara}\\ &\qquad \left ( \overrightarrow{a}+\overrightarrow{b} \right )\: \: \textrm{dan}\: \: \left ( \overrightarrow{a}-\overrightarrow{b} \right )\\\\ &\textbf{Jawab}:\\ &\begin{aligned}\textrm{a}.\quad&\left | \overrightarrow{a}+\overrightarrow{b} \right |^{2}\\ &=\left ( \overrightarrow{a}+\overrightarrow{b} \right )\left ( \overrightarrow{a}+\overrightarrow{b} \right )\\ &=\overrightarrow{a}\overrightarrow{a}+2\overrightarrow{a}\overrightarrow{b}+\overrightarrow{b}\overrightarrow{b}\\ &=\left | \overrightarrow{a} \right |^{2}\cos 0^{\circ}+2\left |\overrightarrow{a} \right |\left |\overrightarrow{b} \right |\cos 60^{\circ}+\left | \overrightarrow{b} \right |^{2}\cos 0^{\circ}\\ &=2^{2}.1+2.2.3.\displaystyle \frac{1}{2}+3^{2}.1\\ &=4+6+9=19\\ &\textrm{Jadi, nilainya adalah}\: \: \left | \overrightarrow{a}+\overrightarrow{b} \right |=\color{red}\sqrt{19} \end{aligned}\\ &\begin{aligned}\textrm{b}.\quad&\left | \overrightarrow{a}-\overrightarrow{b} \right |^{2}\\ &=\left ( \overrightarrow{a}-\overrightarrow{b} \right )\left ( \overrightarrow{a}-\overrightarrow{b} \right )\\ &=\overrightarrow{a}\overrightarrow{a}-2\overrightarrow{a}\overrightarrow{b}+\overrightarrow{b}\overrightarrow{b}\\ &=\left | \overrightarrow{a} \right |^{2}\cos 0^{\circ}-2\left |\overrightarrow{a} \right |\left |\overrightarrow{b} \right |\cos 60^{\circ}+\left | \overrightarrow{b} \right |^{2}\cos 0^{\circ}\\ &=2^{2}.1-2.2.3.\displaystyle \frac{1}{2}+3^{2}.1\\ &=4-6+9=7\\ &\textrm{Jadi, nilainya adalah}\: \: \left | \overrightarrow{a}-\overrightarrow{b} \right |=\color{red}\sqrt{7} \end{aligned}\\ &\begin{aligned}\textrm{c}.\quad \textrm{Untuk menentukan nilai}&\\ \cos \angle \left ( \overrightarrow{a}+\overrightarrow{b},\overrightarrow{a}-\overrightarrow{b} \right )&=\displaystyle \frac{\left (\overrightarrow{a}+\overrightarrow{b} \right ).\left (\overrightarrow{a}-\overrightarrow{b} \right )}{\left | \overrightarrow{a}+\overrightarrow{b} \right |.\left | \overrightarrow{a}-\overrightarrow{b} \right |}\\ &=\displaystyle \frac{\overrightarrow{a}\overrightarrow{a}-\overrightarrow{a}\overrightarrow{b}+\overrightarrow{b}\overrightarrow{a}-\overrightarrow{b}\overrightarrow{b}}{\sqrt{19}.\sqrt{7}}\\ &=\displaystyle \frac{2^{2}-3^{2}}{\sqrt{133}}=-\frac{5}{\sqrt{133}}\\ \angle \left ( \overrightarrow{a}+\overrightarrow{b},\overrightarrow{a}-\overrightarrow{b} \right )&=\color{red}\arccos \left ( -\frac{5}{\sqrt{133}} \right ) \end{aligned} \end{array}$

$\color{blue}\textrm{Berikut contoh untuk bentuk sudutnya}$.

$\begin{array}{ll} 14.&\textrm{Diketahui}\: \: \vec{p} =(x,3,2),\: \: \textrm{dan}\\ & \vec{q}=(2,-6,3)\: .\: \textrm{Tentukan nilai}\: \: x\\ &\textrm{agar kedua vektor}\\ &\textrm{a}\quad \textrm{membentuk sudut lancip}\\ &\textrm{b}\quad \textrm{membentuk sudut siku-siku}\\ &\textrm{c}\quad \textrm{membentuk sudut tumpul}\\ &\textrm{d}\quad \textrm{sama panjang}\\\\ &\textbf{Jawab}\\ &\begin{aligned}\textrm{Dik}&\textrm{etahui bahwa}\\ \triangleright \quad &\vec{p} =(x,3,2)=\begin{pmatrix} x\\ 3\\ 2 \end{pmatrix}\: \: \textrm{dan}\\ \triangleright \quad &\vec{q}=(2,-6,3)=\begin{pmatrix} 2\\ -6\\ 3 \end{pmatrix}\\ \end{aligned}\\ &\textrm{Selanjutnya}\\ &\vec{p}\bullet \vec{q}=\begin{pmatrix} x\\ 3\\ 2 \end{pmatrix}\begin{pmatrix} 2\\ -6\\ 3 \end{pmatrix}\\ &\quad =2x-18+6=2x-12\\ &\textrm{Selanjutnya}\\ &\begin{aligned} \textrm{a}\quad&\textbf{Syarat lancip},\: \textrm{yaitu}:\: \vec{p}\bullet \vec{q}>0\\ &2x-12>0\Leftrightarrow 2x>12\Leftrightarrow x>6\\ \textrm{b}\quad&\textbf{Syarat siku-siku},\: \textrm{yaitu}:\: \vec{p}\bullet \vec{q}=0\\ &2x-12=0\Leftrightarrow 2x=12\Leftrightarrow x=6\\ \textrm{c}\quad&\textbf{Syarat tumpul},\: \textrm{yaitu}:\: \vec{p}\bullet \vec{q}<0\\ &2x-12<0\Leftrightarrow 2x<12\Leftrightarrow x<6\\ \textrm{d}\quad&\textbf{Syarat panjang kedua vektor sama}\\ & \textrm{yaitu}:\: \left |\vec{p} \right |= \left |\vec{q} \right |,\: \textrm{maka}\\ &\begin{aligned}&\sqrt{x^{2}+3^{2}+2^{2}}=\sqrt{2^{2}+(-6)^{2}+3^{2}}\\ &x^{2}+9+4=4+36+9\\ &x^{2}=36\\ &x=\pm \sqrt{36}=\pm 6\\ &\textrm{Jadi},\: \color{red}x=-6\: \: \color{black}\textrm{atau}\: \: \color{red}x=6 \end{aligned} \end{aligned} \end{array}$

DAFTAR PUSTAKA

Johanes, Kastolan, Sulasim. 2006. Kompetensi Matematika Program IPA 3A SMA Kelas XII Semester Pertama. Jakarta: YUDHISTIRA.
Kanginan, M., Nurdiansyah, H., Akhmad, G. 2016. Matematika untuk Siswa SMA/MA Kelas X Kelompok Peminatan Matematika dan Ilmu-Ilmu Alam. Bandung: YRAMA WIDYA.
Noormandiri, Sucipto, E. 2003. Buku Pelajaran Matematika SMU untuk Kelas 3 Program IPA. Jakarta: ERLANGGA.
Yuana, R.A., Indriyastuti. 2017. Perspektif Matematika untuk Kelas X SMA dan MA Kelompok Peminatan Matematika dan Ilmu Alam. Solo: PT. TIGA SERANGKAI PUSTAKA MANDIRI.