Lanjutan Contoh Soal Vektor

\begin{array}{ll}\\ \fbox{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 \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}:\quad \textbf{a}\\ &\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}\\ &=\begin{pmatrix} 12\\ -16 \end{pmatrix} \end{aligned} \end{array}.

\begin{array}{ll}\\ \fbox{12}.&\textrm{Vektor}\: \: \vec{v}=\begin{pmatrix} -2\\ 5 \end{pmatrix}\: \: \textrm{searah dengan vektor}.... \\ &\begin{array}{lll}\\ \textrm{a}.\quad \displaystyle \begin{pmatrix} 2\\ -5 \end{pmatrix}&&\textrm{d}.\quad \displaystyle \begin{pmatrix} -4\\ 5 \end{pmatrix}\\ \textrm{b}.\quad \displaystyle \begin{pmatrix} 2\\ 5 \end{pmatrix}&\textrm{c}.\quad \displaystyle \begin{pmatrix} -6\\ 15 \end{pmatrix}&\textrm{e}.\quad \displaystyle \begin{pmatrix} -3\\ 10 \end{pmatrix} \end{array}\\\\ &\textrm{Jawab}:\: \: \textbf{c}\\ &\begin{aligned}\textrm{Vektor}&\: \: \vec{v}\: \: \: \textrm{searah dengan vektor yang}\: \: k.\vec{v}\\\\ &\begin{array}{|c|c|c|c|c|}\hline \multicolumn{5}{|c|}{k.\vec{v}=k\begin{pmatrix} -2\\ 5 \end{pmatrix}, \textrm{dengan}\: \: k\: \: \textrm{positif}}\\\hline \textrm{a}&\textrm{b}&\textrm{c}&\textrm{d}&\textrm{e}\\\hline \begin{pmatrix} 2\\ -5 \end{pmatrix}=-\begin{pmatrix} -2\\ 5 \end{pmatrix}&\begin{pmatrix} 2\\ 5 \end{pmatrix}=...&\begin{pmatrix} -6\\ 15 \end{pmatrix}=3\begin{pmatrix} -2\\ 5 \end{pmatrix}&\begin{pmatrix} -4\\ 5 \end{pmatrix}=...&\begin{pmatrix} -3\\ 10 \end{pmatrix}=...\\\hline \end{array} \end{aligned} \end{array}.

\begin{array}{ll}\\ \fbox{13}.&\textrm{Vektor satuan}\: \: \vec{v}=\begin{pmatrix} -5\\ -12 \end{pmatrix}\: \: \textrm{adalah}.... \\ &\begin{array}{lll}\\ \textrm{a}.\quad \displaystyle -\frac{1}{13}\begin{pmatrix} 5\\ 12 \end{pmatrix}&&\textrm{d}.\quad \displaystyle -\frac{1}{17}\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{e}.\quad \displaystyle \frac{1}{2} \begin{pmatrix} 5\\ 12 \end{pmatrix} \end{array}\\\\ &\textrm{Jawab}:\: \: \textbf{a}\\ &\begin{aligned}\hat{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}}\\ &=-\displaystyle \frac{1}{13}\begin{pmatrix} 5\\ 12 \end{pmatrix} \end{aligned} \end{array}.

\begin{array}{ll}\\ \fbox{14}.&\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{d}.\quad \displaystyle \frac{1}{7}\sqrt{7}\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{e}.\quad \displaystyle \frac{1}{21}\sqrt{21}\begin{pmatrix} 2\\ -1\\ 4 \end{pmatrix} \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{e}\\ &\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}\\ \fbox{15}.&\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}....\\ &\begin{array}{lll}\\ \textrm{a}.\quad 2&&\textrm{d}.\quad 8\\ \textrm{b}.\quad 4&\textrm{c}.\quad 6&\textrm{e}.\quad 10\end{array}\\\\ &\textrm{Jawab}:\: \: \textbf{e}\\ &\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\\ &=10 \end{aligned} \end{array}.

\begin{array}{ll}\\ \fbox{16}.&\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 =....\\ &\begin{array}{lll}\\ \textrm{a}.\quad \displaystyle \frac{3}{5}&&\textrm{d}.\quad \displaystyle \frac{4}{3}\\\\ \textrm{b}.\quad \displaystyle \frac{9}{16}&\textrm{c}.\quad \displaystyle \frac{3}{4}&\textrm{e}.\quad \displaystyle \frac{16}{9} \end{array}\\\\ &\textrm{Jawab}:\: \: \textbf{c}\\ &\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 \multicolumn{2}{|c|}{\begin{aligned}\tan \theta &=\displaystyle \frac{\sin \theta }{\cos \theta }\\ &=\displaystyle \frac{\frac{6}{10}}{\frac{8}{10}}\\ &=\displaystyle \frac{3}{4} \end{aligned}}\\\hline \end{array} \end{array}.

\begin{array}{ll}\\ \fbox{17}.&\textrm{Jika}\: \: \vec{a}=\begin{pmatrix} 2\\ -5 \end{pmatrix},\: \vec{b}=\begin{pmatrix} 4\\ 3 \end{pmatrix},\: \textrm{maka proyeksi skalar ortogonal vektor}\: \vec{a}\: \: \textrm{pada}\: \: \vec{b}\: \: \textrm{adalah}....\\ &\begin{array}{lll}\\ \textrm{a}.\quad \displaystyle \frac{3}{5}&&\textrm{d}.\quad \displaystyle \frac{9}{5}\\\\ \textrm{b}.\quad \displaystyle \frac{7}{5}&\textrm{c}.\quad \displaystyle \frac{8}{5}&\textrm{e}.\quad \displaystyle \frac{10}{5} \end{array}\\\\ &\textrm{Jawab}:\: \: \textbf{b}\\ &\begin{aligned}\textrm{Misal}&\, \: \textrm{proyeksi ortogonal adalah vektor}\: \: \vec{p},\\ \textrm{maka}&\: \textrm{panjang vektor}\: \: \vec{p}\: \: \textrm{atau proyeksi skalar ortogonalnya adalah}:\\ \left | \vec{p} \right |&=\displaystyle \frac{\vec{a}.\vec{b}}{\left | \vec{b} \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 |\\ &=\displaystyle \frac{7}{5} \end{aligned} \end{array}.

\begin{array}{ll}\\ \fbox{18}.&\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}&\textrm{e}.\quad \displaystyle 150^{\circ} \end{array}\\\\ &\textrm{Jawab}:\quad \textbf{e}\\ &\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,\qquad \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}\\ \fbox{19}.&\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}\\\\ \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}\\\\ &\textrm{Jawab}:\quad \textbf{b}\\ &\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}}\\ &=\displaystyle \frac{1}{6}\sqrt{2} \end{aligned} \end{array}.

\begin{aligned}.\: \: \qquad\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}}\\ &=\displaystyle \frac{1}{6}\sqrt{34} \end{aligned}..

\begin{array}{ll}\\ \fbox{20}.&\textrm{Posisi suatu titik dalam ruang saat waktu}\: \: t\: \: \textrm{ditunjukkan oleh vektor}\\ &\begin{pmatrix} t\\ t^{2}\\ -t \end{pmatrix}.\: \textrm{Jika pada saat}\: \: t=1\: \: \textrm{titik tersebut berada di titik P dan pada}\\ &\textrm{saat}\: \: t=2\: \: \textrm{titik tersebut berada di titik Q, maka jarak titik P dari Q adalah....}\\ &\begin{array}{lll}\\ \textrm{a}.\quad \sqrt{24}-\sqrt{3}&&\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}.

 

Tentang ahmadthohir1089

Nama saya Ahmad Thohir asli orang Purwodadi, Jawa Tengah, lahir di Grobogan 02 Februari 1980. Pendidikan : Tingkat dasar lulus dari MI Nahdlatut Thullab di desa Manggarwetan,kecamatan Godong lulus tahun 1993. dan untuk tingkat menengah saya tempuh di MTs Nahdlatut Thullab Manggar Wetan lulus tahun 1996. Sedang untuk tingkat SMA saya menamatkannya di MA Futuhiyyah-2 Mranggen, Demak lulus tahun 1999. Setelah itu saya Kuliah di IKIP PGRI Semarang pada fakultas FPMIPA Pendidikan Matematika lulus tahun 2004. Pekerjaan : Sebagai guru (PNS DPK Kemenag) mapel matematika di MA Futuhiyah Jeketro, Gubug. Pengalaman mengajar : 1. GTT di MTs Miftahul Mubtadiin Tambakan Gubug tahun 2003 s/d 2005 2. GTT di SMK Negeri 3 Semarang 2005 s/d 2009 3. GT di MA Futuhiyah Jeketro Gubug sejak 1 September 2009
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