PAS MATEMATIKA BLOG

Lingkaran

$\color{blue}\textrm{A. Definisi Lingkaran}$.

Secara definisi lingkaran adalah tempat kedudukam titik-titik yang berjarak sama terhadap satu titik tertentu. Selanjutnya titik tertentu disebut sebagai pusat lingkaran sedangkan jarak yang salalu sama terhadapa titik tertentu tersebut disebut sebagai jari-jari atau radius (r).

Sebagai ilustrasi berikut diberikan gambar berkaitan kedudukan titik-titik tersebut

$\color{blue}\textrm{B. Persamaan Lingkaran Berpusat di O(0,0) }$.

Persamaan sebuah lingkaran dengan dengan jari-jari $r$ dan berpusat di titik pusat koordinat dapat dilustrasikan sebagai berikut

Misalkan sebuah titik $\textrm{P}(x,y)$ terletak pada sebuah lingkaran yang berpusat di O(0,0). Dan titik $\textrm{P}'(x,0)$ adalah proyeksi titik P pada sumbu-X sehingga $\bigtriangleup \textrm{OP}'\textrm{P}$ berupa sebuah segitiga siku-siku di $\textrm{P}'$. Dengan rumus Pythagoras kita mendapatkan

$\begin{aligned}&OP^{2}=(OP')^{2}+(PP')^{2}\\ &\Leftrightarrow \: r^{2}=x^{2}+y^{2}\\ &\Leftrightarrow r=\sqrt{x^{2}+y^{2}} \end{aligned}$

Untuk lebih memudahkan pemahaman Anda, perhatikanlah ilustrasi berikut

Sehingga dapat disimpulkan persamaan lingkaran yang berpusat di O(0,0) adalah:

$\begin{array}{|ccc|}\hline &&\\ &\color{red}x^{2}+y^{2}=r^{2}&\\ &&\\\hline \end{array}$.

$\color{blue}\textrm{C. Persamaan Lingkaran Berpusat di (a,b)}$.

Perhatikanlah ilustrasi berikut

Pada ilustrasi gambar di atas ditunjukkan sebuah lingkaran berpusat di $N(a,b)$ dengan jari-jari $r$, misalkan kita ambil sebuah titik $P(x,y)$ pada keliling lingkaran, maka $NP=r$.

$\begin{aligned}&\sqrt{(x-a)^{2}+(y-b)^{2}}=r^{2}\\ &\color{red}(x-a)^{2}+(y-b)^{2}=r^{2}\\ &\textrm{persamaan di atas adalah}\: \: \textbf{Bentuk Umum}\\ &\textrm{dari}\: \: \textbf{Persamaan Lingkaran}\: \: \textrm{yang}\\ &\textrm{berpusat di}\: \: (a,b) \end{aligned}$

Selanjutnya perhatikanlah rangkuman berikut

$\begin{array}{|l|c|c|}\hline \textrm{Lingkaran} &x^{2}+y^{2}=r^{2}&(x-p)^{2}+(y-q)^{2}=r^{2}\\\hline \textrm{Pusat}&(0,0)&(p,q)\\\hline \textrm{Jari-jari}&r&r\\\hline \begin{aligned}&\textrm{Pesamaan garis}\\ &\textrm{singgung melalui}\\ &\textrm{titik}\: \: (x_{1},y_{1})\\ &\textrm{pada lingkaran} \end{aligned}&x_{1}x+y_{1}y=r^{2}&\begin{aligned}&(x_{1}-p)(x-p)\\ &\: +(y_{1}-q)(y-q)=r^{2} \end{aligned}\\\hline \begin{aligned}&\textrm{Persamaan garis}\\ &\textrm{singgung dengan}\\ &\textrm{gradien}\: \: m \end{aligned}&\begin{aligned}&y=mx\\ &\: \pm r\sqrt{m^{2}+1} \end{aligned}&\begin{aligned}&(y-q)=m(x-a)\\ &\: \pm r\sqrt{m^{2}+1} \end{aligned}\\\hline \end{array}$.

Kusus untuk yang pusat $(a,b)$ adalah:

$\begin{array}{|l|c|}\hline \textrm{Lingkaran} &x^{2}+y^{2}+Ax+By+C=0\\\hline \textrm{Pusat}&\left ( -\frac{1}{2}A,-\frac{1}{2}B \right )\\\hline \textrm{Jari-jari}&r=\sqrt{\displaystyle \frac{1}{4}\left ( A^{2}+B^{2} \right )-C}\\\hline \begin{aligned}&\textrm{Pesamaan garis}\\ &\textrm{singgung melalui}\\ &\textrm{titik}\: \: (x_{1},y_{1})\\ &\textrm{pada lingkaran} \end{aligned}&\begin{aligned}&x_{1}x+y_{1}y\\ &\: +\displaystyle \frac{A}{2}(x_{1}+x)\\ &\: +\displaystyle \frac{B}{2}(y_{1}+y)+C=0 \end{aligned}\\\hline \begin{aligned}&\textrm{Persamaan garis}\\ &\textrm{singgung dengan}\\ &\textrm{gradien}\: \: m \end{aligned}&\begin{aligned}&y+\frac{1}{2}B=m(x+\frac{1}{2}A)\\ &\: \pm \sqrt{\displaystyle \frac{1}{4}\left ( A^{2}+B^{2} \right )-C}.\sqrt{m^{2}+1} \end{aligned}\\\hline \end{array}$

$\color{blue}\textrm{D. Kedudukan Titik Terhadap Lingkaran }$.

Kedudukan sebuah titik terhadap sebuah lingkaran yang berpusat di O(0,0) memiliki 3 kemungkinan, yaitu:

jika titik A(x,y) di dalam lingkaran, maka berlaku $x^{2}+y^{2}<r^{2}$.
jika titik A(x,y) pada lingkaran, maka berlaku $x^{2}+y^{2}=r^{2}$, dan
jika titik A(x,y) di luar lingkaran, maka berlaku $x^{2}+y^{2}>r^{2}$.

Demikian juga kedudukan sebuah titik terhadap sebuah lingkaran yang berpusat di $(a,b)$ memiliki 3 kemungkinan, yaitu:

jika titik A(x,y) di dalam lingkaran, maka berlaku $(x-a)^{2}+(y-b)^{2}<r^{2}$ atau $x^{2}+y^{2}+Ax+By+C<0$.
jika titik A(x,y) pada lingkaran, maka berlaku $(x-a)^{2}+(y-b)^{2}=r^{2}$ atau $x^{2}+y^{2}+Ax+By+C=0$.
jika titik A(x,y) di luar lingkaran, maka berlaku $(x-a)^{2}+(y-b)^{2}>r^{2}$ atau $x^{2}+y^{2}+Ax+By+C>0$.

$\color{blue}\textrm{E. Kedudukan Garis Terhadap Lingkaran }$.

Posisi garis terhadap lingkaran tergantung nilai Diskriminan (D) hasil substitusi persamaan garis ke persamaan lingkaran.

$\begin{cases} \bullet &\textrm{memotong lingkaran di dua titik}\: \: (D>0)\\ & \textrm{ada garis dan titik polar} \\ \bullet &\textrm{menyinggung lingkaran}\: \: (D=0) \\ \bullet &\textrm{tidak memotong ataupun menyinggung}\: \: (D<0) \end{cases}$.

Berikut Ilustrasi gambarnya

$\color{blue}\textrm{F. Jarak Garis ke Pusat Lingkaran}$.

$\begin{array}{|l|l|}\hline \textrm{Jarak titik}\: \: M(p,q)\: \: \textrm{terhadap pusat}&\\ \textrm{lingkaran}\: \: N(a,b)&\left | MN \right |=r\\ \qquad r=\left | \displaystyle \frac{Ap+Bq+C}{\sqrt{A^{2}+B^{2}}} \right |&\\\hline \end{array}$.

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

$\begin{array}{ll}\\ 1.&\textrm{Sebuah lingkaran yang berpusat pada }\\ &\textrm{pangkal koordinat}\\ &\textrm{a}.\quad \textrm{Tentukanlah persamaan lingkaran }\\ &\qquad\textrm{yang berjari-jari 5}\\ &\textrm{b}.\quad \textrm{Gambarlah lingkaran (pada soal a.) }\\ &\qquad\textrm{pada kertas grafiks}\\ &\textrm{c}.\quad \textrm{Lukislah titik-titik dari},\\ &\qquad A(2,3),\: B(4,3),\: \: \textrm{dan}\: \: C(3,6).\\ &\textrm{d}.\quad \textrm{Nyatakan kedudukan titik-titik}\\ &\qquad A,\: B,\: \textrm{dan}\: C\: \textrm{terhadap lingkaran. }\\ &\qquad\textrm{Di dalam, pada, atau}\\ &\qquad\textrm{beradakah di luar lingkaran}\\ &\textbf{Jawab}:\\ &\textrm{Perhatikanlah ilustrasi berikut} \end{array}$.

$\begin{aligned}\textrm{a}.\quad&\textrm{Diketahui}\: \: r=5\\ &\begin{aligned}&x^{2}+y^{2}=5^{2}\\ &\qquad\qquad \updownarrow\\ &x^{2}+y^{2}=25\\ &\textrm{atau}\\ &L\equiv \left \{ (x,y)|x^{2}+y^{2}=25 \right \} \end{aligned}\\ \textrm{b}.\quad&\textrm{Lihat gambar di atas}\\ \textrm{c}.\quad&\textrm{Lihat juga gambar di atas}\\ \textrm{d}.\quad&\textrm{Dari gambar jelas bahwa}:\\ &\begin{matrix} \bullet \quad \textrm{Titik}\: \: A(2,3)\: \textrm{berada di dalam lingkaran}\\ \bullet \quad \textrm{Titik}\: \: A(4,3)\: \textrm{berada pada lingkaran}\: \: \: \: \: \: \: \\ \bullet \quad \textrm{Titik}\: \: A(3,6)\: \textrm{berada di luar lingkaran}\: \: \: \, \end{matrix} \end{aligned}$

$\begin{aligned}\textrm{a}.\quad&\textrm{Diketahui}\: \: r=5\\ &\begin{aligned}&x^{2}+y^{2}=5^{2}\\ &\qquad\qquad \updownarrow\\ &x^{2}+y^{2}=25\\ &\textrm{atau}\\ &L\equiv \left \{ (x,y)|x^{2}+y^{2}=25 \right \} \end{aligned}\\ \textrm{b}.\quad&\textrm{Lihat gambar di atas}\\ \textrm{c}.\quad&\textrm{Lihat juga gambar di atas}\\ \textrm{d}.\quad&\textrm{Dari gambar jelas bahwa}:\\ &\begin{matrix} \bullet \quad \textrm{Titik}\: \: A(2,3)\: \textrm{berada di dalam lingkaran}\\ \textrm{atau}:(2)^{2}+(3)^{2}=4+9=13<\color{red}25\\ \bullet \quad \textrm{Titik}\: \: A(4,3)\: \textrm{berada pada lingkaran}\: \: \: \: \: \: \: \\ \textrm{atau}:(4)^{2}+(3)^{2}=16+9=25=\color{red}25\\ \bullet \quad \textrm{Titik}\: \: A(3,6)\: \textrm{berada di luar lingkaran}\: \: \: \,\\ \textrm{atau}:(3)^{2}+(6)^{2}=9+36=45>\color{red}25\\ \end{matrix} \end{aligned}$.

$\begin{array}{ll}\\ 2.&\textrm{Tentukanlah persamaan lingkaran}\\ &\textrm{yang berpusat di pangkal koordinat}\\ &\textrm{dan melalui titik}\: \: P(5,-3)\\\\ &\textbf{Jawab}:\\ &\begin{aligned}\textrm{Diketahui}&\: \textrm{pusat lingkaran di pangkal }\\ \textrm{koordinat}&\: \: O(0,0)\: \: \textrm{serta lingkaran}\\ \textrm{yang mela}&\textrm{lui titik}\: \: P(5,-3),\: \textrm{maka}\\ r&=\sqrt{(x_{p}-0)^{2}+(y_{p}-0)^{2}}\\ &=\sqrt{5^{2}+(-3)^{2}}\\ &=\sqrt{25+9}\\ &=\sqrt{34}\\ \textrm{Sehingga }&,\: \textrm{persamaan lingkarannya adalah}\\ L&\equiv x^{2}+y^{2}=r^{2}\Leftrightarrow x^{2}+y^{2}=\color{red}34 \end{aligned}\end{array}$.

$\begin{array}{ll}\\ 3.&\textrm{Tentukanlah persamaan lingkaran yang }\\ &\textrm{berpusat di pangkal koordinat dan}\\ &\textrm{menyinggung}\: \: k\equiv 2x+y-5=0\\\\ &\textbf{Jawab}:\\ &\textrm{Perhatikan ilustrasi berikut} \end{array}$.

menjadi

$\begin{aligned}&\textrm{Diketahui}\: \textrm{bahwa titik}\: \: O\: \: \textrm{ke garis}\: \: k\: \: \textrm{adalah}\\ &r=OA=\displaystyle \left |\frac{ax_{1}+by_{1}+c}{\sqrt{a^{2}+b^{2}}} \right |\\ &=\displaystyle \left | \frac{2(0)+(0)-5}{\sqrt{2^{2}+1^{2}}} \right |\\ &=\displaystyle \left | \frac{-5}{\sqrt{5}} \right |\\ &=\left | -\sqrt{5} \right |\\ &=-(-\sqrt{5})=\sqrt{5}\\ &\textrm{(ingat, nilai mutlak bilangan negatif adalah bilngan positif)}\\ &\textrm{Sehingga persamaan lingkarannya adalah}:\\ &\qquad L\equiv x^{2}+y^{2}=r^{2}\Leftrightarrow \color{red}x^{2}+y^{2}=5\end{aligned}$.

$\begin{array}{ll}\\ 4.&\textrm{Tentukanlah pusat dan jari-jari lingkaran berikut?}\\ &\textrm{a}.\quad L\equiv (x+1)^{2}+(y+2)^{2}=9\\ &\textrm{b}.\quad L\equiv (x+1)^{2}+(y-2)^{2}=9\\ &\textrm{c}.\quad L\equiv (x-1)^{2}+(y+2)^{2}=9\\ &\textrm{d}.\quad L\equiv (x-1)^{2}+(y-2)^{2}=9\\ &\textrm{e}.\quad L\equiv (x+3)^{2}+(y-3)^{2}=9\\ &\textrm{f}.\quad L\equiv (x-1)^{2}+(y-2)^{2}=25\\ &\textrm{g}.\quad L\equiv (x-1)^{2}+y^{2}=27\\ &\textrm{h}.\quad L\equiv x^{2}+(y-1)^{2}=27\\\\ &\textrm{Jawab}:\\ &L\equiv (x+1)^{2}+(y+2)^{2}=9,\: \: \textrm{pusat di}\: \: (-1,-2)\\ &\textrm{dan jari-jarinya adalah}\: \: \sqrt{9}=3\\ &\textrm{Soal yang belum dibahas silahkan }\\ &\textrm{diselesaikan sendiri sebagai latihan} \end{array}$.

$\begin{array}{ll}\\ 5.&\textrm{Tentukanlah persamaan lingkaran yang }\\ &\textrm{berpusat di}\: \: A(2,-1)\: \: \textrm{dan menginggung}\\ &\textrm{garis}\: \: 4y+3x-12=0\: \: \textrm{di titik}\: \: P\\\\ &\textbf{Jawab}:\\ &\textrm{Perhatikan ilustrasi berikut} \end{array}$.

$\begin{aligned}&\textrm{Sehingga}\\ &r=AP=\left | \frac{3(2)+4(1)-12}{\sqrt{3^{2}+4^{2}}} \right |\\ &\: \: =\left | \frac{-10}{5} \right |=\left | -2 \right |=2\\ &\textrm{Sehingga persamaan lingkarannya adalah}\\\ &L\equiv (x-2)^{2}+(y+1)^{2}=\color{red}4 \end{aligned}$.

$\begin{array}{ll}\\ 6.&\textrm{Tentukanlah pusat dan jari-jari dari }\\ &\textrm{persamaan lingkaran}\\\ & L\equiv 2x^{2}+2y^{2}-2x+6y-3=0\\\\ &\textbf{Jawab}:\\ &\begin{aligned}&\textrm{Persamaan lingkaran}\\\ & L\equiv 2x^{2}+2y^{2}-2x+6y-3=0\\ &\Leftrightarrow \color{blue}x^{2}+y^{2}-x+3y-\displaystyle \frac{3}{2}=0\color{red}\begin{cases} A & =-1 \\ B & =3 \\ C & =-\displaystyle \frac{3}{2} \end{cases}\\ &\textrm{maka}\: \: \begin{cases} \textrm{Pusat} & =\left ( -\displaystyle \frac{-1}{2},- \frac{3}{2}\right )=\left ( \displaystyle \frac{1}{2},-\frac{3}{2} \right ) \\ \textrm{Jari-jari} & =r=\sqrt{\displaystyle \frac{(-1)^{2}}{4}+\frac{3^{2}}{4}-\left ( -\frac{3}{2} \right )}\\ &=\sqrt{\displaystyle \frac{1}{4}+\frac{9}{4}+\frac{6}{4}}=\sqrt{4}=2 \end{cases}\\ &\textrm{Jadi, lingkaran}\: \: 2x^{2}+2y^{2}-2x+6y-3=0\\\ & \textrm{berpusat di} \: \: \left ( \displaystyle \frac{1}{2},-\frac{3}{2} \right )\: \: \textrm{dan berjari-jari}\: \: 2\end{aligned} \end{array}$.

$\begin{array}{ll}\\ 7.&\textrm{Diketahui persamaan lingkaran}\\\ &L\equiv 2x^{2}+2y^{2}-4x+3py-30=0\\ & \textrm{dan melalui titik}\: \: (-2,1).\: \textrm{Tentukanlah }\\ &\textrm{persamaan lingkaran baru yang} \\ &\textrm{kosentris(sepusat) dan panjang jari-jarinya}\\ &\textrm{dua kali panjang jari-jari lingkaran semula?}\\\\ &\textbf{Jawab}\\ &\begin{aligned}&\textrm{Diketahui persamaan lingkaran}\\ &2x^{2}+2y^{2}-4x+3py-30=0,\: \: \textrm{melalui}\\ &(-2,1), \: \: \textrm{maka}\\ &\begin{aligned}&\textrm{kita tentukan harga}\: \: p\: \: \textrm{dulu, yaitu}:\\ &2(-2)^{2}+2(1)^{2}-4(-2)+3p(1)-30=0\\ &\Leftrightarrow \: \: 8+2+8+3p-30=0\\ &\Leftrightarrow \: \: 3p=12\\ &\Leftrightarrow \: \: \color{red}p=4 \end{aligned}\\ &\textrm{Akibatnya persamaan lingkaran menjadi}\\ &\begin{aligned}&2x^{2}+2y^{2}-4x+12y-30=0\\ &\Leftrightarrow \: \: x^{2}+y^{2}-2x+6y-15=0 \end{aligned}\\ &\begin{aligned}&\begin{cases} \color{blue}\textrm{Pusat}: \\ \left ( -\displaystyle \frac{1}{2}A,-\frac{1}{2}B \right )\\ =\left ( -\displaystyle \frac{1}{2}.(-2),-\frac{1}{2},6 \right )\\ =(1,-3) \\\\ \color{blue}\textrm{Jari-jari }:\\ \begin{aligned}r&=\sqrt{\left ( -\frac{1}{2}A \right )^{2}+\left ( -\frac{1}{2}B \right )-C}\\ &=\sqrt{1^{2}+(-3)^{2}-(-15)}\\ &=\sqrt{1+9+15}=5 \end{aligned} \end{cases} \\ & \end{aligned}\\ &\textrm{Selanjutnya}\\ &\begin{aligned}&\textrm{Persamaan}\: \textrm{lingkaran baru }\\ &\textrm{dengan pusat}\: \: (1,-3)\: \: \textrm{dan jari-jari}\\ & r_{\textrm{baru}}=2r=2.5=10\\ &(x-1)^{2}+(y+3)^{2}=(10)^{2}\\ &\Leftrightarrow \: \: x^{2}-2x+1+y^{2}+6x+9=100\\ &\Leftrightarrow \: \: \color{red}x^{2}+y^{2}-2x+6y-90=0\end{aligned} \end{aligned} \end{array}$

Berikut ilustrasi gambarnya

$\begin{array}{ll}\\ 8.&\textrm{Tentukanlah nilai}\: \: p\: \: \textrm{supaya lingkaran}\\ & x^{2}+y^{2}-px-10y+4=0 \\ &\textrm{a}.\quad \textrm{menyinggung sumbu x}\\ &\textrm{b}.\quad \textrm{memotong sumbu x di dua titik}\\ &\textrm{c}.\quad \textrm{tidak memotong dan tidak menyinggung }\\ &\quad\: \: \: \: \textrm{sumbu x}\\\\ &\textbf{Jawab}:\\ &\begin{aligned}\\ &\textrm{Persamaan lingkaran}:\\ &x^{2}+y^{2}-px-10y+4=0\\ &\textrm{saat menyinggung}\: \: \textrm{sumbu x},\: \: \textrm{maka}\: \: y=0\\ &\textrm{adalah gar}\textrm{is yang sejajar sumbu x, maka}\\ &y=0\Rightarrow \: \: x^{2}+y^{2}-px-10y+4=0\\ &\: \qquad \Leftrightarrow \: \: x^{2}+0^{2}-px-0+4=0\\ &\: \qquad \Leftrightarrow \: \: \color{red}x^{2}-px+4 \end{aligned}\\ &\textrm{Selanjutnya}\\ &\begin{array}{|c|c|c|}\hline \textrm{Menyinggung}&\textrm{memotong}&\textrm{Tidak keduanya}\\\hline \begin{aligned}D&= b^{2}-4ac=0\\ &\Leftrightarrow p^{2}-4.1.4=0\\ &\Leftrightarrow p^{2}=16\\ &\Leftrightarrow p=\pm 4\\ & \end{aligned}&\begin{aligned}D&>0\\ &\Leftrightarrow b^{2}-4ac>0\\ &\Leftrightarrow p^{2}-16>0\\ &\Leftrightarrow (p+4)(p-4)>0\\ &\therefore \quad p<-4\: \: \textrm{atau}\: \: p>4 \end{aligned}&\begin{aligned}D&<0\\ &\Leftrightarrow b^{2}-4ac<0\\ &\Leftrightarrow p^{2}-16<0\\ &\Leftrightarrow (p+4)(p-4)<0\\ &\therefore \quad -4<p<4 \end{aligned}\\\hline \end{array} \end{array}$.

$\begin{array}{ll}\\ 9.&\textrm{Tentukanlah nilai}\: \: a\: \: \textrm{supaya lingkaran}\\ & x^{2}+y^{2}=1\: \: \textrm{dan garis}\: \: y=ax+2 \\ &\textrm{a}.\quad \textrm{bersinggungan}\\ &\textrm{b}.\quad \textrm{berpotongan}\\ &\textrm{c}.\quad \textrm{tidak berpotongan maupun bersinggungan}\\\\ &\textbf{Jawab}:\\ &\begin{aligned}&\textrm{Di sini yang kita bahas adalah yang poin b, }\\ &\textrm{yaitu untuk}\: \: y=ax+2,\: \: \textrm{maka}\\ &x^{2}+y^{2}=1\\ &x^{2}+(ax+2)^{2}=1\\ &x^{2}+a^{2}x^{2}+4ax+4=1\\ &(1+a^{2})x^{2}+4ax+3=0\\ &\textrm{syarat berpotongan}\: \: D=b^{2}-4ac\geq 0\\ &(\textrm{artinya bersinggungan sekaligus berpotongan di 2 titik})\\ &(4a)^{2}-4(1+a^{2})(3)\geq 0\\ &16a^{2}-12a^{2}-12\geq 0\\ &4a^{2}-12\geq 0\\ &a^{2}-3\geq 0\\ &(a+\sqrt{3})(a-\sqrt{3})\geq 0\\ &\therefore \: \: \: \: a\leq -\sqrt{3}\: \: \textrm{atau}\: \: a\geq \sqrt{3} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 10.&\textrm{Tentukanlah persamaan garis singgung lingkaran}\\ & x^{2}+y^{2}=12\: \: \textrm{dan melalui titik}\: \: P(0,4)\\\\ &\textbf{Jawab}:\\ &\begin{aligned}&\textrm{Diketahui persamaan lingkaran}\\ &x^{2}+y^{2}=12\\ &\textrm{Persamaan garis singgung lingkaran}\\ &\textrm{melalui titik}\: \: (x_{1},y_{1})\: \: \textrm{adalah}:\\ &\begin{aligned}x^{2}+y^{2}&=12\\ xx+yy&=12\\ x_{1}x+y_{1}y&=12\\ \textrm{garis ini melalui}&\: \: \textrm{titik}\\ P(0,4)&, \textrm{maka}\\ x_{1}.0+y_{1}.4&=12\\ y_{1}&=3\: ......(1)\end{aligned}\\ &\textrm{Karena titik}\: \: (x_{1},y_{1})\: \: \textrm{pada lingkaran}\\ &\textrm{maka},\\ &x_{1}^{2}+y_{1}^{2}=12\: ......(2)\\ &\textrm{Selanjutnya dari persamaan}\: \: (1)\: \&\: (2)\\ &\textrm{akan diperoleh}\\ &\begin{aligned}&x_{1}^{2}+y_{1}^{2}=12\\ y_{1}=3\Rightarrow \: \: &x_{1}^{2}+(3)^{2}=12\\ \Leftrightarrow \: \: &x_{1}^{2}+9=12\\ \Leftrightarrow \: \: &x_{1}^{2}=3\\ \Leftrightarrow \: \: &x_{1}=\pm \sqrt{3}\end{aligned}\\ &\begin{aligned}& \textrm{Sehingga persa}\textrm{maan garis singgungnya}\\ &\left ( x_{1}x+y_{1}y=12 \right )\: \: \textrm{adalah}:\\ &\begin{cases} \textrm{di titik} & (x_{1},y_{1})=(\sqrt{3},3)\: \: \, \, \Rightarrow \color{red}\sqrt{3}x+3y=12\\ \textrm{di titik} & (x_{1},y_{1})=(-\sqrt{3},3)\Rightarrow \color{red}-\sqrt{3}x+3y=12 \end{cases}\\ &\end{aligned} \end{aligned} \end{array}$

Berikut ilustrasi gambarnya

DAFTAR PUSTAKA

Kartini, Suprapto, Subandi, Setiadi, U. 2005. Matematika Program Studi Ilmu Alam Kelas XI untuk SMA dan MA. Klaten: INTAN PARIWARA.
Sobirin. 2006. Kompas Matematika Strategi Praktis Menguasai Tes Matematika SMA Kelas 2. Jakarta: KAWAN PUSTAKA.
Wirodikromo, S. 2007. Matematika Jilid 2 IPA untuk Kelas XI. Jakarta: ERLANGGA.

Vektor di Dimensi Tiga (Ruang)

$\color{blue}\textrm{A. Letak Titik Dalam Ruang}$

Perhatikan titik P dalam ruang berikut

Kita ambil sistem ortogonal ruang yang terdiri atas tiga bidang yang saling berpotongan tegak lurus menurut tiga potong garis sebagaimana ilustrasi gambar di atas.Ketiga garis potong tersebut berturut-turut adalah sumbu X, sumbu Y, dan sumbu Z.

Letak suatu titik P terhadap bidang OYZ, OXZ, dan OXY. Titik P(a,b,c) berarti P berjarak $\color{blue}\textrm{a}$ terhadap OYZ dan berjarak $\color{blue}\textrm{b}$ terhadap bidang OXZ, serta berjarak $\color{blue}\textrm{c}$ terhadap bidang OXY. Jadi, pasangan tiga buah bilangan riil berurutan menyatakan suatu titik dalam ruang dan demikian sebaliknya.

$\color{blue}\textrm{B. Vektor Dalam Ruang}$

Sebagaimana halnya vektor dalam bidang (dimensi dua), maka vektor dalam ruang atau $\textrm{R}^{3}$ juga dapat dinyatakan dengan 3 bilangan riil yang berbeda.

Sebagai misal

$\begin{aligned}\color{blue}P&(x,y,z), \: \: \textrm{maka vektor posisi}\: \: \overline{OP}\\ &\textrm{dalam ruang adalah}\\ &\overline{OP}=(x,y,z)\: \: \: \color{red}\textrm{atau}\\ &\overline{OP}=\begin{pmatrix} x\\ y\\ z \end{pmatrix}\: \: \: \textrm{ataupun juga dengan}\\ &\textrm{dengan vektor satuannya dalam}\\ &\textrm{hal ini adalah}:\bar{i},\: \bar{j},\: \&\: \bar{k}\: \: \textrm{adalah}:\\ &\overline{OP}=x\bar{i}+y\bar{j}+z\bar{k}\\\\ &\textrm{Selanjutnya}\\ &\textrm{vektor-vektor satuan di atas}\\ &\textrm{dalam ruang adalah}\\ &\bar{i}=\begin{pmatrix} 1\\ 0\\ 0 \end{pmatrix}\: \: \textrm{dan}\\ &\bar{j}=\begin{pmatrix} 0\\ 1\\ 0 \end{pmatrix},\: \: \textrm{serta}\\ &\bar{k}=\begin{pmatrix} 0\\ 0\\ 1 \end{pmatrix} \end{aligned}$.

$\color{blue}\textrm{C. Vektor Basis Dalam Ruang}$

Perhatikan ilustrasi berikut

Dengan penjelasan hampir kurang lebih sama pada poin B di atas, maka vektor basis suatu titik P adalah vektor yang dinyatakan dengan vektor satuan, yaitu: $\overline{OP}=x\bar{i}+y\bar{j}+z\bar{k}$.

Sebagai contoh titik P(3,5,8) jika ilustrasikan adalah sebagai berikut

dan jika dinyakatan dengan vektor basis adalah : $\overline{OP}=3\bar{i}+5\bar{j}+8\bar{k}$.

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

Nyatakan vektor berikut dalam bentuk vektor baris, kolom dan basis

$\begin{aligned}&\textbf{Jawab}:\\ &\textrm{Diktahui vektor adalah}\: \: \vec{w}, \: \: \textrm{boleh juga}\\ &\textrm{disebut sebagai vektor posisi, sehingga}\\ &\textrm{dapat tuliskan juga dengan}:\: \vec{w}=\overline{OW}\\ &\textrm{Dan vektor posisi tersebut dapat dinyatakan}\\ &\textrm{dengan}:\\ &\textrm{a}.\quad \textrm{vektor baris}=(8,-6,13)\\ &\textrm{b}.\quad \textrm{vektor kolom}=\begin{pmatrix} 8\\ -6\\ -13 \end{pmatrix}\\ &\textrm{c}.\quad \textrm{vektor basis}=8\vec{i}-6\vec{j}-13\vec{k}\\ \end{aligned}$

$\color{blue}\textrm{D. Vektor Posisi dan Vektor Bebas dalam Ruang}$

Vektor posisi suatu titik adalah vektor yang titik pangkalnya terletak dititik pusat koordinat dan titik ujungntya pada titik tersebut. Sedangkan vektor bebas di sini adalah sembarang vektor yang titik pangkalnya tidak berada pada pusat koordinat dan berujung pada suatu titik. Perhatikanlah ilustrasi berikut ini

Pada ilustrasi gambar di atas ada dua vektor posisi yaitu $\overrightarrow{OA}=\bar{a}$ dan $\overrightarrow{OB}=\bar{b}$, sedangkan vektor bebas atau vektor sembarangnya adalah $\overrightarrow{AB}$.

$\begin{aligned}&\textrm{Perhatikan bahwa}\\ &\textrm{pada gambar di atas}\\ &\overrightarrow{OA}+\overrightarrow{AB}=\overrightarrow{OB}\\ &\quad \bar{a}+\overrightarrow{AB}=\bar{b}\\ &\: \: \: \, \quad\quad \overrightarrow{AB}=\bar{b}-\bar{a} \end{aligned}$.

$\color{blue}\textrm{E. Modulus Vektor Dalam Ruang}$

Pengertian modulus vektor dalam ruang memiliki pengertian yang sama dalam bidang cuma yang membedakan adalah kondisinya saja. Karena baik di dalam ruang maupun bidang dalam menentukan modulus/bebsar/panjang suatu vektor adalah sama saja.

Jika suatu titik $A(x_{1},y_{1},z_{1})$ dan $B(x_{2},y_{2},z_{2})$, maka modulus dari dari kedua titik itu adalah jarak antara kedua titik tersebut, yaitu:

$\left | \overline{AB} \right |=\sqrt{(x_{1}-x_{2})^{2}+(y_{1}-y_{2})^{2}+(z_{1}-z_{2})^{2}}$.

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

$\begin{array}{ll}\\ 1.&\textrm{Jika di}\: \: \textrm{R}^{3}\: \: \textrm{diketahui titik}\\ &A(0,0,0),\: B(1,2,3),\: \: \textrm{dan}\: \: C(4,5,6)\\ &\textrm{Tentukanlah panjang}\\ &\textrm{a}.\quad \overline{AB}\\ &\textrm{b}.\quad \overline{AC}\\ &\textrm{c}.\quad \overline{BC}\\\\ &\textrm{Jawab}:\\ &\begin{aligned}\textrm{a}.\qquad \overline{AB}&=\begin{pmatrix} 1-0\\ 2-0\\ 3-0 \end{pmatrix}=\begin{pmatrix} 1\\ 2\\ 3 \end{pmatrix}\\ \left | \overline{AB} \right |&=\sqrt{1^{2}+2^{2}+3^{2}}\\ &=\sqrt{1+4+9}\\ &=\color{red}\sqrt{14} \end{aligned}\\ &\begin{aligned}\textrm{b}.\qquad \overline{AC}&=\begin{pmatrix} 4-0\\ 5-0\\ 6-0 \end{pmatrix}=\begin{pmatrix} 4\\ 5\\ 6 \end{pmatrix}\\ \left | \overline{AC} \right |&=\sqrt{4^{2}+5^{2}+6^{2}}\\ &=\sqrt{16+25+36}\\ &=\color{red}\sqrt{77} \end{aligned} \\ &\begin{aligned}\textrm{c}.\qquad \overline{BC}&=\begin{pmatrix} 4-1\\ 5-2\\ 6-3 \end{pmatrix}=\begin{pmatrix} 3\\ 3\\ 3 \end{pmatrix}\\ \left | \overline{BC} \right |&=\sqrt{3^{2}+3^{2}+3^{2}}\\ &=\sqrt{9+9+9}\\ &=\color{red}\sqrt{27} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 2.&\textrm{Tentukanlah panjang dari vektor}\\ &\bar{q}=3\bar{i}-\bar{j}-7\bar{k}\\\\ &\textrm{Jawab}:\\ &\begin{aligned} \bar{q}&=3\bar{i}-\bar{j}-7\bar{k}=\begin{pmatrix} 3\\ -1\\ -7 \end{pmatrix}\\ \left | \bar{q} \right |&=\sqrt{3^{2}+(-1)^{2}+(-7)^{2}}\\ &=\sqrt{9+1+49}\\ &=\color{red}\sqrt{59} \end{aligned} \end{array}$.

DAFTAR PUSTAKA

Koesmartono, Rawuh (editor). 1973. Matematika Pendahuluan (Seri Matematika). Bandung: ITB

Contoh Soal 4 Fungsi Komposisi dan Fungsi Invers

$\begin{array}{ll}\\ 16.&\textrm{Jika}\: \: f\: \: \textrm{dan}\: \: g\: \textrm{adalah fungsi yang }\\ &\textrm{mempunyai invers dan memenuhi}\\ & f(2x)=g(x-3),\: \textrm{maka}\: \: f^{-1}(x)\\ & \textrm{adalah}\: ....\\ &\begin{array}{llllll}\\ \textrm{a}.&g^{-1}\left ( \displaystyle \frac{x}{2}- \frac{2}{3}\right )\\ \textrm{b}.&g^{-1}\left ( \displaystyle \frac{x}{2} \right )-\displaystyle \frac{2}{3}\\ \textrm{c}.&g^{-1}(2x+6)\\ \textrm{d}.&2g^{-1}(x)-6\\ \color{red}\textrm{e}.&2g^{-1}(x)+6 \end{array}\\ & (\textbf{SBMPTN 2016 Mat Das})\\\\ &\color{blue}\textrm{Jawab}:\\ &\begin{aligned} &\textrm{Misalkan bahwa}\\ &f(2x)=g(x-3)=x,\\ & \textrm{maka}\\ &\begin{cases} f^{-1}(x) &=2x \\ g^{-1}(x) &=x-3 \end{cases} \end{aligned}\\ &\begin{array}{|c|c|}\hline \textrm{Sintak}&\textrm{Hasil}\\\hline \begin{aligned}g^{-1}(x) &=x-3\\ x&=g^{-1}(x)+3\\ & \end{aligned}&\begin{aligned}f^{-1}(x) &=2x\\ &=2\left ( g^{-1}(x)+3 \right )\\ &=2g^{-1}(x)+6 \end{aligned}\\\hline \end{array} \end{array}$.

$\begin{array}{ll}\\ 17.&\textrm{Jika}\: \: f^{-1}(x)=\displaystyle \frac{x-1}{5}\: \: \textrm{dan}\: \: g^{-1}(x)=\displaystyle \frac{3-x}{2},\\ & \textrm{maka}\: \: \left (f\circ g \right )^{-1}(6)=....\\ &\begin{array}{llllll}\\ \textrm{a}.&-1&&&\textrm{d}.&2\\ \textrm{b}.&0\qquad&\color{red}\textrm{c}.&1\qquad&\textrm{e}.&3 \end{array}\\ & (\textbf{UMPTN 1995})\\\\ &\color{blue}\textrm{Jawab}:\\ &\begin{aligned}\textrm{Diketahui b}&\textrm{ahwa}:\\ &\begin{cases} f^{-1}(x)&=\displaystyle \frac{x-1}{5} \\ g^{-1}(x)&=\displaystyle \frac{3-x}{2} \end{cases}\\ \left (f\circ g \right )^{-1}(x)&=\left (g^{-1}\circ f^{-1} \right )(x)\\ &=\displaystyle \frac{3-\left ( \displaystyle \frac{x-1}{5} \right )}{2}\\ \left (f\circ g \right )^{-1}(6)&=\displaystyle \frac{3-\left ( \displaystyle \frac{6-1}{5} \right )}{2}\\ &=\displaystyle \frac{3-1}{2}=\frac{2}{2}\\ &=1 \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 18.&\textrm{Invers dari}\: \: f(x)=125^{x}\: \: \textrm{adalah}\: \: f^{-1}(x)\\ &\textrm{Nilai dari}\: \: f^{-1}\left ( 5\sqrt{5} \right )=\: ....\\ &\begin{array}{llllll}\\ \textrm{a}.&1&&&\textrm{d}.&\displaystyle \frac{3}{5}\\ \color{red}\textrm{b}.&\displaystyle \frac{1}{2}&\textrm{c}.&\displaystyle \frac{1}{6}&\textrm{e}.&-\displaystyle \frac{1}{2} \end{array}\\\\ &\color{blue}\textrm{Jawab}:\\ &\begin{aligned} &\textrm{Diketahui bahwa}\: \: f(x)=125^{x},\: \textrm{maka}\\ &f(x)=y=125^{x}\\ &\textrm{Kedua ruas dilogkan masing-masing}\\ &\log y=\log 125^{x}\\ &\Leftrightarrow \log y=x\log 125\\ &\Leftrightarrow x\log 125=\log y\\ &\Leftrightarrow x=\displaystyle \frac{\log y}{\log 125}\\ &\Leftrightarrow x= \, ^{125}\log y\\ &\Leftrightarrow f^{-1}(x)=\, ^{125}\log x\\ &\textrm{Selanjutnya}\\ &f^{-1}(x)=\, ^{125}\log x\\ &\Leftrightarrow f^{-1}(5\sqrt{5})=\, ^{125}\log (5\sqrt{5})\\ &\Leftrightarrow f^{-1}(5\sqrt{5})=\, ^{5^{3}}\log 5^{.^{ \frac{3}{2}}}\\ &\Leftrightarrow f^{-1}(5\sqrt{5})=\displaystyle \frac{\displaystyle \frac{3}{2}}{3}\,^{5}\log 5\\ &\Leftrightarrow f^{-1}(5\sqrt{5})= \displaystyle \frac{1}{2}.1\\ &\Leftrightarrow f^{-1}(5\sqrt{5})= \color{red}\displaystyle \frac{1}{2} \end{aligned} \end{array}$

Contoh Soal 3 Fungsi Komposisi dan Fungsi Invers

$\begin{array}{ll}\\ 11.&\textrm{Diketahui beberapa fungsi memiliki }\\ &\textrm{sifat-sifat sebagaimana berikut ini}:\\ &(i)\quad \Phi (-x)=-\Phi (x)\: \: \textrm{untuk setiap}\: \: x\\ &(ii)\quad \Phi (-x)=\Phi (x)\: \: \textrm{untuk setiap}\: \: x\\ &\textrm{Jika diketahui fungsi}\: \: f\: \: \textrm{dan}\: \: g\\ & \textrm{memiliki sifat}\: \: (i)\: \: \textrm{dan fungsi}\: \: h\: \: \textrm{dan}\: \: k\\ &\textrm{memiliki sifat}\: \: (ii),\: \: \textrm{maka pernyataan }\\ &\textrm{berikut yang salah adalah}\: ....\\ &(1)\quad (f+g)(-x)=-(f+g)(x)\\ &(2)\quad (f.k)(-x)=-(f.k)(x)\\ &(3)\quad (h-k)(-x)=(h-k)(x)\\ &(4)\quad (h-g)(-x)=(h-g)(x)\\ &\begin{array}{llllll}\\ \color{red}\textrm{a}.&(1),(2)\: \textrm{dan}\: (3)\\ \textrm{b}.&(1)\: \textrm{dan}\: (3)\\ \textrm{c}.&(2)\: \textrm{dan}\: (4)\\ \textrm{d}.&(4)\: \textrm{saja}\\ \textrm{e}.&\textrm{semuanya benar} \end{array}\\ & (\textbf{SIMAK UI 2014 Mat Das})\\\\ &\textrm{Jawab}:\\ &\begin{aligned}&\textrm{Diketahui}\: \textrm{bahwa}:\\ &\begin{cases} \Phi (-x)=-\Phi (x) \\ (\textrm{fungsi ganjil})\begin{cases} f & \text{ misal } f(x)=x \\ g & \text{ misal } g(x)=2x \end{cases} \\\\ \Phi (-x)=\Phi (x)\\ (\textrm{fungsi genap})\begin{cases} h & \text{ misal } h(x)=x^{2} \\ k & \text{ misal } k(x)=2x^{2} \end{cases} \end{cases}\\ & \end{aligned}\\ &\begin{array}{|c|}\hline \begin{aligned}(1)\quad (f+g)(-x)&=-(f+g)(x)\\ &\textrm{benar}\\ (2)\qquad (f.k)(-x)&=-(f.k)(x)\\ &\textrm{benar}\\ \end{aligned}\\\hline \begin{aligned} (3)\quad (h-k)(-x)&=(h-k)(x)\\ &\textrm{benar}\\ (4)\quad (h-g)(-x)&=(h-g)(x)\\ &\color{blue}\textrm{salah} \end{aligned}\\\hline \end{array} \end{array}$.

$\begin{array}{ll}\\ 12.&\textrm{Jika}\: \: f(x)=\displaystyle \frac{1}{2x-1}\: \: \textrm{dan}\: \: (f\circ g)(x)=\displaystyle \frac{x}{3x-2}\\ &\textrm{maka}\: \: g(x)=\: ....\\ &(\textbf{UMPTN 1998})\\ &\begin{array}{llllll}\\ \textrm{a}.&x+\displaystyle \frac{1}{2}&&&\textrm{d}.&1-\displaystyle \frac{2}{x}\\\\ \textrm{b}.&x-\displaystyle \frac{1}{2}&\color{red}\textrm{c}.&2-\displaystyle \frac{1}{x}&\textrm{e}.&2-\displaystyle \frac{1}{2x} \end{array}\\\\ &\textrm{Jawab}:\\ &\begin{aligned}&\color{blue}\textbf{Alternatif 1}\\ &\textrm{Diketahui bahwa}\: \: f(x)=\displaystyle \frac{1}{2x-1},\: \textrm{dan}\\ &(f\circ g)(x)=\displaystyle \frac{x}{3x-2},\: \: \textrm{maka}\\ &\Leftrightarrow (f\circ g)(x)=f\left ( g(x) \right )=\displaystyle \frac{x}{3x-2}\\ &\Leftrightarrow \displaystyle \frac{1}{2g(x)-1}=\displaystyle \frac{x}{3x-2}\\ &\Leftrightarrow \displaystyle \frac{1}{2g(x)-1}=\displaystyle \frac{1}{\left (\displaystyle \frac{3x-2}{x} \right )}\\ &\textrm{Dari bentuk di atas didapatkan}\\ &\color{magenta}2g(x)-1=\displaystyle \frac{3x-2}{x}\\ &\Leftrightarrow 2g(x)=1+\left ( 3-\displaystyle \frac{2}{x} \right )\\ &\Leftrightarrow 2g(x)=4-\displaystyle \frac{2}{x}\\ &\Leftrightarrow g(x)=2-\displaystyle \frac{1}{x} \end{aligned}\\ &\begin{aligned}&\color{blue}\textbf{Alternatif 2}\\ &\textrm{Diketahui bahwa}\: \: f(x)=\displaystyle \frac{1}{2x-1},\: \textrm{dengan}\\ &f^{-1}(x)=\displaystyle \frac{x+1}{2x}.........(\textrm{tunjukkan sendiri})\\ &\textrm{serta}\: \: (f\circ g)(x)=\displaystyle \frac{x}{3x-2},\: \: \textrm{maka}\\ &g(x)=(f^{-1}\circ f\circ g)(x)=(I\circ g)(x)=g(x)\\ &\Leftrightarrow g(x)=\displaystyle \frac{f(g(x))+1}{2\left ( f(g(x)) \right )}\\ &\Leftrightarrow g(x)=\displaystyle \frac{\left ( \displaystyle \frac{x}{3x-2} \right )+1}{2\left ( \displaystyle \frac{x}{3x-2} \right )}\\ &\Leftrightarrow g(x)=\displaystyle \frac{\displaystyle \frac{4x-2}{3x-2}}{\displaystyle \frac{2x}{3x-2}}\\ &\Leftrightarrow g(x)=\displaystyle \frac{4x-2}{2x}\\ &\Leftrightarrow g(x)=2-\displaystyle \frac{1}{x} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 13.&\textrm{Jika}\: \: f(x)=x^{2}-9\: \: \textrm{dan}\: \: (f\circ g)(x)=x(x-6)\\ &\textrm{rumus fungsi}\: \: g(x)=\: ....\\ &\begin{array}{llllll}\\ \textrm{a}.&x+3&&&\textrm{d}.&3x+1\\ \textrm{b}.&x-3&\color{red}\textrm{c}.&-x&\textrm{e}.&x \end{array}\\\\ &\textrm{Jawab}:\\ &\begin{aligned}&\color{blue}\textbf{Alternatif 1}\\ &\textrm{Diketahui bahwa}\: \: f(x)=x^{2}-9,\: \textrm{dan}\\ &(f\circ g)(x)=x(x-6)=x^{2}-6x,\: \: \textrm{maka}\\ &\Leftrightarrow (f\circ g)(x)=f\left ( g(x) \right )=x^{2}-6x\\ &\Leftrightarrow \left (g(x) \right )^{2}-9=x^{2}-6x\\ &\Leftrightarrow \left (g(x) \right )^{2}-9=x^{2}-6x+9-9\\ &\Leftrightarrow \left (g(x) \right )^{2}-9=(x-3)^{2}-9\\ &\textrm{Dari bentuk di atas didapatkan}\\ &\qquad g(x)=x-3 \end{aligned}\\ &\begin{aligned}&\color{blue}\textbf{Alternatif 2}\\ &\textrm{Diketahui bahwa}\: \: f(x)=x^{2}-9,\: \textrm{dengan}\\ &f^{-1}(x)=\sqrt{x+9}\: .......(\textrm{tunjukkan sendiri})\\ &\textrm{serta}\: \: (f\circ g)(x)=x^{2}-6x,\: \: \textrm{maka}\\ &g(x)=(f^{-1}\circ f\circ g)(x)=(I\circ g)(x)=g(x)\\ &\Leftrightarrow g(x)=\sqrt{\left ( f(g(x)) \right )+9}\\ &\Leftrightarrow g(x)=\sqrt{x^{2}-6x+9}\\ &\Leftrightarrow g(x)=\sqrt{(x-3)^{2}}\\ &\Leftrightarrow g(x)=x-3 \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 14.&\textrm{Jika}\: \: f(x)=\displaystyle \frac{1}{\sqrt{x^{2}-2}}\: \: \textrm{dan}\\ &\left ( f\circ g \right )(x)=\displaystyle \frac{1}{\sqrt{x^{2}+6x+7}},\\ & \textrm{maka}\: \: g(x+2)=....\\ &\begin{array}{llllll}\\ \textrm{a}.&\displaystyle \frac{1}{x+3}&&&\textrm{d}.&x+3\\ \textrm{b}.&\displaystyle \frac{1}{x-2}\qquad&\textrm{c}.&x-2\qquad&\color{red}\textrm{e}.&x+5 \end{array}\\ & (\textbf{UM UGM 2010 Mat Das})\\\\ &\textrm{Jawab}:\\ &\color{blue}\textbf{Alternatif 1}\\ &\begin{aligned}\left ( f\circ g \right )(x)&=\displaystyle \frac{1}{\sqrt{x^{2}+6x+7}}\\ f\left ( g(x) \right )&=\displaystyle \frac{1}{\sqrt{x^{2}+6x+7}}\\ \displaystyle \frac{1}{\sqrt{\left (g(x) \right )^{2}-2}}&=\displaystyle \frac{1}{\sqrt{x^{2}+6x+7}}\\ \left (g(x) \right )^{2}-2&=x^{2}+6x+7\\ \left (g(x) \right )^{2}&=x^{2}+6x+9\\ g(x)&=\sqrt{x^{2}+6x+9}=\sqrt{(x+3)^{2}}\\ g(x)&=x+3\\ g(x+2)&=(x+2)+3\\ &=x+5 \end{aligned}\\ &\begin{aligned}&\color{blue}\textbf{Alternatif 2}\\ &\textrm{Diketahui bahwa}\: \: f(x)=\displaystyle \frac{1}{\sqrt{x^{2}-2}},\: \textrm{dengan}\\ &f^{-1}(x)=\sqrt{\displaystyle \frac{1}{x^{2}}+2}\: .......(\textrm{akan ditunjukkan})\\ &\textrm{serta}\: \: (f\circ g)(x)=\displaystyle \frac{1}{\sqrt{x^{2}+6x+7}},\: \: \textrm{maka}\\ &\begin{array}{|c|c|}\hline \begin{aligned}f(x) =y&= \displaystyle \frac{1}{\sqrt{x^{2}-2}}\\ y^{2}&=\displaystyle \frac{1}{x^{2}-2}\\ x^{2}-2&=\displaystyle \frac{1}{y^{2}}\\ x^{2}&=\displaystyle \frac{1}{y^{2}}+2\\ x&=\sqrt{\displaystyle \frac{1}{y^{2}}+2}\\ f^{-1}(y)&=\sqrt{\displaystyle \frac{1}{y^{2}}+2}\\ f^{-1}(x)&=\sqrt{\displaystyle \frac{1}{x^{2}}+2} \end{aligned}&\begin{aligned}g(x)&=\left (f^{-1}\circ f\circ g \right )(x)\\ &=\sqrt{\displaystyle \frac{1}{\left ( \displaystyle \frac{1}{\sqrt{x^{2}+6x+7}} \right )^{2}}+2}\\ &=\sqrt{\left ( x^{2}+6x+7 \right )+2}\\ &=\sqrt{\left ( x^{2}+6x+9 \right )}\\ &=\sqrt{(x+3)^{2}}\\ &=x+3\\ g(x+2)&=(x+2)+3\\ &=x+5\\ & \end{aligned}\\\hline \end{array} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 15.&\textrm{Jika}\: \: g(x)=2x+4\: \: \textrm{dan}\: \: \left ( f\circ g \right )(x)=4x^{2}+8x-3,\\ & \textrm{maka}\: \: f^{-1}(x)=....\\ &\begin{array}{llllll}\\ \textrm{a}.&x+9\\ \textrm{b}.&\sqrt{x}+2\\ \textrm{c}.&x^{2}-4x-3\\ \textrm{d}.&\sqrt{x+1}+2\\ \color{red}\textrm{e}.&\sqrt{x+7}+2 \end{array}\\\\ &\textrm{Jawab}:\\ &\begin{array}{|c|c|c|}\hline \textrm{Sintak 1}&\textrm{ Sintak 2}&\textrm{Hasil Invers}\\\hline \begin{aligned}g(x)=y&=2x+4\\ y-4&=2x\\ x&=\displaystyle \frac{y-4}{2}\\ f^{-1}(y)&=\displaystyle \frac{y-4}{2}\\ f^{-1}(x)&=\displaystyle \frac{x-4}{2}\\ & \end{aligned}&\begin{aligned}f(x)&=\left (f\circ g\circ g^{-1} \right )(x)\\ &=4\left ( g^{^{-1}}(x) \right )^{2}+8\left ( g^{-1}(x) \right )-3\\ &=4\left ( \displaystyle \frac{x-4}{2} \right )^{2}+8\left ( \displaystyle \frac{x-4}{2} \right )-3\\ &=\left ( \displaystyle x^{2}-8x+16 \right )+4x-16-3\\ &=x^{2}-4x-3\\ &=x^{2}-4x+4-7\\ &=(x-2)^{2}-7 \end{aligned}&\begin{aligned}f(x)=y&=(x-2)^{2}-7\\ y+7&=(x-2)^{2}\\ \sqrt{y+7}&=(x-2)\\ (x-2)&=\sqrt{y+7}\\ x&=\sqrt{y+7}+2\\ f^{-1}(y)&=\sqrt{y+7}+2\\ f^{-1}(x)&=\sqrt{x+7}+2 \end{aligned} \\\hline \end{array} \end{array}$

Contoh Soal 2 Fungsi Komposisi dan Fungsi Invers

$\begin{array}{ll}\\ 6.&\textrm{Fungsi}\: \: g:\mathbb{R}\rightarrow \mathbb{R}\: \: \textrm{ditentukan oleh}\\ &g(x)=x^{2}-x+3\: \: \textrm{dan}\: \: f:\mathbb{R}\rightarrow \mathbb{R}\\ &\textrm{sehingga}\: \: (f\circ g)(x)=3x^{2}-3x+4 \:, \\ &\textrm{maka fungsi}\: \: f(x-2)=\: ....\\ &\begin{array}{llllll}\\ \textrm{a}.&2x-11&&&\textrm{d}.&3x-7\\ \textrm{b}.&2x-7&\textrm{c}.&3x+1&\color{red}\textrm{e}.&3x-11 \end{array}\\\\ &\textrm{Jawab}:\\ &\begin{aligned}&\textrm{Diketahui}\: \textrm{bahwa}:\\ &\begin{cases} &\color{blue}f(x) =f(x)\\ &\color{blue}g(x) =x^{2}-x+3\end{cases}\\ &\left ( f\circ g \right )(x)=3x^{2}-3x+4\\ &\begin{aligned}f(g(x))&=3x^{2}-3x+4\\ f(x^{2}-&x+3)=3(x^{2}-x+3)-5\\ \textrm{Sehing}&\textrm{ga}\\ f(x)&=3x-5\\ f(x-2)&=3(x-2)-5\\ &=3x-11 \end{aligned} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 7.&\textrm{Fungsi}\: \: f:\mathbb{R}\rightarrow \mathbb{R}\: \: \textrm{dan}\: \: g:\mathbb{R}\rightarrow \mathbb{R}\\ &\textrm{dengan}\: \: f(x)=x-3\: \: \textrm{dan}\\ & g(x)=x^{2}+5.\: \textrm{Jika}\: \: (f\circ g)(x)=(g\circ f)(x)\\ &\textrm{maka nilai}\: \: x\: \: \textrm{adalah}\: ....\\ &\begin{array}{llllll}\\ \textrm{a}.&1&&&\textrm{d}.&4\\ \color{red}\textrm{b}.&2&\textrm{c}.&3&\textrm{e}.&5 \end{array}\\\\ &\textrm{Jawab}:\\ &\begin{aligned}&\textrm{Diketahui}\: \textrm{bahwa}:\\ &\begin{cases} &\color{blue}f(x) =x-3\\ &\color{blue}g(x) =x^{2}+5\end{cases}\\ &\left ( f\circ g \right )(x)=(g\circ f)(x)\\ &\begin{aligned}f(g(x))&=g(f(x))\\ (x^{2}+5)-3&=(x-3)^{2}+5\\ x^{2}+2&=x^{2}-6x+14\\ 6x&=14-2\\ x&=\displaystyle \frac{12}{6}\\ x&=2 \end{aligned} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 8.&\textrm{Fungsi}\: \: f:\mathbb{R}\rightarrow \mathbb{R}\: \: \textrm{dan}\: \: g:\mathbb{R}\rightarrow \mathbb{R}\\ &\textrm{dengan}\: \: f(x)=3x-10\: \: \textrm{dan}\\ & g(x)=4x+n.\: \textrm{Jika}\\ & (g\circ f)(x)-(f\circ g)(x)=0\\ &\textrm{maka nilai}\: \: n\: \: \textrm{adalah}\: ....\\ &\begin{array}{llllll}\\ \textrm{a}.&15&&&\textrm{d}.&-10\\ \textrm{b}.&10&\textrm{c}.&5&\color{red}\textrm{e}.&-15 \end{array}\\\\ &\textrm{Jawab}:\\ &\begin{aligned}&\textrm{Diketahui}\: \textrm{bahwa}:\\ &\begin{cases} &\color{blue}f(x) =3x-10\\ &\color{blue}g(x) =4x+n\end{cases}\\ &\left ( g\circ f \right )(x)-(f\circ g)(x)=0\\ &\begin{aligned}g(f(x))&=f(g(x))\\ 3(4x+n)-10&=4(3x-10)+n\\ 12x+3n-10&=12x-40+n\\ 2n&=-30\\ x&=\displaystyle \frac{-30}{2}\\ x&=-15 \end{aligned} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 9.&\textrm{Jika}\: \: f(x)=\sqrt{2x+3}\\ & \textrm{dan}\: \: g(x)=x^{2}+1,\: \textrm{maka}\: \: \left ( f\circ g \right )(2)=....\\\\ &\begin{array}{llllll}\\ \textrm{a}.&2,24&&&\textrm{d}.&6\\ \textrm{b}.&3\qquad&\color{red}\textrm{c}.&3,61\qquad&\textrm{e}.&6,16 \end{array}\\ & (\textbf{SAT Subject Test})\\\\\\ &\textrm{Jawab}:\quad \textbf{c}\\ &\begin{aligned}\left ( f\circ g \right )(x)&=f\left ( g(x) \right )\\ &=\sqrt{2g(x)+3}\\ &=\sqrt{2\left ( x^{2}+1 \right )+3}\\ \left ( f\circ g \right )(2)&=\sqrt{2\left ( 2^{2}+1 \right )+3}\\ &=\sqrt{2(5)+3}\\ &=\sqrt{13}\\ &\color{blue}\approx 3,61 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 10.&\textrm{Misalkan}\: \: f(x)=x^{2},\: \: g(x)=2x\\ &\textrm{dan}\: \: h(x)=1-x.\\ & \textrm{Fungsi}\: \: (f\circ g\circ h)(x)=\: ....\\ &\begin{array}{llllll}\\ \color{red}\textrm{a}.&4x^{2}-8x+4\\ \textrm{b}.&4x^{2}+8x-4\\ \textrm{c}.&2x^{2}-4x+1\\ \textrm{d}.&x^{2}-2x+1\\ \textrm{e}.&4-2x+x^{2} \end{array}\\\\\\ &\textrm{Jawab}:\\ &\begin{aligned}\textrm{Dike}&\textrm{tahui bahwa}\\ \bullet \quad&f(x)=x^{2}\\ \bullet \quad &g(x)=2x\\ \bullet \quad&h(x)=1-x\\ \textrm{mak}&\textrm{a}\\ &(f\circ g\circ h)=f\left ( g(h(x)) \right )\\ &\: =\left ( 2(1-x) \right )^{2}\\ &\: =(2-2x)^{2}\\ &\: =4x^{2}-8x+4 \end{aligned} \end{array}$

Contoh 7 Vektor

$\begin{array}{ll}\\ 31.&\textrm{Jika}\: \: \vec{p}=\begin{pmatrix} 2\\ -5 \end{pmatrix},\: \vec{q}=\begin{pmatrix} 4\\ 3 \end{pmatrix},\: \textrm{maka }\\ &\textrm{proyeksi skalar ortogonal vektor}\: \vec{p}\\ &\textrm{pada}\: \: \vec{q}\: \: \textrm{adalah}....\\ &\textrm{a}.\quad \displaystyle \frac{3}{5}\\\\ &\textrm{b}.\quad \color{red}\displaystyle \frac{7}{5}\\\\ &\textrm{c}.\quad \displaystyle \frac{8}{5}\\\\ &\textrm{d}.\quad \displaystyle \frac{9}{5}\\\\ &\textrm{e}.\quad \displaystyle 2\\\\ &\textrm{Jawab}\\ &\begin{aligned}\left | \vec{r} \right |&=\displaystyle \frac{\vec{p}.\vec{q}}{\left | \vec{q} \right |}\\ &=\displaystyle \frac{\begin{pmatrix} 2\\ -5 \end{pmatrix}.\begin{pmatrix} 4\\ 3 \end{pmatrix}}{\sqrt{4^{2}+3^{2}}}\\ &=\displaystyle \frac{8+(-15)}{\sqrt{25}}\\ &=\left | \displaystyle \frac{-7}{5} \right |\\ &=\color{red}\displaystyle \frac{7}{5} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 32.&\textrm{Panjang Proyeksi vektor}\: \: \vec{a}=\begin{pmatrix} 5\\ 1 \end{pmatrix}\\ & \textrm{pada}\: \: \vec{b}=\begin{pmatrix} 0\\ 4 \end{pmatrix}\: \: \textrm{adalah}....\\ &\textrm{a}.\quad \displaystyle -1\\ &\textrm{b}.\quad -\displaystyle \frac{1}{2}\\ &\textrm{c}.\quad \color{red}1\\ &\textrm{d}.\quad \displaystyle 2\\ &\textrm{e}.\quad 4\\\\ &\textrm{Jawab}\\ &\begin{aligned}\left |\vec{c} \right |&=\left |\displaystyle \frac{\vec{a}.\vec{b}}{\left | \vec{b} \right |} \right |\\ &=\left |\displaystyle \frac{\begin{pmatrix} 5\\ 1 \end{pmatrix}\bullet \begin{pmatrix} 0\\ -4 \end{pmatrix}}{\left | \sqrt{0^{2}+(-4)^{2}} \right |} \right |\\ &=\left |\displaystyle \frac{0-4}{4} \right |=\left |-1 \right |=\color{red}1 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 33.&\textrm{Proyeksi vektor ortogonal}\: \: \vec{a}=\begin{pmatrix} 2\\ -4 \end{pmatrix}\\ & \textrm{pada}\: \: \vec{b}=\begin{pmatrix} -1\\ 2 \end{pmatrix}\: \: \textrm{adalah}....\\ &\textrm{a}.\quad \begin{pmatrix} 2\\ -1 \end{pmatrix}\\ &\textrm{b}.\quad \begin{pmatrix} 2\\ -2 \end{pmatrix}\\ &\textrm{c}.\quad \color{red}\begin{pmatrix} 2\\ -4 \end{pmatrix}\\ &\textrm{d}.\quad \begin{pmatrix} -1\\ 2 \end{pmatrix}\\ &\textrm{e}.\quad \begin{pmatrix} -2\\ 4 \end{pmatrix}\\\\ &\textrm{Jawab}\\ &\begin{aligned}\vec{c}&=\left ( \displaystyle \frac{\vec{a}\bullet \vec{b}}{\left |\vec{b} \right |^{2}} \right ).\vec{b}\\ &=\left (\displaystyle \frac{\begin{pmatrix} 2\\ -4 \end{pmatrix}\bullet \begin{pmatrix} -1\\ 2 \end{pmatrix}}{(-1)^{2}+2^{2}} \right ).\begin{pmatrix} -1\\ 2 \end{pmatrix}\\ &=\left (\displaystyle \frac{-2-8}{1+4} \right ).\begin{pmatrix} -1\\ 2 \end{pmatrix}\\ &=-2\begin{pmatrix} -1\\ 2 \end{pmatrix}\\ &=\color{red}\begin{pmatrix} 2\\ -4 \end{pmatrix} \end{aligned} \end{array}$

Contoh 6 Vektor

$\begin{array}{ll}\\ 26.&\textrm{Jika}\: \: \vec{p}=\begin{pmatrix} -2\\ 4 \end{pmatrix}\: \: \textrm{dan}\: \: \vec{q}=\begin{pmatrix} 8\\ 4 \end{pmatrix},\: \: \textrm{maka}\\ &\textrm{sudut yang dibentuk vektor}\: \: \vec{p}\: \: \textrm{dan}\: \: \vec{q}\\ &\textrm{adalah}....\\ &\textrm{a}.\quad 0^{\circ}\\ &\textrm{b}.\quad 60^{\circ}\\ &\textrm{c}.\quad 45^{\circ}\\ &\textrm{d}.\quad 60^{\circ}\\ &\textrm{e}.\quad \color{red}90^{\circ}\\\\ &\textrm{Jawab}\\ & \begin{aligned}\vec{p}.\vec{q}&=\displaystyle \begin{vmatrix} \vec{p} \end{vmatrix}.\begin{vmatrix} \vec{q} \end{vmatrix}.\cos \angle \left (\vec{p},\, \vec{q} \right )\\ \cos \angle \left (\vec{p},\, \vec{q} \right )&=\displaystyle \frac{\vec{p}.\vec{q}}{\left | \vec{p} \right |.\left | \vec{q} \right |}\\ &=\displaystyle \frac{\begin{pmatrix} -2\\ 1 \end{pmatrix}.\begin{pmatrix} 8\\ 4 \end{pmatrix}}{\sqrt{(-2)^{2}+1^{2}}.\sqrt{8^{2}+4^{2}}}\\ &=\displaystyle \frac{-16+16}{\sqrt{20}.\sqrt{80}}\\ &=\displaystyle \frac{0}{40}\\ &=0\\ \cos \angle \left (\vec{p},\, \vec{q} \right )&=\cos 90^{\circ}\\ \angle \left (\vec{p},\, \vec{q} \right )&=\color{red}90^{\circ} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 27.&\textrm{Jika}\: \: \overline{OA}=\begin{pmatrix} 1\\ 2 \end{pmatrix},\: \overline{OB}=\begin{pmatrix} 4\\ 2 \end{pmatrix},\: \textrm{dan}\\ &\theta =\angle \left ( \overline{OA},\: \overline{OB} \right ),\: \textrm{maka}\: \: \tan \theta =....\\\\ &\textrm{a}.\quad \displaystyle \frac{3}{5}\\\\ &\textrm{b}.\quad \displaystyle \frac{9}{16}\\\\ &\textrm{c}.\quad \color{red}\displaystyle \frac{3}{4}\\\\ &\textrm{d}.\quad \displaystyle \frac{4}{3}\\\\ &\textrm{e}.\quad \displaystyle \frac{16}{9}\\\\ &\textrm{Jawab}\\ &\begin{array}{|c|c|}\hline \begin{aligned}\cos \theta &=\displaystyle \frac{\vec{a}.\vec{b}}{\left | \vec{a} \right |\left | \vec{b} \right |}\\ &=\displaystyle \frac{\begin{pmatrix} 1\\ 2 \end{pmatrix}.\begin{pmatrix} 4\\ 2 \end{pmatrix}}{\sqrt{1^{2}+2^{2}}\sqrt{4^{2}+2^{2}}}\\ &=\displaystyle \frac{4+4}{\sqrt{5}.\sqrt{20}}\\ &=\displaystyle \frac{8}{10} \end{aligned}&\begin{aligned}\sin \theta &=\sqrt{1-\cos ^{2}\theta }\\ &=\sqrt{1-\left ( \displaystyle \frac{8}{10} \right )^{2}}\\ &=\sqrt{\displaystyle \frac{36}{100}}\\ &=\displaystyle \frac{6}{10}\\ & \end{aligned}\\\hline \end{array} \\ &\textrm{Selanjutnya}\\ &\begin{aligned}\tan \theta &=\displaystyle \frac{\sin \theta }{\cos \theta }\\ &=\displaystyle \frac{\displaystyle \frac{6}{10}}{\displaystyle \frac{8}{10}}\\ &=\color{red}\displaystyle \frac{3}{4} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 28.&\textrm{Jika}\: \: \vec{a},\: \vec{b}\: \: \textrm{dan}\: \: \vec{c}\: \: \textrm{adalah vektor satuan dengan}\\ & \vec{a}+\vec{b}+\vec{c}=0.\: \textrm{Nilai dari}\: \: \vec{a}.\vec{b}+\vec{a}.\vec{c}+\vec{b}.\vec{c}\: \: \textrm{adalah}....\\ &\textrm{a}.\quad \displaystyle -3\\ &\textrm{b}.\quad \color{red}\displaystyle -\frac{3}{2}\\ &\textrm{c}.\quad \displaystyle 0\\ &\textrm{d}.\quad \displaystyle \frac{3}{2}\\ &\textrm{e}.\quad \displaystyle 3\\\\ &\textrm{Jawab}\\ & \begin{aligned}\textrm{Karena}&\left\{\begin{matrix} \vec{a},\vec{b},\vec{c}\: \: \textrm{adalah vektor satuan, dan}\\ \vec{a}+\vec{b}+\vec{c}=0\qquad\qquad\qquad\qquad . \end{matrix}\right.\\ \textrm{segitig}&\textrm{a ABC adalah segitiga sama sisi}\\ \vec{a}.\vec{b}=&\left | \vec{a} \right |\left | \vec{b} \right |\cos 120^{0}=1.1.\left ( -\frac{1}{2} \right )=-\frac{1}{2}\\ \vec{a}.\vec{c}=&\left | \vec{a} \right |\left | \vec{c} \right |\cos 120^{0}=1.1.\left ( -\frac{1}{2} \right )=-\frac{1}{2}\\ \vec{b}.\vec{c}=&\left | \vec{b} \right |\left | \vec{c} \right |\cos 120^{0}=1.1.\left ( -\frac{1}{2} \right )=-\frac{1}{2}\\ \textrm{Jadi, ni}&\textrm{lai dari}\: \: \vec{a}.\vec{b}+\vec{a}.\vec{c}+\vec{b}.\vec{c}\\ &=\left ( -\frac{1}{2} \right )+\left ( -\frac{1}{2} \right )+\left ( -\frac{1}{2} \right )=\color{red}-\frac{3}{2} \end{aligned} \end{array}$

$.\: \qquad \color{blue}\textrm{berikut ilustrasinya}$

$\begin{array}{ll}\\ 29.&\textrm{Jika}\: \: \angle \left ( \vec{a},\vec{b} \right )=60^{\circ},\: \: \left |\vec{a} \right |=4\: \: \textrm{dan}\\ &\left |\vec{b} \right |=3,\: \: \textrm{maka}\: \: \vec{a}(\vec{a}-\vec{b})\: \: \textrm{adalah}....\\ &\textrm{a}.\quad 2\\ &\textrm{b}.\quad 4\\ &\textrm{c}.\quad 6\\ &\textrm{d}.\quad 8\\ &\textrm{e}.\quad \color{red}10\\\\ &\textrm{Jawab}\\ &\begin{aligned}\vec{a}(\vec{a}-\vec{b})&=\vec{a}.\vec{a}-\vec{a}.\vec{b}\\ &=\left | \vec{a} \right |\left | \vec{a} \right |\cos 0^{\circ}-\left | \vec{a} \right |\left | \vec{b} \right |\cos 60^{\circ}\\ &=\left | \vec{a} \right |^{2}-\left | \vec{a} \right |\left | \vec{b} \right |.\displaystyle \frac{1}{2}\\ &=4^{2}-4.3.\displaystyle \frac{1}{2}\\ &=16-6\\ &=\color{red}10 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 30.&\textrm{Diketahui}\: \: \left | \vec{a} \right |=\sqrt{3}\: ,\left | \vec{b} \right |=1,\: \: \textrm{dan}\: \: \left | \vec{a}-\vec{b} \right |=1\\ &\textrm{maka panjang vektor}\: \: \vec{a}+\vec{b}\: \: \textrm{adalah}\: ....\\ &\begin{array}{llllllll}\\ \textrm{a}.&\sqrt{3}&&&\textrm{d}.&2\sqrt{2}\\ \textrm{b}.&\sqrt{5}&\textrm{c}.&\color{red}\sqrt{7}&\textrm{e}.&3 \end{array}\\\\ &\textbf{Jawab}\\ &\begin{aligned}\textrm{Diketahui}&\: \, \textrm{sebagaimana pada soal}\\ \left | \vec{a}-\vec{b} \right |^{2}&=\left | \vec{a} \right |^{2}+\left | \vec{b} \right |^{2}-2\left | \vec{a} \right |\left | \vec{b} \right |\cos \theta \\ 1^{2}&=\left ( \sqrt{3} \right )^{2}+1^{2}-2.\sqrt{3}.1.\cos \theta \\ 2\sqrt{3}\cos \theta &=3\\ \textrm{maka pan}&\textrm{jang vektor}\: \: \vec{a}+\vec{b}\: \: \textrm{adalah}\\ \left | \vec{a}+\vec{b} \right |&=\sqrt{\left | \vec{a} \right |^{2}+\left | \vec{b} \right |^{2}+2\left | \vec{a} \right |\left | \vec{b} \right |\cos \theta}\\ &=\sqrt{\left ( \sqrt{3} \right )^{2}+1^{2}+3}\\ &=\sqrt{3+1+3}\\ &=\color{red}\sqrt{7} \end{aligned} \end{array}$

Contoh 5 Vektor

$\begin{array}{ll}\\ 21.&\textrm{Jika}\: \: \vec{g}=\begin{pmatrix} 3^{x+y}\\ 5 \end{pmatrix}\: \: \textrm{dan}\: \: \vec{h}=\begin{pmatrix} 81\\ \displaystyle \frac{y+7}{2} \end{pmatrix}\\ & \textrm{sehingga}\: \: \vec{g}=\vec{h}\: \: \: \textrm{nilai dari}\: \: 4x-3y=.... \\ &\textrm{a}.\quad \color{red}-5\\ &\textrm{b}.\quad -1\\ &\textrm{c}.\quad 0\\ &\textrm{d}.\quad 5\\ &\textrm{e}.\quad 10\\\\ &\textrm{Jawab}\\ &\begin{aligned}\textrm{Diketahui}\: &\: \textrm{bahwa}:\quad \vec{g}=\vec{h}\\ \begin{pmatrix} 3^{x+y}\\ 5 \end{pmatrix}&=\begin{pmatrix} 81\\ \displaystyle \frac{y+7}{2} \end{pmatrix}\\ 3^{x+y}&=81=3^{4}\Leftrightarrow x+y=4\\ \displaystyle \frac{y+7}{2}&=5\Leftrightarrow y=10-7=3,\quad \textrm{sehingga}\\ x+y&=4\Leftrightarrow x+3=4\Leftrightarrow x=4-3=1,\\ &\textrm{maka}\\ 4x-3y&=4(1)-3(3)\\ &=4-9=\color{red}-5 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 22.&\textrm{Vektor}\: \: \vec{m}=\begin{pmatrix} -2\\ 5 \end{pmatrix}\: \: \textrm{searah }\\ &\textrm{dengan vektor}\: .... \\ &\textrm{a}.\quad \displaystyle \begin{pmatrix} 2\\ -5 \end{pmatrix}\\ &\textrm{b}.\quad \displaystyle \begin{pmatrix} 2\\ 5 \end{pmatrix}\\ &\textrm{c}.\quad \color{red}\displaystyle \begin{pmatrix} -6\\ 15 \end{pmatrix}\\ &\textrm{d}.\quad \displaystyle \begin{pmatrix} -4\\ 5 \end{pmatrix}\\ &\textrm{e}.\quad \displaystyle \begin{pmatrix} -3\\ 10 \end{pmatrix} \\\\ &\textrm{Jawab}\\ &\begin{aligned}&\textrm{Vektor}\quad \vec{m}\: \: \: \textrm{searah dengan vektor}\: \: k.\vec{m}\\ &\color{blue}k.\vec{m}=k\begin{pmatrix} -2\\ 5 \end{pmatrix}, \color{black}\textrm{dengan}\: \: k\: \: \textrm{skalar positif}\\ &\begin{array}{|c|c|c|}\hline \textrm{a}&\textrm{b}\\\hline \begin{pmatrix} 2\\ -5 \end{pmatrix}=-\begin{pmatrix} -2\\ 5 \end{pmatrix}&\begin{pmatrix} 2\\ 5 \end{pmatrix}=...\\\hline \textrm{c}&\textrm{d}\\\hline \color{red}\begin{pmatrix} -6\\ 15 \end{pmatrix}=3\begin{pmatrix} -2\\ 5 \end{pmatrix}&\begin{pmatrix} -4\\ 5 \end{pmatrix}=...\\\hline \textrm{e}&\\\hline \begin{pmatrix} -3\\ 10 \end{pmatrix}=...&\\\hline \end{array} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 23.&\textrm{Jika vektor}\: \: \overline{AC}=\vec{p},\: \overline{BC}=\vec{q}\: \: \textrm{dan}\\ &\overline{AD}:\overline{DC}=1:2,\: \: \textrm{maka vektor}\\ & \overline{BD}\: \: \textrm{bila dinyatakan}\\ &\textrm{dalam}\: \: \vec{p}\: \: \textrm{dan}\: \: \vec{q}\: \: \textrm{adalah}....\\ &\begin{array}{lll}\\ \textrm{a}.\quad \displaystyle \frac{1}{3}\left ( 3\vec{p}-2\vec{q} \right )\\ \textrm{b}.\quad \color{red}\displaystyle \frac{1}{3}\left ( 3\vec{q}-2\vec{p} \right )\\ \textrm{c}.\quad \displaystyle \frac{1}{3}\left ( \vec{p}-2\vec{q} \right )\\ \textrm{d}.\quad \displaystyle \left ( \vec{p}-\displaystyle \frac{1}{3}\vec{q} \right )\\ \textrm{e}.\quad \displaystyle \frac{1}{3}\left ( \vec{p}-\vec{q} \right ) \end{array}\\\\ &\textbf{Jawab}\\ &\begin{aligned}\overline{AC}:\overline{CD}&=3:-2\\ \overline{CD}&=-\displaystyle \frac{2}{3}\, \overline{AC}\\ &=\displaystyle -\frac{2}{3}\vec{p}\\ \textrm{maka}\, ,\qquad&\\ \overline{BD}&=\overline{BC}+\overline{CD}\\ &=\vec{q}+\left ( -\displaystyle \frac{2}{3}\vec{p} \right )\\ &=\color{red}\displaystyle \frac{1}{3}\left ( 3\vec{q}-2\vec{p} \right ) \end{aligned} \end{array}$

$.\: \quad \color{blue}\textrm{Gambar berikut untuk soal 24}$

$\begin{array}{ll}\\ 24.&\textrm{Jika vektor}\: \: \overline{AC}=\vec{p},\: \overline{BC}=\vec{q}\: \: \textrm{dan}\\ & \overline{AD}:\overline{DC}=1:2,\: \textrm{maka vektor}\: \: \overline{BD}\\ &\textrm{bila dinyatakan dalam}\: \: \vec{p}\: \: \textrm{dan}\: \: \vec{q}\: \: \textrm{adalah}.... \\ &\textrm{a}.\quad \displaystyle \frac{1}{3}\left ( 3\vec{p}-2\vec{q} \right )\\ &\textrm{b}.\quad \color{red}\displaystyle \frac{1}{3}\left ( 3\vec{q}-2\vec{p} \right )\\ &\textrm{c}.\quad \displaystyle \left ( \vec{p}-\displaystyle \frac{1}{3}\vec{q} \right )\\ &\textrm{d}.\quad \displaystyle \frac{1}{3}\left ( \vec{p}-2\vec{q} \right )\\ &\textrm{e}.\quad \displaystyle \frac{1}{3}\left ( \vec{p}-\vec{q} \right )\\\\ &\textrm{Jawab}\\ &\begin{aligned}\overline{AC}:\overline{CD}&=3:-2\\ \overline{CD}&=-\displaystyle \frac{2}{3}\, \overline{AC}\\ &=\displaystyle -\frac{2}{3}\vec{p}\\ \textrm{maka}\, ,\qquad&\\ \overline{BD}&=\overline{BC}+\overline{CD}\\ &=\vec{q}+\left ( -\displaystyle \frac{2}{3}\vec{p} \right )\\ &=\color{red}\displaystyle \frac{1}{3}\left ( 3\vec{q}-2\vec{p} \right ) \end{aligned} \end{array}$

$\begin{array}{ll}\\ 25.&\textrm{Jika titik}\: \: A(2,6)\: \: \textrm{dan}\: \: B(5,3)\: \: \textrm{demikian }\\ &\textrm{juga titik}\: \: P\: \: \textrm{terletak pada}\: \: \overline{AB}\\ &\textrm{dengan}\: \: \overline{AP}:\overline{PB}=2:1,\: \textrm{maka vektor }\\ &\textrm{posisi}\: \: \vec{p}\: \: \textrm{adalah}....\\ &\textrm{a}.\quad \color{red}\begin{pmatrix} 4\\ 4 \end{pmatrix}\\ &\textrm{b}.\quad \begin{pmatrix} 4\\ 5 \end{pmatrix}\\ &\textrm{c}.\quad \begin{pmatrix} -4\\ 4 \end{pmatrix}\\ &\textrm{d}.\quad \begin{pmatrix} 4\\ 2 \end{pmatrix}\\ &\textrm{e}.\quad \begin{pmatrix} -4\\ 6 \end{pmatrix}\\\\ &\textrm{Jawab}\\ &\begin{aligned}\overline{AP}:\overline{PB}&=2:1\\ \overline{AP}&=2\, \overline{PB}\\ \vec{p}-\vec{a}&=2\left ( \vec{b}-\vec{p} \right )\\ \vec{p}+2\vec{p}&=\vec{a}+2\vec{b}\\ 3\vec{p}&=\vec{a}+2\vec{b}\\ \vec{p}&=\displaystyle \frac{1}{3}\left ( \vec{a}+2\vec{b} \right )\\ &=\displaystyle \frac{1}{3}\begin{pmatrix} 2+2.5\\ 6+2.3 \end{pmatrix}\\ &=\displaystyle \frac{1}{3}\begin{pmatrix} 12\\ 12 \end{pmatrix}\\ &=\color{red}\begin{pmatrix} 4\\ 4 \end{pmatrix} \end{aligned} \end{array}$

Contoh 4 Vektor

$\begin{array}{ll}\\ 16.&\textrm{Diketahui titik A(-1,1,0) dan titik B(1,-2,2)}\\ &\textrm{maka panjang vektor}\: \: \overrightarrow{BA}\: \: \textrm{adalah}\: ....\\ &\begin{array}{llllllll}\\ \textrm{a}.&\sqrt{2}&&&\textrm{d}.&\color{red}\sqrt{17}\\ \textrm{b}.&\sqrt{5}&\textrm{c}.&\sqrt{9}&\textrm{e}.&\sqrt{21} \end{array}\\\\ &\textbf{Jawab}\\ &\begin{aligned}&\textrm{Diketahui}\: \, \textrm{sebagaimana pada soal}\\ &\textrm{maka pan}\textrm{jang vektor}\: \: \overrightarrow{BA}\: \: \textrm{adalah}\\ &\left | \overrightarrow{BA} \right |=\sqrt{(1-(-1))^{2}+(-2-1)^{2}+(2-0)^{2}}\\ &=\sqrt{2^{2}+(-3)^{2}+2^{2}}\\ &=\sqrt{4+9+4}\\ &=\color{red}\sqrt{17} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 17.&\textrm{Vektor satuan untuk vektor}\: \: \vec{a}=\begin{pmatrix} 2, & 1, &-2 \end{pmatrix}=\: ....\\ &\begin{array}{llllllll}\\ \textrm{a}.&\begin{pmatrix} -\displaystyle \frac{2}{3}, & -\displaystyle \frac{1}{3}, &\displaystyle \frac{2}{3} \end{pmatrix}&\\\\ \textrm{b}.&\color{red}\begin{pmatrix} \displaystyle \frac{2}{3}, & \displaystyle \frac{1}{3}, &-\displaystyle \frac{2}{3} \end{pmatrix}&\\\\ \textrm{c}.&\begin{pmatrix} \displaystyle \frac{2}{4}, & \displaystyle \frac{1}{4}, &-\displaystyle \frac{2}{4} \end{pmatrix}&\\\\ \textrm{d}.&\begin{pmatrix} -\displaystyle \frac{2}{4}, & -\displaystyle \frac{1}{4}, &\displaystyle \frac{2}{4} \end{pmatrix}\\\\ \textrm{e}.&\begin{pmatrix} -\displaystyle \frac{2}{9}, &-\displaystyle \frac{1}{9}, & \displaystyle \frac{2}{9} \end{pmatrix} \end{array}\\\\ &\textbf{Jawab}\\ &\begin{aligned}\textrm{Vektor }&\: \textrm{satuan dari vektor}\: \: \vec{a}\: \: \textrm{adalah}:\\ \hat{a}&=\displaystyle \frac{\vec{a}}{\left | \vec{a} \right |}\\ &=\displaystyle \frac{\begin{pmatrix} 2, & 1, & -2 \end{pmatrix}}{\sqrt{2^{2}+1^{2}+(-2)^{2}}}\\ &=\displaystyle \frac{\begin{pmatrix} 2, & 1, & -2 \end{pmatrix}}{\sqrt{9}}\\ &=\displaystyle \frac{\begin{pmatrix} 2, & 1, & -2 \end{pmatrix}}{3}\\ &=\color{red}\begin{pmatrix} \displaystyle \frac{2}{3}, & \displaystyle \frac{1}{3}, &-\displaystyle \frac{2}{3} \end{pmatrix} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 18.&\textrm{Jika titik A(-2,3,5) dan B(4,1,-3)},\\ & \textrm{maka vektor posisi AB adalah}\: ....\\ &\begin{array}{llllllll}\\ \textrm{a}.&\begin{pmatrix} -6, & 2, &8 \end{pmatrix}&\\ \textrm{b}.&\begin{pmatrix} 8, & 2, &-6 \end{pmatrix}&\\ \textrm{c}.&\color{red}\begin{pmatrix} 6, & -2, &-8 \end{pmatrix}&\\ \textrm{d}.&\begin{pmatrix} -8 & -2, &6 \end{pmatrix}\\ \textrm{e}.&\begin{pmatrix} 2, &4, & 2 \end{pmatrix} \end{array}\\\\ &\textbf{Jawab}\\ &\begin{aligned}\textrm{Vektor posisi}&\: \textrm{dari}\: \: \overrightarrow{AB}\: \: \textrm{adalah}:\\ \overrightarrow{AB}&=\overrightarrow{OB}-\overrightarrow{OA}\\ &=\begin{pmatrix} 4\\ 1\\ -3 \end{pmatrix}-\begin{pmatrix} -2\\ 3\\ 5 \end{pmatrix}\\ &=\begin{pmatrix} 4+2\\ 1-3\\ -3-5 \end{pmatrix}\\ &=\begin{pmatrix} 6\\ -2\\ -8 \end{pmatrix}\quad\: \textbf{atau}\\ &=\color{red}\begin{pmatrix} 6, & -2, & -8 \end{pmatrix} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 19.&\textrm{Jika}\: \: \vec{p}=\begin{pmatrix} ^{2}\log 8x\\ \left ( ^{2}\log x \right )^{y} \end{pmatrix}\: \: \textrm{dan}\: \: \vec{q}=\begin{pmatrix} 5\\ 8 \end{pmatrix}\\ &\textrm{sehingga}\: \: \vec{p}=\vec{q}\: \: \: \textrm{nilai dari}\: \: x.y=.... \\ &\textrm{a}.\quad 6\\ &\textrm{b}.\quad \color{red}12\\ &\textrm{c}.\quad 18\\ &\textrm{d}.\quad 24\\ &\textrm{e}.\quad 30\\\\ &\textrm{Jawab}\\ &\begin{aligned}\textrm{Diketahui}\: &\: \textrm{bahwa}:\quad \vec{p}=\vec{q}\\ \begin{pmatrix} ^{2}\log 8x\\ \left ( ^{2}\log x \right )^{y} \end{pmatrix}&=\begin{pmatrix} 5\\ 8 \end{pmatrix},\quad\: \: \textrm{maka}\\ 8x&=2^{5}=32\\ \Leftrightarrow x&=\displaystyle \frac{32}{8}=4\\ \left (^{2}\log 4 \right )^{y}&=8\\ \Leftrightarrow 2^{y}&=8=2^{3}\\ \Leftrightarrow y&=3\\ \textrm{Sehingga}&\\ x.y&=4\times 3=\color{red}12 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 20.&\textrm{Jika}\: \: \vec{p}=\begin{pmatrix} 3x\\ 4x+y \end{pmatrix}\: \: \textrm{dan}\: \: \vec{q}=\begin{pmatrix} \displaystyle \frac{2x-4}{2}\\ 6 \end{pmatrix}\\ & \textrm{sehingga}\: \: \vec{p}=\vec{q}\: \: \: \textrm{nilai dari}\: \: 2x+y=.... \\ &\textrm{a}.\quad -12\\ &\textrm{b}.\quad 0\\ &\textrm{c}.\quad 8\\ &\textrm{d}.\quad \color{red}9\\ &\textrm{e}.\quad 19\\\\ &\textrm{Jawab}\\ &\begin{aligned}\textrm{Diketahui}\: &\: \textrm{bahwa}:\quad \vec{p}=\vec{q}\\ \begin{pmatrix} 3x\\ 4x+y \end{pmatrix}&=\begin{pmatrix} \displaystyle \frac{2x-4}{2}\\ 6 \end{pmatrix}\\ 3x&=\displaystyle \frac{2x-4}{2}\Leftrightarrow 6x=2x-4\\ \Leftrightarrow x&=-1\\ 4(-1)+y&=6\Leftrightarrow -4+y=6\\ \Leftrightarrow y&=6+4\\ y&=10\\ x+y&=(-1)+10=\color{red}9 \end{aligned} \end{array}$

Contoh 3 Vektor

$\begin{array}{ll}\\ 11.&\textrm{Jika vektor}\: \: \vec{a}=\begin{pmatrix} 6\\ -4 \end{pmatrix}\: \: \textrm{dan}\: \: \vec{b}=\begin{pmatrix} 3\\ 2 \end{pmatrix},\\ &\textrm{maka}\: \: 3\vec{a}-2\vec{b}\: \: \textrm{adalah}....\\ &\begin{array}{lll}\\ \textrm{a}.\quad \color{red}\begin{pmatrix} 12\\ -16 \end{pmatrix}&&\textrm{d}.\quad \begin{pmatrix} 24\\ 16 \end{pmatrix}\\ \textrm{b}.\quad \begin{pmatrix} 24\\ -16 \end{pmatrix}&\textrm{c}.\quad \begin{pmatrix} 12\\ 16 \end{pmatrix}&\textrm{e}.\quad \begin{pmatrix} -12\\ -16 \end{pmatrix} \end{array}\\\\ &\textbf{Jawab}\\ &\begin{aligned}3\vec{a}-2\vec{b}&=3\begin{pmatrix} 6\\ -4 \end{pmatrix}-2\begin{pmatrix} 3\\ 2 \end{pmatrix}\\ &=\begin{pmatrix} 18-6\\ -12-4 \end{pmatrix}\\ &=\color{red}\begin{pmatrix} 12\\ -16 \end{pmatrix} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 12.&\textrm{Diketahui jajar genjang ABCD }\\ &\textrm{dengan titik E adalah perpotongan }\\ &\textrm{diagonal jajar genjang}. \end{array}$

$\begin{array}{ll}\\ .\, \quad&\textrm{Jika}\: \: \overline{AB}=\vec{b}\: \: \textrm{dan}\: \: \overline{AD}=\vec{a},\: \textrm{maka}\: \: \overline{CE}\\ & \textrm{bila dinyatakan dalam}\: \: \vec{a}\: \: \textrm{dan}\: \: \vec{b}\: \: \textrm{adalah}....\\ &\begin{array}{lll}\\ \textrm{a}.\quad \displaystyle \frac{1}{2}\left ( \vec{a}+\vec{b} \right )&\\ \textrm{b}.\quad \displaystyle \frac{1}{2}\left ( \vec{a}-\vec{b} \right )&\\ \textrm{c}.\quad \displaystyle \frac{1}{2}\left ( \vec{b}-\vec{a} \right )&\\ \textrm{d}.\quad \color{red}\displaystyle -\frac{1}{2}\left ( \vec{a}+\vec{b} \right )\\ \textrm{e}.\quad -\displaystyle \frac{1}{2}\left ( 2\vec{a}+\vec{b} \right ) \end{array}\\\\ &\textbf{Jawab}\\ &\begin{aligned} \overline{AC}&=\overline{AD}+\overline{DC}\\ \overline{CA}&=\overline{CD}+\overline{DA}\\ \overline{CE}&=\displaystyle \frac{1}{2}\, \overline{CA}\\ &=\displaystyle \frac{1}{2}\left ( -\vec{b}-\vec{a} \right )\\ &=\color{red}-\displaystyle \frac{1}{2}\left ( \vec{a}+\vec{b} \right ) \end{aligned} \end{array}$

$\begin{array}{ll}\\ 13.&\textrm{Pada segi enam beraturan ABCDEF},\\ & \textrm{jika}\: \: \overrightarrow{AB}=\vec{u}\: \: \textrm{dan}\: \: \overrightarrow{AF}=\vec{v}\: \: \textrm{maka vektor}\\ &\overrightarrow{AB}+\overrightarrow{AC}+\overrightarrow{AD}+\overrightarrow{AE}+\overrightarrow{AF}=....\\ &\begin{array}{llllllll}\\ \textrm{a}.&2\vec{u}+2\vec{v}&&&\textrm{d}.&\color{red}6\vec{u}+6\vec{v}\\ \textrm{b}.&4\vec{u}+4\vec{v}&\textrm{c}.&5\vec{u}+5\vec{v}&\textrm{e}.&8\vec{u}+8\vec{v} \end{array}\\\\ &\textbf{Jawab}\\ &\textrm{Perhatikanlah ilustrasi gambar berikut} \end{array}$

$\begin{aligned}.\: \, \qquad &\overrightarrow{AB}+\overrightarrow{AC}+\overrightarrow{AD}+\overrightarrow{AE}+\overrightarrow{AF}\\ &=\overrightarrow{AB}+\left (\overrightarrow{AO}+\overrightarrow{OC} \right )+2\overrightarrow{AO}+\left (\overrightarrow{AO}+\overrightarrow{OE} \right )+\overrightarrow{AF}\\ &=\vec{u}+\left (2\vec{u}+\vec{v} \right )+2\left ( \vec{v}+\vec{u} \right )+\left ( 2\vec{v}+\vec{u} \right )+\vec{v}\\ &=\color{red}6\vec{u}+6\vec{v} \end{aligned}$

$\begin{array}{ll}\\ 14.&\textrm{Perhatikanlah juga ilustrasi gambar berikut} \end{array}$

$\begin{array}{ll}\\ .\, \quad&\textrm{maka vektor}\: \: \vec{w}\: \: \textrm{adalah}\: ....\\ &\begin{array}{llllllll}\\ \textrm{a}.&\color{red}8\vec{i}-6\vec{j}-13\vec{k}&\\ \textrm{b}.&8\vec{i}-13\vec{j}-6\vec{k}\\ \textrm{c}.&6\vec{i}-8\vec{j}-13\vec{k}&\\ \textrm{d}.&-6\vec{i}+8\vec{j}-13\vec{k}\\ \textrm{e}.&-6\vec{i}-13\vec{j}+8\vec{k} \end{array}\\\\ &\textbf{Jawab}\\ &\textrm{Kita perhatikan juga ilustrasi }\\ &\textrm{gambarnya semisal dengan soal No.1}\\ &\textrm{Misalkan titiknya adalah titik }\\ &\textrm{W dengan koordinat (8,-6,-13)},\\ &\textrm{maka vektor posisi titik }\\ &\textrm{W tersebut adalah}\: \: \overrightarrow{OW}=\vec{w}\\ & \textrm{di mana}\\ &\begin{aligned}&\textrm{Vektor}\: \: \vec{w}\: \textrm{jika dinyatakan }\\ &\textrm{dalam kombinasi linear adalah}\\ &\vec{w}=\color{red}8\vec{i}-6\vec{j}-13\vec{k} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 15.&\textrm{Jika titik Z(4,-5,2)},\: \textrm{maka panjang }\\ &\textrm{vektor posisi titik Z adalah}\: ....\\ &\begin{array}{llllllll}\\ \textrm{a}.&1&&&\textrm{d}.&5\sqrt{2}\\ \textrm{b}.&2\sqrt{5}&\textrm{c}.&\color{red}3\sqrt{5}&\textrm{e}.&5\sqrt{3} \end{array}\\\\ &\textbf{Jawab}\\ &\begin{aligned}\textrm{Vektor posisi}&\: \, \textrm{titik Z tersebut adalah}\\ \overrightarrow{OZ}=\vec{z}&=\begin{pmatrix} 4, & -5, & 2 \end{pmatrix},\\ \textrm{Dan panjang}&\: \, \textrm{vektor}\: \: \vec{z}\: \: \textrm{ini adalah}\\ \left | \vec{z} \right |&=\sqrt{4^{2}+(-5)^{2}+2^{2}}\\ &=\sqrt{16+25+4}\\ &=\sqrt{45}=\sqrt{9\times 5}\\ &=\color{red}3\sqrt{5} \end{aligned} \end{array}$.

Contoh 2 Vektor

$\begin{array}{ll}\\ 6.&\textrm{Diketahui titik}\: \: P(n,2),\: Q(1,-2),\: n>0\\ &\textrm{dan panjang}\: \: \overline{PQ}=5,\: \: \textrm{maka nilai}\: \: n\\ & \textrm{adalah}....\\ &\textrm{a}.\quad 1\\ &\textrm{b}.\quad 2\\ &\textrm{c}.\quad 3\\ &\textrm{d}.\quad \color{red}4\\ &\textrm{e}.\quad 5\\\\ &\textrm{Jawab}\\ &\begin{aligned}\overline{PQ}&=5\\ \sqrt{(x_{Q}-x_{P})^{2}+(y_{Q}-y_{P})^{2}}&=5\\ (x_{Q}-x_{P})^{2}+(y_{Q}-y_{P})^{2}&=25\\ (1-n)^{2}+(-2-2)^{2}&=25\\ (1-n)^{2}+16&=25\\ (1-n)^{2}-3^{2}&=0\\ (1+3-n)(1-3-n)&=0\\ \color{red}n=4\: \: \color{black}\textrm{atau}\: \: \color{red}n=-2& \end{aligned} \end{array}$

$\begin{array}{ll}\\ 7.&\textrm{Diketahui vektor}\: \: \vec{u}=\begin{pmatrix} 3\\ 4 \end{pmatrix}\: \: \textrm{dan}\: \: \vec{v}=\begin{pmatrix} 2\\ -1 \end{pmatrix}.\\ & \textrm{Nilai}\: \: \left | \vec{u}+\vec{v} \right |\: \: \textrm{adalah}....\\ &\textrm{a}.\quad \sqrt{28}\\ &\textrm{b}.\quad \sqrt{30}\\ &\textrm{c}.\quad \color{red}\sqrt{34}\\ &\textrm{d}.\quad \sqrt{44}\\ &\textrm{e}.\quad \sqrt{50}\\\\ &\textrm{Jawab}\\ &\begin{aligned}\vec{u}+\vec{v}&=\begin{pmatrix} 3\\ 4 \end{pmatrix} + \begin{pmatrix} 2\\ -1 \end{pmatrix}\\ &=\begin{pmatrix} 3+2\\ 4+(-1) \end{pmatrix}\\ &=\begin{pmatrix} 5\\ 3 \end{pmatrix}\\ \left | \vec{u}+\vec{v} \right |&=\sqrt{5^{2}+3^{2}}\\ &=\sqrt{25+9}\\ &=\color{red}\sqrt{34} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 8.&\textrm{Vektor satuan}\: \: \vec{u}=\begin{pmatrix} -5\\ -12 \end{pmatrix}\: \: \textrm{adalah}.... \\ &\textrm{a}.\quad \color{red}\displaystyle -\frac{1}{13}\begin{pmatrix} 5\\ 12 \end{pmatrix}\\ &\textrm{b}.\quad \displaystyle \frac{1}{15}\begin{pmatrix} -5\\ -12 \end{pmatrix}\\ &\textrm{c}.\quad \displaystyle -\frac{1}{17}\begin{pmatrix} 5\\ 12 \end{pmatrix}\\ &\textrm{d}.\quad \displaystyle -\frac{1}{17}\begin{pmatrix} 5\\ 12 \end{pmatrix}\\ &\textrm{e}.\quad \displaystyle \frac{1}{2} \begin{pmatrix} 5\\ 12 \end{pmatrix}\\\\ &\textrm{Jawab}\\ &\begin{aligned}\vec{e}_{\vec{u}}&=\displaystyle \frac{\vec{u}}{\left | \vec{u} \right |},\quad \textrm{maka}\\ \vec{e}_{\begin{pmatrix} -5\\ -12 \end{pmatrix}}&=\displaystyle \frac{\begin{pmatrix} -5\\ -12 \end{pmatrix}}{\left | \begin{pmatrix} -5\\ -12 \end{pmatrix} \right |}\\ &=\displaystyle \frac{\begin{pmatrix} -5\\ -12 \end{pmatrix}}{\sqrt{(-5)^{2}+(-12)^{2}}}\\ &=\displaystyle \frac{-\begin{pmatrix} 5\\ 12 \end{pmatrix}}{\sqrt{169}}\\ &=\color{red}-\displaystyle \frac{1}{13}\begin{pmatrix} 5\\ 12 \end{pmatrix} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 9.&\textrm{Jika vektor}\: \: \vec{p}=\begin{pmatrix} 8\\ 7 \end{pmatrix}\: \: \textrm{dan}\: \: \vec{q}=\begin{pmatrix} -3\\ 9 \end{pmatrix}\\ & \textrm{Hasil dari}\: \: \vec{p}+\vec{q}\: \: \textrm{adalah}....\\ &\textrm{a}.\quad \begin{pmatrix} 6\\ 14 \end{pmatrix}\\ &\textrm{b}.\quad \begin{pmatrix} 6\\ 13 \end{pmatrix}\\ &\textrm{c}.\quad \begin{pmatrix} 6\\ 15 \end{pmatrix}\\ &\textrm{d}.\quad \color{red}\begin{pmatrix} 5\\ 16 \end{pmatrix}\\ &\textrm{e}.\quad \begin{pmatrix} 5\\ 48 \end{pmatrix}\\\\ &\textrm{Jawab}\\ &\begin{aligned}\vec{p}+\vec{q}&=\begin{pmatrix} 8\\ 7 \end{pmatrix}+\begin{pmatrix} -3\\ 9 \end{pmatrix}\\ &=\begin{pmatrix} 8-3\\ 7+9 \end{pmatrix}\\ &=\color{red}\begin{pmatrix} 5\\ 16 \end{pmatrix} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 10.&\textrm{Jika vektor}\: \: \vec{p}=\begin{pmatrix} ^{2}\log 32\\ -\, ^{3}\log \displaystyle \frac{1}{81} \end{pmatrix}\: \: \textrm{dan}\: \: \vec{q}=\begin{pmatrix} -9\\ -21 \end{pmatrix}\\ & \textrm{Hasil dari}\: \: \vec{p}+\vec{q}\: \: \textrm{adalah}....\\ &\textrm{a}.\quad \begin{pmatrix} 6\\ 17 \end{pmatrix}\\ &\textrm{b}.\quad \begin{pmatrix} -6\\ -17 \end{pmatrix}\\ &\textrm{c}.\quad \begin{pmatrix} 4\\ 17 \end{pmatrix}\\ &\textrm{d}.\quad \color{red}\begin{pmatrix} -4\\ -17 \end{pmatrix}\\ &\textrm{e}.\quad \begin{pmatrix} 5\\ -16 \end{pmatrix}\\\\ &\textrm{Jawab}\\ &\begin{aligned}\vec{p}+\vec{q}&=\begin{pmatrix} ^{2}\log 32\\ -\, ^{3}\log \displaystyle \frac{1}{81} \end{pmatrix}+\begin{pmatrix} -9\\ -21 \end{pmatrix}\\ &=\begin{pmatrix} 5-9\\ 4-21 \end{pmatrix}\\ &=\color{red}\begin{pmatrix} -4\\ -17 \end{pmatrix} \end{aligned} \end{array}$

Contoh 1 Vektor

$\begin{array}{ll}\\ 1.&\textrm{Perhatikanlah ilustrasi gambar berikut} \end{array}$

$\begin{array}{ll}\\ .\, \quad&\textrm{maka vektor}\: \: \vec{u}\: \: \textrm{adalah}\: ....\\ &\begin{array}{llllllll}\\ \textrm{a}.&3\vec{i}+5\vec{j}&&&\textrm{d}.&-3\vec{i}-5\vec{j}\\ \textrm{b}.&5\vec{i}+3\vec{j}&\textrm{c}.&\color{red}-3\vec{i}+5\vec{j}&\textrm{e}.&-5\vec{i}+3\vec{j} \end{array}\\\\ &\textbf{Jawab}\\ &\textrm{Kita perhatikan lagi gambarnya}\\ &\begin{aligned}&\textrm{Vektor}\: \: \vec{u}\: \: \textrm{jika dinyatakan sebagai }\\ &\textrm{kombinasi linear adalah}\: \: \vec{u}=\color{red}-3\vec{i}+5\vec{j} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 2.&\textrm{Panjang vektor}\: \: \vec{p}=\begin{pmatrix} 4\\ -8 \end{pmatrix}\: \: \textrm{adalah}....\\ &\textrm{a}.\quad \displaystyle \sqrt{4}\\ &\textrm{b}.\quad \sqrt{12}\\ &\textrm{b}.\quad \sqrt{20}\\ &\textrm{d}.\quad \color{red}\displaystyle \sqrt{80}\\ &\textrm{e}.\quad \sqrt{100}\\\\ &\textrm{Jawab}\\ &\begin{aligned}\vec{p}&=\begin{pmatrix} 4\\ -8 \end{pmatrix},\: \: \textrm{maka besar dari vektor}\\ \vec{p}&\: \: \textrm{adalah}=\left | \vec{p} \right |=\sqrt{x^{2}+y^{2}}\\ &\textrm{Yaitu}\\ \left | \vec{p} \right |&=\sqrt{4^{2}+(-8)^{2}}\\ &=\sqrt{16+64}=\color{red}\sqrt{80} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 3.&\textrm{Perhatikanlah gambar berikut}\\ & \end{array}$.

$\begin{array}{ll}\\ .\quad &\textrm{Panjang vektor}\: \: \vec{h}\: \: \textrm{tersebut di atas adalah}....\\ &\textrm{a}.\quad 5\\ &\textrm{b}.\quad 7\\ &\textrm{c}.\quad \color{red}10\\ &\textrm{d}.\quad 12\\ &\textrm{e}.\quad 15\\\\ &\textrm{Jawab}\\ &\begin{aligned}\textrm{Diketahui}&\quad \vec{h}=\overline{OH}=8\vec{i}+6\vec{j}\\ \vec{h}&=\sqrt{x_{H}^{2}+y_{H}^{2}}\\ &=\sqrt{8^{2}+6^{2}}\qquad \textbf{(ingat tripel Pythagoras)}\\ &=\sqrt{10^{2}}=\color{red}10 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 4.&\textrm{Vektor satuan dari}\: \: \vec{q}=3\vec{i}-4\vec{j}\: \: \textrm{adalah}....\\ &\textrm{a}.\quad \displaystyle \frac{4}{5}\vec{i}-\frac{3}{5}\vec{j}\\ &\textrm{b}.\quad \color{red}\displaystyle \frac{3}{5}\vec{i}-\frac{4}{5}\vec{j}\\ &\textrm{c}.\quad 3\vec{i}-4\vec{j}\\ &\textrm{d}.\quad 4\vec{i}-3\vec{j}\\ &\textrm{e}.\quad 15\vec{i}-20\vec{j}\\\\ &\textrm{Jawab}\\ &\begin{aligned}\vec{q}&=3\vec{i}-4\vec{j}\\ &\textrm{Vektor satuan dari vektor}\: \: \vec{q}\: \: \textrm{adalah}:\\ &\: \hat{e}_{_{\vec{q}}}=\displaystyle \frac{1}{\left | \vec{q} \right |}. \vec{q}\\ &\textrm{Sehingga}\\ \hat{e}_{_{\vec{q}}}&=\displaystyle \frac{1}{\sqrt{3^{2}+(-4)^{2}}}.\begin{pmatrix} 3\\ -4 \end{pmatrix}\\ &=\displaystyle \frac{1}{5}\begin{pmatrix} 3\\ -4 \end{pmatrix}\\ &\textrm{atau dalam vektor basis}\\ &=\color{red}\displaystyle \frac{3}{5}\vec{i}-\frac{4}{5}\vec{j} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 5.&\textrm{Vektor berikut yang memiliki panjang}\\ & 29\: \: \textrm{satuan adalah}....\\ &\textrm{a}.\quad \displaystyle 18\vec{i}-19\vec{j}\\ &\textrm{b}.\quad \displaystyle 19\vec{i}-20\vec{j}\\ &\textrm{c}.\quad 20\vec{i}-21\vec{j}\\ &\textrm{d}.\quad \color{red}21\vec{i}-22\vec{j}\\ &\textrm{e}.\quad 22\vec{i}-23\vec{j}\\\\ &\textrm{Jawab}\\ &\begin{aligned}\textrm{Ingat}&\textrm{lah akan tigaan Pythagoras}\\ &\begin{cases} (3,4,5) &\Rightarrow 3^{2}+4^{2}=5^{2} \\ (5,12,13) & \Rightarrow 5^{2}+12^{2}=13^{2} \\ (8,15,17) &\Rightarrow \cdots \\ (20,21,29)&\Rightarrow \cdots \\ \vdots & dll \end{cases}\\ \textrm{Sehin}&\textrm{gga}\\ \textrm{yang}&\: \: \textrm{paling mungkin adalah}\: : \: \\ &=\sqrt{20^{2}+21^{2}}=\sqrt{400+441}\\ &=\sqrt{841}\\ &=\color{red}29 \end{aligned} \end{array}$

Contoh Soal 2 Limit Fungsi Aljabar

$\begin{array}{ll}\\ 6.&\textrm{Diketahui bahwa}\: \: f(x)=x^{2}-2,\\ & \textrm{maka nilai}\: \: \: \underset{h\rightarrow 0 }{\textrm{Lim}}\: \: \displaystyle \frac{f(x+h)-f(x)}{h}=....\\ &\begin{array}{lll}\\ \textrm{a}.\quad \displaystyle x^{2}-2&&\textrm{d}.\quad \displaystyle x\\\\ \textrm{b}.\quad \displaystyle x^{2}\quad &\textrm{c}.\quad \displaystyle 2x\quad &\textrm{e}.\quad \displaystyle 2x-2 \end{array}\\\\ &\textrm{Jawab}:\\ &\begin{aligned}\textrm{Diketahui}&\: \textrm{bahwa}\: f(x)=x^{2}-2,\\ \textrm{maka nila}&\textrm{i untuk}\\ \underset{h\rightarrow 0 }{\textrm{Lim}}\: &\: \displaystyle \frac{f(x+h)-f(x)}{h}\, \, \, \, \\ &=\underset{h\rightarrow 0 }{\textrm{Lim}}\: \: \displaystyle \frac{\left ((x+h)^{2}-2 \right )-\left ( x^{2}-2 \right )}{h}\\ &=\underset{h\rightarrow 0 }{\textrm{Lim}}\: \: \displaystyle \displaystyle \frac{x^{2}+2xh+h^{2}-2-x^{2}+2}{h}\\ &=\underset{h\rightarrow 0 }{\textrm{Lim}}\: \: \displaystyle \displaystyle \frac{2xh+h^{2}}{h}\\ &=\underset{h\rightarrow 0 }{\textrm{Lim}}\: \: \displaystyle \left (2x+h \right )\\ &=\color{red}2x \end{aligned} \end{array}$

$\begin{array}{ll}\\ 7.&\textrm{Diketahui}\: \: f(x)=\sqrt{x-1},\\ & \textrm{maka nilai}\: \: \: \underset{h\rightarrow 0 }{\textrm{Lim}}\: \:\displaystyle \frac{f(2+h)-f(2)}{h}=....\\ &\begin{array}{lll}\\ \textrm{a}.\quad \color{red}\displaystyle \frac{1}{2}&&\textrm{d}.\quad \displaystyle 1\\\\ \textrm{b}.\quad -\displaystyle \frac{1}{2}\quad &\textrm{c}.\quad \displaystyle 0\quad &\textrm{e}.\quad \displaystyle -1\end{array}\\\\ &\textrm{Jawab}:\\ &\begin{aligned}\textrm{Diketa}&\textrm{hui bahwa}\: f(x)=\sqrt{x-1},\\ \textrm{maka}\: \: & \textrm{nilai untuk}\\ \underset{h\rightarrow 0 }{\textrm{Lim}}\: &\: \displaystyle \frac{f(2+h)-f(2)}{h}\\ &=\underset{h\rightarrow 0 }{\textrm{Lim}}\: \: \displaystyle \frac{\sqrt{(2+h)-1}-\sqrt{2-1}}{h}\\ &=\underset{h\rightarrow 0 }{\textrm{Lim}}\: \: \displaystyle \displaystyle \frac{\sqrt{h+1}-1}{h}\\ &=\underset{h\rightarrow 0 }{\textrm{Lim}}\: \: \displaystyle \displaystyle \frac{\sqrt{h+1}-1}{h}\times \displaystyle \frac{\sqrt{h+1}+1}{\sqrt{h+1}+1}\\ &=\underset{h\rightarrow 0 }{\textrm{Lim}}\: \: \displaystyle \frac{(h+1)-1}{h\times \left ( \sqrt{h+1}+1 \right )}\\ &=\underset{h\rightarrow 0 }{\textrm{Lim}}\: \: \displaystyle \frac{1}{\left ( \sqrt{h+1}+1 \right )}\\ &=\displaystyle \frac{1}{\sqrt{0+1}+1}\\ &=\color{red}\displaystyle \frac{1}{2} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 8.&\textbf{(Mat Das SIMAK UI 2013)}\\ &\textrm{Nilai}\: \: \underset{x\rightarrow 5}{\textrm{Lim}}\: \: \displaystyle \frac{\sqrt{x+2\sqrt{x+1}}}{\sqrt{x-2\sqrt{x+1}}}=....\\ &\begin{array}{lll}\\ \textrm{a}.\quad \sqrt{3}+\sqrt{2}&&\textrm{d}.\quad 5\\ \textrm{b}.\quad 5-2\sqrt{6}\quad &\textrm{c}.\quad 2\sqrt{6}\quad &\textrm{e}.\quad \color{red}5+2\sqrt{6}\end{array}\\\\ &\textrm{Jawab}:\\ & \begin{aligned}\underset{x\rightarrow 5}{\textrm{Lim}}\: &\: \displaystyle \frac{\sqrt{x+2\sqrt{x+1}}}{\sqrt{x-2\sqrt{x+1}}}\\ &=\displaystyle \frac{\sqrt{5+2\sqrt{5+1}}}{\sqrt{5-2\sqrt{5+1}}}\\ &=\displaystyle \frac{\sqrt{5+2\sqrt{6}}}{\sqrt{5-2\sqrt{6}}}\\ &=\displaystyle \frac{\sqrt{3+2+2\sqrt{3.2}}}{\sqrt{3+2-2\sqrt{3.2}}}\\ &=\displaystyle \frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}\\ &=\displaystyle \frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}\times \displaystyle \frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}+\sqrt{2}}\\ &=\displaystyle \frac{3+2+2\sqrt{6}}{3-2}\\ &=\color{red}5+2\sqrt{6} \end{aligned}\end{array}$

$\begin{array}{ll}\\ 9.&\textbf{(Mat IPA SBMPTN 2014)}\\\\ &\textrm{Jika}\: \: \underset{x\rightarrow a}{\textrm{Lim}}\: \: \displaystyle \left ( f(x)+\frac{1}{g(x)} \right )=4\\ & \textrm{dan}\: \: \underset{x\rightarrow a}{\textrm{Lim}}\: \: \displaystyle \left ( f(x)-\frac{1}{g(x)} \right )=-3,\\\\ &\textrm{maka nilai}\: \: \underset{x\rightarrow a}{\textrm{Lim}}\: \: \displaystyle f(x).g(x)=....\\ &\begin{array}{lll}\\ \textrm{a}.\quad \displaystyle \frac{1}{14}&&\textrm{d}.\quad \displaystyle \frac{4}{14}\\\\ \textrm{b}.\quad \color{red}\displaystyle \frac{2}{14}\quad &\textrm{c}.\quad \displaystyle \frac{3}{14}\quad &\textrm{e}.\quad \displaystyle \frac{5}{14}\end{array}\\\\ &\textrm{Jawab}:\\ &\begin{aligned}&\textrm{Perhatikan bahwa}\: ,\\ &\begin{array}{lll}\\ \underset{x\rightarrow a}{\textrm{Lim}}\: \: \displaystyle \left ( f(x)+\frac{1}{g(x)} \right )=4&&\\ &&\\ \underset{x\rightarrow a}{\textrm{Lim}}\: \: \displaystyle \left ( f(x)-\frac{1}{g(x)} \right )=-3&+&\\ &&\\\hline &&\\ 2\underset{x\rightarrow a}{\textrm{Lim}}\: \: \displaystyle f(x)\qquad\qquad\: \: =1&&\\ &&\\ \qquad\qquad\: \: \: \: \underset{x\rightarrow a}{\textrm{Lim}}\: \: f(x)=\displaystyle \frac{1}{2},&&\\\end{array}\\ &\textrm{sehingga}\: \: \underset{x\rightarrow a}{\textrm{Lim}}\: \: f(g)=\displaystyle \frac{2}{7}\\ &&\\ &\textrm{maka},\\ &\underset{x\rightarrow a}{\textrm{Lim}}\: \: f(x).g(x)=\displaystyle \frac{1}{2}\times \frac{2}{7}=\color{red}\displaystyle \frac{2}{14} \end{aligned} \end{array}$