PAS MATEMATIKA BLOG

Persamaan Trigonometri

$\Large\textrm{A. 1 Identitas Trigonometri}$.

A. 1. 1 Nilai Trigonometri Sudut

$\textrm{a. Perbandingan Trigonometri dalam Segitiga Siku-Siku}$.

Perhatikanlah ilustrasi sebuah segitiga siku-siku sama kaki berikut

Diketahui pula bahwa :

$\begin{matrix} \bullet \quad \sin 45^{\circ}=\displaystyle \frac{1}{\sqrt{2}}=\frac{1}{2}\sqrt{2}\\ \bullet \quad \cos 45^{\circ}=\displaystyle \frac{1}{\sqrt{2}}=\frac{1}{2}\sqrt{2}\\ \bullet \quad \tan 45^{\circ}=1 \qquad\qquad\: \: \end{matrix}$.

$\begin{matrix} \bullet \quad \csc 45^{\circ}=\displaystyle \sqrt{2}\\ \bullet \quad \sec 45^{\circ}=\displaystyle \sqrt{2}\\ \bullet \quad \cot 45^{\circ}=1 \: \: \, \end{matrix}$.

Berikut ilustrasi segitiga dengan sudut istimewa yang lain yaitu $30^{\circ}$ dan $60^{\circ}$.

$\begin{array}{|c|c|}\hline \begin{matrix} \bullet \quad \color{purple}\sin 30^{\circ}=\displaystyle \frac{1}{2}\\ \bullet \quad \cos 30^{\circ}=\displaystyle \frac{1}{2}\sqrt{3}\\ \bullet \quad \tan 30^{\circ}=\displaystyle \frac{1}{\sqrt{3}}=\frac{1}{3}\sqrt{3} \: \: \,\\ \bullet \quad \color{blue}\sin 60^{\circ}=\displaystyle \frac{1}{2}\sqrt{3}\\ \bullet \quad \cos 60^{\circ}=\displaystyle \frac{1}{2}\\ \bullet \quad \tan 30^{\circ}=\displaystyle \sqrt{3}\\ \end{matrix} &\begin{matrix} \bullet \quad \csc 30^{\circ}=\displaystyle 2\\ \bullet \quad \sec 30^{\circ}=\displaystyle \frac{2}{\sqrt{3}}=\frac{2}{3}\sqrt{3}\\ \bullet \quad \cot 30^{\circ}=\displaystyle \sqrt{3} \: \: \,\\ \bullet \quad \color{red}\csc 60^{\circ}=\displaystyle \frac{2}{\sqrt{3}}=\frac{2}{3}\sqrt{3}\\ \bullet \quad \sec 60^{\circ}=\displaystyle 2\\ \bullet \quad \color{purple}\cot 30^{\circ}=\displaystyle \frac{1}{3}\sqrt{3}\\ \end{matrix} \\\hline \end{array}$

Perhatikan segitiga ABC siku-siku di C berikut

Perhatikanlah segitiga OAB berikut

$\begin{aligned}\textrm{a}.\quad&\color{purple}\sin \alpha =\displaystyle \frac{y}{r}\\ \textrm{b}.\quad&\cos \alpha =\displaystyle \frac{x}{r}\\ \textrm{c}.\quad&\color{blue}\tan \alpha =\displaystyle \frac{y}{x}\\ \textrm{d}.\quad&\csc \alpha =\displaystyle \frac{r}{y}\\ \textrm{e}.\quad&\sec \alpha =\displaystyle \frac{r}{x}\\ \textrm{f}.\quad&\color{red}\cot \alpha =\displaystyle \frac{x}{y}\\ \end{aligned}$.

A. 1. 2 Identitas Trigonometri Dasar

$\textrm{a. Dalil Pythagoras Segitiga Siku-Siku}$.

$\begin{array}{c|c}\\ \begin{aligned}&\textrm{Dalil/rumus Pythagoras}\\ &a^{2}+b^{2} =c^{2}\\ &\color{red}\textrm{atau}\\ &c=\sqrt{a^{2}+b^{2}} \end{aligned}&\begin{aligned}&\sin \angle ACB=\displaystyle \frac{a}{c}\\ &\cos \angle ACB=\displaystyle \frac{b}{c}\\ &\tan \angle ACB=\displaystyle \frac{a}{b}=\displaystyle \frac{\sin \angle ACB}{\cos \angle ACB}\\ &\csc \angle ACB=\displaystyle \frac{c}{a}\\ &\sec \angle ACB=\displaystyle \frac{c}{b}\\ &\cot \angle ACB=\displaystyle \frac{b}{a}=\displaystyle \frac{\cos \angle ACB}{\sin \angle ACB} \end{aligned} \end{array}$

$\color{purple}\textrm{b. Identitas trigonometri pada segitiga siku-siku}$.

$\begin{aligned}&\textrm{Dalil/rumus Pythagoras}\\ &a^{2}+b^{2} =c^{2}\\ &\textrm{Perhatikan lagi gambar di poin c di atas}\\ &\begin{array}{|c|l|}\hline 1.&\textrm{Rumus saat dibagi dengan}\: \: c^{2}\\ &\displaystyle \frac{a^{2}}{c^{2}}+\displaystyle \frac{b^{2}}{c^{2}}=\displaystyle \frac{c^{2}}{c^{2}}\Leftrightarrow \color{red}\displaystyle \frac{a^{2}}{c^{2}}+\displaystyle \frac{b^{2}}{c^{2}}=1\\&\\ &\textrm{menjadi}\: \: \: \left ( \displaystyle \frac{a}{c} \right )^{2}+\left ( \displaystyle \frac{b}{c} \right )^{2}=1\\ &\color{blue}\sin ^{2}\angle ACB+\cos ^{2}\angle ACB=1\\\hline 2&\textrm{Rumus saat dibagi dengan}\: \: b^{2}\\ &\displaystyle \frac{a^{2}}{b^{2}}+\displaystyle \frac{b^{2}}{b^{2}}=\displaystyle \frac{c^{2}}{b^{2}}\Leftrightarrow \color{red}\displaystyle \frac{a^{2}}{b^{2}}+1=\displaystyle \frac{c^{2}}{b^{2}}\\&\\ &\textrm{menjadi}\: \: \: \left ( \displaystyle \frac{a}{b} \right )^{2}+1=\left ( \displaystyle \frac{c}{b} \right )^{2}\\ &\color{blue}\tan ^{2}\angle ACB+1=\sec ^{2}\angle ACB\\\hline 3&\textrm{Rumus saat dibagi dengan}\: \: a^{2}\\ &\displaystyle \frac{a^{2}}{a^{2}}+\displaystyle \frac{b^{2}}{a^{2}}=\displaystyle \frac{c^{2}}{a^{2}}\Leftrightarrow \color{red}1+\displaystyle \frac{b^{2}}{a^{2}}=\displaystyle \frac{c^{2}}{a^{2}}\\&\\ &\textrm{menjadi}\: \: \: 1+\left ( \displaystyle \frac{b}{a} \right )^{2}=\left ( \displaystyle \frac{c}{a} \right )^{2}\\ &\color{blue}1+\cot ^{2}\angle ACB=\csc ^{2}\angle ACB\\\hline \end{array} \end{aligned}$

$\color{purple}\textrm{c. Tabel trigonometri nilai sudut istimewa}$.

$\begin{array}{|c|c|c|c|c|c|c|}\hline \alpha &0^{\circ}&30&45^{\circ}&60^{\circ}&90^{\circ}&180^{\circ}\\\hline \sin \alpha &0&\displaystyle \frac{1}{2}&\displaystyle \frac{1}{2}\sqrt{2}&\displaystyle \frac{1}{2}\sqrt{3}&1&0\\\hline \cos \alpha &1&\displaystyle \frac{1}{2}\sqrt{3}&\displaystyle \frac{1}{2}\sqrt{2}&\displaystyle \frac{1}{2}&0&-1\\\hline \tan \alpha &0&\displaystyle \frac{1}{3}\sqrt{3}&1&\sqrt{3}&\color{red}\textrm{TD}&0\\\hline \end{array}$.

$\color{purple}\textrm{d. Aturan sinus pada segitiga sebarang}$.

$\displaystyle \frac{BC}{\sin \angle A}=\displaystyle \frac{AC}{\sin \angle B}=\displaystyle \frac{AB}{\sin \angle C}$

$\color{purple}\textrm{e. Aturan cosinus pada segitiga sebarang}$.

Perhatikanlah gmabar pada poin e di atas, aturan cosinusnya adalah:

$\begin{aligned}\bullet \: \: &\cos \angle A=\displaystyle \frac{b^{2}+c^{2}-a^{2}}{2bc}\\ \bullet \: \: &\cos \angle B=\displaystyle \frac{a^{2}+c^{2}-a^{2}}{2ac}\\ \bullet \: \: &\cos \angle C=\displaystyle \frac{a^{2}+b^{2}-a^{2}}{2ab} \end{aligned}$.

$\begin{aligned}&\color{red}\textrm{Macam-Macam Identitas Trigonometri Dasar}\\ &1.\quad \csc \alpha =\displaystyle \frac{1}{\sin \alpha }\qquad\qquad 5.\quad \tan \alpha =\displaystyle \frac{\sin \alpha }{\cos \alpha }\\ &2.\quad \sec \alpha =\displaystyle \frac{1}{\cos \alpha }\qquad\qquad 6.\quad \tan^{2} \alpha +1=\sec ^{2}\alpha \\ &3.\quad \cot \alpha =\displaystyle \frac{1}{\tan \alpha }\qquad\qquad 7.\quad \cot^{2} \alpha +1=\csc ^{2}\alpha \\ &4.\quad \cot \alpha =\displaystyle \frac{\cos \alpha }{\sin \alpha }\qquad\qquad 8.\quad \sin^{2} \alpha +\cos ^{2}=1\\ \end{aligned}$.

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

$\begin{array}{ll}\\ 1.&\textrm{Tunjukkan bahwa} \\ &\qquad\qquad\tan \alpha =\displaystyle \frac{\sin \alpha \cos \alpha }{1-\sin ^{2}\alpha }\\\\ &\color{blue}\textrm{Bukti}:\\ &\begin{aligned}\tan \alpha &=\displaystyle \frac{\sin \alpha }{\cos \alpha }\\ &=\displaystyle \frac{\sin \alpha }{\cos \alpha }\times \frac{\cos \alpha }{\cos \alpha }\\ &=\displaystyle \frac{\sin \alpha \cos \alpha }{\cos^{2} \alpha }\\ &=\displaystyle \frac{\sin \alpha \cos \alpha }{1-\sin^{2} \alpha }\qquad \blacksquare \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 2.&\textrm{Tunjukkan bahwa} \\ &\qquad\qquad \displaystyle \frac{1}{\sqrt{\tan ^{2}\beta }}\times \sin \beta =\cos \beta \\\\ &\color{blue}\textrm{Bukti}:\\ &\begin{aligned}\displaystyle \frac{1}{\sqrt{\tan ^{2}\beta }}\times \sin \beta &=\displaystyle \frac{1}{\tan \beta }\times \sin \beta \\ &=\displaystyle \frac{\cos \beta }{\sin \beta }\times \sin \beta \\ &=\cos \beta \qquad\blacksquare \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 3.&\textrm{Tunjukkan bahwa} \\ &\qquad\qquad \displaystyle \frac{\cos ^{2}\gamma }{1-\sin \gamma } =1+\sin \gamma \\\\ &\color{blue}\textrm{Bukti}:\\ &\begin{aligned}\displaystyle \frac{\cos ^{2}\gamma }{1-\sin \gamma } &=\displaystyle \frac{1-\sin^{2} \gamma }{1-\sin \gamma }\\ &=\displaystyle \frac{(1-\sin \gamma )(1+\sin \gamma )}{1-\sin \gamma }\\ &=1+\sin \gamma \qquad\quad \blacksquare \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 4.&\textrm{Tunjukkan bahwa} \\ &\qquad\qquad \displaystyle \frac{1-\tan ^{2}\theta }{1+\tan ^{2}\theta } =\cos ^{2}\theta -\sin ^{2}\theta \\\\ &\color{blue}\textrm{Bukti}:\\ &\begin{aligned}\displaystyle \frac{1-\tan ^{2}\theta }{1+\tan ^{2}\theta } &=\displaystyle \frac{1-\tan ^{2}\theta }{\sec ^{2}\theta }=\displaystyle \frac{1-\displaystyle \frac{\sin ^{2}\theta }{\cos ^{2}\theta }}{\displaystyle \frac{1}{\cos ^{2}\theta }}\\ &=\displaystyle \frac{\displaystyle \frac{\cos ^{2}\theta -\sin ^{2}\theta }{\cos ^{2}\theta }}{\displaystyle \frac{1}{\cos ^{2}\theta }} \\ &=\cos ^{2}\theta -\sin ^{2}\theta\qquad \blacksquare \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 5.&\textrm{Tunjukkan bahwa} \\ &\qquad\qquad \cos ^{4}\alpha -\sin ^{4}\alpha =1-2\sin ^{2}\alpha \\\\ &\color{blue}\textrm{Bukti}:\\ &\begin{aligned}\cos ^{4}\alpha -\sin ^{4}\alpha &=\left ( \cos ^{2}\alpha \right )^{2} -\left (\sin ^{2}\alpha \right )^{2}\\ &=\left (\cos ^{2}\alpha -\sin ^{2}\alpha \right )\left ( \cos ^{2}\alpha +\sin ^{2}\alpha \right )\\ &=\left (\cos ^{2}\alpha -\sin ^{2}\alpha \right )\times 1\\ &=\cos ^{2}\alpha -\sin ^{2}\alpha \\ &=\left ( 1-\sin ^{2}\alpha \right )-\sin ^{2}\alpha \\ &=1-2\sin ^{2}\alpha \qquad\quad \blacksquare \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 6.&\textrm{Tunjukkan bahwa} \\ &\qquad\qquad \displaystyle \frac{\sin \beta \sec \beta }{\sin ^{2}\beta -\tan ^{2}\beta } =-\displaystyle \frac{\cos \beta }{\sin ^{3}\beta } \\\\ &\color{blue}\textrm{Bukti}:\\ &\begin{aligned}\displaystyle \frac{\sin \beta \sec \beta }{\sin ^{2}\beta -\tan ^{2}\beta }&=\displaystyle \frac{\sin \beta \left ( \displaystyle \frac{1}{\cos \beta } \right ) }{\sin ^{2}\beta -\displaystyle \frac{\sin ^{2}\beta }{\cos ^{2}\beta } } \\ &=\displaystyle \frac{\left ( \displaystyle \frac{\sin \beta }{\cos \beta } \right )}{\sin ^{2}\beta \left ( 1-\displaystyle \frac{1}{\cos ^{2}\beta } \right )}\times \frac{\cos ^{2}\beta }{\cos ^{2}\beta }\\ &=\displaystyle \frac{\sin \beta \cos \beta }{\sin ^{2}\beta \left ( \cos ^{2}\beta -1 \right )}\\ &=\displaystyle \frac{\cos \beta }{\sin \beta \left ( -\sin ^{2}\beta \right )}\\ &=-\displaystyle \frac{\cos \beta }{\sin ^{3}\beta } \qquad\quad \blacksquare \end{aligned} \end{array}$.

DAFTAR PUSTAKA

Noormandiri, B. K. 2016. Matematika untuk SMA/MA Kelas XI Kelompok Peminatan Matematika dan Ilmu-Ilmu Alam. Jakarta: ERLANGGA.
Sukino. 2016. Matematika untuk SMA/MA Kelas XI Kelompok Peminatan Matematika dan Ilmu-Ilmu Alam. Jakarta: ERLANGGA

Lanjutan Limit Fungsi Trigonometri

$\color{blue}\textrm{C. Menentukan Nilai Limit Fungsi Trigonometri}$

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:

$\color{blue}\textrm{C. 1 dengan substitusi langsung}$

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

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

$\color{blue}\textrm{C. 2 dengan menyederhanakan}$

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:

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

Demikian juga

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

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

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

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

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

$\color{blue}\textrm{C. 3 dengan rumus limit fungsi trigonometri}$

Berikut rumus limit fungsi trigonometri yang akan kita gunakan

$\begin{aligned}\bullet \: \: \: \underset{x\rightarrow 0 }{\textrm{lim}}\: \displaystyle \frac{\sin ax}{ax}&=\underset{x\rightarrow 0 }{\textrm{lim}}\: \displaystyle \frac{ax}{\sin ax}=\frac{a}{a}=1\\ \bullet \: \: \: \underset{x\rightarrow 0 }{\textrm{lim}}\: \displaystyle \frac{\tan ax}{ax}&=\underset{x\rightarrow 0 }{\textrm{lim}}\: \displaystyle \frac{ax}{\tan ax}=\frac{a}{a}=1 \end{aligned}$.

BUKTINYA ada di sini

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

$\begin{array}{ll}\\ 4.&\textrm{Tentukanlah nilai limit berikut}\\ &\begin{array}{lll} \textrm{a}&\underset{x\rightarrow 0 }{\textrm{lim}}\: \displaystyle \frac{\sin 4x}{7x}\\ \textrm{b}&\underset{x\rightarrow 0 }{\textrm{lim}}\: \displaystyle \frac{ 2x}{\tan 9x}\\ \textrm{c}&\underset{x\rightarrow 0 }{\textrm{lim}}\: \displaystyle \frac{\tan 8x}{\sin 3x} \end{array}\\\\ &\color{blue}\textrm{Jawab}:\\ &\begin{aligned}\textrm{a}.&\underset{x\rightarrow 0 }{\textrm{lim}}\: \displaystyle \frac{\sin 4x}{7x}=\underset{x\rightarrow 0 }{\textrm{lim}}\: \displaystyle \frac{4}{7}\times \displaystyle \frac{\sin 4x}{4x}=\displaystyle \frac{4}{7}\\ \textrm{b}.&\underset{x\rightarrow 0 }{\textrm{lim}}\: \displaystyle \frac{2x}{\tan 9x}=\underset{x\rightarrow 0 }{\textrm{lim}}\: \displaystyle \frac{2}{9}\times \displaystyle \frac{9x}{\tan 9x}=\displaystyle \frac{2}{9}\\ \textrm{c}.&\underset{x\rightarrow 0 }{\textrm{lim}}\: \displaystyle \frac{\tan 8x}{\sin 3x}=\underset{x\rightarrow 0 }{\textrm{lim}}\: \displaystyle \frac{\tan 8x}{8x}\times \underset{x\rightarrow 0 }{\textrm{lim}}\: \displaystyle \frac{3x}{\sin 3x}\times \frac{8}{3}=\displaystyle \frac{8}{3} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 5.&\textrm{Tentukanlah nilai limit berikut}\\ &\begin{array}{lll} \textrm{a}&\underset{x\rightarrow 0 }{\textrm{lim}}\: \displaystyle \frac{24x^{2}}{8\sin ^{2}x}\\ \textrm{b}&\underset{x\rightarrow 0 }{\textrm{lim}}\: \displaystyle \frac{ 5x^{2}}{15\tan (-9x)\sin 3x} \end{array}\\\\ &\color{blue}\textrm{Jawab}:\\ &\begin{aligned}\textrm{a}.\: \: &\underset{x\rightarrow 0 }{\textrm{lim}}\: \displaystyle \frac{24x^{2}}{8\sin ^{2}x}=\displaystyle \frac{24}{8}\times \underset{x\rightarrow 0 }{\textrm{lim}}\: \displaystyle \frac{x}{\sin x}\times \underset{x\rightarrow 0 }{\textrm{lim}}\: \displaystyle \frac{x}{\sin x}\\ &=3\times 1\times 1\\ &=3\\ \textrm{b}.\: \: &\underset{x\rightarrow 0 }{\textrm{lim}}\: \displaystyle \frac{ 5x^{2}}{15\tan (-9x)\sin 3x}\\ &=\displaystyle \frac{5}{15}\times \underset{x\rightarrow 0 }{\textrm{lim}}\: \displaystyle \frac{ x}{\tan (-9x)}\times \underset{x\rightarrow 0 }{\textrm{lim}}\: \displaystyle \frac{ x}{\tan 3x}\\ &=\displaystyle \frac{1}{3}\times \left ( -\times \underset{x\rightarrow 0 }{\textrm{lim}}\: \displaystyle \frac{ 9x}{\tan 9x}\times \frac{1}{9} \right )\times \times \underset{x\rightarrow 0 }{\textrm{lim}}\: \displaystyle \frac{ 3x}{\tan 3x}\times \frac{1}{3}\\ &=-\displaystyle \frac{1}{3}\times \frac{1}{9}\times \frac{1}{3}=-\frac{1}{81} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 6.&\textrm{Tentukanlah nilai limit berikut}\\ &\begin{array}{lll} \textrm{a}&\underset{x\rightarrow 0 }{\textrm{lim}}\: \displaystyle \frac{1-\cos 6x}{x^{2}}\\ \textrm{b}&\underset{x\rightarrow 0 }{\textrm{lim}}\: \displaystyle \frac{ \cos 4x-\cos 2x}{x^{2}}\\ \textrm{c}&\underset{x\rightarrow 3 }{\textrm{lim}}\: \displaystyle \frac{ \left (x^{2}-5x+6 \right )\sin \left ( x-3 \right )}{\left ( x^{2}-7x+12 \right )^{2}} \end{array}\\\\ &\color{blue}\textrm{Jawab}:\\ &\begin{aligned}\textrm{a}.\: \: &\underset{x\rightarrow 0 }{\textrm{lim}}\: \displaystyle \frac{1-\cos 6x}{x^{2}}=\underset{x\rightarrow 0 }{\textrm{lim}}\: \displaystyle \frac{1-\left ( 1-2\sin ^{2}3x \right )}{x^{2}}\\ &=\underset{x\rightarrow 0 }{\textrm{lim}}\: \displaystyle \frac{2\sin ^{2}3x}{x^{2}}=2\times \underset{x\rightarrow 0 }{\textrm{lim}}\: \displaystyle \frac{\sin 3x}{x}\times \underset{x\rightarrow 0 }{\textrm{lim}}\: \displaystyle \frac{\sin 3x}{x}\\ &=2\times 3\times 3=18\\ \textrm{b}.\: \: &\underset{x\rightarrow 0 }{\textrm{lim}}\: \displaystyle \frac{ \cos 4x-\cos 2x}{x^{2}}\\ &=\underset{x\rightarrow 0 }{\textrm{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\rightarrow 0 }{\textrm{lim}}\: \displaystyle \frac{\sin 3x.\sin x}{x^{2}}\\ &=-2\times \underset{x\rightarrow 0 }{\textrm{lim}}\: \displaystyle \frac{\sin 3x}{x}\times \underset{x\rightarrow 0 }{\textrm{lim}}\: \displaystyle \frac{x}{x}=-2\times 3\times 1=-6\\ \textrm{c}.\: \: &\underset{x\rightarrow \color{red}3 }{\textrm{lim}}\: \displaystyle \frac{ \left (x^{2}-5x+6 \right )\sin \left ( x-3 \right )}{\left ( x^{2}-7x+12 \right )^{2}}\\ &=\underset{x\rightarrow \color{red}3 }{\textrm{lim}}\: \displaystyle \frac{ (x-2)(x-3)\sin \left ( x-3 \right )}{\left ( (x-3)(x-4) \right )^{2}}\\ &=\underset{x\rightarrow \color{red}3 }{\textrm{lim}}\: \displaystyle \frac{ (x-2)(x-3)\sin \left ( x-3 \right )}{(x-3)(x-3)(x-4)^{2}}\\ &=\underset{x\rightarrow \color{red}3 }{\textrm{lim}}\: \displaystyle \displaystyle \frac{x-2}{(x-4)^{2}}\times \underset{x\rightarrow \color{red}3 }{\textrm{lim}}\: \displaystyle \frac{ \sin \left ( x-3 \right )}{(x-3)} \\ &=\displaystyle \frac{(\color{red}3\color{black}-2)}{(\color{red}3\color{black}-4)^{2}}\times 1=\displaystyle \frac{1}{(-1)^{2}}=\frac{1}{1}=1 \end{aligned} \end{array}$.

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

$\LARGE\colorbox{magenta}{LATIHAN SOAL}$.

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

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

DAFTAR PUSTAKA

Kanginan, M., Nurdiansyah, H., Akhmad, G. 2016. Matematika untuk Siswa SMA/MA Kelas XII Kelompok Peminatan Matematika dan Ilmu-Ilmu Alam. Bandung: YRAMA WIDYA.
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.
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

$\color{blue}\textrm{A. Pendahuluan}$

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

$\begin{aligned}\underset{x\rightarrow a }{\textrm{lim}}\: f(x)=L&\: \: \textbf{adalah}\, \: \textrm{Jika}\: \: x\: \: \textrm{mendekati}\: \: a\\ &\textrm{dengan tidak sama dengan}\: \: a,\\ &\textrm{maka nilai}\: \: f(x)\: \: \textrm{mendekati}\: \: L \end{aligned}$.

Perhatikan definisi di atas istilah $x\: \: \textrm{mendekati}\: \: a$ dituliskan dengan simbol $(x\rightarrow a)$. 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

$\begin{aligned}\underset{x\rightarrow \color{blue}a^{-} }{\textrm{lim}}\: f(x)=\underset{x\rightarrow \color{blue}a^{+} }{\textrm{lim}}\: f(x)=\underset{x\rightarrow \color{blue}a }{\textrm{lim}}\: f(x)=L&\: \: \ \end{aligned}$.

$\begin{aligned}\textrm{Perlu di}&\textrm{perhatikan bahwa didekati dari}\\ \bullet \: \: \textbf{kiri}\: &\textrm{disimbolkan dengan}\: \: \underset{x\rightarrow \color{blue}a^{-} }{\textrm{lim}}\: f(x),\: \: \textrm{dan}\\ \bullet \: \: \, \textbf{kan}&\textbf{an}\: \: \textrm{disimbolkan dengan}\: \: \underset{x\rightarrow \color{blue}a^{+} }{\textrm{lim}}\: f(x) \end{aligned}$.

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

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

$\begin{aligned}.\qquad&\textrm{Jadi, nilai}\: \: \underset{x\rightarrow \color{blue}2}{\textrm{lim}} \: \displaystyle \frac{x^{2}-4}{x-2}=2 \: \: \textrm{atau dapat dikatakan}\\ &\textrm{nilai} \: \: \underset{x\rightarrow \color{blue}2}{\textrm{lim}} \: \displaystyle \frac{x^{2}-4}{x-2}\: \: \textbf{ada}\\ &\textrm{meskipun nilai substitusi langsung}\: \: x=2\: \: \textrm{yaitu}\\ &f(0)=\displaystyle \frac{0^{2}-0}{0-0}=\frac{0}{0}\: \: \textrm{berupa bentuk tak}\\ &\textrm{tentu. Berikut ilustrasinya} \end{aligned}$

$\begin{array}{ll}\\ 2.&\textrm{Selidikilah limit fungsi berikut apakah}\\ &\textrm{memiliki harga limit}\\ &\underset{x\rightarrow \color{red}5}{\textrm{lim}} \:f(x),\: \: \textrm{untuk}\: \: f(x)=\begin{cases} x &\textrm{saat}\: \: x<5 \\ 5-x &\textrm{saat}\: \: x\geq 5 \end{cases}\\\\ &\color{blue}\textrm{Jawab}:\\ &\begin{aligned}&\textrm{Perhatikanlah ketika didekati dari kiri}\\\ &\textrm{yaitu}\: \: x\rightarrow \color{red}5^{-},\: \: \color{black}\textrm{maka}\: \: \underset{x\rightarrow \color{red}5^{-}}{\textrm{lim}} \:f(x)=x\\ & \textrm{atau}\: \: \underset{x\rightarrow \color{red}5^{-}}{\textrm{lim}} \:f(x)=5\\ &\textrm{boleh juga dituliskan dengan}\\ &\underset{x\rightarrow \color{red}5^{-}}{\textrm{lim}} \:f(x)=x=5.\: \: \textrm{Sedangkan ketika}\\ &\textrm{didekati dari arah kanan yaitu}\: \: x\rightarrow \color{red}5^{+},\\ &\textrm{maka}\: \: \underset{x\rightarrow \color{red}5^{+}}{\textrm{lim}} \:f(x)=\underset{x\rightarrow \color{red}5^{+}}{\textrm{lim}} \: (5-x)=5-5=0.\\ &\textrm{Karena nilai}\: \: \underset{x\rightarrow \color{red}5^{-}}{\textrm{lim}} \:f(x)\neq \underset{x\rightarrow \color{red}5^{+}}{\textrm{lim}} \:f(x),\: \: \textrm{maka}\\ &\textrm{nilai atau harga}\: \: \underset{x\rightarrow \color{red}5}{\textrm{lim}} \:f(x)\: \: \textbf{tidak ada}\\ &\textrm{Berikut ilustrasi gambarnya} \end{aligned}\end{array}$.

$\begin{array}{ll}\\ 3.&\textrm{Selidikilah limit fungsi berikut apakah}\\ &\textrm{memiliki harga limit}\\ &\underset{x\rightarrow \color{red}0}{\textrm{lim}} \:f(x),\: \: \textrm{untuk}\: \: f(x)=\cos x\\\\ &\color{blue}\textrm{Jawab}:\\ &\begin{aligned}&\textrm{Perhatikanlah ketika didekati dari kiri}\\\ &\textrm{yaitu}\: \: x\rightarrow \color{red}0^{-},\: \: \color{black}\textrm{maka}\: \: \underset{x\rightarrow \color{red}0^{-}}{\textrm{lim}} \: \cos x\\ &\begin{array}{|c|c|c|c|c|c|c|}\hline x&-0,5&-0,4&-0,3&-0,2&-0,1&0\\\hline \cos x&...&...&0,999986&0,999994&0,9999985&1\\\hline \end{array}\\ &\textrm{Sedangkan ketika}\\ &\textrm{didekati dari arah kanan yaitu}\: \: x\rightarrow \color{red}0^{+},\\ &\textrm{maka}\: \: \underset{x\rightarrow \color{red}0^{+}}{\textrm{lim}} \:\cos x\\ &\begin{array}{|c|c|c|c|c|c|c|}\hline x&0&0,1&0,2&0,3&0,4&0,5\\\hline \cos x&1&0,9999985&0,999994&0,999986&...&...\\\hline \end{array}\\ &\textrm{Karena nilai}\: \: \underset{x\rightarrow \color{red}0^{-}}{\textrm{lim}} \:f(x)= \underset{x\rightarrow \color{red}0^{+}}{\textrm{lim}} \:f(x)=1,\: \: \textrm{maka}\\ &\textrm{nilai}\: \: \underset{x\rightarrow \color{red}0}{\textrm{lim}} \:\cos x\: \: \textbf{ ada}\\ &\textrm{Berikut ilustrasi gambarnya} \end{aligned} \end{array}$.

$\color{blue}\textrm{B. Sifat-Sifat Limit Fungsi}$

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

Fungsi Eksponen

$\Large\textrm{A. Bilangan Pangkat Positif}$

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

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

$\color{purple}\begin{aligned}\textrm{Bilangan}&:\\ \color{blue}a&\: \: \textrm{disebut basis atau bilangan pokok}\\ \color{blue}n&\: \: \textrm{disebut sebagai bilangan pangkat/eksponen} \end{aligned}$.

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

$\color{purple}(1).\quad 3^{4}=3\times 3\times 3\times 3=81$

$\color{purple}(2).\quad 5^{4}=5\times 5\times 5\times 5=625$

$\color{purple}(3).\quad 2^{6}=2\times 2\times 2\times 2\times 2\times 2=64$

$\color{purple}(4).\quad 6^{7}=6\times 6\times 6\times 6\times 6\times 6\times 6=279936$

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

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

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

$\color{purple}(8).\quad \left ( -\displaystyle \frac{1}{2} \right )^{3}=\left ( -\displaystyle \frac{1}{2} \right )\times \left (- \displaystyle \frac{1}{2} \right )\times \left ( -\displaystyle \frac{1}{2} \right )=- \displaystyle \frac{1}{8}$

$\Large\textrm{B. Sifat-Sifat Bilangan Pangkat Positif}$

$\begin{aligned}\\ 1.\quad&a^{m}.a^{n}=a^{m+n}\\ 2.\quad&a^{m}:a^{n}=a^{m-n}\\ 3.\quad&\left ( a^{m} \right )^{n}=a^{m.n},\: \: \textrm{syarat}\: \: a\neq 0\\ 4.\quad&\left ( ab \right )^{n}=a^{n}.b^{n}\\ 5.\quad&\left ( \frac{a}{b} \right )^{n}=\frac{a^{n}}{b^{n}},\: \: \textrm{syarat}\: \: b\neq 0 \end{aligned}$

Beberpa hal yang perlu diketahui juga, yaitu

$\color{blue}\begin{aligned}\\ 1.\quad&(a+b)^{2}=a^{2}+2ab+b^2\\ 2.\quad&(a+b)^{3}=a^{3}+3a^{2}b+3ab^{2}+b^{3}\\ 3.\quad&\left ( a+\frac{1}{a} \right )^{2}=a^{2}+2+\displaystyle \frac{1}{a^{2}},\: \: \textrm{syarat}\: \: a\neq 0\\ \end{aligned}$

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

$\color{blue}\begin{aligned}\\ (1).\quad&2^{6} \times 2^{4} \times 2^{7} = 2^{6+4+7}=2^{17}\\ (2).\quad&2^{5} \times 3^{5} \times 7^{5} = \left ( 2 . 3 . 7 \right )^{5}=\left ( 42 \right )^{5}\\ (3).\quad&\displaystyle \frac{a^{3}.a^{7}.a^{6}}{a^{9}}=\displaystyle \frac{a^{3+7+6}}{a^{9}}=\frac{a^{16}}{a^{9}}=a^{16-9}=a^{7},\: \: \textrm{syarat}\: \: a\neq 0\\ \end{aligned}$

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

$\color{blue}\begin{aligned}(5).\quad\displaystyle \frac{2^{2020}+2^{2021}+2^{2022}}{7}&=\displaystyle \frac{1.2^{2020}+2^{1}.2^{2020}+2^{2}.2^{2020}}{7}\\ &=\frac{\left ( 1+2+4 \right ).2^{2020}}{7}\\ &=\frac{7.2^{2020}}{7}\\ &=2^{2020} \end{aligned}$

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

$\LARGE\textrm{C. Bentuk Akar}$

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.

$\color{purple}\begin{aligned}&\\ 1.\quad&a^{ \frac{1}{n}}=\sqrt[n]{a}\\ 2.\quad&a^{\frac{m}{n}}=\sqrt[n]{a^{m}}\\ 3.\quad&a^{\frac{1}{2}}=\sqrt[2]{a^{1}}=\sqrt{a} \end{aligned}$.

$\textrm{Cara membaca}$.

$\begin{aligned}1.\quad&\sqrt[n]{p}\: \: \: \textbf{dibaca}\: \: \: \color{red}\textrm{akar pangkat n dari p}\\ 2.\quad&\sqrt[n]{p^{2}}\: \: \: \textbf{dibaca}\: \: \: \color{red}\textrm{akar pangkat n dari p kuadrat}\\ 3.\quad&\sqrt[n]{p^{3}}\: \: \: \textbf{dibaca}\: \: \: \color{red}\textrm{akar pangkat n dari p pangkat tiga}\\ 4.\quad&\sqrt{p}\: \: \: \textbf{dibaca}\: \: \: \color{red}\textrm{akar dari p}\: \: \: \color{black}\textrm{atau}\\ &\qquad\qquad\qquad \color{red}\textrm{akar kuadrat dari p}\\ &\qquad\qquad\qquad \textrm{ingat bahwa}:\: \: \sqrt{p}=\sqrt[2]{p} \end{aligned}$.

$\begin{aligned}\color{blue}\textrm{Defini}&\color{blue}\textrm{si}\\ \textrm{Jika}\: &\: a\: \: \textrm{dan}\: \: b\: \: \textrm{bilangan real dan}\\ &n\: \: \textrm{bilangan bulat positif, maka}:\\ &a^{n}=b\Leftrightarrow \sqrt[n]{b}=a\\ \textrm{keter}&\textrm{angan}:\\ \sqrt[n]{b}&\quad \textrm{disebut}\: \: \textbf{akar (radikal)}\\ b&\quad \textrm{disebut}\: \: \textbf{radikan}\\ &\quad \textrm{(bilangan pokok yang ditarik akarnya)}\\ n&\quad \textrm{disebut}\: \: \textbf{indeks}\\ &\quad (\textrm{pangkat akar}) \end{aligned}$.

$\Large\textrm{C.1 Bilangan Pangkat Pecahan}$.

Operasi Bilangan pangkat pecahan sama dengan operasi pangkat bilangan bulat.

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

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

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

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

$\begin{array}{ll}\\ 8.&\textrm{Jabarkanlah bentuk}\\ &\qquad\qquad\quad \left ( 2m^{.^{\frac{3}{2}}}+n^{.^{\frac{3}{4}}} \right )^{2}\\\\ &\color{blue}\textrm{Jawab}:\\ &\begin{aligned}&\left ( 2m^{.^{\frac{3}{2}}}+n^{.^{\frac{3}{4}}} \right )^{2}\\ &=\left ( 2m^{.^{\frac{3}{2}}} \right )^{2}+2\left ( 2m^{.^{\frac{3}{2}}} \right )\left ( n^{.^{\frac{3}{4}}} \right )+\left ( n^{.^{\frac{3}{4}}} \right )^{2}\\ &\color{red}\textrm{INGAT}\: :\: \: \color{black}\left ( A+B \right )^{2}=A^{2}+2AB+B^{2}\\ &=2^{2}m^{.^{\frac{3.2}{2}}}+2.2.m^{.^{\frac{3}{2}}}n^{.^{\frac{3}{4}}}+n^{.^{\frac{3.2}{4}}}\\ &=4m^{3}+4m^{.^{\frac{3}{2}}}n^{.^{\frac{3}{4}}}+n^{.^{\frac{3}{2}}} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 9.&\textrm{Jabarkanlah bentuk}\\ &\qquad\qquad\quad \left ( 2m^{.^{\frac{3}{2}}}-n^{.^{\frac{3}{4}}} \right )^{3}\\\\ &\color{blue}\textrm{Jawab}:\\ &\begin{aligned}&\left ( 2m^{.^{\frac{3}{2}}}-n^{.^{\frac{3}{4}}} \right )^{3}\\ &=\left ( 2m^{.^{\frac{3}{2}}} \right )^{3}-3\left ( 2m^{.^{\frac{3}{2}}} \right )^{2}\left ( n^{.^{\frac{3}{4}}} \right )+3\left ( 2m^{.^{\frac{3}{2}}} \right )\left ( n^{.^{\frac{3}{4}}} \right )^{2}-\left ( n^{.^{\frac{3}{4}}} \right )^{3}\\ &\color{red}\textrm{INGAT}\: :\: \: \color{black}\left ( A-B \right )^{3}=A^{3}-3A^{2}B+3AB^{2}-B^{3}\\ &=2^{3}m^{.^{\frac{3.3}{2}}}-3.2^{2}.m^{.^{\frac{3.2}{2}}}n^{.^{\frac{3}{4}}}+3.2.m^{.^{\frac{3}{2}}}n^{.^{\frac{3.2}{4}}}-n^{.^{\frac{3.3}{4}}}\\ &=8m^{.^{\frac{9}{2}}}-12m^{3}n^{.^{\frac{3}{4}}}+6m^{.^{\frac{3}{2}}}n^{.^{\frac{3}{2}}}-n^{.^{\frac{9}{2}}} \end{aligned} \end{array}$.

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

DAFTAR PUSTAKA

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)

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

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