Lanjutan Limit Fungsi Trigonometri

 $\text{C. Menentukan Nilai Limit Fungsi Trigonometri}$

Dalam bahasan ini yang akan dibahas adalah nilai limit mendekati $a$ atau nilai $x$ di sekitar $a$. Ada 3 cara yang populer digunakan untuk menentukan nilai limit fungsi trigonometri ini dengan salah satunya yang paling sering digunakan adalah substitusi langsung di antara cara-cara penyelesaian lainnya. Jika dengan cara substitusi langsung nantinya mendfapatkan nilai bentuk tak tentu yaitu ${\displaystyle \frac{0}{0}}$, maka cara Anda harus menggunakan cara yang lainnya sampai Anda temukan nilai limitnya. Selanjutnya 3 cara yang dimaksud di atas adalah sebagai berikut:

$\text{C. 1 dengan substitusi langsung}$

$\text{CONTOH SOAL}$.

$\begin{array}{l}\\ 1.& \text{Tentukanlah nilai limit dari fungsi-sungsi}\\ & \text{berikut}\\ & \begin{array}{ll}\text{a}& \underset{x\to 0}{\text{lim}}(\sin x+\tan x)\\ \text{b}& \underset{x\to \pi }{\text{lim}}\left({\displaystyle \sin x+\cos x}\right)\\ \text{c}& \underset{x\to {\displaystyle \frac{\pi }{4}}}{\text{lim}}\left({\displaystyle \frac{\sin x+\cos x}{\tan x}}\right)\\ \text{d}& \underset{x\to {\displaystyle 0}}{\text{lim}}\left({\displaystyle \frac{1+\cos 2x}{1+2\cos x}}\right)\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}\text{De}& \text{ngan substitusi langsung didapatkan}\\ \text{a}.& \underset{x\to 0}{\text{lim}}(\sin x+\tan x)\\ & =\sin 0+\tan 0=0+0=0\\ \text{b}.& \underset{x\to \pi }{\text{lim}}\left({\displaystyle \sin x+\cos x}\right)\\ & =\sin \pi +\cos \pi =0+(-1)=-1\\ \text{c}.& \underset{x\to {\displaystyle \frac{\pi }{4}}}{\text{lim}}\left({\displaystyle \frac{\sin x+\cos x}{\tan x}}\right)\\ & =\left({\displaystyle \frac{\sin {\displaystyle \frac{\pi }{4}+\cos {\displaystyle \frac{\pi }{4}}}}{\tan {\displaystyle \frac{\pi }{4}}}}\right)={\displaystyle \frac{{\displaystyle \frac{1}{2}\sqrt{2}+{\displaystyle \frac{1}{2}\sqrt{2}}}}{1}}\\ & ={\displaystyle \frac{\sqrt{2}}{1}=\sqrt{2}}\\ \text{d}.& \underset{x\to 0}{\text{lim}}\left({\displaystyle \frac{1+\cos 2x}{1+2\cos x}}\right)\\ & =\left({\displaystyle \frac{1+\cos 2(0)}{1+2\cos (0)}}\right)={\displaystyle \frac{1+1}{1+2.1}={\displaystyle \frac{2}{3}}}\end{array}\end{array}$

$\text{C. 2 dengan menyederhanakan}$

Langkah ini ditempuh setelah langkah substitusi langsung tidak memungkinkan atau ketemu bentuk tak tentu  ${\displaystyle \frac{0}{0}}$.

Gunakanlah identitas-identitas trigonometri yang Anda dapatkan di kelas XI  dan akan sering digunakan nantinya di antaranya, yaitu:

$\begin{array}{rl}\bullet \sin 2x& =2\sin x\cos x\\ \bullet \cos 2x& ={\cos }^{2}x-{\sin }^{2}x\\ & =1-2{\sin }^{2}x\\ & =2{\cos }^{2}x-1\\ \bullet \tan 2x& ={\displaystyle \frac{2\tan x}{1-{\tan }^{2}x}}\end{array}$

Demikian juga

$\begin{array}{rl}\bullet \sin A+\sin B& =2\sin \left({\displaystyle \frac{A+B}{2}}\right)\cos \left({\displaystyle \frac{A-B}{2}}\right)\\ \bullet \sin A-\sin B& =2\cos \left({\displaystyle \frac{A+B}{2}}\right)\sin \left({\displaystyle \frac{A-B}{2}}\right)\\ \bullet \cos A+\cos B& =2\cos \left({\displaystyle \frac{A+B}{2}}\right)\cos \left({\displaystyle \frac{A-B}{2}}\right)\\ \bullet \cos A-\cos B& =-2\sin \left({\displaystyle \frac{A+B}{2}}\right)\sin \left({\displaystyle \frac{A-B}{2}}\right)\end{array}$.

Masih banyak bentuk identitas trigonometri selain di atas, karenanya sekiranya perlu maka hafalkanlah

$\text{CONTOH SOAL}$.

$\begin{array}{l}\\ 2.& \text{Tentukanlah nilai limit dari fungsi-sungsi}\\ & \text{berikut}\\ & \begin{array}{ll}\text{a}& \underset{x\to 0}{\text{lim}}{\displaystyle \frac{\sin 2x}{\sin x}}\\ \text{b}& \underset{x\to \frac{\pi }{2}}{\text{lim}}{\displaystyle \frac{\sin 4x}{\sin x\cos x}}\\ \text{c}& \underset{x\to {\displaystyle 0}}{\text{lim}}{\displaystyle \frac{1-{\cos }^{2}x}{{\tan }^{2}x}}\\ \text{d}& \underset{x\to {\displaystyle 0}}{\text{lim}}{\displaystyle \frac{{\sin }^{2}x}{1-\cos x}}\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}\text{De}& \text{ngan menyederhankan bentuk trigonometri}\\ \text{ak}& \text{an didapatkan bentuk yang lebih sederhana}\\ \text{a}.& \underset{x\to 0}{\text{lim}}{\displaystyle \frac{\sin 2x}{\sin x}}\\ & \text{saat substitusi langsung menghsilkan bentuk}\\ & ={\displaystyle \frac{\sin 0}{\sin 0}={\displaystyle \frac{0}{0},\text{maka perlu disederhanakan}}}\\ & =\underset{x\to 0}{\text{lim}}{\displaystyle \frac{\sin 2x}{\sin x}=\underset{x\to 0}{\text{lim}}{\displaystyle \frac{2\sin x\cos x}{\sin x}=\underset{x\to 0}{\text{lim}}{\displaystyle 2\cos x}}}\\ & \text{baru digunakan substitusinya}\\ & =2\cos 0=2.1=2\\ \text{b}.& \underset{x\to \frac{\pi }{2}}{\text{lim}}{\displaystyle \frac{\sin 4x}{\sin x\cos x}}\\ & \text{Sama seperti langkah di atas, yaitu}:\\ & =\underset{x\to \frac{\pi }{2}}{\text{lim}}{\displaystyle \frac{\sin 4x}{\sin x\cos x}=\underset{x\to \frac{\pi }{2}}{\text{lim}}{\displaystyle \frac{2\sin 2x\cos 2x}{\sin x\cos x}}}\\ & =\underset{x\to \frac{\pi }{2}}{\text{lim}}{\displaystyle \frac{4\sin x\cos x\cos 2x}{\sin x\cos x}=\underset{x\to \frac{\pi }{2}}{\text{lim}}{\displaystyle 4\cos 2x}}\\ & =4\cos 2\left({\displaystyle \frac{\pi }{2}}\right)=4\cos \pi =4.(-1)=-4\\ \text{c}.& \underset{x\to {\displaystyle 0}}{\text{lim}}{\displaystyle \frac{1-{\cos }^{2}x}{{\tan }^{2}x}}\\ & \text{saat substitusi langsung menghsilkan bentuk}\\ & ={\displaystyle \frac{1-{\cos }^{2}0}{{\tan }^{2}0}={\displaystyle \frac{1-1}{0}=\frac{0}{0},}}\\ & \text{maka perlu disederhanakan}\\ & =\underset{x\to {\displaystyle 0}}{\text{lim}}{\displaystyle \frac{1-{\cos }^{2}x}{{\tan }^{2}x}=\underset{x\to {\displaystyle 0}}{\text{lim}}{\displaystyle \frac{{\sin }^{2}x}{\left({\displaystyle \frac{{\sin }^{2}x}{{\cos }^{2}x}}\right)}}}\\ & =\underset{x\to {\displaystyle 0}}{\text{lim}}{\displaystyle {\cos }^{2}x={\cos }^{2}0=1^2=1}\\ \text{d}.& \underset{x\to {\displaystyle 0}}{\text{lim}}{\displaystyle \frac{{\sin }^{2}x}{1-\cos x}}\\ & \text{saat substitusi langsung menghsilkan bentuk}\\ & ={\displaystyle \frac{{\sin }^{2}0}{1-\cos 0}={\displaystyle \frac{0}{1-1}=\frac{0}{0},}}\\ & \text{maka perlu disederhanakan}\\ & =\underset{x\to {\displaystyle 0}}{\text{lim}}{\displaystyle \frac{{\sin }^{2}x}{1-\cos x}=\underset{x\to {\displaystyle 0}}{\text{lim}}{\displaystyle \frac{1-{\cos }^{2}x}{1-\cos x}}}\\ & =\underset{x\to {\displaystyle 0}}{\text{lim}}{\displaystyle \frac{(1+\cos x)(1-\cos x)}{1-\cos x}}\\ & =\underset{x\to {\displaystyle 0}}{\text{lim}}{\displaystyle (1+\cos x)}\\ & =1+\cos 0=1+1=2\end{array}\end{array}$.

$\begin{array}{l}\\ 3.& \text{Tentukanlah nilai limit dari fungsi-sungsi}\\ & \text{berikut}\\ & \begin{array}{ll}\text{a}& \underset{x\to 0}{\text{lim}}{\displaystyle \frac{\cos x-\cos 3x}{1-\cos 2x}}\\ \text{b}& \underset{x\to \frac{\pi }{4}}{\text{lim}}{\displaystyle \frac{1-\tan x}{\sin x-\cos x}}\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}\text{De}& \text{ngan menyederhankan bentuk trigonometri}\\ \text{ak}& \text{an didapatkan bentuk yang lebih sederhana}\\ \text{a}.& \underset{x\to 0}{\text{lim}}{\displaystyle \frac{\cos x-\cos 3x}{1-\cos 2x}}\\ & \text{saat substitusi langsung menghsilkan bentuk}\\ & ={\displaystyle \frac{\cos 0-\cos 0}{1-\cos 0}={\displaystyle \frac{1-1}{1-1}={\displaystyle \frac{0}{0},}}}\\ & \text{maka perlu disederhanakan}\\ & =\underset{x\to 0}{\text{lim}}{\displaystyle \frac{\cos x-\cos 3x}{1-\cos 2x}}\\ & =\underset{x\to 0}{\text{lim}}{\displaystyle \frac{-2\sin \left({\displaystyle \frac{x+3x}{2}}\right)\sin \left({\displaystyle \frac{x-3x}{2}}\right)}{1-(1-2{\sin }^{2}x)}}\\ & =\underset{x\to 0}{\text{lim}}{\displaystyle \frac{-2\sin 2x\sin (-x)}{2{\sin }^{2}x}}\\ & =\underset{x\to 0}{\text{lim}}{\displaystyle \frac{2\sin 2x\sin x}{2\sin x.\sin x}}\\ & =\underset{x\to 0}{\text{lim}}{\displaystyle \frac{2(2\sin x\cos x)\sin x}{2\sin x.\sin x}}\\ & =\underset{x\to 0}{\text{lim}}{\displaystyle 2\cos x}\\ & \text{baru digunakan substitusinya}\\ & ={\displaystyle 2\cos 0=2.1=2}\\ \text{b}.& \underset{x\to \frac{\pi }{4}}{\text{lim}}{\displaystyle \frac{1-\tan x}{\sin x-\cos x}}\\ & \text{saat substitusi langsung menghsilkan bentuk}\\ & ={\displaystyle \frac{1-\tan x}{\sin x-\cos x}=\frac{1-1}{\frac{1}{2}\sqrt{2}-\frac{1}{2}\sqrt{2}}={\displaystyle \frac{0}{0},}}\\ & \text{maka perlu disederhanakan}\\ & =\underset{x\to \frac{\pi }{4}}{\text{lim}}{\displaystyle \frac{1-\tan x}{\sin x-\cos x}=\underset{x\to \frac{\pi }{4}}{\text{lim}}{\displaystyle \frac{1-\left({\displaystyle \frac{\sin x}{\cos x}}\right)}{\sin x-\cos x}}}\\ & =\underset{x\to \frac{\pi }{4}}{\text{lim}}{\displaystyle \frac{-(\sin x-\cos x)}{\cos x(\sin x-\cos x)}}\\ & =\underset{x\to \frac{\pi }{4}}{\text{lim}}{\displaystyle \frac{-1}{\cos x}={\displaystyle \frac{-1}{\cos \left({\displaystyle \frac{\pi }{4}}\right)}={\displaystyle \frac{-1}{\frac{1}{\sqrt{2}}}=-\sqrt{2}}}}\end{array}\end{array}$

$\text{C. 3 dengan rumus limit fungsi trigonometri}$

Berikut rumus limit fungsi trigonometri yang akan kita gunakan

$\begin{array}{rl}\bullet \underset{x\to 0}{\text{lim}}{\displaystyle \frac{\sin ax}{ax}}& =\underset{x\to 0}{\text{lim}}{\displaystyle \frac{ax}{\sin ax}=\frac{a}{a}=1}\\ \bullet \underset{x\to 0}{\text{lim}}{\displaystyle \frac{\tan ax}{ax}}& =\underset{x\to 0}{\text{lim}}{\displaystyle \frac{ax}{\tan ax}=\frac{a}{a}=1}\end{array}$.

BUKTINYA ada di sini

$\text{CONTOH SOAL}$.

$\begin{array}{l}\\ 4.& \text{Tentukanlah nilai limit berikut}\\ & \begin{array}{ll}\text{a}& \underset{x\to 0}{\text{lim}}{\displaystyle \frac{\sin 4x}{7x}}\\ \text{b}& \underset{x\to 0}{\text{lim}}{\displaystyle \frac{2x}{\tan 9x}}\\ \text{c}& \underset{x\to 0}{\text{lim}}{\displaystyle \frac{\tan 8x}{\sin 3x}}\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}\text{a}.& \underset{x\to 0}{\text{lim}}{\displaystyle \frac{\sin 4x}{7x}=\underset{x\to 0}{\text{lim}}{\displaystyle \frac{4}{7}\times {\displaystyle \frac{\sin 4x}{4x}={\displaystyle \frac{4}{7}}}}}\\ \text{b}.& \underset{x\to 0}{\text{lim}}{\displaystyle \frac{2x}{\tan 9x}=\underset{x\to 0}{\text{lim}}{\displaystyle \frac{2}{9}\times {\displaystyle \frac{9x}{\tan 9x}={\displaystyle \frac{2}{9}}}}}\\ \text{c}.& \underset{x\to 0}{\text{lim}}{\displaystyle \frac{\tan 8x}{\sin 3x}=\underset{x\to 0}{\text{lim}}{\displaystyle \frac{\tan 8x}{8x}\times \underset{x\to 0}{\text{lim}}{\displaystyle \frac{3x}{\sin 3x}\times \frac{8}{3}={\displaystyle \frac{8}{3}}}}}\end{array}\end{array}$.

$\begin{array}{l}\\ 5.& \text{Tentukanlah nilai limit berikut}\\ & \begin{array}{ll}\text{a}& \underset{x\to 0}{\text{lim}}{\displaystyle \frac{24x^2}{8{\sin }^{2}x}}\\ \text{b}& \underset{x\to 0}{\text{lim}}{\displaystyle \frac{5x^2}{15\tan (-9x)\sin 3x}}\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}\text{a}.& \underset{x\to 0}{\text{lim}}{\displaystyle \frac{24x^2}{8{\sin }^{2}x}={\displaystyle \frac{24}{8}\times \underset{x\to 0}{\text{lim}}{\displaystyle \frac{x}{\sin x}\times \underset{x\to 0}{\text{lim}}{\displaystyle \frac{x}{\sin x}}}}}\\ & =3\times 1\times 1\\ & =3\\ \text{b}.& \underset{x\to 0}{\text{lim}}{\displaystyle \frac{5x^2}{15\tan (-9x)\sin 3x}}\\ & ={\displaystyle \frac{5}{15}\times \underset{x\to 0}{\text{lim}}{\displaystyle \frac{x}{\tan (-9x)}\times \underset{x\to 0}{\text{lim}}{\displaystyle \frac{x}{\tan 3x}}}}\\ & ={\displaystyle \frac{1}{3}\times (-\times \underset{x\to 0}{\text{lim}}{\displaystyle \frac{9x}{\tan 9x}\times \frac{1}{9}})\times \times \underset{x\to 0}{\text{lim}}{\displaystyle \frac{3x}{\tan 3x}\times \frac{1}{3}}}\\ & =-{\displaystyle \frac{1}{3}\times \frac{1}{9}\times \frac{1}{3}=-\frac{1}{81}}\end{array}\end{array}$.

$\begin{array}{l}\\ 6.& \text{Tentukanlah nilai limit berikut}\\ & \begin{array}{ll}\text{a}& \underset{x\to 0}{\text{lim}}{\displaystyle \frac{1-\cos 6x}{x^2}}\\ \text{b}& \underset{x\to 0}{\text{lim}}{\displaystyle \frac{\cos 4x-\cos 2x}{x^2}}\\ \text{c}& \underset{x\to 3}{\text{lim}}{\displaystyle \frac{(x^2-5x+6)\sin (x-3)}{{(x^2-7x+12)}^2}}\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}\text{a}.& \underset{x\to 0}{\text{lim}}{\displaystyle \frac{1-\cos 6x}{x^2}=\underset{x\to 0}{\text{lim}}{\displaystyle \frac{1-(1-2{\sin }^{2}3x)}{x^2}}}\\ & =\underset{x\to 0}{\text{lim}}{\displaystyle \frac{2{\sin }^{2}3x}{x^2}=2\times \underset{x\to 0}{\text{lim}}{\displaystyle \frac{\sin 3x}{x}\times \underset{x\to 0}{\text{lim}}{\displaystyle \frac{\sin 3x}{x}}}}\\ & =2\times 3\times 3=18\\ \text{b}.& \underset{x\to 0}{\text{lim}}{\displaystyle \frac{\cos 4x-\cos 2x}{x^2}}\\ & =\underset{x\to 0}{\text{lim}}{\displaystyle \frac{-2\sin \left({\displaystyle \frac{4x+2x}{2}}\right)\sin \left({\displaystyle \frac{4x-2x}{2}}\right)}{x^2}}\\ & =-2\times \underset{x\to 0}{\text{lim}}{\displaystyle \frac{\sin 3x.\sin x}{x^2}}\\ & =-2\times \underset{x\to 0}{\text{lim}}{\displaystyle \frac{\sin 3x}{x}\times \underset{x\to 0}{\text{lim}}{\displaystyle \frac{x}{x}=-2\times 3\times 1=-6}}\\ \text{c}.& \underset{x\to 3}{\text{lim}}{\displaystyle \frac{(x^2-5x+6)\sin (x-3)}{{(x^2-7x+12)}^2}}\\ & =\underset{x\to 3}{\text{lim}}{\displaystyle \frac{(x-2)(x-3)\sin (x-3)}{{((x-3)(x-4))}^2}}\\ & =\underset{x\to 3}{\text{lim}}{\displaystyle \frac{(x-2)(x-3)\sin (x-3)}{(x-3)(x-3)(x-4{)}^2}}\\ & =\underset{x\to 3}{\text{lim}}{\displaystyle {\displaystyle \frac{x-2}{(x-4{)}^2}\times \underset{x\to 3}{\text{lim}}{\displaystyle \frac{\sin (x-3)}{(x-3)}}}}\\ & ={\displaystyle \frac{(3-2)}{(3-4{)}^2}\times 1={\displaystyle \frac{1}{(-1{)}^2}=\frac{1}{1}=1}}\end{array}\end{array}$.

$\begin{array}{l}\\ 7.& \text{Tentukanlah nilai limit dari}\underset{x\to \frac{1}{2}}{\text{lim}}{\displaystyle \frac{\sin (4x-2)}{\tan 2x-1}}\\ \\ & \text{Jawab}:\\ & \text{Kita misalkan}a=2x-1,\\ & \text{ketika}x\to {\displaystyle \frac{1}{2},\text{maka akan didapatkan}a\to 0}\\ & (\text{textbf{dibaca}: saat nilai}x\text{mendekati}{\displaystyle \frac{1}{2},\text{maka nilai}}\\ & a\text{akan mendekati nilai}0).\\ & \text{Selanjutnya kita buatkan penyesuaian, yaitu}:\\ & \begin{array}{rl}\underset{x\to \frac{1}{2}}{\text{lim}}{\displaystyle \frac{\sin (4x-2)}{\tan 2x-1}}& =\underset{x\to \frac{1}{2}}{\text{lim}}{\displaystyle \frac{\sin 2(2x-1)}{\tan 2x-1}}\\ & =\underset{x\to 0}{\text{lim}}{\displaystyle \frac{\sin 2a}{\tan a}}\\ & =\underset{x\to 0}{\text{lim}}{\displaystyle \frac{\sin 2a}{2a}\times 2\times \underset{x\to 0}{\text{lim}}{\displaystyle \frac{a}{\tan a}}}\\ & =2\times 1=2\end{array}\end{array}$.

$\text{LATIHAN SOAL}$.

$\begin{array}{l}\\ 1.& \text{Selidikilah limit fungsi berikut, apakah}\\ & \text{memiliki nilai limit atau tidak}\\ & \text{a}.\underset{x\to 0}{\text{lim}}2x-1\\ & \text{b}.\underset{x\to 0}{\text{lim}}{x}^2-x-2\\ & \text{c}.\underset{x\to 1}{\text{lim}}f(x),\text{dengan}f(x)=\{\begin{array}{ll}2x& ;x<1\\ 4x-1& ;x\ge 1\end{array}\\ & \text{ d}.\underset{x\to 2}{\text{lim}}\sqrt{f(x)},\text{dengan}f(x)=\{\begin{array}{ll}4x-1& ;x<2\\ 2x+5& ;x\ge 2\end{array}\\ & \text{e}.\underset{x\to 0}{\text{lim}}(x+\cos x)\\ & \text{f}.\underset{x\to 0}{\text{lim}}x\tan x\\ & \text{g}.\underset{x\to {\displaystyle \frac{\pi }{2}}}{\text{lim}}(\sin x+2\cos x)\\ & \text{h}.\underset{x\to {\displaystyle \frac{\pi }{2}}}{\text{lim}}(2\tan x-\sin 2x)\end{array}$.

$\begin{array}{l}\\ 2.& \text{Tentukanlah nilai limit berikut}\\ & \begin{array}{ll}\text{a}& \underset{x\to 0}{\text{lim}}{\displaystyle \frac{\sin 5x}{x}}\\ \text{b}& \underset{x\to 0}{\text{lim}}{\displaystyle \frac{4x}{\sin x}}\\ \text{c}& \underset{x\to 0}{\text{lim}}{\displaystyle \frac{\tan 6x}{8x}}\\ \text{d}& \underset{x\to 0}{\text{lim}}{\displaystyle \frac{2x}{\tan 7x}}\\ \text{e}& \underset{x\to 0}{\text{lim}}{\displaystyle \frac{\sin 3x}{\sin 2x}}\\ \text{f}& \underset{x\to 0}{\text{lim}}{\displaystyle \frac{\sin 4x}{\tan 8x}}\\ \text{9}& \underset{x\to 0}{\text{lim}}{\displaystyle \frac{{\sin }^{2}5x}{2x^2}}\\ \text{h}& \underset{x\to 0}{\text{lim}}{\displaystyle \frac{\sin 5x\tan 6x}{x\tan 7x}}\\ \text{i}& \underset{x\to y}{\text{lim}}{\displaystyle \frac{\sin x-\sin y}{x-y}}\\ \text{j}& \underset{x\to 2}{\text{lim}}{\displaystyle \frac{\sin (x^2-4)}{x-2}}\\ \text{k}& \underset{x\to 0}{\text{lim}}{\displaystyle \frac{\sin mx-\sin nx}{\cos mx-\cos nx}}\\ \text{l}& \underset{x\to 0}{\text{lim}}{\displaystyle \frac{1-\cos 3x\cos x}{x^2}}\end{array}\\ \end{array}$

DAFTAR PUSTAKA

1. Kanginan, M., Nurdiansyah, H., Akhmad, G. 2016. Matematika untuk Siswa SMA/MA Kelas XII Kelompok Peminatan Matematika dan Ilmu-Ilmu Alam. Bandung: YRAMA WIDYA.
2. Sembiring, S., Zulkifli, M., Marsito, Rusdi, I. 2016. Matematika untuk Siswa SMA/MA Kelas XII Kelompok Peminatan Matematika dan Ilmu-Ilmu Alam. Bandung: SRIKANDI EMPAT WIDYA UTAMA.
3. Yuana, A.R., Indriyastuti. 2017. Perspektif Matematika untuk Kelas XII SMA dan MA Kelompok Peminatan Matematika dan Ilmu Alam. Solo: PT TIGA SERANGKAI PUSTAKA MANDIRI.

Limit Fungsi Trigonometri

 $\text{A. Pendahuluan}$

Mengingat kembali definisi limit yang telah dipelajari sebelumnya di kelas XI, yaitu limit fungsi aljabar $f(x)$ yang didefinisikan dengan:

$\begin{array}{rl}\underset{x\to a}{\text{lim}}f(x)=L& \mathbf{\text{adalah}}\text{Jika}x\text{mendekati}a\\ & \text{dengan tidak sama dengan}a,\\ & \text{maka nilai}f(x)\text{mendekati}L\end{array}$.

Perhatikan definisi di atas istilah  $x\text{mendekati}a$ dituliskan dengan simbol  $(x\to a)$. Suatu nilai limit dianggap ada jika nilai $f(x)$ mendekati  $a$ dari arah kiri sama dengan nilai $f(x)$ mendekati  $a$ dari arah kanan dengan nilai yang sama misalnya $L$. Jika disimbolkan pernyataan ini menjadi berikut

$\begin{array}{rl}\underset{x\to a^{-}}{\text{lim}}f(x)=\underset{x\to a^{+}}{\text{lim}}f(x)=\underset{x\to a}{\text{lim}}f(x)=L& \end{array}$.

$\begin{array}{rl}\text{Perlu di}& \text{perhatikan bahwa didekati dari}\\ \bullet \mathbf{\text{kiri}}& \text{disimbolkan dengan}\underset{x\to a^{-}}{\text{lim}}f(x),\text{dan}\\ \bullet \mathbf{\text{kan}}& \mathbf{\text{an}}\text{disimbolkan dengan}\underset{x\to a^{+}}{\text{lim}}f(x)\end{array}$.

$\text{CONTOH SOAL}$.

$\begin{array}{l}\\ 1.& \text{Tentukanlah nilai limit dari}f(x)={\displaystyle \frac{x^2-4}{x-2}}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}& \text{Perhatikanlah ketika fungsi}{\displaystyle \frac{x^2-4}{x-2}}\\ & \text{di sekitar}x=2\text{sebagaimana dalam tabel}\\ & \text{berikut}\end{array}\end{array}$.

$\begin{array}{rl}.& \text{Jadi, nilai}\underset{x\to 2}{\text{lim}}{\displaystyle \frac{x^2-4}{x-2}=2\text{atau dapat dikatakan}}\\ & \text{nilai}\underset{x\to 2}{\text{lim}}{\displaystyle \frac{x^2-4}{x-2}\mathbf{\text{ada}}}\\ & \text{meskipun nilai substitusi langsung}x=2\text{yaitu}\\ & f(0)={\displaystyle \frac{0^2-0}{0-0}=\frac{0}{0}\text{berupa bentuk tak}}\\ & \text{tentu. Berikut ilustrasinya}\end{array}$

$\begin{array}{l}\\ 2.& \text{Selidikilah limit fungsi berikut apakah}\\ & \text{memiliki harga limit}\\ & \underset{x\to 5}{\text{lim}}f(x),\text{untuk}f(x)=\{\begin{array}{ll}x& \text{saat}x<5\\ 5-x& \text{saat}x\ge 5\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}& \text{Perhatikanlah ketika didekati dari kiri}\\ & \text{yaitu}x\to 5^{-},\text{maka}\underset{x\to 5^{-}}{\text{lim}}f(x)=x\\ & \text{atau}\underset{x\to 5^{-}}{\text{lim}}f(x)=5\\ & \text{boleh juga dituliskan dengan}\\ & \underset{x\to 5^{-}}{\text{lim}}f(x)=x=5.\text{Sedangkan ketika}\\ & \text{didekati dari arah kanan yaitu}x\to 5^{+},\\ & \text{maka}\underset{x\to 5^{+}}{\text{lim}}f(x)=\underset{x\to 5^{+}}{\text{lim}}(5-x)=5-5=0.\\ & \text{Karena nilai}\underset{x\to 5^{-}}{\text{lim}}f(x)\ne \underset{x\to 5^{+}}{\text{lim}}f(x),\text{maka}\\ & \text{nilai atau harga}\underset{x\to 5}{\text{lim}}f(x)\mathbf{\text{tidak ada}}\\ & \text{Berikut ilustrasi gambarnya}\end{array}\end{array}$.

$\begin{array}{l}\\ 3.& \text{Selidikilah limit fungsi berikut apakah}\\ & \text{memiliki harga limit}\\ & \underset{x\to 0}{\text{lim}}f(x),\text{untuk}f(x)=\cos x\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}& \text{Perhatikanlah ketika didekati dari kiri}\\ & \text{yaitu}x\to 0^{-},\text{maka}\underset{x\to 0^{-}}{\text{lim}}\cos x\\ & \begin{array}{|ccccccc|}\hline x& -0,5& -0,4& -0,3& -0,2& -0,1& 0\\ \cos x& ...& ...& 0,999986& 0,999994& 0,9999985& 1\\ \hline\end{array}\\ & \text{Sedangkan ketika}\\ & \text{didekati dari arah kanan yaitu}x\to 0^{+},\\ & \text{maka}\underset{x\to 0^{+}}{\text{lim}}\cos x\\ & \begin{array}{|ccccccc|}\hline x& 0& 0,1& 0,2& 0,3& 0,4& 0,5\\ \cos x& 1& 0,9999985& 0,999994& 0,999986& ...& ...\\ \hline\end{array}\\ & \text{Karena nilai}\underset{x\to 0^{-}}{\text{lim}}f(x)=\underset{x\to 0^{+}}{\text{lim}}f(x)=1,\text{maka}\\ & \text{nilai}\underset{x\to 0}{\text{lim}}\cos x\mathbf{\text{ ada}}\\ & \text{Berikut ilustrasi gambarnya}\end{array}\end{array}$.

$\text{B. Sifat-Sifat Limit Fungsi}$

$\begin{array}{rl}& \text{Misalkan}f\text{dan}g\text{adalah fungsi-fungsi yang}\\ & \text{mempunyai nilai limit di titik sekitar}x=a\\ & \text{atau}(x\to a)\text{dan}c\text{adalah suatu konstanta}\\ & \text{serta}n\text{adalah suatu bilangan bulat positif},\\ & \text{maka berlaku sifat-sifat berikut}:\\ & \begin{array}{l}\\ 1.& \underset{x\to a}{\text{lim}}{\displaystyle c=c}\\ 2.& \underset{x\to a}{\text{lim}}{\displaystyle x^n=a^n}\\ 3.& \underset{x\to a}{\text{lim}}c.f(x)=c.\underset{x\to a}{\text{lim}}f(x)\\ 4.& \underset{x\to a}{\text{lim}}(f(x)\pm g(x))=\underset{x\to a}{\text{lim}}f(x)\pm \underset{x\to a}{\text{lim}}g(x)\\ 5.& \underset{x\to a}{\text{lim}}(f(x)\times g(x))=\underset{x\to a}{\text{lim}}f(x)\times \underset{x\to a}{\text{lim}}g(x)\\ 6.& \underset{x\to a}{\text{lim}}{\displaystyle \frac{f(x)}{g(x)}={\displaystyle \frac{\underset{x\to a}{\text{lim}}f(x)}{\underset{x\to a}{\text{lim}}g(x)}}}\\ 7.& \underset{x\to a}{\text{lim}}{(f(x))}^n={[\underset{x\to a}{\text{lim}}f(x))]}^n\\ 8.& \underset{x\to a}{\text{lim}}\sqrt[n]{f(x)}=\sqrt[n]{\underset{x\to \mathrm{\infty }}{\text{lim}}f(x)},\text{dengan}\underset{x\to a}{\text{lim}}f(x)\ge 0\\ & \text{dan}n\text{genap}\end{array}\end{array}$

Fungsi Eksponen

  $\text{A. Bilangan Pangkat Positif}$

Misalkan diketahui bahwa $a$ adalah suatu bilangan tidak nol dan $m$ adalah bilangan asli, maka bilangan ekponen atau bilangan berpangkat dedefinisikan dengan:

$a^m=\underset{m}{\underbrace{a\times a\times \times a\times ...\times a}}$

$\begin{array}{rl}\text{Bilangan}& :\\ a& \text{disebut basis atau bilangan pokok}\\ n& \text{disebut sebagai bilangan pangkat/eksponen}\end{array}$.

$\text{ CONTOH SOAL}$.

$(1).3^4=3\times 3\times 3\times 3=81$

$(2).5^4=5\times 5\times 5\times 5=625$

$(3).2^6=2\times 2\times 2\times 2\times 2\times 2=64$

$(4).6^7=6\times 6\times 6\times 6\times 6\times 6\times 6=279936$

$(5).(-3{)}^3=(-3)\times (-3)\times (-3)=-27$

$(6).(-2{)}^4=(-2)\times (-2)\times (-2)\times (-2)=16$

$(7).{\left({\displaystyle \frac{1}{5}}\right)}^3=\left({\displaystyle \frac{1}{5}}\right)\times \left({\displaystyle \frac{1}{5}}\right)\times \left({\displaystyle \frac{1}{5}}\right)={\displaystyle \frac{1}{125}}$

$(8).{(-{\displaystyle \frac{1}{2}})}^3=(-{\displaystyle \frac{1}{2}})\times (-{\displaystyle \frac{1}{2}})\times (-{\displaystyle \frac{1}{2}})=-{\displaystyle \frac{1}{8}}$

$\text{B. Sifat-Sifat Bilangan Pangkat Positif}$

$\begin{array}{r}\\ 1.& a^m.a^n=a^{m+n}\\ 2.& a^m:a^n=a^{m-n}\\ 3.& {\left(a^m\right)}^n=a^{m.n},\text{syarat}a\ne 0\\ 4.& {\left(ab\right)}^n=a^n.b^n\\ 5.& {\left(\frac{a}{b}\right)}^n=\frac{a^n}{b^n},\text{syarat}b\ne 0\end{array}$

Beberpa hal yang perlu diketahui juga, yaitu

$\begin{array}{r}\\ 1.& (a+b{)}^2=a^2+2ab+b^2\\ 2.& (a+b{)}^3=a^3+3a^{2}b+3a{b}^2+b^3\\ 3.& {(a+\frac{1}{a})}^2=a^2+2+{\displaystyle \frac{1}{a^2},\text{syarat}a\ne 0}\end{array}$

$\text{ CONTOH SOAL}$

$\begin{array}{r}\\ (1).& 2^6\times 2^4\times 2^7=2^{6+4+7}=2^{17}\\ (2).& 2^5\times 3^5\times 7^5={(2.3.7)}^5={\left(42\right)}^5\\ (3).& {\displaystyle \frac{a^3.a^7.a^6}{a^9}={\displaystyle \frac{a^{3+7+6}}{a^9}=\frac{a^{16}}{a^9}=a^{16-9}=a^7,\text{syarat}a\ne 0}}\end{array}$

$\begin{array}{rl}(4).{\displaystyle \frac{3^7{.7}^3.2}{{\left(42\right)}^3}}& =\frac{2^1{.3}^7{.7}^3}{{\left(2.3.7\right)}^3}=\frac{2^1{.3}^7{.7}^3}{2^3{.3}^3{.7}^3}\\ & =2^{1-3}{.3}^{7-3}{.7}^{3-3}=2^{-2}{.3}^4{.7}^0\\ & =\frac{1}{2^2}{.3}^4.1=\frac{3^4}{2^2}\end{array}$

$\begin{array}{rl}(5).{\displaystyle \frac{2^{2020}+2^{2021}+2^{2022}}{7}}& ={\displaystyle \frac{{1.2}^{2020}+2^1{.2}^{2020}+2^2{.2}^{2020}}{7}}\\ & =\frac{(1+2+4){.2}^{2020}}{7}\\ & =\frac{{7.2}^{2020}}{7}\\ & =2^{2020}\end{array}$

$\begin{array}{rl}(6){\displaystyle \frac{{\left(2^{n+2}\right)}^2-2^2{.2}^{2n}}{2^n{.2}^{n+2}}}& ={\displaystyle \frac{2^{2(n+2)}-2^2{.2}^{2n}}{2^n{.2}^n{.2}^2}}\\ & ={\displaystyle \frac{2^{2n}{.2}^{2.2}-2^2{.2}^{2n}}{2^{n+n}{.2}^2}}\\ & ={\displaystyle \frac{2^{2n}(2^4-2^2)}{2^{2n}{.2}^2}}\\ & ={\displaystyle \frac{(2^4-2^2)}{2^2}=\frac{16-4}{4}}\\ & ={\displaystyle \frac{12}{4}=3}\end{array}$

$\text{C. Bentuk Akar}$

Bilangan bentuk akar di sini adalah kebalikan dari bilangan bentuk pangkat. Bilangan bentuk akar selanjutnya disebut bilangan irasional. Sebagai contoh $\sqrt{2}$, $\sqrt{3}$, $\sqrt{8}$, $\sqrt[3]{3}$, $\sqrt[3]{4}$, $\sqrt[3]{7}$ dan tapi ingat $\sqrt{4}$ dan  $\sqrt[3]{8}$ serta  $\sqrt[3]{27}$ adalah bukan bentuk akar, karena nantinya akan menghasilkan masing-masing 2 dan 3 serta 3.

$\begin{array}{rl}& \\ 1.& a^{\frac{1}{n}}=\sqrt[n]{a}\\ 2.& a^{\frac{m}{n}}=\sqrt[n]{a^m}\\ 3.& a^{\frac{1}{2}}=\sqrt[2]{a^1}=\sqrt{a}\end{array}$.

$\text{Cara membaca}$.

$\begin{array}{rl}1.& \sqrt[n]{p}\mathbf{\text{dibaca}}\text{akar pangkat n dari p}\\ 2.& \sqrt[n]{p^2}\mathbf{\text{dibaca}}\text{akar pangkat n dari p kuadrat}\\ 3.& \sqrt[n]{p^3}\mathbf{\text{dibaca}}\text{akar pangkat n dari p pangkat tiga}\\ 4.& \sqrt{p}\mathbf{\text{dibaca}}\text{akar dari p}\text{atau}\\ & \text{akar kuadrat dari p}\\ & \text{ingat bahwa}:\sqrt{p}=\sqrt[2]{p}\end{array}$.

$\begin{array}{rl}\text{Defini}& \text{si}\\ \text{Jika}& a\text{dan}b\text{bilangan real dan}\\ & n\text{bilangan bulat positif, maka}:\\ & a^n=b\iff \sqrt[n]{b}=a\\ \text{keter}& \text{angan}:\\ \sqrt[n]{b}& \text{disebut}\mathbf{\text{akar (radikal)}}\\ b& \text{disebut}\mathbf{\text{radikan}}\\ & \text{(bilangan pokok yang ditarik akarnya)}\\ n& \text{disebut}\mathbf{\text{indeks}}\\ & (\text{pangkat akar})\end{array}$.

$\text{C.1 Bilangan Pangkat Pecahan}$.

Operasi Bilangan pangkat pecahan sama dengan operasi pangkat bilangan bulat.

$\text{ CONTOH SOAL}$.

$\begin{array}{l}\\ 1.& a^{{.}^{\frac{1}{2}}}\times a^{{.}^{\frac{1}{3}}}=a^{{.}^{\frac{1}{2}+\frac{1}{3}}}=a^{{.}^{\frac{5}{6}}}\\ 2.& a^{{.}^{\frac{1}{5}}}:a^{{.}^{\frac{1}{3}}}=a^{{.}^{\frac{1}{5}-\frac{1}{3}}}=a^{{.}^{-\frac{2}{15}}}\\ 3.& {\left(a^{{.}^{\frac{2}{5}}}\right)}^{\frac{4}{7}}=a^{{.}^{\frac{8}{35}}}\\ 4.& {81}^{{.}^{\frac{1}{2}}}={\left(9^2\right)}^{{.}^{\frac{1}{2}}}=9^1=9\\ 5.& {27}^{{.}^{-\frac{2}{3}}}={\left(3^3\right)}^{{.}^{-\frac{2}{3}}}={\left(3\right)}^{-2}={\displaystyle \frac{1}{3^2}=\frac{1}{9}}\end{array}$.

$\begin{array}{l}\\ 6.& \text{Sederhanakanlah bentuk berikut dan}\\ & \text{nyatakan hasilnya dalam pangkat positif}\\ \\ & \text{a}.\left(3p^{{.}^{\frac{5}{3}}}{q}^{{.}^{-\frac{3}{4}}}\right)\left(2p^{{.}^{-\frac{2}{3}}}{q}^{{.}^{\frac{5}{4}}}\right)\\ \\ & \text{b}.{\displaystyle \frac{\left(8p^{{.}^{\frac{2}{3}}}{q}^0{r}^{{.}^{-\frac{1}{2}}}\right)}{\left(4p^{{.}^{-\frac{1}{2}}}{q}^{{.}^{-\frac{1}{3}}}r\right)}}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}\text{a}.& \left(3p^{{.}^{\frac{5}{3}}}{q}^{{.}^{-\frac{3}{4}}}\right)\left(2p^{{.}^{-\frac{2}{3}}}{q}^{{.}^{\frac{5}{4}}}\right)\\ & =3.2.p^{{.}^{\frac{5}{3}+(-\frac{2}{3})}}.q^{{.}^{-\frac{3}{4}+\frac{5}{4}}}\\ & =6.p^{{.}^{\frac{3}{3}}}{q}^{{.}^{\frac{2}{4}}}\\ & =6p{q}^{{.}^{\frac{1}{2}}}\end{array}\\ & \begin{array}{rl}\text{b}.& {\displaystyle \frac{\left(8p^{{.}^{\frac{2}{3}}}{q}^0{r}^{{.}^{-\frac{1}{2}}}\right)}{\left(4p^{{.}^{-\frac{1}{2}}}{q}^{{.}^{-\frac{1}{3}}}r\right)}}\\ & =2.p^{{.}^{\frac{2}{3}-(-\frac{1}{2})}}{q}^{{.}^0-(-\frac{1}{3})}{r}^{{.}^{-\frac{1}{2}-1}}\\ & =2p^{{.}^{\frac{2}{3}+\frac{1}{2}}}{q}^{{.}^{\frac{1}{3}}}{r}^{{.}^{-\frac{3}{2}}}\\ & =2p^{{.}^{\frac{4+3}{6}}}{q}^{{.}^{\frac{1}{3}}}.r^{{.}^{-\frac{3}{2}}}\\ & =2p^{{.}^{\frac{7}{6}}}{q}^{{.}^{\frac{1}{3}}}.r^{{.}^{-\frac{3}{2}}}\\ & ={\displaystyle \frac{2p^{{.}^{\frac{7}{6}}}{q}^{{.}^{\frac{1}{3}}}}{r^{{.}^{\frac{3}{2}}}}}\end{array}\end{array}$.

$\begin{array}{l}\\ 7.& \text{Sederhanakanlah bentuk berikut dan}\\ & \text{nyatakan hasilnya dalam pangkat positif}\\ & \text{a}.{\left({\displaystyle \frac{p^{3n+1}{q}^n}{p^{3n+4}{q}^{4n}}}\right)}^{\frac{1}{3}}\\ & \text{b}.{\left({\displaystyle \frac{p^{-2}{q}^3}{p^4{q}^{-3}}}\right)}^{-\frac{1}{2}}{\left({\displaystyle \frac{p^4{q}^{-5}}{pq}}\right)}^{-\frac{1}{3}}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}\text{a}.& {\left({\displaystyle \frac{p^{3n+1}{q}^n}{p^{3n+4}{q}^{4n}}}\right)}^{\frac{1}{3}}\\ & ={\left(p^{(3n+1)-(3n+4)}{q}^{n-4n}\right)}^{\frac{1}{3}}\\ & ={\left(p^{-3}{q}^{-3n}\right)}^{\frac{1}{3}}\\ & =p^{-3.\frac{1}{3}}{q}^{-3n.\frac{1}{3}}\\ & =p^{-1}{q}^{-n}\\ & ={\displaystyle \frac{1}{p{q}^n}}\end{array}\\ & \begin{array}{rl}\text{b}.& {\left({\displaystyle \frac{p^{-2}{q}^3}{p^4{q}^{-3}}}\right)}^{-\frac{1}{2}}{\left({\displaystyle \frac{p^4{q}^{-5}}{pq}}\right)}^{-\frac{1}{3}}\\ & =\left({\displaystyle \frac{p^{-2.(-\frac{1}{2})}{q}^{3.(-\frac{1}{2})}}{p^{4.(-\frac{1}{2})}{q}^{-3.(-\frac{1}{2})}}}\right)\left({\displaystyle \frac{p^{4.(-\frac{1}{3})}{q}^{-5.(-\frac{1}{3})}}{p^{{.}^{-\frac{1}{3}}}{q}^{{.}^{-\frac{1}{3}}}}}\right)\\ & ={\displaystyle \frac{p^1{q}^{{.}^{-\frac{3}{2}}}}{p^{-2}{q}^{{.}^{\frac{3}{2}}}}\times \frac{p^{{.}^{-\frac{4}{3}}}{q}^{{.}^{\frac{5}{3}}}}{p^{{.}^{-\frac{1}{3}}}{q}^{{.}^{-\frac{1}{3}}}}}\\ & =p^{1-(-2)+(-\frac{4}{3})-(-\frac{1}{3})}{q}^{-\frac{3}{2}-\frac{3}{2}+\frac{5}{3}-(-\frac{1}{3})}\\ & =p^{3-\frac{3}{3}}{q}^{-\frac{6}{2}+\frac{6}{3}}\\ & =p^{3-1}{q}^{-3+2}\\ & =p^2{q}^{-1}\\ & ={\displaystyle \frac{p^2}{q}}\end{array}\end{array}$

$\begin{array}{l}\\ 8.& \text{Jabarkanlah bentuk}\\ & {(2m^{{.}^{\frac{3}{2}}}+n^{{.}^{\frac{3}{4}}})}^2\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}& {(2m^{{.}^{\frac{3}{2}}}+n^{{.}^{\frac{3}{4}}})}^2\\ & ={\left(2m^{{.}^{\frac{3}{2}}}\right)}^2+2\left(2m^{{.}^{\frac{3}{2}}}\right)\left(n^{{.}^{\frac{3}{4}}}\right)+{\left(n^{{.}^{\frac{3}{4}}}\right)}^2\\ & \text{INGAT}:{(A+B)}^2=A^2+2AB+B^2\\ & =2^2{m}^{{.}^{\frac{3.2}{2}}}+2.2.m^{{.}^{\frac{3}{2}}}{n}^{{.}^{\frac{3}{4}}}+n^{{.}^{\frac{3.2}{4}}}\\ & =4m^3+4m^{{.}^{\frac{3}{2}}}{n}^{{.}^{\frac{3}{4}}}+n^{{.}^{\frac{3}{2}}}\end{array}\end{array}$.

$\begin{array}{l}\\ 9.& \text{Jabarkanlah bentuk}\\ & {(2m^{{.}^{\frac{3}{2}}}-n^{{.}^{\frac{3}{4}}})}^3\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}& {(2m^{{.}^{\frac{3}{2}}}-n^{{.}^{\frac{3}{4}}})}^3\\ & ={\left(2m^{{.}^{\frac{3}{2}}}\right)}^3-3{\left(2m^{{.}^{\frac{3}{2}}}\right)}^2\left(n^{{.}^{\frac{3}{4}}}\right)+3\left(2m^{{.}^{\frac{3}{2}}}\right){\left(n^{{.}^{\frac{3}{4}}}\right)}^2-{\left(n^{{.}^{\frac{3}{4}}}\right)}^3\\ & \text{INGAT}:{(A-B)}^3=A^3-3A^{2}B+3A{B}^2-B^3\\ & =2^3{m}^{{.}^{\frac{3.3}{2}}}-{3.2}^2.m^{{.}^{\frac{3.2}{2}}}{n}^{{.}^{\frac{3}{4}}}+3.2.m^{{.}^{\frac{3}{2}}}{n}^{{.}^{\frac{3.2}{4}}}-n^{{.}^{\frac{3.3}{4}}}\\ & =8m^{{.}^{\frac{9}{2}}}-12m^3{n}^{{.}^{\frac{3}{4}}}+6m^{{.}^{\frac{3}{2}}}{n}^{{.}^{\frac{3}{2}}}-n^{{.}^{\frac{9}{2}}}\end{array}\end{array}$.

$\begin{array}{l}\\ 10.& \text{Jabarkanlah bentuk berikut}\\ \\ & \text{a}.(2p^{{.}^{\frac{1}{2}}}-3q^{{.}^{\frac{1}{2}}})(p^{{.}^{\frac{1}{2}}}+4q^{{.}^{\frac{1}{2}}})\\ \\ & \text{b}.(p^{{.}^{\frac{1}{3}}}-q^{{.}^{\frac{1}{3}}})(p^{{.}^{\frac{2}{3}}}+p^{{.}^{\frac{1}{3}}}{q}^{{.}^{\frac{1}{3}}}+q^{{.}^{\frac{2}{3}}})\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}\text{a}.& (2p^{{.}^{\frac{1}{2}}}-3q^{{.}^{\frac{1}{2}}})(p^{{.}^{\frac{1}{2}}}+4q^{{.}^{\frac{1}{2}}})\\ & =2{\left(p^{{.}^{\frac{1}{2}}}\right)}^2+2.4.p^{{.}^{\frac{1}{2}}}{q}^{{.}^{\frac{1}{2}}}-3q^{{.}^{\frac{1}{2}}}.p^{{.}^{\frac{1}{2}}}-3.4.{\left(q^{{.}^{\frac{1}{2}}}\right)}^2\\ & =2p^1+8q^{{.}^{\frac{1}{2}}}{q}^{{.}^{\frac{1}{2}}}-3p^{{.}^{\frac{1}{2}}}{q}^{{.}^{\frac{1}{2}}}-12.q^1\\ & =2p+5(pq{)}^{{.}^{\frac{1}{2}}}-12q\end{array}\\ & \begin{array}{rl}\text{b}.& (p^{{.}^{\frac{1}{3}}}-q^{{.}^{\frac{1}{3}}})(p^{{.}^{\frac{2}{3}}}+p^{{.}^{\frac{1}{3}}}{q}^{{.}^{\frac{1}{3}}}+q^{{.}^{\frac{2}{3}}})\\ & =p^{{.}^{\frac{1+2}{3}}}+{\left(p^{{.}^{\frac{1}{3}}}\right)}^2{q}^{{.}^{\frac{1}{3}}}+p^{{.}^{\frac{1}{3}}}{q}^{{.}^{\frac{2}{3}}}-p^{{.}^{\frac{2}{3}}}{q}^{{.}^{\frac{1}{3}}}-p^{{.}^{\frac{1}{3}}}{\left(q^{{.}^{\frac{1}{3}}}\right)}^2-q^{{.}^{\frac{1+2}{3}}}\\ & =p^1+0+0-q^1\\ & =p-q\end{array}\end{array}$

DAFTAR PUSTAKA

1. Sembiring, S., Zulkifli, M., Marsito, Rusdi, I. 2016. Matematika untuk Siswa SMA/MA Kelas X kelompok Peminatan Matematika dan Ilmu-Ilmu Alam. Bandung: SRIKANDI EMPAT WIDYA UTAMA.

Persamaan Garis Singgung Lingkaran (PGSL)

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

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

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

Kurang lebih dengan penjelasan yang sama dan persamaan garisnya di rumuskan 

$\{\begin{array}{ll}(p,q)& \equiv (x_1-p)(x-p)+(y_1-q)(y-q)=r^2\\ (p,q)& \equiv px+qy+{\displaystyle \frac{1}{2}A(p+x)+{\displaystyle \frac{1}{2}B(q+y)+C=0}}\end{array}$.

$\text{C. PGSL dengan gradien m}$

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

Contoh 9 Vektor

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

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

$\begin{array}{l}\\ 42.& \text{Supaya vektor}\overrightarrow{a}=\left(\begin{array}{c}x\\ 4\\ 7\end{array}\right),\text{dan}\overrightarrow{b}=\left(\begin{array}{c}6\\ y\\ 14\end{array}\right),\\ & \text{segaris, harga}x-y=....\\ & \begin{array}{l}\\ \text{a}.-5& & \text{d}.4\\ \text{b}.-2& \text{c}.3& \text{e}.6\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}\text{Dikethui}& \text{bahwa}:\\ \text{vektor}& \overrightarrow{a}\text{dan}\overrightarrow{b}\text{segaris},\text{maka}\\ m\overrightarrow{a}& =\overrightarrow{b}\\ \text{dengan}& m\text{adalah skalar/faktor pengali}\\ m\left(\begin{array}{c}x\\ 4\\ 7\end{array}\right)& =\left(\begin{array}{c}6\\ y\\ 14\end{array}\right)\\ \text{di}\text{dapat}& \text{kan}\{\begin{array}{ll}mx& =6...................(1)\\ 4m& =y....................(2)\\ 7m& =14...................(3)\end{array}\\ \text{Dari per}& \text{samaan}(3)\text{akan}\\ \text{didapat}& \text{kan nilai}m=2\\ \text{maka}& \text{akan didapatkan juga}\{\begin{array}{ll}x& =3\\ y& =8\end{array}\\ \text{sehin}& \text{gga nilai dari}\\ x-y& =3-8\\ & =-5\end{array}\end{array}$.

$\begin{array}{l}\\ 43.& \text{Diketahui}O\text{titik pangkal}A(0,1,2)\\ & \text{dan}B(3,4,5),\text{maka luas segitiga}\\ & OAB\text{sama dengan}....\\ & \begin{array}{l}\\ \text{a}.3\sqrt{6}\\ \text{b}.{\displaystyle \frac{2}{3}\sqrt{6}}\\ \text{c}.{\displaystyle \frac{4}{3}\sqrt{6}}& \\ \text{d}.{\displaystyle \frac{3}{2}\sqrt{6}}\\ \text{e}.2\sqrt{6}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & \text{Coba perhatikanlah ilustrasi berikut}\\ & \begin{array}{rl}& \text{Misalkan luas segitiga}{\displaystyle \frac{1}{2}|\overrightarrow{p}\times \overrightarrow{q}|,}\\ & \text{dengan}\\ & \{\begin{array}{ll}\overrightarrow{p}& =\overline{OA}=\left(\begin{array}{c}0\\ 1\\ 2\end{array}\right)\\ \overrightarrow{q}& =\overline{OB}=\left(\begin{array}{c}3\\ 4\\ 5\end{array}\right)\end{array}\end{array}\\ & \begin{array}{rl}\overrightarrow{p}& \times \overrightarrow{q}=\left|\begin{array}{ccc}\overrightarrow{i}& \overrightarrow{j}& \overrightarrow{k}\\ x_1& y_1& z_1\\ x_2& y_2& z_2\end{array}\right|\\ & =(y_1{z}_2-z_1{y}_2)\overrightarrow{i}-(x_1{z}_2-z_1{x}_2)\overrightarrow{j}+(x_1{y}_2-y_1{x}_2)\overrightarrow{k}\\ & =\left|\begin{array}{ccc}\overrightarrow{i}& \overrightarrow{j}& \overrightarrow{k}\\ 0& 1& 2\\ 3& 4& 5\end{array}\right|\\ & =(5-8)\overrightarrow{i}-(0-6)\overrightarrow{j}+(0-3)\overrightarrow{k}\\ & =-3\overrightarrow{i}+6\overrightarrow{j}-3\overrightarrow{k}\\ & \text{Sehingga}\\ & |\overrightarrow{p}\times \overrightarrow{q}|=\sqrt{(-3{)}^2+6^2+(-3{)}^2}\\ & =\sqrt{9+36+9}=\sqrt{54}=3\sqrt{6}\\ & \text{Maka luas segi tiganya adalah}:\\ & \text{luas}\mathrm{△}ABC={\displaystyle \frac{1}{2}|\overrightarrow{p}\times \overrightarrow{q}|={\displaystyle \frac{1}{2}\left(3\sqrt{6}\right)={\displaystyle \frac{3}{2}\sqrt{6}}}}\end{array}\end{array}$.

$\begin{array}{l}\\ 44.& \text{Proyeksi skalar ortogonal}\\ & \overrightarrow{a}=2\overrightarrow{i}-3\overrightarrow{j}+6\overrightarrow{k},\text{pada}\\ & \overrightarrow{b}=\overrightarrow{i}+2\overrightarrow{j}+2\overrightarrow{k}\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.{\displaystyle \frac{4}{3}}& & & & & \text{d}.& {\displaystyle \frac{16}{3}}\\ \\ \text{b}.{\displaystyle \frac{8}{3}}& & \text{c}& {\displaystyle \frac{10}{3}}& & \text{e}.& {\displaystyle \frac{20}{3}}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & \text{Diketahui bahwa}\\ & \{\begin{array}{ll}\overrightarrow{a}& =(2,-3,6)\\ \overrightarrow{b}& =(1,2,2)\end{array}\\ & \text{Selanjutnya}\\ & \begin{array}{r}\\ \left|\overrightarrow{c}\right|& =\left|{\displaystyle \frac{\overrightarrow{a}\overrightarrow{b}}{\left|\overrightarrow{b}\right|}}\right|\\ & ={\displaystyle \frac{\left(\begin{array}{c}2\\ -3\\ 6\end{array}\right)\left(\begin{array}{c}1\\ 2\\ 2\end{array}\right)}{\sqrt{1^2+2^2+2^2}}}\\ & ={\displaystyle \frac{2-6+12}{\sqrt{9}}=\frac{8}{3}}\end{array}\end{array}$

$\begin{array}{l}\\ 45.& \text{Diketahui vektor}\overrightarrow{a}=3\overrightarrow{i}-4\overrightarrow{j}+p\overrightarrow{k}\\ & \text{dan}\overrightarrow{b}=2\overrightarrow{i}+2\overrightarrow{j}-3\overrightarrow{k}.\text{Jika panjang}\\ & \text{proyeksi vektor}\overrightarrow{a}\text{pada}\overrightarrow{b}\text{adalah}\\ & {\displaystyle \frac{4}{\sqrt{17}},\text{maka nilai}p\text{adalah}....}\\ & \begin{array}{l}\\ \text{a}.{\displaystyle -2}& & & & & \text{d}.& {\displaystyle 2}\\ \\ \text{b}.{\displaystyle -1}& & \text{c}& {\displaystyle 1}& & \text{e}.& {\displaystyle 3}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & \text{Panjang proyeksi skalar vektor}\overrightarrow{a}\text{pada}\overrightarrow{b}\\ & \begin{array}{r}\\ \left|\overrightarrow{c}\right|& ={\displaystyle \frac{\overrightarrow{a}\overrightarrow{b}}{\left|\overrightarrow{b}\right|}}\\ {\displaystyle \frac{4}{\sqrt{17}}}& ={\displaystyle \frac{\left(\begin{array}{c}3\\ -4\\ p\end{array}\right)\left(\begin{array}{c}2\\ 2\\ -3\end{array}\right)}{\sqrt{2^2+2^2+(-3{)}^2}}}\\ {\displaystyle \frac{4}{\sqrt{17}}}& ={\displaystyle \frac{6-8-3p}{\sqrt{17}}}\\ 4& =-2-3p\\ 6& =-3p\\ -3p& =6\\ p& =-2\end{array}\end{array}$

Contoh 8 Vektor

Contoh soal sebelumnya di sini Contoh Soal 7

$\begin{array}{l}\\ 34.& \text{Vektor satuan untuk}\overrightarrow{a}=\left(\begin{array}{c}2\\ -1\\ 4\end{array}\right)\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.{\displaystyle \frac{1}{2}\sqrt{3}\left(\begin{array}{c}2\\ -1\\ 4\end{array}\right)}& \\ \text{b}.{\displaystyle \frac{1}{3}\sqrt{3}\left(\begin{array}{c}2\\ -1\\ 4\end{array}\right)}& \\ \text{c}.{\displaystyle \frac{1}{5}\sqrt{5}\left(\begin{array}{c}2\\ -1\\ 4\end{array}\right)}& \\ \text{d}.{\displaystyle \frac{1}{7}\sqrt{7}\left(\begin{array}{c}2\\ -1\\ 4\end{array}\right)}\\ \text{e}.{\displaystyle \frac{1}{21}\sqrt{21}\left(\begin{array}{c}2\\ -1\\ 4\end{array}\right)}\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}\text{Vektor satuan}\overrightarrow{a}& \text{adalah}{\overrightarrow{e}}_{\overrightarrow{a}},\text{yaitu}:\\ {\overrightarrow{e}}_{\overrightarrow{a}}& ={\displaystyle \frac{\overrightarrow{a}}{\left|\overrightarrow{a}\right|}}\\ & ={\displaystyle \frac{\left(\begin{array}{c}2\\ -1\\ 4\end{array}\right)}{\sqrt{2^2+(-1{)}^2+4^2}}}\\ & ={\displaystyle \frac{1}{\sqrt{21}}\left(\begin{array}{c}2\\ -1\\ 4\end{array}\right)}\\ & ={\displaystyle \frac{1}{21}\sqrt{21}\left(\begin{array}{c}2\\ -1\\ 4\end{array}\right)}\end{array}\end{array}$.

$\begin{array}{l}\\ 35.& \text{Posisi suatu titik dalam ruang saat }\\ & \text{waktu}t\text{ditunjukkan oleh vektor}\\ & \left(\begin{array}{c}t\\ t^2\\ -t\end{array}\right).\text{Jika pada saat}t=1\text{titik }\\ & \text{tersebut berada di titik P dan pada}\\ & \text{saat}t=2\text{titik tersebut berada }\\ & \text{di titik Q, maka jarak titik P dari Q}\\ & \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.\sqrt{24}-\sqrt{3}& & \text{d}.\sqrt{11}\\ \text{b}.2-\sqrt{2}& \text{c}.3& \text{e}.\sqrt{43}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{a}}\\ & \begin{array}{rl}\left|\overrightarrow{PQ}\right|& =\sqrt{(x_q-x_p{)}^2+(y_q-y_p{)}^2+(y_q-y_p{)}^2}\\ & =\sqrt{(2-1{)}^2+(2^2-1^2{)}^2+((-2)-(-1){)}^2}\\ & =\sqrt{1^2+3^2+(-1{)}^2}\\ & =\sqrt{1+9+1}\\ & =\sqrt{11}\end{array}\end{array}$

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

$\begin{array}{l}\\ 37.& \text{Jika diketahui titik}A(2,-1,4),B(4,1,3),\\ & \text{ dan}C(2,0,5),\text{maka}\sin \mathrm{\angle }(\overline{AB},\overline{AC})=....\\ & \begin{array}{l}\\ \text{a}.{\displaystyle \frac{1}{7}\sqrt{5}}& & \text{d}.{\displaystyle \frac{1}{6}\sqrt{3}}\\ \\ \text{b}.{\displaystyle \frac{1}{6}\sqrt{34}}& \text{c}.{\displaystyle \frac{2}{3}\sqrt{2}}& \text{e}.{\displaystyle \frac{1}{6}\sqrt{2}}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& \cos \mathrm{\angle }(\overline{AB},\overline{AC})={\displaystyle \frac{\overline{AB}.\overline{AC}}{\left|\overline{AB}\right|.\left|\overline{AC}\right|}}\\ & ={\displaystyle \frac{(\overrightarrow{b}-\overrightarrow{a}).(\overrightarrow{c}-\overrightarrow{a})}{\sqrt{x_{(\overrightarrow{b}-\overrightarrow{a})}^2+y_{(\overrightarrow{b}-\overrightarrow{a})}^2+z_{(\overrightarrow{b}-\overrightarrow{a})}^2}.\sqrt{x_{(\overrightarrow{c}-\overrightarrow{a})}^2+y_{(\overrightarrow{c}-\overrightarrow{a})}^2+z_{(\overrightarrow{c}-\overrightarrow{a})}^2}}}\\ & ={\displaystyle \frac{\left(\begin{array}{c}4-2\\ 1+1\\ 3-4\end{array}\right).\left(\begin{array}{c}2-2\\ 0+1\\ 5-4\end{array}\right)}{\sqrt{(4-2{)}^2+(1+1{)}^2+(3-4{)}^2}.\sqrt{(2-2{)}^2+(0+1{)}^2+(5-4{)}^2}}}\\ & ={\displaystyle \frac{2.0+2.1+-1.1}{\sqrt{4+4+1}.\sqrt{0+1+1}}}\\ & ={\displaystyle \frac{1}{3\sqrt{2}}}\\ & ={\displaystyle \frac{1}{6}\sqrt{2}}\end{array}\\ & \begin{array}{rl}& \text{Sehingga},\\ & \sin \mathrm{\angle }(\overline{AB},\overline{AC})\\ & =\sqrt{1-{\cos }^2\mathrm{\angle }(\overline{AB},\overline{AC})}\\ & =\sqrt{1-{\left({\displaystyle \frac{1}{6}\sqrt{2}}\right)}^2}\\ & =\sqrt{1-{\displaystyle \frac{2}{36}}}\\ & =\sqrt{{\displaystyle \frac{34}{36}}}\\ & ={\displaystyle \frac{1}{6}\sqrt{34}}\end{array}\end{array}$.

$\begin{array}{l}\\ 38.& \text{Diketahui segitiga ABC. Titik M }\\ & \text{di tengah AC, dan titik N pada BC}\\ & \text{Jika}\overrightarrow{AB}=\overrightarrow{c},\overrightarrow{AC}=\overrightarrow{b},\overrightarrow{BC}=\overrightarrow{a},\\ & \text{maka}\overrightarrow{MN}=....\\ & \begin{array}{l}\\ \text{a}.{\displaystyle \frac{1}{2}(\overrightarrow{b}-\overrightarrow{c})}& \\ \text{b}.{\displaystyle \frac{1}{2}(-\overrightarrow{b}+\overrightarrow{c})}& \\ \text{c}.{\displaystyle \frac{1}{2}(-\overrightarrow{a}+\overrightarrow{c})}\\ \text{d}.{\displaystyle \frac{1}{2}(\overrightarrow{a}-\overrightarrow{b})}\\ \text{e}.{\displaystyle \frac{1}{2}(-\overrightarrow{a}+\overrightarrow{b})}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}\overrightarrow{MN}& =\overrightarrow{MC}+\overrightarrow{CN}\\ & ={\displaystyle \frac{1}{2}\overrightarrow{AC}+{\displaystyle \frac{1}{2}(-\overrightarrow{BC})}}\\ & ={\displaystyle \frac{1}{2}(\overrightarrow{AC}-\overrightarrow{BC})}\\ & ={\displaystyle \frac{1}{2}(\overrightarrow{b}-\overrightarrow{a})\mathbf{\text{atau}}}\\ & ={\displaystyle \frac{1}{2}(-\overrightarrow{a}+\overrightarrow{b})}\end{array}\end{array}$.

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

$.\begin{array}{r}\\ & \begin{array}{rl}\text{koordinat}& {\text{titik A}}^{\prime }\\ & ={\displaystyle \frac{1}{2}(5+0,4-1,10+6)}\\ & =(\frac{5}{2},\frac{3}{2},8)\end{array}\\ & \begin{array}{rl}\text{Dalam se}& \text{gitiga ABC untuk titik berat Z }\\ \text{berlaku}& \text{ketentuan sebagai berikut}\\ AZ:Z{A}^{\prime }& =2:1\\ \overrightarrow{AZ}:\overrightarrow{Z{A}^{\prime }}& =2:1\\ \overrightarrow{OZ}& ={\displaystyle \frac{\overrightarrow{OA}+2\overrightarrow{O{A}^{\prime }}}{3}}\\ & ={\displaystyle \frac{\left(\begin{array}{c}1\\ 0\\ 2\end{array}\right)+2\left(\begin{array}{c}\frac{5}{2}\\ \frac{3}{2}\\ 8\end{array}\right)}{3}}\\ & =\left(\begin{array}{c}2\\ 1\\ 6\end{array}\right)\\ \text{Jadi},& \text{koordinat titik Z adalah}(2,1,6)\end{array}\end{array}$.

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

$.\begin{array}{|cc|}\hline \text{Diketahui}& \text{Proses Penyelesaian}\\ \begin{array}{rl}\overrightarrow{AM}:\overrightarrow{MB}& =3:2\\ \overrightarrow{m}& ={\displaystyle \frac{2\overrightarrow{a}+3\overrightarrow{b}}{5}}\\ & \\ & \\ & \\ & \\ & \\ & \\ & \\ & \\ & \\ & \\ & \end{array}& \begin{array}{rl}\overrightarrow{MC}& =\overrightarrow{c}-\overrightarrow{m}\\ & =\overrightarrow{c}-{\displaystyle \frac{2\overrightarrow{a}+3\overrightarrow{b}}{5}}\\ & ={\displaystyle \frac{5\overrightarrow{c}-2\overrightarrow{a}-3\overrightarrow{b}}{5}}\\ & ={\displaystyle \frac{5\left(\begin{array}{c}2\\ 3\\ 5\end{array}\right)-2\left(\begin{array}{c}4\\ -1\\ -2\end{array}\right)-3\left(\begin{array}{c}-6\\ 4\\ 3\end{array}\right)}{5}}\\ & ={\displaystyle \frac{\left(\begin{array}{c}20\\ 5\\ 20\end{array}\right)}{5}}\\ & =\left(\begin{array}{c}4\\ 1\\ 4\end{array}\right)\end{array}\\ \hline\end{array}$

Proyeksi Ortoganal Suatu Vektor di Dimensi Tiga

 Materinya sama dengan proyeksi ortogonal pada dimensi dua klik di sini

Uraian berikut sebagai pengingat saja

$\begin{array}{rl}▹& \mathbf{\text{Proyeksi skalar vektor}}\\ & \left|\overrightarrow{c}\right|={\displaystyle \frac{\overrightarrow{a}\bullet \overrightarrow{b}}{\left|\overrightarrow{b}\right|}}\\ ▹& \mathbf{\text{Vektor proyeksi ortogonal}}\\ & \overrightarrow{c}={\displaystyle \frac{\overrightarrow{a}\bullet \overrightarrow{b}}{{\left|\overrightarrow{b}\right|}^2}.\overrightarrow{b}}\end{array}$

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!

$\begin{array}{|cc|}\hline \mathrm{△}\text{OAC}& \mathrm{\angle }(\overrightarrow{a},\overrightarrow{b})\\ \begin{array}{rl}\cos \theta & ={\displaystyle \frac{\left|\overrightarrow{c}\right|}{\left|\overrightarrow{a}\right|}}\\ \iff \left|\overrightarrow{c}\right|& =\left|\overrightarrow{a}\right|\cos \theta ........(1)\end{array}& \begin{array}{rl}\cos \theta & ={\displaystyle \frac{\overrightarrow{a}\overrightarrow{b}}{\left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right|}........(2)}\end{array}\\ \hline\end{array}$. 

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

Dan penjelasan kedua berkaitan dengan vektor proyeksi ortogonalnya, yaitu:

$\begin{array}{rl}& \text{Perhatikan pula misal}\hat{c}\\ & \text{adalah vektor satuan dari}\overrightarrow{c}\text{dan}\overrightarrow{b},\\ & \text{maka}\\ & \begin{array}{rl}\overrightarrow{c}& =\left|\overrightarrow{c}\right|\hat{c}\end{array},\text{dan}\\ & \begin{array}{rl}\overrightarrow{b}& =\left|\overrightarrow{b}\right|\hat{b}=\left|\overrightarrow{b}\right|\hat{c}\end{array}\end{array}$.

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

$\text{CONTOH SOAL}$.

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

$\begin{array}{l}\\ 2.& \text{Diketahui}\overrightarrow{a}=\left(\begin{array}{c}-3\\ -2\\ m\end{array}\right)\text{dan}\overrightarrow{b}=\left(\begin{array}{c}2\\ -1\\ -2\end{array}\right).\\ & \text{Jika proyeksi skalar}\overrightarrow{a}\text{pada}\overrightarrow{b}\text{adalah}\\ & \text{bernilai}-{\displaystyle \frac{2}{3},\text{maka tentukan nilai}m}\\ \\ & \mathbf{\text{Jawab}}:\\ & \text{Misalkan proyeksi skalar}\overrightarrow{a}\text{pada}\overrightarrow{b}\text{adalah}\left|\overrightarrow{f}\right|,\\ & \text{maka}\\ & \begin{array}{rl}\left|\overrightarrow{f}\right|& ={\displaystyle \frac{\overrightarrow{a}\bullet \overrightarrow{b}}{\left|\overrightarrow{b}\right|}}\\ \iff -{\displaystyle \frac{2}{3}}& ={\displaystyle \frac{\left(\begin{array}{c}-3\\ -2\\ m\end{array}\right)\left(\begin{array}{c}2\\ -1\\ -2\end{array}\right)}{\sqrt{2^2+(-1{)}^2+(-2{)}^2}}}\\ \iff -{\displaystyle \frac{2}{3}}& ={\displaystyle \frac{-6+2-2m}{\sqrt{9}}=\frac{-4-2m}{3}}\\ \iff -2& =-4-2m\\ \iff 1& =2+m\\ \iff -m& =2-1=1\\ \iff m& =-1\end{array}\end{array}$.

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

DAFTAR PUSTAKA

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

 

Lanjutan 2 Materi Operasi Vektor Berdimensi Tiga

 $\text{G. Perbandingan Vektor }$.

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  $\overrightarrow{a}$  dan  $\overrightarrow{b}$ sebagai vektor posisi dari masing-masing titik A dan B dan vektor posisi titik T dapat ditentukan, yaitu:

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

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:

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

$\text{CONTOH SOAL}$.

$\begin{array}{l}\\ 1.& \text{Jika titik}A(12,12,0)\text{dan}B(6,6,12)\\ & \text{serta}P\text{membagi garis dengan}\\ & \overline{AP}:\overline{PB}=1:2.\text{Tentukanlah koordinat}\\ & \text{titik}P\text{jika},\\ & P\text{membagi di dalam}\\ & P\text{membagi di luar}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}\text{a}.& P\text{membagi di dalam}\\ & \overrightarrow{p}={\displaystyle \frac{n\overrightarrow{a}+m\overrightarrow{b}}{m+n}}\\ & ={\displaystyle \frac{2\overrightarrow{a}+1\overrightarrow{p}}{1+2}=\frac{2\left(\begin{array}{c}12\\ 12\\ 0\end{array}\right)+\left(\begin{array}{c}6\\ 6\\ 12\end{array}\right)}{3}}\\ & ={\displaystyle \frac{\left(\begin{array}{c}30\\ 30\\ 12\end{array}\right)}{3}=\left(\begin{array}{c}10\\ 10\\ 4\end{array}\right)}\\ \text{b}.& P\text{membagi di luar}\\ & \overrightarrow{p}={\displaystyle \frac{-n\overrightarrow{a}+m\overrightarrow{b}}{m-n}}\\ & ={\displaystyle \frac{-2\left(\begin{array}{c}12\\ 12\\ 0\end{array}\right)+\left(\begin{array}{c}6\\ 6\\ 12\end{array}\right)}{1-2}}\\ & =-\left(\begin{array}{c}-18\\ -18\\ 12\end{array}\right)=\left(\begin{array}{c}18\\ 18\\ -12\end{array}\right)\end{array}\end{array}$.

$\begin{array}{l}\\ 2.& \text{Jika titik}A(3,2,-1)\text{dan}B(1,-2,1)\\ & \text{dan}C(7,m-1,-5).\text{Tentukan nilai}m\\ & \text{agar ketiga titik itu segaris}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& \text{Diketahui}\\ & \overrightarrow{a}=\left(\begin{array}{c}3\\ 2\\ -1\end{array}\right),\overrightarrow{b}=\left(\begin{array}{c}1\\ -2\\ 1\end{array}\right),\left(\begin{array}{c}7\\ m-1\\ -5\end{array}\right)\\ & \text{Agar ketiga titik segaris, maka}\overline{AB}=k\overline{BC}\\ & \overrightarrow{b}-\overrightarrow{a}=k(\overrightarrow{c}-\overrightarrow{b})\\ & \left(\begin{array}{c}1\\ -2\\ 1\end{array}\right)-\left(\begin{array}{c}3\\ 2\\ -1\end{array}\right)=k(\begin{array}{c}7\\ m-1\\ -5\end{array}-\left(\begin{array}{c}1\\ -2\\ 1\end{array}\right))\\ & \iff \left(\begin{array}{c}-2\\ -4\\ 2\end{array}\right)=k\left(\begin{array}{c}6\\ m+1\\ -6\end{array}\right)\\ & \iff \{\begin{array}{ll}-2& =6k\\ -4& =k(m+1)\\ 2& =-6k\end{array}\iff -2=6k\\ & \iff k=-{\displaystyle \frac{1}{3}}\\ & \text{maka}\\ & \iff -4=k(m+1)\iff -4=-{\displaystyle \frac{1}{3}(m+1)}\\ & \iff 12=m+1\\ & \iff m=11\end{array}\end{array}$

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

 $\text{F. 4. Perkalian Silang Vektor (Pengayaan)}$.

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

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

$\begin{array}{rl}\overrightarrow{u}& \times \overrightarrow{v}=\overrightarrow{c}=\left|\begin{array}{ccc}\overrightarrow{i}& \overrightarrow{j}& \overrightarrow{k}\\ x_1& y_1& z_1\\ x_2& y_2& z_2\end{array}\right|\\ & =(y_1{z}_2-z_1{y}_2)\overrightarrow{i}-(x_1{z}_2-z_1{x}_2)\overrightarrow{j}+(x_1{y}_2-y_1{x}_2)\overrightarrow{k}\end{array}$

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

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

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

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

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

Pada hasil kali silang dua vektor berlaku

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

$\text{CONTOH SOAL}$.

$\begin{array}{l}\\ 1.& \text{Diketahui}\overrightarrow{a}=4\overrightarrow{i}+3\overrightarrow{j}\text{dan}\overrightarrow{b}=4\overrightarrow{i}-3\overrightarrow{k}\\ & \text{Tentukanlah hasil}\overrightarrow{a}\times \overrightarrow{b}\text{dan}\overrightarrow{b}\times \overrightarrow{a}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}\overrightarrow{a}& \times \overrightarrow{b}=\left|\begin{array}{ccc}\overrightarrow{i}& \overrightarrow{j}& \overrightarrow{k}\\ x_1& y_1& z_1\\ x_2& y_2& z_2\end{array}\right|\\ & =(y_1{z}_2-z_1{y}_2)\overrightarrow{i}-(x_1{z}_2-z_1{x}_2)\overrightarrow{j}+(x_1{y}_2-y_1{x}_2)\overrightarrow{k}\\ & =\left|\begin{array}{ccc}\overrightarrow{i}& \overrightarrow{j}& \overrightarrow{k}\\ 4& 3& 0\\ 4& 0& -3\end{array}\right|\\ & =(-9-0)\overrightarrow{i}-(-12-0)\overrightarrow{j}+(0-12)\overrightarrow{k}\\ & =-9\overrightarrow{i}+12\overrightarrow{j}-12\overrightarrow{k}\end{array}\\ & \begin{array}{rl}\overrightarrow{b}& \times \overrightarrow{a}=\left|\begin{array}{ccc}\overrightarrow{i}& \overrightarrow{j}& \overrightarrow{k}\\ x_2& y_2& z_2\\ x_1& y_1& z_1\end{array}\right|\\ & =\left|\begin{array}{ccc}\overrightarrow{i}& \overrightarrow{j}& \overrightarrow{k}\\ 4& 0& -3\\ 4& 3& 0\end{array}\right|\\ & =(0-(-9))\overrightarrow{i}-(0-(-12))\overrightarrow{j}+(12-0)\overrightarrow{k}\\ & =9\overrightarrow{i}-12\overrightarrow{j}+12\overrightarrow{k}\end{array}\end{array}$.

$\begin{array}{l}\\ 2.& \text{Diketahui}\overrightarrow{a}=6\overrightarrow{i}+2\overrightarrow{j}+10\overrightarrow{k}\text{dan}\overrightarrow{b}=4\overrightarrow{i}+\overrightarrow{j}+9\overrightarrow{k}\\ & \text{Tentukanlah hasil}\overrightarrow{a}\times \overrightarrow{b}\text{dan}\overrightarrow{b}\times \overrightarrow{a}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}\overrightarrow{a}& \times \overrightarrow{b}=\left|\begin{array}{ccc}\overrightarrow{i}& \overrightarrow{j}& \overrightarrow{k}\\ x_1& y_1& z_1\\ x_2& y_2& z_2\end{array}\right|\\ & =(y_1{z}_2-z_1{y}_2)\overrightarrow{i}-(x_1{z}_2-z_1{x}_2)\overrightarrow{j}+(x_1{y}_2-y_1{x}_2)\overrightarrow{k}\\ & =\left|\begin{array}{ccc}\overrightarrow{i}& \overrightarrow{j}& \overrightarrow{k}\\ 6& 2& 10\\ 4& 1& 9\end{array}\right|\\ & =(18-10)\overrightarrow{i}-(54-40)\overrightarrow{j}+(6-8)\overrightarrow{k}\\ & =8\overrightarrow{i}-14\overrightarrow{j}-2\overrightarrow{k}\end{array}\\ & \begin{array}{rl}\overrightarrow{b}& \times \overrightarrow{a}=\left|\begin{array}{ccc}\overrightarrow{i}& \overrightarrow{j}& \overrightarrow{k}\\ x_2& y_2& z_2\\ x_1& y_1& z_1\end{array}\right|\\ & =\left|\begin{array}{ccc}\overrightarrow{i}& \overrightarrow{j}& \overrightarrow{k}\\ 4& 1& 9\\ 6& 2& 10\end{array}\right|\\ & =(10-18)\overrightarrow{i}-(40-54)\overrightarrow{j}+(8-6)\overrightarrow{k}\\ & =-8\overrightarrow{i}+14\overrightarrow{j}+2\overrightarrow{k}\end{array}\end{array}$.

$\begin{array}{l}\\ 3.& \text{Tentukanlah luas segitiga}ABC\text{jika}\\ & \text{diketahui}A(2,1,-2),B(0,-1,0),\text{dan}\\ & C(-1,2,-1)\\ \\ & \mathbf{\text{Jawab}}:\\ & \text{Misalkan luas segitiga}{\displaystyle \frac{1}{2}|\overrightarrow{p}\times \overrightarrow{q}|,\text{dengan}}\\ & \{\begin{array}{ll}\overrightarrow{p}& =\overline{AB}=\overline{OB}-\overline{OA}=\left(\begin{array}{c}0\\ -1\\ 0\end{array}\right)-\left(\begin{array}{c}2\\ 1\\ -2\end{array}\right)=\left(\begin{array}{c}-2\\ -2\\ 2\end{array}\right)\\ \overrightarrow{q}& =\overline{AC}=\overline{OC}-\overline{OA}=\left(\begin{array}{c}-1\\ 2\\ -1\end{array}\right)-\left(\begin{array}{c}2\\ 1\\ -2\end{array}\right)=\left(\begin{array}{c}-3\\ 1\\ 1\end{array}\right)\end{array}\\ & \begin{array}{rl}\overrightarrow{p}& \times \overrightarrow{q}=\left|\begin{array}{ccc}\overrightarrow{i}& \overrightarrow{j}& \overrightarrow{k}\\ x_1& y_1& z_1\\ x_2& y_2& z_2\end{array}\right|\\ & =(y_1{z}_2-z_1{y}_2)\overrightarrow{i}-(x_1{z}_2-z_1{x}_2)\overrightarrow{j}+(x_1{y}_2-y_1{x}_2)\overrightarrow{k}\\ & =\left|\begin{array}{ccc}\overrightarrow{i}& \overrightarrow{j}& \overrightarrow{k}\\ -2& -2& 2\\ -3& 1& 1\end{array}\right|\\ & =(-2-2)\overrightarrow{i}-(-2-(-6))\overrightarrow{j}+(-2-6)\overrightarrow{k}\\ & =-4\overrightarrow{i}-4\overrightarrow{j}-8\overrightarrow{k}\\ & \text{Sehingga}\\ & |\overrightarrow{p}\times \overrightarrow{q}|=\sqrt{(-4{)}^2+(-4{)}^2+(-8{)}^2}\\ & =\sqrt{16+16+64}=\sqrt{96}=4\sqrt{6}\\ & \text{Maka luas segi tiganya adalah}:\\ & \text{luas}\mathrm{△}ABC={\displaystyle \frac{1}{2}|\overrightarrow{p}\times \overrightarrow{q}|={\displaystyle \frac{1}{2}\left(4\sqrt{6}\right)=2\sqrt{6}}}\end{array}\end{array}$

DAFTAR PUSTAKA

1. 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

 $\text{F. Operasi Vektor Dalam Ruang}$

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

$\text{F. 1. Penjumlahan dan Pengurangan}$.

$\begin{array}{rl}\text{Jika}& \text{diketahui sebagai misal}\\ \overline{u}& =a\overline{i}+b\overline{j}+c\overline{k}\text{dan}\\ \overline{v}& =p\overline{i}+q\overline{j}+r\overline{k}\\ \text{mak}& \text{a}\\ \mathbf{\text{Pen}}& \mathbf{\text{jumlahan dua vektor di atas adalah}}\\ \overline{u}+\overline{v}& =(a+p)\overline{i}+(b+q)\overline{j}+(c+r)\overline{k}\\ \mathbf{\text{dem}}& \mathbf{\text{ikian juga untuk pengurangan}}\\ \overline{u}-\overline{v}& =(a-p)\overline{i}+(b-q)\overline{j}+(c-r)\overline{k}\end{array}$.

$\text{CONTOH SOAL}$.

$\begin{array}{l}\\ 1.& \text{Jika diketahui}\overline{a}=\left(\begin{array}{c}1\\ 3\\ 7\end{array}\right)\text{dan}\overline{b}=\left(\begin{array}{c}8\\ -2\\ 0\end{array}\right)\\ & \text{Tentukanlah hasil dari}\\ & \text{a}.\overline{a}+\overline{b}\\ & \text{b}.\overline{a}-\overline{b}\\ \\ & \mathbf{\text{Jawab}}\\ & \begin{array}{rl}& \text{Diketahui bahwa}\\ & \overline{a}=\left(\begin{array}{c}1\\ 3\\ 7\end{array}\right)\text{dan}\overline{b}=\left(\begin{array}{c}8\\ -2\\ 0\end{array}\right),\\ & \text{maka}\\ & \overline{a}+\overline{b}=\left(\begin{array}{c}1\\ 3\\ 7\end{array}\right)+\left(\begin{array}{c}8\\ -2\\ 0\end{array}\right)\\ & =\left(\begin{array}{c}1+8\\ 3+(-2)\\ 7+0\end{array}\right)=\left(\begin{array}{c}9\\ 1\\ 7\end{array}\right)\\ & \text{Dan untuk}\overline{a}-\overline{b}\text{adalah}:\\ & \overline{a}-\overline{b}=\left(\begin{array}{c}1\\ 3\\ 7\end{array}\right)-\left(\begin{array}{c}8\\ -2\\ 0\end{array}\right)\\ & =\left(\begin{array}{c}1-8\\ 3-(-2)\\ 7-0\end{array}\right)=\left(\begin{array}{c}-7\\ 5\\ 7\end{array}\right)\end{array}\end{array}$.

$\text{F. 2. Perkalian Skalar dengan Vektor}$.

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

$\text{CONTOH SOAL}$.

$\begin{array}{ll}2.& \text{Jika}\overline{a}=\left(\begin{array}{c}2022\\ 2021\\ 2020\end{array}\right),\text{tentukanlah nilai}\\ & \text{dari}2\overline{a}\text{dan}-3\overline{a}\\ \\ & \mathbf{\text{Jawab}}\\ & \begin{array}{rl}2\overline{a}& =2\left(\begin{array}{c}2022\\ 2021\\ 2020\end{array}\right)=\left(\begin{array}{c}4044\\ 4042\\ 4040\end{array}\right),\text{dan}\\ -3\overline{a}& =-3\left(\begin{array}{c}2022\\ 2021\\ 2020\end{array}\right)=\left(\begin{array}{c}-6066\\ -6063\\ -6060\end{array}\right)\end{array}\end{array}$

$\text{F. 3. Perkalian Skalar Dua Vektor}$.

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

Dengan

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

$\begin{array}{rl}& \text{Tanda dari hasil skalar ini adalah}\\ & \begin{array}{|lll|}\hline \mathbf{\text{Besar sudut}}\theta & \mathbf{\text{Tanda}}& \text{Bentuk}\\ 0^{\circ }\le \theta <{90}^{\circ }& \text{Positif}& \text{Lancip}\\ \theta ={90}^{\circ }& \text{Nol}& \text{Siku-siku}\\ {90}^{\circ }<\theta \le {180}^{\circ }& \text{Negatif}& \text{Tumpul}\\ \hline\end{array}\\ & \text{Untuk}\theta \text{berupa sudut istimewa}:\\ & \begin{array}{|ccccccc|}\hline \theta & 0^{\circ }& {30}^{\circ }& {45}^{\circ }& {60}^{\circ }& {90}^{\circ }& {180}^{\circ }\\ \cos \theta & 1& {\displaystyle \frac{1}{2}\sqrt{3}}& {\displaystyle \frac{1}{2}\sqrt{2}}& {\displaystyle \frac{1}{2}}& 0& -1\\ \hline\end{array}\end{array}$

Adapun secara rumus untuk menentukan besar sudutnya adalah:

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

Sebagai ilustrasinya perhatikanlah gambar berikut

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

$\begin{array}{rl}\text{Jika}& \text{diketahui sebagai misal}\\ \overline{u}& =a\overline{i}+b\overline{j}+c\overline{k}\text{dan}\\ \overline{v}& =p\overline{i}+q\overline{j}+r\overline{k}\\ \text{mak}& \text{a}\\ \mathbf{\text{Per}}& \mathbf{\text{kalian skalar dua vektor adalah}}:\\ \overline{u}\bullet & \overline{v}=(a\overline{i}+b\overline{j}+c\overline{k})(p\overline{i}+q\overline{j}+r\overline{k})\\ & =ap.\overline{i}\bullet \overline{i}+aq.\overline{i}\bullet \overline{j}+ar.\overline{i}\bullet \overline{k}\\ & +bp.\overline{j}\bullet \overline{i}+bq.\overline{j}\bullet \overline{j}+br.\overline{j}\bullet \overline{k}\\ & +cp.\overline{k}\bullet \overline{i}+cq.\overline{k}\bullet \overline{j}+cr.\overline{k}\bullet \overline{k}\\ & =ap+0+0+0+bq+0+0+0+cr\\ & =ap+bq+cr\end{array}$

$\begin{array}{rl}& \text{Sebagai penjelasannya adalah}:\\ & ▹\overline{i}\bullet \overline{i}=\left|\overline{i}\right|\left|\overline{i}\right|\cos 0^{\circ }=1.1.1=1\\ & ▹\overline{i}\bullet \overline{j}=\left|\overline{i}\right|\left|\overline{j}\right|\cos {90}^{\circ }=1.1.0=0\\ & ▹\overline{i}\bullet \overline{k}=\left|\overline{i}\right|\left|\overline{k}\right|\cos {90}^{\circ }=1.1.0=0\\ & ▹\overline{j}\bullet \overline{i}=\overline{i}\bullet \overline{j}=0\\ & ▹\overline{j}\bullet \overline{j}=\left|\overline{j}\right|\left|\overline{j}\right|\cos 0^{\circ }=1.1.1=1\\ & ▹\overline{j}\bullet \overline{k}=\left|\overline{j}\right|\left|\overline{k}\right|\cos {90}^{\circ }=1.1.0=0\\ & ▹\overline{k}\bullet \overline{i}=\overline{i}\bullet \overline{k}=0\\ & ▹\overline{k}\bullet \overline{j}=\overline{j}\bullet \overline{k}=0\\ & ▹\overline{k}\bullet \overline{k}=\left|\overline{k}\right|\left|\overline{k}\right|\cos {90}^{\circ }=1.1.1=1\end{array}$

$\begin{array}{rl}& \text{Atau jika ditabelkan nilainya}\\ & \begin{array}{|cccc|}\hline \overline{u}\bullet \overline{v}& p\overline{i}& q\overline{j}& r\overline{k}\\ a\overline{k}& ap& 0& 0\\ b\overline{j}& 0& bq& 0\\ c\overline{k}& 0& 0& cr\\ \hline\end{array}\end{array}$

$\text{CONTOH SOAL}$.

$\begin{array}{ll}3.& \text{Jika}\overrightarrow{a}=\left(\begin{array}{c}1\\ 2\\ 4\end{array}\right),\text{dan}\overrightarrow{b}=\left(\begin{array}{c}5\\ 4\\ 0\end{array}\right)\\ & \text{tentukanlah nilai}\text{dari}\overrightarrow{a}\bullet \overrightarrow{b}\\ \\ & \mathbf{\text{Jawab}}\\ & \begin{array}{rl}\overrightarrow{a}\bullet \overrightarrow{b}& =1.5+2.4+4.0=5+8+0=13\end{array}\end{array}$

$\begin{array}{ll}4.& \text{Jika diketahui}\overrightarrow{a}=\overrightarrow{i}-2\overrightarrow{j}+3\overrightarrow{k},\\ & \text{dan}\overrightarrow{b}=3\overrightarrow{i}-4\overrightarrow{j}+m\overrightarrow{k}\text{serta}\\ & \text{nilai}\overrightarrow{a}\bullet \overrightarrow{b}=-4,\text{maka tentukan}\\ & \text{nilai}m\\ \\ & \mathbf{\text{Jawab}}\\ & \text{Diketahui bahwa}\\ & ▹\overrightarrow{a}=\overrightarrow{i}-2\overrightarrow{j}+3\overrightarrow{k}=\left(\begin{array}{c}1\\ -2\\ 3\end{array}\right),\text{dan}\\ & ▹\overrightarrow{b}=3\overrightarrow{i}-4\overrightarrow{j}+m\overrightarrow{k}=\left(\begin{array}{c}3\\ -4\\ m\end{array}\right)\\ & \begin{array}{rl}\overrightarrow{a}\bullet \overrightarrow{b}& =1.3+3.(-4)+3.m\\ -4& =3+8+3m\\ -3m& =11+4\\ m& =-{\displaystyle \frac{15}{3}}\\ & =-5\end{array}\end{array}$

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

$\begin{array}{l}\\ 6.& \text{Diketahui}\overrightarrow{a}=\left(\begin{array}{c}-2\\ 1\\ 3\end{array}\right)\text{dan}\overrightarrow{b}=\left(\begin{array}{c}4\\ -1\\ t\end{array}\right),\\ & \text{jika}\overrightarrow{p}\text{tegak lurus}\overrightarrow{q},\text{maka tentukanlah}\\ & \text{nilai}t\text{adalah}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}\text{Karena}& \text{kedua vektor tersebut saling }\\ \text{tegak l}& \text{urus maka}\\ \overrightarrow{a}.\overrightarrow{b}& =0\\ \left(\begin{array}{c}-2\\ 1\\ 3\end{array}\right)& \left(\begin{array}{c}4\\ -1\\ t\end{array}\right)=0\\ (-2).4& +1.(-1)+3.t=0\\ -8-1& +3t=0\\ 3t& =9\\ t& =3\end{array}\end{array}$

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

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

$\begin{array}{l}\\ 9.& \text{Tentukanlah besar sudut antara vektor}\\ & \overrightarrow{a}=\left(\begin{array}{c}-1\\ 1\\ 0\end{array}\right)\text{dan}\overrightarrow{b}=\left(\begin{array}{c}1\\ -2\\ 2\end{array}\right)\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}\cos \theta & ={\displaystyle \frac{\overrightarrow{a}.\overrightarrow{b}}{\left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right|}}\\ & ={\displaystyle \frac{\left(\begin{array}{c}-1\\ 1\\ 0\end{array}\right)\left(\begin{array}{c}1\\ -2\\ 2\end{array}\right)}{\sqrt{(-1{)}^2+1^2}\sqrt{1^2+(-2{)}^2+2^2}}}\\ & ={\displaystyle \frac{-1-2+0}{\sqrt{2}\sqrt{9}}}\\ & =-{\displaystyle \frac{1}{\sqrt{2}}=-{\displaystyle \frac{1}{2}\sqrt{2}}}\\ & =-\cos {45}^{\circ }\\ & =\cos ({180}^{\circ }-{45}^{\circ })\\ \cos \theta & =\cos {135}^{\circ }\\ \therefore \theta & ={135}^{\circ }\end{array}\end{array}$.

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

$\text{Berikut dua contoh untuk sudut tidak istimewa}$.

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

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

$\begin{array}{l}\\ 13.& \text{Diketahui vektor}\overrightarrow{a}\text{dan}\overrightarrow{b}\text{memiliki }\\ & \text{panjang masing-masing adalah 2 dan 3}\\ & \text{serta}\mathrm{\angle }(\overrightarrow{a},\overrightarrow{b})={60}^{\circ }.\text{Carilah nilai}\\ & \text{a}.|\overrightarrow{a}+\overrightarrow{b}|\\ \\ & \text{b}.|\overrightarrow{a}-\overrightarrow{b}|\\ & \text{b}\text{besar sudut antara}\\ & (\overrightarrow{a}+\overrightarrow{b})\text{dan}(\overrightarrow{a}-\overrightarrow{b})\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}\text{a}.& {|\overrightarrow{a}+\overrightarrow{b}|}^2\\ & =(\overrightarrow{a}+\overrightarrow{b})(\overrightarrow{a}+\overrightarrow{b})\\ & =\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\\ & \text{Jadi, nilainya adalah}|\overrightarrow{a}+\overrightarrow{b}|=\sqrt{19}\end{array}\\ & \begin{array}{rl}\text{b}.& {|\overrightarrow{a}-\overrightarrow{b}|}^2\\ & =(\overrightarrow{a}-\overrightarrow{b})(\overrightarrow{a}-\overrightarrow{b})\\ & =\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\\ & \text{Jadi, nilainya adalah}|\overrightarrow{a}-\overrightarrow{b}|=\sqrt{7}\end{array}\\ & \begin{array}{rl}\text{c}.\text{Untuk menentukan nilai}& \\ \cos \mathrm{\angle }(\overrightarrow{a}+\overrightarrow{b},\overrightarrow{a}-\overrightarrow{b})& ={\displaystyle \frac{(\overrightarrow{a}+\overrightarrow{b}).(\overrightarrow{a}-\overrightarrow{b})}{|\overrightarrow{a}+\overrightarrow{b}|.|\overrightarrow{a}-\overrightarrow{b}|}}\\ & ={\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}}}\\ \mathrm{\angle }(\overrightarrow{a}+\overrightarrow{b},\overrightarrow{a}-\overrightarrow{b})& =\arccos (-\frac{5}{\sqrt{133}})\end{array}\end{array}$

$\text{Berikut contoh untuk bentuk sudutnya}$.

$\begin{array}{ll}14.& \text{Diketahui}\overrightarrow{p}=(x,3,2),\text{dan}\\ & \overrightarrow{q}=(2,-6,3).\text{Tentukan nilai}x\\ & \text{agar kedua vektor}\\ & \text{a}\text{membentuk sudut lancip}\\ & \text{b}\text{membentuk sudut siku-siku}\\ & \text{c}\text{membentuk sudut tumpul}\\ & \text{d}\text{sama panjang}\\ \\ & \mathbf{\text{Jawab}}\\ & \begin{array}{rl}\text{Dik}& \text{etahui bahwa}\\ ▹& \overrightarrow{p}=(x,3,2)=\left(\begin{array}{c}x\\ 3\\ 2\end{array}\right)\text{dan}\\ ▹& \overrightarrow{q}=(2,-6,3)=\left(\begin{array}{c}2\\ -6\\ 3\end{array}\right)\end{array}\\ & \text{Selanjutnya}\\ & \overrightarrow{p}\bullet \overrightarrow{q}=\left(\begin{array}{c}x\\ 3\\ 2\end{array}\right)\left(\begin{array}{c}2\\ -6\\ 3\end{array}\right)\\ & =2x-18+6=2x-12\\ & \text{Selanjutnya}\\ & \begin{array}{rl}\text{a}& \mathbf{\text{Syarat lancip}},\text{yaitu}:\overrightarrow{p}\bullet \overrightarrow{q}>0\\ & 2x-12>0\iff 2x>12\iff x>6\\ \text{b}& \mathbf{\text{Syarat siku-siku}},\text{yaitu}:\overrightarrow{p}\bullet \overrightarrow{q}=0\\ & 2x-12=0\iff 2x=12\iff x=6\\ \text{c}& \mathbf{\text{Syarat tumpul}},\text{yaitu}:\overrightarrow{p}\bullet \overrightarrow{q}<0\\ & 2x-12<0\iff 2x<12\iff x<6\\ \text{d}& \mathbf{\text{Syarat panjang kedua vektor sama}}\\ & \text{yaitu}:\left|\overrightarrow{p}\right|=\left|\overrightarrow{q}\right|,\text{maka}\\ & \begin{array}{rl}& \sqrt{x^2+3^2+2^2}=\sqrt{2^2+(-6{)}^2+3^2}\\ & x^2+9+4=4+36+9\\ & x^2=36\\ & x=\pm \sqrt{36}=\pm 6\\ & \text{Jadi},x=-6\text{atau}x=6\end{array}\end{array}\end{array}$

DAFTAR PUSTAKA

1. Johanes, Kastolan, Sulasim. 2006. Kompetensi Matematika Program IPA 3A SMA Kelas XII Semester Pertama. Jakarta: YUDHISTIRA.
2. Kanginan, M., Nurdiansyah, H., Akhmad, G. 2016. Matematika untuk Siswa SMA/MA Kelas X Kelompok Peminatan Matematika dan Ilmu-Ilmu Alam. Bandung: YRAMA WIDYA.
3. Noormandiri, Sucipto, E. 2003. Buku Pelajaran Matematika SMU untuk Kelas 3 Program IPA. Jakarta: ERLANGGA.
4. 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.