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Penulis: Mr. Benz (Teacher)
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Partikel Penyusun Atom
Atom tersusun atas inti atom atau sering disebut juga nukleon yang terdiri dari proton dan neutron, serta terdapat elektron-elektron yang mengelilingi inti atom tersebut.
- Proton adalah partikel subatomik yang bermuatan listrik positif \(+1,6\times 10^{-19}\text{ Coulomb}\)
- Neutron adalah partikel subatomik yang tidak bermuatan
- Elektron adalah partikel subatomik yang bermuatan listrik negatif \(-1,6\times 10^{-19}\text{ Coulomb}\)
Satuan Internasional muatan listrik adalah Coulomb atau disingkat \(\text{C}\)
Di sekitar kita banyak kita jumpai material yang mudah untuk menghantarkan listrik (disebut sebagai konduktor). Contoh konduktor adalah logam-logam seperti besi, tembaga, nikel, aluminium dan masih banyak lagi. Ada juga material yang tidak dapat/sulit sekali untuk menghantarkan listrik (disebut sebagai isolator). Contoh isolator adalah kain, kayu, plastik, kaca, dan karet.
Elektron-elektron dalam atom konduktor bebas bergerak sehingga mampu mengalirkan listrik, sedangkan elektron-elektron dalam atom isolator cenderung stabil (tidak bebas bergerak), sehingga tidak mampu menghantarkan listrik.
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Menulis Pecahan
Tuliskan pecahan dari daerah yang diarsir pada setiap gambar di bawah ini
Gambar Persegi
Gambar 01
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Gambar 02
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Gambar 03
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Gambar 04
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Gambar Segi Enam
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Gambar Lingkaran
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Gambar 03
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Gambar 05
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Gambar 06
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Gambar 07
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Gambar 08
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Gambar 09
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Gambar 10
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Gambar 11
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Gambar 12
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Gambar Segitiga
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Gambar 02
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Gambar 03
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Gambar 04
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Gambar 05
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Gambar 06
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Kuis Vektor 02
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Dear Students,
Welcome to today’s quiz! This is your opportunity to demonstrate what you’ve learned so far, so do your best. Please keep in mind that you have a maximum of 40 minutes to complete all the questions. Make sure to manage your time wisely and answer each question thoughtfully.
Good luck!
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Pertanyaan 1 dari 10
1. Pertanyaan
1 pointsDiketahui vektor \(\textbf{a} = 2\textbf{i}\:-\:\textbf{j} + 3\textbf{k}\) dan \(\textbf{b} =5\textbf{j} \:-\:\textbf{k}\). Hasil dari \(\textbf{a}\cdot \textbf{b} = \dotso\)
Benar
\(\textbf{a} = \left(\begin{array}{c}2\\ -1\\3\end{array}\right)\)
\(\textbf{b} = \left(\begin{array}{c}0\\ 5\\-1\end{array}\right)\)
\(\textbf{a}\cdot \textbf{b} = \left(\begin{array}{c}2\\ -1\\3\end{array}\right) \cdot \left(\begin{array}{c}0\\ 5\\-1\end{array}\right)\)
\(\textbf{a}\cdot \textbf{b} = 2(0) + (-1)(5) + 3(-1)\)
\(\textbf{a}\cdot \textbf{b} = 0\:-\:5\:-3\)
\(\textbf{a}\cdot \textbf{b} = -8\)
Salah
\(\textbf{a} = \left(\begin{array}{c}2\\ -1\\3\end{array}\right)\)
\(\textbf{b} = \left(\begin{array}{c}0\\ 5\\-1\end{array}\right)\)
\(\textbf{a}\cdot \textbf{b} = \left(\begin{array}{c}2\\ -1\\3\end{array}\right) \cdot \left(\begin{array}{c}0\\ 5\\-1\end{array}\right)\)
\(\textbf{a}\cdot \textbf{b} = 2(0) + (-1)(5) + 3(-1)\)
\(\textbf{a}\cdot \textbf{b} = 0\:-\:5\:-3\)
\(\textbf{a}\cdot \textbf{b} = -8\)
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Pertanyaan 2 dari 10
2. Pertanyaan
1 pointsDiketahui \(\textbf{a} = \left(\begin{array}{c}-4\\ 2\\-1\end{array}\right)\) dan \(\textbf{b} = \left(\begin{array}{c}1\\ 3\\6\end{array}\right)\), Hasil dari \((3\textbf{a} + \textbf{b})(2\textbf{a} \:-\:\textbf{b}) = \dotso\)
Benar
\(3\textbf{a} + \textbf{b} = 3\left(\begin{array}{c}-4\\ 2\\-1\end{array}\right) + \left(\begin{array}{c}1\\ 3\\6\end{array}\right)\)
\(3\textbf{a} + \textbf{b} = \left(\begin{array}{c}-12\\ 6\\-3\end{array}\right) + \left(\begin{array}{c}1\\ 3\\6\end{array}\right)\)
\(3\textbf{a} + \textbf{b} = \left(\begin{array}{c}-11\\ 9\\3\end{array}\right)\)
\(2\textbf{a} \:-\:\textbf{b} = 2\left(\begin{array}{c}-4\\ 2\\-1\end{array}\right)\:-\:\left(\begin{array}{c}1\\ 3\\6\end{array}\right)\)
\(2\textbf{a} \:-\:\textbf{b} = \left(\begin{array}{c}-8\\ 4\\-2\end{array}\right)\:-\:\left(\begin{array}{c}1\\ 3\\6\end{array}\right)\)
\(2\textbf{a} \:-\:\textbf{b} = \left(\begin{array}{c}-9\\1\\-8\end{array}\right)\)
\((3\textbf{a} + \textbf{b})(2\textbf{a} \:-\:\textbf{b}) =\left(\begin{array}{c}-11\\ 9\\3\end{array}\right)\cdot \left(\begin{array}{c}-9\\1\\-8\end{array}\right)\)
\((3\textbf{a} + \textbf{b})(2\textbf{a} \:-\:\textbf{b}) = -11(-9) + 9(1) + 3(-8)\)
\((3\textbf{a} + \textbf{b})(2\textbf{a} \:-\:\textbf{b}) = 99 + 9 \:-\:24 = 84\)
Salah
\(3\textbf{a} + \textbf{b} = 3\left(\begin{array}{c}-4\\ 2\\-1\end{array}\right) + \left(\begin{array}{c}1\\ 3\\6\end{array}\right)\)
\(3\textbf{a} + \textbf{b} = \left(\begin{array}{c}-12\\ 6\\-3\end{array}\right) + \left(\begin{array}{c}1\\ 3\\6\end{array}\right)\)
\(3\textbf{a} + \textbf{b} = \left(\begin{array}{c}-11\\ 9\\3\end{array}\right)\)
\(2\textbf{a} \:-\:\textbf{b} = 2\left(\begin{array}{c}-4\\ 2\\-1\end{array}\right)\:-\:\left(\begin{array}{c}1\\ 3\\6\end{array}\right)\)
\(2\textbf{a} \:-\:\textbf{b} = \left(\begin{array}{c}-8\\ 4\\-2\end{array}\right)\:-\:\left(\begin{array}{c}1\\ 3\\6\end{array}\right)\)
\(2\textbf{a} \:-\:\textbf{b} = \left(\begin{array}{c}-9\\1\\-8\end{array}\right)\)
\((3\textbf{a} + \textbf{b})(2\textbf{a} \:-\:\textbf{b}) =\left(\begin{array}{c}-11\\ 9\\3\end{array}\right)\cdot \left(\begin{array}{c}-9\\1\\-8\end{array}\right)\)
\((3\textbf{a} + \textbf{b})(2\textbf{a} \:-\:\textbf{b}) = -11(-9) + 9(1) + 3(-8)\)
\((3\textbf{a} + \textbf{b})(2\textbf{a} \:-\:\textbf{b}) = 99 + 9 \:-\:24 = 84\)
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Pertanyaan 3 dari 10
3. Pertanyaan
1 pointsDiketahui vektor \(\textbf{u} = \left(\begin{array}{c}m\\ -1\\-2\end{array}\right)\) dan \(\textbf{v} = \left(\begin{array}{c}4\\ 3\\m\end{array}\right)\). Jika \(\textbf{u}\cdot \textbf{v} = 5\) maka nilai \(m = \dotso\)
Benar
\(\textbf{u}\cdot \textbf{v} = 5\)
\(\left(\begin{array}{c}m\\ -1\\-2\end{array}\right) \cdot \left(\begin{array}{c}4\\ 3\\m\end{array}\right) = 5\)
\(4m + (-1)(3) + (-2)(m) = 5\)
\(4m\:-\:3\:-\:2m = 5\)
\(2m = 5 + 3\)
\(2m = 8\)
\(m = 4\)
Salah
\(\textbf{u}\cdot \textbf{v} = 5\)
\(\left(\begin{array}{c}m\\ -1\\-2\end{array}\right) \cdot \left(\begin{array}{c}4\\ 3\\m\end{array}\right) = 5\)
\(4m + (-1)(3) + (-2)(m) = 5\)
\(4m\:-\:3\:-\:2m = 5\)
\(2m = 5 + 3\)
\(2m = 8\)
\(m = 4\)
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Pertanyaan 4 dari 10
4. Pertanyaan
1 pointsJika vektor \(\textbf{u} = \left(\begin{array}{c}-1\\ 0\\2m + 2\end{array}\right)\) dan \(\textbf{v} = \left(\begin{array}{c}8\\ 3\\m + 1\end{array}\right)\) saling tegak lurus, maka nilai \(m = \dotso\)
Benar
Karena vektor \(\textbf{u}\) dan \(\textbf{v}\) saling tegak lurus maka berlaku \(\textbf{u}\cdot \textbf{v} = 0\)
\(\left(\begin{array}{c}-1\\ 0\\2m + 2\end{array}\right)\cdot \left(\begin{array}{c}8\\ 3\\m + 1\end{array}\right) = 0\)
\(-8 + 0 + (2m + 2)(m + 1) = 0\)
\(-8 + 2m^2 + 4m + 2 = 0\)
\(2m^2 + 4m \:-\:6 = 0\)
\(m^2 + 2m\:-\:3 = 0\)
\((m + 3)(m\:-\:1) = 0\)
\(m + 3 = 0 \rightarrow m = -3\)
\(m – 1 = 0 \rightarrow m = 1\)
Jadi nilai \(m\) yang memenuhi adalah \(-3\) atau \(1\)
Salah
Karena vektor \(\textbf{u}\) dan \(\textbf{v}\) saling tegak lurus maka berlaku \(\textbf{u}\cdot \textbf{v} = 0\)
\(\left(\begin{array}{c}-1\\ 0\\2m + 2\end{array}\right)\cdot \left(\begin{array}{c}8\\ 3\\m + 1\end{array}\right) = 0\)
\(-8 + 0 + (2m + 2)(m + 1) = 0\)
\(-8 + 2m^2 + 4m + 2 = 0\)
\(2m^2 + 4m \:-\:6 = 0\)
\(m^2 + 2m\:-\:3 = 0\)
\((m + 3)(m\:-\:1) = 0\)
\(m + 3 = 0 \rightarrow m = -3\)
\(m – 1 = 0 \rightarrow m = 1\)
Jadi nilai \(m\) yang memenuhi adalah \(-3\) atau \(1\)
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Pertanyaan 5 dari 10
5. Pertanyaan
1 pointsDiketahui vektor \(\textbf{u} = \left(\begin{array}{c}2\\ -2\\1\end{array}\right)\) dan \(\textbf{v} = \left(\begin{array}{c}4\\ 4\\-2\end{array}\right)\). Jika \(\alpha\) adalah sudut yang dibentuk oleh vektor \(\textbf{u}\) dan \(\textbf{v}\), maka nilai \(\cos \alpha = \dotso\)
Benar
\(\cos \alpha = \dfrac{\textbf{u}\cdot \textbf{v}}{||\textbf{u}||\cdot ||\textbf{v}||}\)
\(\cos \alpha = \dfrac{\left(\begin{array}{c}2\\ -2\\1\end{array}\right)\cdot \left(\begin{array}{c}4\\ 4\\-2\end{array}\right)}{\sqrt{2^2 + (-2)^2 + 1^2}\cdot \sqrt{4^2 + 4^2 + (-2)^2}}\)
\(\cos \alpha = \dfrac{2(4) + (-2)(4) + 1(-2)}{\sqrt{4 + 4 + 1}\cdot \sqrt{16 + 16 +4}}\)
\(\cos \alpha = \dfrac{8\:-\:8 \:-\:2}{\sqrt{9}\cdot \sqrt{36}}\)
\(\cos \alpha = \dfrac{-2}{3\times 6}\)
\(\cos \alpha = \dfrac{-2}{18}\)
\(\cos \alpha = -\dfrac{1}{9}\)
Salah
\(\cos \alpha = \dfrac{\textbf{u}\cdot \textbf{v}}{||\textbf{u}||\cdot ||\textbf{v}||}\)
\(\cos \alpha = \dfrac{\left(\begin{array}{c}2\\ -2\\1\end{array}\right)\cdot \left(\begin{array}{c}4\\ 4\\-2\end{array}\right)}{\sqrt{2^2 + (-2)^2 + 1^2}\cdot \sqrt{4^2 + 4^2 + (-2)^2}}\)
\(\cos \alpha = \dfrac{2(4) + (-2)(4) + 1(-2)}{\sqrt{4 + 4 + 1}\cdot \sqrt{16 + 16 +4}}\)
\(\cos \alpha = \dfrac{8\:-\:8 \:-\:2}{\sqrt{9}\cdot \sqrt{36}}\)
\(\cos \alpha = \dfrac{-2}{3\times 6}\)
\(\cos \alpha = \dfrac{-2}{18}\)
\(\cos \alpha = -\dfrac{1}{9}\)
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Pertanyaan 6 dari 10
6. Pertanyaan
1 pointsDiketahui \(||\textbf{a}|| = 2\) dan \(||\textbf{b}|| = 3\). Jika besar sudut antara vektor \(\textbf{a}\) dan \(\textbf{b}\) adalah \(120^{\circ}\), maka \(||\textbf{a} + \textbf{b}|| = \dotso\)
Benar
\(||\textbf{a} + \textbf{b}|| = \sqrt{||\textbf{a}||^2 +||\textbf{b}||^2 + 2\cdot ||\textbf{a}||\cdot ||\textbf{b}||\cdot \cos \alpha}\)
\(||\textbf{a} + \textbf{b}|| = \sqrt{2^2 + 3^2 + 2\cdot 2\cdot 3\cdot \cos 120^{\circ}}\)
\(||\textbf{a} + \textbf{b}|| = \sqrt{4 + 9 + 2\cdot 2\cdot 3\cdot (-\frac{1}{2})}\)
\(||\textbf{a} + \textbf{b}|| = \sqrt{13 \:-\: 6}\)
\(||\textbf{a} + \textbf{b}|| = \sqrt{7}\)
Salah
\(||\textbf{a} + \textbf{b}|| = \sqrt{||\textbf{a}||^2 +||\textbf{b}||^2 + 2\cdot ||\textbf{a}||\cdot ||\textbf{b}||\cdot \cos \alpha}\)
\(||\textbf{a} + \textbf{b}|| = \sqrt{2^2 + 3^2 + 2\cdot 2\cdot 3\cdot \cos 120^{\circ}}\)
\(||\textbf{a} + \textbf{b}|| = \sqrt{4 + 9 + 2\cdot 2\cdot 3\cdot (-\frac{1}{2})}\)
\(||\textbf{a} + \textbf{b}|| = \sqrt{13 \:-\: 6}\)
\(||\textbf{a} + \textbf{b}|| = \sqrt{7}\)
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Pertanyaan 7 dari 10
7. Pertanyaan
1 pointsDiketahui \(||\textbf{a}|| = 4\) dan \(||\textbf{b}|| = 5\). Jika besar sudut antara vektor \(\textbf{a}\) dan \(\textbf{b}\) adalah \(60^{\circ}\), maka \(||\textbf{a} \:-\: \textbf{b}|| = \dotso\)
Benar
\(||\textbf{a} \:-\: \textbf{b}|| = \sqrt{||\textbf{a}||^2 +||\textbf{b}||^2 \:-\:2\cdot ||\textbf{a}||\cdot ||\textbf{b}||\cdot \cos \alpha}\)
\(||\textbf{a} \:-\: \textbf{b}|| = \sqrt{4^2 + 5^2 \:-\:2\cdot 4\cdot 5\cdot \cos 60^{\circ}}\)
\(||\textbf{a} \:-\: \textbf{b}|| = \sqrt{16 + 25 \:-\:2\cdot 4\cdot 5\cdot \dfrac{1}{2}}\)
\(||\textbf{a} \:-\: \textbf{b}|| = \sqrt{41 \:-\:20}\)
\(||\textbf{a} \:-\: \textbf{b}|| = \sqrt{21}\)
Salah
\(||\textbf{a} \:-\: \textbf{b}|| = \sqrt{||\textbf{a}||^2 +||\textbf{b}||^2 \:-\:2\cdot ||\textbf{a}||\cdot ||\textbf{b}||\cdot \cos \alpha}\)
\(||\textbf{a} \:-\: \textbf{b}|| = \sqrt{4^2 + 5^2 \:-\:2\cdot 4\cdot 5\cdot \cos 60^{\circ}}\)
\(||\textbf{a} \:-\: \textbf{b}|| = \sqrt{16 + 25 \:-\:2\cdot 4\cdot 5\cdot \dfrac{1}{2}}\)
\(||\textbf{a} \:-\: \textbf{b}|| = \sqrt{41 \:-\:20}\)
\(||\textbf{a} \:-\: \textbf{b}|| = \sqrt{21}\)
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Pertanyaan 8 dari 10
8. Pertanyaan
1 pointsJika \(||\textbf{a}|| = 5\), \(||\textbf{b}|| = 2\), dan \(||\textbf{a} + \textbf{b}|| = \sqrt{31}\) maka nilai dari \(\textbf{a}\cdot \textbf{b} = \dotso\)
Benar
\(||\textbf{a} + \textbf{b}|| = \sqrt{||\textbf{a}||^2 +||\textbf{b}||^2 + 2\cdot ||\textbf{a}||\cdot ||\textbf{b}||\cdot \cos \alpha}\)
\(||\textbf{a} + \textbf{b}|| = \sqrt{||\textbf{a}||^2 +||\textbf{b}||^2 + 2\cdot \textbf{a}\cdot \textbf{b}}\)
\(\sqrt{31} = \sqrt{5^2 + 2^2 + 2\cdot \textbf{a}\cdot \textbf{b}}\)
\(31 = 25 + 4 + 2\cdot \textbf{a}\cdot \textbf{b}\)
\(31 \:-\:29 = 2\cdot \textbf{a}\cdot \textbf{b}\)
\(2 = 2\cdot \textbf{a}\cdot \textbf{b}\)
\(\textbf{a}\cdot \textbf{b} = 1\)
Salah
\(||\textbf{a} + \textbf{b}|| = \sqrt{||\textbf{a}||^2 +||\textbf{b}||^2 + 2\cdot ||\textbf{a}||\cdot ||\textbf{b}||\cdot \cos \alpha}\)
\(||\textbf{a} + \textbf{b}|| = \sqrt{||\textbf{a}||^2 +||\textbf{b}||^2 + 2\cdot \textbf{a}\cdot \textbf{b}}\)
\(\sqrt{31} = \sqrt{5^2 + 2^2 + 2\cdot \textbf{a}\cdot \textbf{b}}\)
\(31 = 25 + 4 + 2\cdot \textbf{a}\cdot \textbf{b}\)
\(31 \:-\:29 = 2\cdot \textbf{a}\cdot \textbf{b}\)
\(2 = 2\cdot \textbf{a}\cdot \textbf{b}\)
\(\textbf{a}\cdot \textbf{b} = 1\)
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Pertanyaan 9 dari 10
9. Pertanyaan
1 pointsDiketahui vektor \(||\textbf{a}|| = 3\) dan \(||\textbf{b}|| = 5\). Jika besar sudut antara vektor \(||\textbf{a}||\) dan \(||\textbf{b}||\) adalah \(60^{\circ}\), maka nilai dari \(||2\textbf{a} + \textbf{b}|| = \dotso\)
Benar
\(||\textbf{a} + \textbf{b}|| = \sqrt{||\textbf{a}||^2 +||\textbf{b}||^2 + 2\cdot ||\textbf{a}||\cdot ||\textbf{b}||\cdot \cos \alpha}\)
\(||\textbf{2a} + \textbf{b}|| = \sqrt{||2\textbf{a}||^2 +||\textbf{b}||^2 + 2\cdot ||2\textbf{a}||\cdot ||\textbf{b}||\cdot \cos \alpha}\)
\(||\textbf{2a} + \textbf{b}|| = \sqrt{4||\textbf{a}||^2 +||\textbf{b}||^2 + 4\cdot ||\textbf{a}||\cdot ||\textbf{b}||\cdot \cos \alpha}\)
\(||\textbf{2a} + \textbf{b}|| = \sqrt{4(3)^2 + (5)^2 + 4\cdot 3\cdot 5\cdot \cos 60^{\circ}}\)
\(||\textbf{2a} + \textbf{b}|| = \sqrt{36 + 25 + 4\cdot 3\cdot 5\cdot \dfrac{1}{2}}\)
\(||\textbf{2a} + \textbf{b}|| = \sqrt{36 + 25 + 30}\)
\(||\textbf{2a} + \textbf{b}|| = \sqrt{91}\)
Salah
\(||\textbf{a} + \textbf{b}|| = \sqrt{||\textbf{a}||^2 +||\textbf{b}||^2 + 2\cdot ||\textbf{a}||\cdot ||\textbf{b}||\cdot \cos \alpha}\)
\(||\textbf{2a} + \textbf{b}|| = \sqrt{||2\textbf{a}||^2 +||\textbf{b}||^2 + 2\cdot ||2\textbf{a}||\cdot ||\textbf{b}||\cdot \cos \alpha}\)
\(||\textbf{2a} + \textbf{b}|| = \sqrt{4||\textbf{a}||^2 +||\textbf{b}||^2 + 4\cdot ||\textbf{a}||\cdot ||\textbf{b}||\cdot \cos \alpha}\)
\(||\textbf{2a} + \textbf{b}|| = \sqrt{4(3)^2 + (5)^2 + 4\cdot 3\cdot 5\cdot \cos 60^{\circ}}\)
\(||\textbf{2a} + \textbf{b}|| = \sqrt{36 + 25 + 4\cdot 3\cdot 5\cdot \dfrac{1}{2}}\)
\(||\textbf{2a} + \textbf{b}|| = \sqrt{36 + 25 + 30}\)
\(||\textbf{2a} + \textbf{b}|| = \sqrt{91}\)
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Pertanyaan 10 dari 10
10. Pertanyaan
1 pointsJika \(||\textbf{a}|| = 5\) dan \(\textbf{a}\cdot \textbf{b} = 2\), maka \(\textbf{a}\cdot (\textbf{a} \:-\:\textbf{b}) = \dotso\)
Benar
\(\textbf{a}\cdot (\textbf{a} \:-\:\textbf{b}) = \textbf{a}\cdot \textbf{a}\:-\:\textbf{a}\cdot \textbf{b}\)
\(\textbf{a}\cdot (\textbf{a} \:-\:\textbf{b}) = ||\textbf{a}||^2\:-\:\textbf{a}\cdot \textbf{b}\)
\(\textbf{a}\cdot (\textbf{a} \:-\:\textbf{b}) = 5^2 \:-\:2 \)
\(\textbf{a}\cdot (\textbf{a} \:-\:\textbf{b}) = 25 \:-\:2 = 23\)
Salah
\(\textbf{a}\cdot (\textbf{a} \:-\:\textbf{b}) = \textbf{a}\cdot \textbf{a}\:-\:\textbf{a}\cdot \textbf{b}\)
\(\textbf{a}\cdot (\textbf{a} \:-\:\textbf{b}) = ||\textbf{a}||^2\:-\:\textbf{a}\cdot \textbf{b}\)
\(\textbf{a}\cdot (\textbf{a} \:-\:\textbf{b}) = 5^2 \:-\:2 \)
\(\textbf{a}\cdot (\textbf{a} \:-\:\textbf{b}) = 25 \:-\:2 = 23\)
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Kuis Vektor 01
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Information
Dear Students,
Welcome to today’s quiz! This is your opportunity to demonstrate what you’ve learned so far, so do your best. Please keep in mind that you have a maximum of 45 minutes to complete all the questions. Make sure to manage your time wisely and answer each question thoughtfully.
Good luck!
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Pertanyaan 1 dari 13
1. Pertanyaan
1 pointsDiketahui koordinat titik A\((-1, 2, 0)\) dan titik B\((3, -4, 2)\). Vektor satuan dari \(\overrightarrow{\text{BA}}\) adalah …
Benar
\(\widehat{\text{BA}}=\dfrac{\textbf{BA}}{||\textbf{BA}||}\)
\(\overrightarrow{\text{BA}} = \textbf{a}\:-\:\textbf{b}\)
\(\overrightarrow{\text{BA}} = \left(\begin{array}{c}-1\\ 2\\0\end{array}\right)\:-\:\left(\begin{array}{c}3\\-4\\2\end{array}\right)\)
\(\overrightarrow{\text{BA}} = \left(\begin{array}{c}-4\\ 6\\-2\end{array}\right)\)
\(\widehat{\text{BA}}=\dfrac{\left(\begin{array}{c}-4\\ 6\\-2\end{array}\right)}{\sqrt{(-4)^2 + 6^2 + (-2)^2}}\)
\(\widehat{\text{BA}}=\dfrac{\left(\begin{array}{c}-4\\ 6\\-2\end{array}\right)}{\sqrt{16 + 36 + 4}}\)
\(\widehat{\text{BA}}=\dfrac{\left(\begin{array}{c}-4\\ 6\\-2\end{array}\right)}{\sqrt{56}}\)
\(\widehat{\text{BA}}=\dfrac{\left(\begin{array}{c}-4\\ 6\\-2\end{array}\right)}{2\sqrt{14}}\)
\(\widehat{\text{BA}}=\dfrac{1}{\sqrt{14}}\left(\begin{array}{c}-2\\ 3\\-1\end{array}\right)\)
Salah
\(\widehat{\text{BA}}=\dfrac{\textbf{BA}}{||\textbf{BA}||}\)
\(\overrightarrow{\text{BA}} = \textbf{a}\:-\:\textbf{b}\)
\(\overrightarrow{\text{BA}} = \left(\begin{array}{c}-1\\ 2\\0\end{array}\right)\:-\:\left(\begin{array}{c}3\\-4\\2\end{array}\right)\)
\(\overrightarrow{\text{BA}} = \left(\begin{array}{c}-4\\ 6\\-2\end{array}\right)\)
\(\widehat{\text{BA}}=\dfrac{\left(\begin{array}{c}-4\\ 6\\-2\end{array}\right)}{\sqrt{(-4)^2 + 6^2 + (-2)^2}}\)
\(\widehat{\text{BA}}=\dfrac{\left(\begin{array}{c}-4\\ 6\\-2\end{array}\right)}{\sqrt{16 + 36 + 4}}\)
\(\widehat{\text{BA}}=\dfrac{\left(\begin{array}{c}-4\\ 6\\-2\end{array}\right)}{\sqrt{56}}\)
\(\widehat{\text{BA}}=\dfrac{\left(\begin{array}{c}-4\\ 6\\-2\end{array}\right)}{2\sqrt{14}}\)
\(\widehat{\text{BA}}=\dfrac{1}{\sqrt{14}}\left(\begin{array}{c}-2\\ 3\\-1\end{array}\right)\)
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Pertanyaan 2 dari 13
2. Pertanyaan
1 pointsDiketahui \(\textbf{u} = 2\textbf{i} \:-\:5\textbf{j} + \textbf{k}\) dan \(\textbf{v} = \textbf{j} + 3\textbf{k}\). Pernyataan di bawah ini yang benar adalah …
Benar
\(\textbf{u} = 2\textbf{i} \:-\:5\textbf{j} + \textbf{k}\)
\(||\textbf{u}|| = \sqrt{2^2 + (-5)^2 + 1^2}\)
\(||\textbf{u}|| = \sqrt{4 + 25 + 1}\)
\(||\textbf{u}|| = \sqrt{30}\)
\(\textbf{v} = \textbf{j} + 3\textbf{k}\)
\(||\textbf{v}|| = \sqrt{1^2 + 3^2}\)
\(||\textbf{v}|| = \sqrt{1 + 9}\)
\(||\textbf{v}|| = \sqrt{10}\)
\(||\textbf{u}|| \cdot ||\textbf{v}|| =\sqrt{30}\cdot \sqrt{10} = 10\sqrt{3}\)
\(||\textbf{u}||\:-\: ||\textbf{v}|| =\sqrt{30}\:-\: \sqrt{10}\)
\(\textbf{u}\:-\:2\textbf{v} = 2\textbf{i} \:-\:5\textbf{j} + \textbf{k}\:-\:(\textbf{j} + 3\textbf{k})\)
\(\textbf{u}\:-\:2\textbf{v} = 2\textbf{i} \:-\:7\textbf{j} \:-\:5\textbf{k})\)
\(||\textbf{u}\:-\:2\textbf{v} || = \sqrt{2^2 + (-7)^2 + (-5)^2}\)
\(||\textbf{u}\:-\:2\textbf{v} || = \sqrt{4 + 49 + 25}\)
\(||\textbf{u}\:-\:2\textbf{v} || = \sqrt{78}\)
Salah
\(\textbf{u} = 2\textbf{i} \:-\:5\textbf{j} + \textbf{k}\)
\(||\textbf{u}|| = \sqrt{2^2 + (-5)^2 + 1^2}\)
\(||\textbf{u}|| = \sqrt{4 + 25 + 1}\)
\(||\textbf{u}|| = \sqrt{30}\)
\(\textbf{v} = \textbf{j} + 3\textbf{k}\)
\(||\textbf{v}|| = \sqrt{1^2 + 3^2}\)
\(||\textbf{v}|| = \sqrt{1 + 9}\)
\(||\textbf{v}|| = \sqrt{10}\)
\(||\textbf{u}|| \cdot ||\textbf{v}|| =\sqrt{30}\cdot \sqrt{10} = 10\sqrt{3}\)
\(||\textbf{u}||\:-\: ||\textbf{v}|| =\sqrt{30}\:-\: \sqrt{10}\)
\(\textbf{u}\:-\:2\textbf{v} = 2\textbf{i} \:-\:5\textbf{j} + \textbf{k}\:-\:(\textbf{j} + 3\textbf{k})\)
\(\textbf{u}\:-\:2\textbf{v} = 2\textbf{i} \:-\:7\textbf{j} \:-\:5\textbf{k})\)
\(||\textbf{u}\:-\:2\textbf{v} || = \sqrt{2^2 + (-7)^2 + (-5)^2}\)
\(||\textbf{u}\:-\:2\textbf{v} || = \sqrt{4 + 49 + 25}\)
\(||\textbf{u}\:-\:2\textbf{v} || = \sqrt{78}\)
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Pertanyaan 3 dari 13
3. Pertanyaan
1 pointsJika \(\textbf{p} = \left(\begin{array}{c}1\\ 3\\0\end{array}\right)\), \(\textbf{q} = \left(\begin{array}{c}0\\ -1\\2\end{array}\right)\) dan \(\textbf{r} = \left(\begin{array}{c}1\\ 3\\-3\end{array}\right)\), maka nilai \(||(2\textbf{p} + \textbf{q})\:-\:(\textbf{p}\:-\:3\textbf{r})|| = \dotso\)
Benar
\(2\textbf{p} + \textbf{q}\:-\:(\textbf{p}\:-\:3\textbf{r})\)
\(2\textbf{p} + \textbf{q}\:-\:\textbf{p}+ 3\textbf{r}\)
\(\textbf{p} + \textbf{q}+ 3\textbf{r}\)
\(\left(\begin{array}{c}1\\ 3\\0\end{array}\right) + \left(\begin{array}{c}0\\ -1\\2\end{array}\right) + 3\left(\begin{array}{c}1\\ 3\\-3\end{array}\right)\)
\(\textbf{p} + \textbf{q}+ 3\textbf{r} = \left(\begin{array}{c}1 + 0 + 3\\3\:-\:1 + 9\\0 + 2 \:-\:9\end{array}\right)\)
\(\textbf{p} + \textbf{q}+ 3\textbf{r} = \left(\begin{array}{c}4\\11\\-7\end{array}\right)\)
\(||\textbf{p} + \textbf{q}+ 3\textbf{r} ||= \sqrt{4^2 + 11^2 + (-7)^2}\)
\(\sqrt{16 + 121 +49}\)
\(\sqrt{186}\)
Salah
\(2\textbf{p} + \textbf{q}\:-\:(\textbf{p}\:-\:3\textbf{r})\)
\(2\textbf{p} + \textbf{q}\:-\:\textbf{p}+ 3\textbf{r}\)
\(\textbf{p} + \textbf{q}+ 3\textbf{r}\)
\(\left(\begin{array}{c}1\\ 3\\0\end{array}\right) + \left(\begin{array}{c}0\\ -1\\2\end{array}\right) + 3\left(\begin{array}{c}1\\ 3\\-3\end{array}\right)\)
\(\textbf{p} + \textbf{q}+ 3\textbf{r} = \left(\begin{array}{c}1 + 0 + 3\\3\:-\:1 + 9\\0 + 2 \:-\:9\end{array}\right)\)
\(\textbf{p} + \textbf{q}+ 3\textbf{r} = \left(\begin{array}{c}4\\11\\-7\end{array}\right)\)
\(||\textbf{p} + \textbf{q}+ 3\textbf{r} ||= \sqrt{4^2 + 11^2 + (-7)^2}\)
\(\sqrt{16 + 121 +49}\)
\(\sqrt{186}\)
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Pertanyaan 4 dari 13
4. Pertanyaan
1 pointsDiketahui titik A\((2, 0, -6)\) dan titik B\((-4, 2, 4)\). Titik P terletak di tengah-tengah AB. Vektor posisi dari P adalah …
Benar
\(\textbf{p} = \dfrac{\textbf{a} + \textbf{b}}{2}\)
\(\textbf{p} = \dfrac{\left(\begin{array}{c}2\\ 0\\-6\end{array}\right) + \left(\begin{array}{c}-4\\ 2\\4\end{array}\right)}{2}\)
\(\textbf{p} = \dfrac{\left(\begin{array}{c}2\:-\:4\\ 0 + 2\\-6 + 4\end{array}\right) }{2}\)
\(\textbf{p} = \dfrac{\left(\begin{array}{c}-2\\2\\-2\end{array}\right) }{2}\)
\(\textbf{p} = \left(\begin{array}{c}-1\\1\\-1\end{array}\right)\)
Salah
\(\textbf{p} = \dfrac{\textbf{a} + \textbf{b}}{2}\)
\(\textbf{p} = \dfrac{\left(\begin{array}{c}2\\ 0\\-6\end{array}\right) + \left(\begin{array}{c}-4\\ 2\\4\end{array}\right)}{2}\)
\(\textbf{p} = \dfrac{\left(\begin{array}{c}2\:-\:4\\ 0 + 2\\-6 + 4\end{array}\right) }{2}\)
\(\textbf{p} = \dfrac{\left(\begin{array}{c}-2\\2\\-2\end{array}\right) }{2}\)
\(\textbf{p} = \left(\begin{array}{c}-1\\1\\-1\end{array}\right)\)
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Pertanyaan 5 dari 13
5. Pertanyaan
1 pointsDiketahui A\((3, 7)\) dan titik B\((3, 3)\). Titik P terletak pada ruas garis AB, dengan perbandingan AP : PB = 1 : 3. Vektor posisi titik P tersebut ditulis dalam vektor basis adalah …
Benar
\(\textbf{p} = \dfrac{1\cdot \textbf{b} + 3\cdot \textbf{a}}{1 + 3}\)
\(\textbf{p} = \dfrac{\left(\begin{array}{c}3\\3\end{array}\right)+ 3\left(\begin{array}{c}3\\7\end{array}\right)}{4}\)
\(\textbf{p} = \dfrac{\left(\begin{array}{c}3 + 9\\3 + 21\end{array}\right)}{4}\)
\(\textbf{p} = \dfrac{\left(\begin{array}{c}12\\24\end{array}\right)}{4}\)
\(\textbf{p} = \left(\begin{array}{c}3\\6\end{array}\right)\)
\(\textbf{p} =3\textbf{i} + 6\textbf{j}\)
Salah
\(\textbf{p} = \dfrac{1\cdot \textbf{b} + 3\cdot \textbf{a}}{1 + 3}\)
\(\textbf{p} = \dfrac{\left(\begin{array}{c}3\\3\end{array}\right)+ 3\left(\begin{array}{c}3\\7\end{array}\right)}{4}\)
\(\textbf{p} = \dfrac{\left(\begin{array}{c}3 + 9\\3 + 21\end{array}\right)}{4}\)
\(\textbf{p} = \dfrac{\left(\begin{array}{c}12\\24\end{array}\right)}{4}\)
\(\textbf{p} = \left(\begin{array}{c}3\\6\end{array}\right)\)
\(\textbf{p} =3\textbf{i} + 6\textbf{j}\)
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Pertanyaan 6 dari 13
6. Pertanyaan
1 pointsTitik A\((3, 0, -6)\), B\((3, -12, 0)\) dan P kolinear dengan perbandingan AP : PB = – 2 : 5. Jika \(\textbf{p}\) adalah vektor posisi dari titik P, maka \(\textbf{p} = \dotso\)
Benar
\(\dfrac{\overrightarrow{\text{AP}}}{\overrightarrow{\text{PB}}} = \dfrac{- 2}{5}\)
\(\dfrac{\textbf{p}\:-\:\textbf{a}}{\textbf{b}\:-\:\textbf{p}} = -2 : 5\)
\(5\textbf{p}\:-\:5\textbf{a} = -2\textbf{b} + 2\textbf{p}\)
\(5\textbf{p}\:-\:2\textbf{p} = 5\textbf{a}\:-\:2\textbf{b}\)
\(3\textbf{p} = 5\textbf{a}\:-\:2\textbf{b}\)
\(\textbf{p} = \dfrac{5}{3}\textbf{a}\:-\:\dfrac{2}{3}\textbf{b}\)
\(\textbf{p} = \dfrac{5}{3}\left(\begin{array}{c}3\\0\\-6\end{array}\right)\:-\:\dfrac{2}{3}\left(\begin{array}{c}3\\-12\\0\end{array}\right)\)
\(\textbf{p} = \left(\begin{array}{c}5\\0\\-10\end{array}\right)\:-\:\left(\begin{array}{c}2\\-8\\0\end{array}\right)\)
\(\textbf{p} = \left(\begin{array}{c}3\\8\\-10\end{array}\right)\)
Salah
\(\dfrac{\overrightarrow{\text{AP}}}{\overrightarrow{\text{PB}}} = \dfrac{- 2}{5}\)
\(\dfrac{\textbf{p}\:-\:\textbf{a}}{\textbf{b}\:-\:\textbf{p}} = -2 : 5\)
\(5\textbf{p}\:-\:5\textbf{a} = -2\textbf{b} + 2\textbf{p}\)
\(5\textbf{p}\:-\:2\textbf{p} = 5\textbf{a}\:-\:2\textbf{b}\)
\(3\textbf{p} = 5\textbf{a}\:-\:2\textbf{b}\)
\(\textbf{p} = \dfrac{5}{3}\textbf{a}\:-\:\dfrac{2}{3}\textbf{b}\)
\(\textbf{p} = \dfrac{5}{3}\left(\begin{array}{c}3\\0\\-6\end{array}\right)\:-\:\dfrac{2}{3}\left(\begin{array}{c}3\\-12\\0\end{array}\right)\)
\(\textbf{p} = \left(\begin{array}{c}5\\0\\-10\end{array}\right)\:-\:\left(\begin{array}{c}2\\-8\\0\end{array}\right)\)
\(\textbf{p} = \left(\begin{array}{c}3\\8\\-10\end{array}\right)\)
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Pertanyaan 7 dari 13
7. Pertanyaan
1 points
Besar sudut antara vektor \(\textbf{a}\) dan \(\textbf{b}\) di atas adalah …
Benar
Salah
-
Pertanyaan 8 dari 13
8. Pertanyaan
1 pointsMisalkan ABCD adalah segi empat dengan diagonal AC dan BD berpotongan di titik P. Jika DP = PB dan AP = 2PC, \(\overrightarrow{\text{AB}} = \textbf{a}\) dan \(\overrightarrow{\text{AD}} = \textbf{b}\), maka \(\overrightarrow{\text{PC}} = \dotso\)
Benar
Langkah 1 Menentukan \(\overrightarrow{\text{BD}}\) dan \(\overrightarrow{\text{BP}}\)
\(\overrightarrow{\text{BD}} = \overrightarrow{\text{BA}} + \overrightarrow{\text{AD}}\)
\(\overrightarrow{\text{BD}} = -\textbf{a} + \textbf{b}\)
\(\overrightarrow{\text{BP}} = \dfrac{1}{2}\overrightarrow{\text{BD}}\)
\(\overrightarrow{\text{BP}} = -\dfrac{1}{2}\textbf{a} + \dfrac{1}{2}\textbf{b}\)
Langkah 2 Menentukan \(\overrightarrow{\text{AP}}\)
\(\overrightarrow{\text{AP}} = \overrightarrow{\text{AB}} +\overrightarrow{\text{BP}}\)
\(\overrightarrow{\text{AP}} = \textbf{a} \:-\:\dfrac{1}{2}\textbf{a} + \dfrac{1}{2}\textbf{b}\)
\(\overrightarrow{\text{AP}} = \dfrac{1}{2}\textbf{a} + \dfrac{1}{2}\textbf{b}\)
Langkah 3 Menentukan \(\overrightarrow{\text{PC}}\)
\(\overrightarrow{\text{PC}} = \dfrac{1}{2}\overrightarrow{\text{AP}}\)
\(\overrightarrow{\text{PC}} = \dfrac{1}{2}\left[\dfrac{1}{2}\textbf{a} + \dfrac{1}{2}\textbf{b}\right]\)
\(\overrightarrow{\text{PC}} = \dfrac{1}{4}\textbf{a} + \dfrac{1}{4}\textbf{b}\)
Salah
Langkah 1 Menentukan \(\overrightarrow{\text{BD}}\) dan \(\overrightarrow{\text{BP}}\)
\(\overrightarrow{\text{BD}} = \overrightarrow{\text{BA}} + \overrightarrow{\text{AD}}\)
\(\overrightarrow{\text{BD}} = -\textbf{a} + \textbf{b}\)
\(\overrightarrow{\text{BP}} = \dfrac{1}{2}\overrightarrow{\text{BD}}\)
\(\overrightarrow{\text{BP}} = -\dfrac{1}{2}\textbf{a} + \dfrac{1}{2}\textbf{b}\)
Langkah 2 Menentukan \(\overrightarrow{\text{AP}}\)
\(\overrightarrow{\text{AP}} = \overrightarrow{\text{AB}} +\overrightarrow{\text{BP}}\)
\(\overrightarrow{\text{AP}} = \textbf{a} \:-\:\dfrac{1}{2}\textbf{a} + \dfrac{1}{2}\textbf{b}\)
\(\overrightarrow{\text{AP}} = \dfrac{1}{2}\textbf{a} + \dfrac{1}{2}\textbf{b}\)
Langkah 3 Menentukan \(\overrightarrow{\text{PC}}\)
\(\overrightarrow{\text{PC}} = \dfrac{1}{2}\overrightarrow{\text{AP}}\)
\(\overrightarrow{\text{PC}} = \dfrac{1}{2}\left[\dfrac{1}{2}\textbf{a} + \dfrac{1}{2}\textbf{b}\right]\)
\(\overrightarrow{\text{PC}} = \dfrac{1}{4}\textbf{a} + \dfrac{1}{4}\textbf{b}\)
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Pertanyaan 9 dari 13
9. Pertanyaan
1 pointsPerhatikan gambar berikut:
ABCD adalah sebuah persegi panjang dan ABEC adalah sebuah jajargenjang. Jika \(\overrightarrow{\text{AB}} = \textbf{u}\) dan \(\overrightarrow{\text{AD}} = \textbf{v}\), maka \(\overrightarrow{\text{AE}} +\overrightarrow{\text{OD}} = \dotso\)
Benar
Langkah 1: Menentukan \(\overrightarrow{\text{BD}}\) dan \(\overrightarrow{\text{OD}}\)
\(\overrightarrow{\text{BD}} = \overrightarrow{\text{BA}} + \overrightarrow{\text{AD}}\)
\(\overrightarrow{\text{BD}} = -\textbf{u} + \textbf{v}\)
\(\overrightarrow{\text{OD}} = \dfrac{1}{2}\overrightarrow{\text{BD}}\)
\(\overrightarrow{\text{OD}} = \dfrac{1}{2}(-\textbf{u} + \textbf{v})\)
\(\overrightarrow{\text{OD}} = -\dfrac{1}{2}\textbf{u} + \dfrac{1}{2}\textbf{v}\)
Langkah 2: Menentukan \(\overrightarrow{\text{AC}}\) dan \(\overrightarrow{\text{BE}}\)
\(\overrightarrow{\text{AC}} = \overrightarrow{\text{AB}} + \overrightarrow{\text{BC}}\)
\(\overrightarrow{\text{AC}} = \textbf{u} + \textbf{v}\)
Karena ABEC adalah jajargenjang maka:
\(\overrightarrow{\text{AC}} = \overrightarrow{\text{BE}} = \textbf{u} + \textbf{v}\)
Langkah 3: Menentukan \(\overrightarrow{\text{AE}}\)
\(\overrightarrow{\text{AE}} = \overrightarrow{\text{AB}} + \overrightarrow{\text{BE}}\)
\(\overrightarrow{\text{AE}} = \textbf{u} + \textbf{u} + \textbf{v}\)
\(\overrightarrow{\text{AE}} = 2\textbf{u} + \textbf{v}\)
Langkah 4: Menentukan \(\overrightarrow{\text{AE}} + \overrightarrow{\text{OD}}\)
\(\overrightarrow{\text{AE}} + \overrightarrow{\text{OD}} = 2\textbf{u} + \textbf{v} + -\dfrac{1}{2}\textbf{u} + \dfrac{1}{2}\textbf{v}\)
\(\overrightarrow{\text{AE}} + \overrightarrow{\text{OD}} = \dfrac{3}{2}\textbf{u} + \dfrac{3}{2}\textbf{v}\)
Salah
Langkah 1: Menentukan \(\overrightarrow{\text{BD}}\) dan \(\overrightarrow{\text{OD}}\)
\(\overrightarrow{\text{BD}} = \overrightarrow{\text{BA}} + \overrightarrow{\text{AD}}\)
\(\overrightarrow{\text{BD}} = -\textbf{u} + \textbf{v}\)
\(\overrightarrow{\text{OD}} = \dfrac{1}{2}\overrightarrow{\text{BD}}\)
\(\overrightarrow{\text{OD}} = \dfrac{1}{2}(-\textbf{u} + \textbf{v})\)
\(\overrightarrow{\text{OD}} = -\dfrac{1}{2}\textbf{u} + \dfrac{1}{2}\textbf{v}\)
Langkah 2: Menentukan \(\overrightarrow{\text{AC}}\) dan \(\overrightarrow{\text{BE}}\)
\(\overrightarrow{\text{AC}} = \overrightarrow{\text{AB}} + \overrightarrow{\text{BC}}\)
\(\overrightarrow{\text{AC}} = \textbf{u} + \textbf{v}\)
Karena ABEC adalah jajargenjang maka:
\(\overrightarrow{\text{AC}} = \overrightarrow{\text{BE}} = \textbf{u} + \textbf{v}\)
Langkah 3: Menentukan \(\overrightarrow{\text{AE}}\)
\(\overrightarrow{\text{AE}} = \overrightarrow{\text{AB}} + \overrightarrow{\text{BE}}\)
\(\overrightarrow{\text{AE}} = \textbf{u} + \textbf{u} + \textbf{v}\)
\(\overrightarrow{\text{AE}} = 2\textbf{u} + \textbf{v}\)
Langkah 4: Menentukan \(\overrightarrow{\text{AE}} + \overrightarrow{\text{OD}}\)
\(\overrightarrow{\text{AE}} + \overrightarrow{\text{OD}} = 2\textbf{u} + \textbf{v} + -\dfrac{1}{2}\textbf{u} + \dfrac{1}{2}\textbf{v}\)
\(\overrightarrow{\text{AE}} + \overrightarrow{\text{OD}} = \dfrac{3}{2}\textbf{u} + \dfrac{3}{2}\textbf{v}\)
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Pertanyaan 10 dari 13
10. Pertanyaan
1 pointsDiketahui \(\textbf{p} = \left(\begin{array}{c}9\\4\\5\end{array}\right)\), \(\textbf{q} = \left(\begin{array}{c}3\\-2\\3\end{array}\right)\) dan \(\textbf{r} = \left(\begin{array}{c}0\\5\\-2\end{array}\right)\). Jika ketiga vektor tersebut satu bidang sehingga dapat dinyatakan \(\textbf{p} = m\textbf{q} + n\textbf{r}\), maka nilai \(m + n = \dotso\)
Benar
\(\textbf{p} = m\textbf{q} + n\textbf{r}\)
\(\left(\begin{array}{c}9\\4\\5\end{array}\right) = m\left(\begin{array}{c}3\\-2\\3\end{array}\right) + n\left(\begin{array}{c}0\\5\\-2\end{array}\right)\)
\(\left(\begin{array}{c}9\\4\\5\end{array}\right) = \left(\begin{array}{c}3m\\-2m\\3m\end{array}\right) + \left(\begin{array}{c}0\\5n\\-2n\end{array}\right)\)
\(\left(\begin{array}{c}9\\4\\5\end{array}\right) = \left(\begin{array}{c}3m\\-2m + 5n\\3m\:-\:2n\end{array}\right)\)
\(\left(\begin{array}{c}9\\4\\5\end{array}\right) = \left(\begin{array}{c}3m\\-2m + 5n\\3m\:-\:2n\end{array}\right)\)
Dengan prinsip kesamaan dua vektor:
\(9 = 3m\rightarrow m = 3\)
\(4 = -2m + 5n\)
\(4 = -2(3) + 5n\)
\(4 = -6 + 5n\)
\(4 + 6 = 5n\)
\(10 = 5n\)
\(n = 2\)
Jadi, \(m + n = 3 + 2 = 5\)
Salah
\(\textbf{p} = m\textbf{q} + n\textbf{r}\)
\(\left(\begin{array}{c}9\\4\\5\end{array}\right) = m\left(\begin{array}{c}3\\-2\\3\end{array}\right) + n\left(\begin{array}{c}0\\5\\-2\end{array}\right)\)
\(\left(\begin{array}{c}9\\4\\5\end{array}\right) = \left(\begin{array}{c}3m\\-2m\\3m\end{array}\right) + \left(\begin{array}{c}0\\5n\\-2n\end{array}\right)\)
\(\left(\begin{array}{c}9\\4\\5\end{array}\right) = \left(\begin{array}{c}3m\\-2m + 5n\\3m\:-\:2n\end{array}\right)\)
\(\left(\begin{array}{c}9\\4\\5\end{array}\right) = \left(\begin{array}{c}3m\\-2m + 5n\\3m\:-\:2n\end{array}\right)\)
Dengan prinsip kesamaan dua vektor:
\(9 = 3m\rightarrow m = 3\)
\(4 = -2m + 5n\)
\(4 = -2(3) + 5n\)
\(4 = -6 + 5n\)
\(4 + 6 = 5n\)
\(10 = 5n\)
\(n = 2\)
Jadi, \(m + n = 3 + 2 = 5\)
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Pertanyaan 11 dari 13
11. Pertanyaan
1 pointsPerhatikan jajargenjang ABCD di bawah ini:
Jika koordinat titik A\((-2, 0, 3)\), B\((-2, 1, 2)\) dan D\((-3, 1, 1)\), maka \(\overrightarrow{\text{AC}}=\dotso\)
Benar
Langkah 1: Menentukan \(\overrightarrow{\text{AD}}\) dan \(\overrightarrow{\text{BC}}\)
\(\overrightarrow{\text{AD}}= \textbf{d}\:-\:\textbf{a}\)
\(\overrightarrow{\text{AD}}= \left(\begin{array}{c}-3\\1\\1\end{array}\right)\:-\:\left(\begin{array}{c}-2\\0\\3\end{array}\right)\)
\(\overrightarrow{\text{AD}}= \left(\begin{array}{c}-3\:-\:(-2)\\1\:-\:0\\1\:-\:3\end{array}\right)\)
\(\overrightarrow{\text{AD}}= \left(\begin{array}{c}-1\\1\\-2\end{array}\right)\)
Karena ABCD jajargenjang, maka:
\(\overrightarrow{\text{AD}}= \overrightarrow{\text{BC}} = \left(\begin{array}{c}-1\\1\\-2\end{array}\right)\)
Langkah 2: Menentukan \(\overrightarrow{\text{AB}}\)
\(\overrightarrow{\text{AB}}= \textbf{b}\:-\:\textbf{a}\)
\(\overrightarrow{\text{AB}}= \left(\begin{array}{c}-2\\1\\2\end{array}\right)\:-\:\left(\begin{array}{c}-2\\0\\3\end{array}\right)\)
\(\overrightarrow{\text{AB}}= \left(\begin{array}{c}0\\1\\-1\end{array}\right)\)
Langkah 3: Menentukan \(\overrightarrow{\text{AC}}\)
\(\overrightarrow{\text{AC}}= \overrightarrow{\text{AB}} + \overrightarrow{\text{BC}}\)
\(\overrightarrow{\text{AC}}= \left(\begin{array}{c}0\\1\\-1\end{array}\right) + \left(\begin{array}{c}-1\\1\\-2\end{array}\right)\)
\(\overrightarrow{\text{AC}}= \left(\begin{array}{c}-1\\2\\-3\end{array}\right) \)
\(\overrightarrow{\text{AC}}= -\textbf{i} + 2\textbf{j} \:-\: 3\textbf{k}\)
Salah
Langkah 1: Menentukan \(\overrightarrow{\text{AD}}\) dan \(\overrightarrow{\text{BC}}\)
\(\overrightarrow{\text{AD}}= \textbf{d}\:-\:\textbf{a}\)
\(\overrightarrow{\text{AD}}= \left(\begin{array}{c}-3\\1\\1\end{array}\right)\:-\:\left(\begin{array}{c}-2\\0\\3\end{array}\right)\)
\(\overrightarrow{\text{AD}}= \left(\begin{array}{c}-3\:-\:(-2)\\1\:-\:0\\1\:-\:3\end{array}\right)\)
\(\overrightarrow{\text{AD}}= \left(\begin{array}{c}-1\\1\\-2\end{array}\right)\)
Karena ABCD jajargenjang, maka:
\(\overrightarrow{\text{AD}}= \overrightarrow{\text{BC}} = \left(\begin{array}{c}-1\\1\\-2\end{array}\right)\)
Langkah 2: Menentukan \(\overrightarrow{\text{AB}}\)
\(\overrightarrow{\text{AB}}= \textbf{b}\:-\:\textbf{a}\)
\(\overrightarrow{\text{AB}}= \left(\begin{array}{c}-2\\1\\2\end{array}\right)\:-\:\left(\begin{array}{c}-2\\0\\3\end{array}\right)\)
\(\overrightarrow{\text{AB}}= \left(\begin{array}{c}0\\1\\-1\end{array}\right)\)
Langkah 3: Menentukan \(\overrightarrow{\text{AC}}\)
\(\overrightarrow{\text{AC}}= \overrightarrow{\text{AB}} + \overrightarrow{\text{BC}}\)
\(\overrightarrow{\text{AC}}= \left(\begin{array}{c}0\\1\\-1\end{array}\right) + \left(\begin{array}{c}-1\\1\\-2\end{array}\right)\)
\(\overrightarrow{\text{AC}}= \left(\begin{array}{c}-1\\2\\-3\end{array}\right) \)
\(\overrightarrow{\text{AC}}= -\textbf{i} + 2\textbf{j} \:-\: 3\textbf{k}\)
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Pertanyaan 12 dari 13
12. Pertanyaan
1 points
ABCD adalah sebuah persegi panjang. AC dan BD merupakan diagonal persegi panjang yang berpotongan di titik O. Titik R terletak pada AC dengan perbandingan AR : RC = 5 : 3 dan titik P berada pada OB dengan perbandingan BP : PO = 1 : 3. Jika \(\overrightarrow{\text{AB}} = \textbf{u}\) dan \(\overrightarrow{\text{AD}} = \textbf{v}\), maka \(\overrightarrow{\text{PR}}\) dapat dinyatakan sebagai …
Benar
Langkah 1: Menentukan \(\overrightarrow{\text{AC}}\) dan \(\overrightarrow{\text{OR}}\)
\(\overrightarrow{\text{AC}} = \overrightarrow{\text{AB}} + \overrightarrow{\text{BC}}\)
\(\overrightarrow{\text{AC}} = \textbf{u} + \textbf{v}\)
\(\overrightarrow{\text{OR}} = \dfrac{1}{8}\overrightarrow{\text{AC}}\)
\(\overrightarrow{\text{OR}} = \dfrac{1}{8}(\textbf{u} + \textbf{v})\)
\(\overrightarrow{\text{OR}} = \dfrac{1}{8}\textbf{u} + \dfrac{1}{8}\textbf{v}\)
Langkah 2: Menentukan \(\overrightarrow{\text{BD}}\) dan \(\overrightarrow{\text{PO}}\)
\(\overrightarrow{\text{BD}} = \overrightarrow{\text{BA}} + \overrightarrow{\text{AD}}\)
\(\overrightarrow{\text{BD}} = -\textbf{u} + \textbf{v}\)
\(\overrightarrow{\text{PO}} = \dfrac{3}{8}\overrightarrow{\text{BD}}\)
\(\overrightarrow{\text{PO}} = \dfrac{3}{8}(-\textbf{u} + \textbf{v})\)
\(\overrightarrow{\text{PO}} = -\dfrac{3}{8}\textbf{u} + \dfrac{3}{8}\textbf{v}\)
Langkah 3: Menentukan \(\overrightarrow{\text{PR}}\)
\(\overrightarrow{\text{PR}} = \overrightarrow{\text{PO}} + \overrightarrow{\text{OR}}\)
\(\overrightarrow{\text{PR}} = -\dfrac{3}{8}\textbf{u} + \dfrac{3}{8}\textbf{v} + \dfrac{1}{8}\textbf{u} + \dfrac{1}{8}\textbf{v}\)
\(\overrightarrow{\text{PR}} = -\dfrac{2}{8}\textbf{u} + \dfrac{4}{8}\textbf{v}\)
\(\overrightarrow{\text{PR}} = -\dfrac{1}{4}\textbf{u} + \dfrac{1}{2}\textbf{v}\)
Salah
Langkah 1: Menentukan \(\overrightarrow{\text{AC}}\) dan \(\overrightarrow{\text{OR}}\)
\(\overrightarrow{\text{AC}} = \overrightarrow{\text{AB}} + \overrightarrow{\text{BC}}\)
\(\overrightarrow{\text{AC}} = \textbf{u} + \textbf{v}\)
\(\overrightarrow{\text{OR}} = \dfrac{1}{8}\overrightarrow{\text{AC}}\)
\(\overrightarrow{\text{OR}} = \dfrac{1}{8}(\textbf{u} + \textbf{v})\)
\(\overrightarrow{\text{OR}} = \dfrac{1}{8}\textbf{u} + \dfrac{1}{8}\textbf{v}\)
Langkah 2: Menentukan \(\overrightarrow{\text{BD}}\) dan \(\overrightarrow{\text{PO}}\)
\(\overrightarrow{\text{BD}} = \overrightarrow{\text{BA}} + \overrightarrow{\text{AD}}\)
\(\overrightarrow{\text{BD}} = -\textbf{u} + \textbf{v}\)
\(\overrightarrow{\text{PO}} = \dfrac{3}{8}\overrightarrow{\text{BD}}\)
\(\overrightarrow{\text{PO}} = \dfrac{3}{8}(-\textbf{u} + \textbf{v})\)
\(\overrightarrow{\text{PO}} = -\dfrac{3}{8}\textbf{u} + \dfrac{3}{8}\textbf{v}\)
Langkah 3: Menentukan \(\overrightarrow{\text{PR}}\)
\(\overrightarrow{\text{PR}} = \overrightarrow{\text{PO}} + \overrightarrow{\text{OR}}\)
\(\overrightarrow{\text{PR}} = -\dfrac{3}{8}\textbf{u} + \dfrac{3}{8}\textbf{v} + \dfrac{1}{8}\textbf{u} + \dfrac{1}{8}\textbf{v}\)
\(\overrightarrow{\text{PR}} = -\dfrac{2}{8}\textbf{u} + \dfrac{4}{8}\textbf{v}\)
\(\overrightarrow{\text{PR}} = -\dfrac{1}{4}\textbf{u} + \dfrac{1}{2}\textbf{v}\)
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Pertanyaan 13 dari 13
13. Pertanyaan
1 pointsPerhatikan gambar kubus OABC.DEFG di bawah ini:
Titik P terletak pada rusuk FG, sehingga FP : PG = 1 : 2. Jika \(\overrightarrow{\text{OA}} = \textbf{u}\), \(\overrightarrow{\text{OC}} = \textbf{v}\) dan \(\overrightarrow{\text{OD}} = \textbf{w}\), maka \(\overrightarrow{\text{OP}} = \dotso\)
Benar
\(\overrightarrow{\text{OP}} = \overrightarrow{\text{OA}} + \overrightarrow{\text{AB}} + \overrightarrow{\text{BQ}} + \overrightarrow{\text{QP}}\)
\(\overrightarrow{\text{OP}} = \textbf{u} + \textbf{v} \:-\: \dfrac{1}{3}\textbf{u} + \textbf{w}\)
\(\overrightarrow{\text{OP}} = \dfrac{2}{3}\textbf{u} + \textbf{v} + \textbf{w}\)
Salah
\(\overrightarrow{\text{OP}} = \overrightarrow{\text{OA}} + \overrightarrow{\text{AB}} + \overrightarrow{\text{BQ}} + \overrightarrow{\text{QP}}\)
\(\overrightarrow{\text{OP}} = \textbf{u} + \textbf{v} \:-\: \dfrac{1}{3}\textbf{u} + \textbf{w}\)
\(\overrightarrow{\text{OP}} = \dfrac{2}{3}\textbf{u} + \textbf{v} + \textbf{w}\)
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Vektor 3 Dimensi
Perhatikan gambar kubus ABCD.EFGH berikut:
Diketahui kubus ABCD.EFGH memiliki rusuk dengan panjang 10 satuan.
- \(\overrightarrow{\text{DA}}= 10\textbf{i} \text{ dan } \overrightarrow{\text{AD}}= -10\textbf{i} \)
- \(\overrightarrow{\text{DC}}= 10\textbf{j} \text{ dan } \overrightarrow{\text{CD}}= -10\textbf{j} \)
- \(\overrightarrow{\text{DH}}= 10\textbf{k} \text{ dan } \overrightarrow{\text{HD}}= -10\textbf{k} \)
Dari informasi di atas, nyatakan:
A. Vektor \(\overrightarrow{\text{DB}}\) dan panjang DB
B. Vektor\(\overrightarrow{\text{DF}}\) dan panjang DF
C. Vektor \(\overrightarrow{\text{AC}}\) dan panjang AC
D. Vektor \(\overrightarrow{\text{AG}}\) dan panjang AG
Penyelesaian:
Menyatakan vektor \(\overrightarrow{\text{DB}}\)
\(\overrightarrow{\text{DB}} = \overrightarrow{\text{DA}} + \overrightarrow{\text{AB}}\)
\(\overrightarrow{\text{DB}} = 10\textbf{i} + 10\textbf{j}\)
\(\text{DB} = \sqrt{10^2 + 10^2} = \sqrt{200} = 10\sqrt{2}\text{ satuan}\)
Menyatakan vektor \(\overrightarrow{\text{DF}}\)
\(\overrightarrow{\text{DF}} = \overrightarrow{\text{DB}} + \overrightarrow{\text{BF}}\)
\(\overrightarrow{\text{DF}} = 10\textbf{i} + 10\textbf{j} + 10\textbf{k}\)
\(\text{DF} = \sqrt{10^2 + 10^2 + 10^2} = \sqrt{300} = 10\sqrt{3}\text{ satuan}\)
Menyatakan vektor \(\overrightarrow{\text{AC}}\)
\(\overrightarrow{\text{AC}} = \overrightarrow{\text{AB}} + \overrightarrow{\text{BC}}\)
\(\overrightarrow{\text{AC}} = 10\textbf{j} \:-\: 10\textbf{i}\)
\(\text{AC} = \sqrt{10^2 + (-10)^2} = \sqrt{200} = 10\sqrt{2}\text{ satuan}\)
Menyatakan vektor \(\overrightarrow{\text{AG}}\)
\(\overrightarrow{\text{AG}} = \overrightarrow{\text{AC}} + \overrightarrow{\text{CG}}\)
\(\overrightarrow{\text{AG}} = 10\textbf{j} \:-\: 10\textbf{i} + 10\textbf{k}\)
\(\text{AG} = \sqrt{10^2 + (-10)^2 + 10^2} = \sqrt{300} = 10\sqrt{3}\text{ satuan}\)
Contoh Soal
Soal 1
Kubus ABCD.EFGH memiliki panjang rusuk 10 satuan. Titik P terletak di tengah HG. Tentukan panjang AP.
Cara 1:
Sebelum menentukan panjang AP, terlebih dahulu kita nyatakan vektor \(\overrightarrow{\text{AP}}\)
\(\overrightarrow{\text{AP}} = \overrightarrow{\text{AB}} + \overrightarrow{\text{BC}} + \overrightarrow{\text{CG}} + \overrightarrow{\text{GP}}\)
\(\overrightarrow{\text{AP}} = 10\textbf{j} \:-\: 10\textbf{i} + 10\textbf{k} + \frac{1}{2}\cdot \overrightarrow{\text{GH}}\)
\(\overrightarrow{\text{AP}} = 10\textbf{j} \:-\: 10\textbf{i} + 10\textbf{k} + \frac{1}{2}\cdot (-10)\textbf{j} \)
\(\overrightarrow{\text{AP}} = 10\textbf{j} \:-\: 10\textbf{i} + 10\textbf{k} \:-\:5\textbf{j} \)
\(\overrightarrow{\text{AP}} = -10\textbf{i} + 5\textbf{j} + 10\textbf{k} \)
Menghitung panjang AP
\(||\overrightarrow{\text{AP}}|| = \sqrt{(-10)^2 + 5^2 + 10^2}\)
\(||\overrightarrow{\text{AP}}|| = \sqrt{100 + 25 + 100}\)
\(||\overrightarrow{\text{AP}}|| = \sqrt{225}\)
\(||\overrightarrow{\text{AP}}|| = 15\text { satuan}\)
Cara 2:
Tentukan koordinat titik A dan P terlebih dahulu
Koordinat titik A adalah \((10, 0, 0)\) dan koordinat titik P adalah \((0, 5, 10)\)
\(\overrightarrow{\text{AP}} = \textbf{p}\:-\:\textbf{a}\)
\(\overrightarrow{\text{AP}} = \left(\begin{array}{c}0\\ 5\\10\end{array}\right)\:-\:\left(\begin{array}{c}10\\ 0\\0\end{array}\right) = \left(\begin{array}{c}-10\\ 5\\10\end{array}\right)\)
\(||\overrightarrow{\text{AP}}|| =\sqrt{(-10)^2 + 5^2 + 10^2}\)
\(||\overrightarrow{\text{AP}}|| =\sqrt{100 + 25 + 100}\)
\(||\overrightarrow{\text{AP}}|| =\sqrt{225}\)
\(||\overrightarrow{\text{AP}}|| = 15\text { satuan}\)
Soal 2
Kubus ABCD.EFGH memiliki panjang rusuk 10 satuan. Titik P terletak pada EF dengan perbandingan EP : PF = 2 : 3. Titik Q terletak pada CG dengan perbandingan CQ : QG = 1 : 4. Tentukan panjang PQ.
Cara 1:
Sebelum menentukan panjang PQ, terlebih dahulu kita nyatakan vektor \(\overrightarrow{\text{PQ}}\)
\(\overrightarrow{\text{PQ}} = \overrightarrow{\text{PF}} + \overrightarrow{\text{FB}} + \overrightarrow{\text{BC}} + \overrightarrow{\text{CQ}}\)
\(\overrightarrow{\text{PQ}} = \frac{3}{5}\cdot \overrightarrow{\text{EF}} + \overrightarrow{\text{FB}} + \overrightarrow{\text{BC}} + \frac{1}{5}\cdot \overrightarrow{\text{CG}}\)
\(\overrightarrow{\text{PQ}} = \frac{3}{5}\cdot 10\textbf{j} \:-\:10\textbf{k}\:-\:10\textbf{i} + \frac{1}{5}\cdot 10\textbf{k}\)
\(\overrightarrow{\text{PQ}} = 6\textbf{j} \:-\:10\textbf{k}\:-\:10\textbf{i} + 2\textbf{k}\)
\(\overrightarrow{\text{PQ}} = -10\textbf{i} + 6\textbf{j} \:-\:8\textbf{k}\)
Menghitung panjang PQ
\(||\overrightarrow{\text{PQ}}|| = \sqrt{(-10)^2 + 6^2 + (-8)^2}\)
\(||\overrightarrow{\text{PQ}}|| = \sqrt{100 + 36 + 64}\)
\(||\overrightarrow{\text{PQ}}|| = \sqrt{200}\)
\(||\overrightarrow{\text{PQ}}|| = 10\sqrt{2}\text { satuan}\)
Cara 2:
Tentukan koordinat titik P dan Q terlebih dahulu
Koordinat titik P adalah \((10, 4, 10)\) dan koordinat titik Q adalah \((0, 10, 2)\)
\(\overrightarrow{\text{PQ}} = \textbf{q}\:-\:\textbf{p}\)
\(\overrightarrow{\text{AP}} = \left(\begin{array}{c}0\\10\\2\end{array}\right)\:-\:\left(\begin{array}{c}10\\ 4\\10\end{array}\right) = \left(\begin{array}{c}-10\\6\\-8\end{array}\right)\)
\(||\overrightarrow{\text{PQ}}|| =\sqrt{(-10)^2 + 6^2 + (-8)^2}\)
\(||\overrightarrow{\text{PQ}}|| =\sqrt{100 + 36 + 64}\)
\(||\overrightarrow{\text{PQ}}|| =\sqrt{200}\)
\(||\overrightarrow{\text{PQ}}|| = 10\sqrt{2}\text { satuan}\)
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Perkalian Silang (Cross Product)
Perkalian silang antara vektor \(\textbf{a} = a_1\textbf{i} + a_2\textbf{j} + a_3\textbf{k}\) dengan \(\textbf{b} = b_1\textbf{i} + b_2\textbf{j} + b_3\textbf{k}\) dihitung sebagai berikut:
\(\textbf{a} \times \textbf{b} = \begin{vmatrix}\textbf{i}&\textbf{j}&\textbf{k}\\a_1 & a_2 & a_3\\b_1 & b_2 & b_3\end{vmatrix}\)
\(\textbf{a} \times \textbf{b} =\begin{vmatrix}a_2 & a_3\\ b_2 & b_3\end{vmatrix}\textbf{i}\:-\:\begin{vmatrix}a_1 & a_3\\ b_1 & b_3\end{vmatrix}\textbf{j} + \begin{vmatrix}a_1 & a_2\\ b_1 & b_2\end{vmatrix}\textbf{k}\)
\(\textbf{a} \times \textbf{b} =(a_2\cdot b_3\:-\:a_3\cdot b_2)\textbf{i}\:-\:(a_1\cdot b_3\:-\:a_3\cdot b_1)\textbf{j} + (a_1\cdot b_2\:-\:a_2\cdot b_1)\textbf{k}\)
Hasil perkalian silang antara vektor \(\textbf{a}\) dengan vektor \(\textbf{b}\) akan menghasilkan suatu vektor \(\textbf{c}\) yang tegak lurus terhadap kedua vektor tersebut.
Menentukan Luas Segitiga
\(\color{blue} \text{Luas Segitiga } = \dfrac{1}{2} \left|\overrightarrow{\text{AB}} \times \overrightarrow{\text{AC}}\right|\)
\(\overrightarrow{\text{AB}}=\textbf{b}\:-\:\textbf{a}\)
\(\overrightarrow{\text{AB}}=\left(\begin{array}{c}x_2\\ y_2\\z_2\end{array}\right) \:-\: \left(\begin{array}{c}x_1\\y_1\\z_1\end{array}\right)\)
\(\overrightarrow{\text{AC}}=\textbf{c}\:-\:\textbf{a}\)
\(\overrightarrow{\text{AC}}=\left(\begin{array}{c}x_3\\ y_3\\z_3\end{array}\right) \:-\: \left(\begin{array}{c}x_1\\y_1\\z_1\end{array}\right)\)
Contoh Soal
Soal 1
Tentukan perkalian silang antara vektor \(\textbf{a} = -\textbf{i} + 2\textbf{j} + 3\textbf{k}\) dengan \(\textbf{b} = 4\textbf{i} \:-\:\textbf{j} + 5\textbf{k}\)
\(\textbf{a} \times \textbf{b} = \begin{vmatrix}\textbf{i}&\textbf{j}&\textbf{k}\\-1 & 2 & 3\\4 & -1 & 5\end{vmatrix}\)
\(\textbf{a} \times \textbf{b} =\begin{vmatrix}2 & 3\\ -1 & 5\end{vmatrix}\textbf{i}\:-\:\begin{vmatrix}-1 & 3\\ 4 & 5\end{vmatrix}\textbf{j} + \begin{vmatrix}-1 & 2\\ 4 & -1\end{vmatrix}\textbf{k}\)
\(\textbf{a} \times \textbf{b} =(2\cdot 5\:-\:3\cdot (-1))\textbf{i}\:-\:(-1\cdot 5\:-\:3\cdot 4)\textbf{j} + (-1\cdot (-1)\:-\:2\cdot 4)\textbf{k}\)
\(\textbf{a} \times \textbf{b} =13\textbf{i}\:-\:(-17)\textbf{j} + (-7)\textbf{k}\)
\(\textbf{a} \times \textbf{b} =13\textbf{i} + 17\textbf{j} \:-\:7\textbf{k}\)
Jadi, hasil perkalian silang antara vektor \(\textbf{a}\) dengan \(\textbf{b}\) adalah \(\textbf{c} = 13\textbf{i} + 17\textbf{j} \:-\:7\textbf{k}\)
Vektor \(\textbf{c}\) ini tegak lurus terhadap vektor \(\textbf{a}\) dan juga terhadap vektor \(\textbf{b}\)
Soal 2
Find the area of the triangle with vertices:
A (2, 0, -1)
B (3, -2, 0)
C (4, 4, 1)
\(\overrightarrow{\text{AB}}=\textbf{b}\:-\:\textbf{a}\)
\(\overrightarrow{\text{AB}}=\left(\begin{array}{c}3\\ -2\\0\end{array}\right) \:-\: \left(\begin{array}{c}2\\0\\-1\end{array}\right) = \left(\begin{array}{c}1\\-2\\1\end{array}\right)\)
\(\overrightarrow{\text{AC}}=\textbf{c}\:-\:\textbf{a}\)
\(\overrightarrow{\text{AC}}=\left(\begin{array}{c}4\\ 4\\1\end{array}\right) \:-\: \left(\begin{array}{c}2\\0\\-1\end{array}\right) = \left(\begin{array}{c}2\\4\\2\end{array}\right)\)
\(\color{blue} \text{Area of triangle } = \dfrac{1}{2} \left|\overrightarrow{\text{AB}} \times \overrightarrow{\text{AC}}\right|\)
\(\overrightarrow{\text{AB}} \times \overrightarrow{\text{AC}} = \left(\begin{array}{c}1\\-2\\1\end{array}\right)\times \left(\begin{array}{c}2\\4\\2\end{array}\right)\)
\(\overrightarrow{\text{AB}} \times \overrightarrow{\text{AC}} =\begin{vmatrix}\textbf{i}&\textbf{j}&\textbf{k}\\1 & -2 & 1\\2 & 4 & 2\end{vmatrix}\)
\(\overrightarrow{\text{AB}}\times \overrightarrow{\text{AC}} =\begin{vmatrix}-2 & 1\\ 4& 2\end{vmatrix}\textbf{i}\:-\:\begin{vmatrix}1&1\\ 2& 2\end{vmatrix}\textbf{j} + \begin{vmatrix}1 & -2\\ 2 & 4\end{vmatrix}\textbf{k}\)
\(\overrightarrow{\text{AB}}\times \overrightarrow{\text{AC}} =(-2\cdot 2\:-\:1\cdot 4)\textbf{i}\:-\:(1\cdot 2\:-\:1\cdot 2)\textbf{j} + (1\cdot 4\:-\:(-2)\cdot 2)\textbf{k}\)
\(\overrightarrow{\text{AB}} \times \overrightarrow{\text{AC}} =-8\textbf{i}\:-\:0\textbf{j} +8\textbf{k}\)
\(\overrightarrow{\text{AB}} \times \overrightarrow{\text{AC}} = -8\textbf{i} + 8\textbf{k}\)
\(\text{Area of triangle } = \dfrac{1}{2}\sqrt{(-8)^2 + 8^2}\)
\(\text{Area of triangle } = \dfrac{1}{2}\sqrt{64 + 64}\)
\(\text{Area of triangle } = \dfrac{1}{2}\sqrt{2\times 64}\)
\(\text{Area of triangle } = \dfrac{8}{2}\sqrt{2}\)
\(\text{Area of triangle } = 4\sqrt{2} \text{ square units}\)
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Proyeksi Vektor
Proyeksi vektor ortogonal \(\textbf{u}\) pada \(\textbf{v}\) menghasilkan vektor \(\textbf{w}\)
\(\textbf{w} = \dfrac{\textbf{u} \cdot \textbf{v}}{(||\textbf{v}||)^2} \cdot \textbf{v}\)
Sedangkan panjang proyeksi vektor \(\textbf{u}\) pada \(\textbf{v}\) adalah \(||\textbf{w}||\)
\(||\textbf{w}|| = \dfrac{|\textbf{u} \cdot \textbf{v}|}{||\textbf{v}||}\)
Contoh Soal
Soal 1
Tentukan proyeksi vektor ortogonal \(\textbf{u} = -\textbf{i} + 2\textbf{j}\) pada \(\textbf{v} = 3\textbf{i} \:-\: \textbf{j}\).
Misal hasil vektor proyeksi \(\textbf{u}\) pada \(\textbf{v}\) adalah \(\textbf{w}\)
\(\textbf{w} = \dfrac{\textbf{u} \cdot \textbf{v}}{(||\textbf{v}||)^2} \cdot \textbf{v}\)
\(\textbf{w} = \dfrac{\left(\begin{array}{c}-1\\ 2\end{array}\right) \cdot \left(\begin{array}{c}3\\ -1\end{array}\right)}{(\sqrt{3^2 + (-1)^2})^2} \cdot \left(\begin{array}{c}3\\ -1\end{array}\right)\)
\(\textbf{w} = \dfrac{(-1)(3) + (2)(-1)}{3^2 + (-1)^2} \cdot \left(\begin{array}{c}3\\ -1\end{array}\right)\)
\(\textbf{w} = \dfrac{-5}{10} \cdot \left(\begin{array}{c}3\\ -1\end{array}\right)\)
\(\textbf{w} = -\dfrac{1}{2} \cdot \left(\begin{array}{c}3\\ -1\end{array}\right)\)
\(\textbf{w} = \left(\begin{array}{c}-\dfrac{3}{2}\\ \dfrac{1}{2}\end{array}\right)\)
\(\textbf{w} = -\dfrac{3}{2}\textbf{i}+ \dfrac{1}{2}\textbf{j} \)
Jadi hasil vektor proyeksi \(\textbf{u}\) pada \(\textbf{v}\) adalah \(\textbf{w} = -\frac{3}{2}\textbf{i} + \frac{1}{2}\textbf{j} \)
Soal 2
Tentukan proyeksi skalar vektor \(\textbf{v} = 4\textbf{i} + \textbf{j}\) pada \(\textbf{u} = -2\textbf{i} \:-\: 3\textbf{j}\).
Proyeksi skalar (panjang proyeksi) vektor \(\textbf{v}\) pada \(\textbf{u}\) adalah \(||\textbf{w}||\)
\(||\textbf{w}|| = \dfrac{|\textbf{v} \cdot \textbf{u}|}{||\textbf{u}||}\)
\(||\textbf{w}|| = \dfrac{\left|\left(\begin{array}{c}4\\ 1\end{array}\right) \cdot \left(\begin{array}{c}-2\\ -3\end{array}\right)\right|}{\sqrt{(-2)^2 + (-3)^2}}\)
\(||\textbf{w}|| = \dfrac{|(4)(-2) + (1)(-3)|}{\sqrt{4 + 9}}\)
\(||\textbf{w}|| = \dfrac{11}{\sqrt{13}}\times \color{red} \dfrac{\sqrt{13}}{\sqrt{13}}\)
\(||\textbf{w}|| = \dfrac{11}{13}\sqrt{13}\)
Jadi proyeksi skalar \(\textbf{v}\) pada \(\textbf{u}\) adalah \(||\textbf{w}|| = \dfrac{11}{13}\sqrt{13}\) satuan
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Sudut Antara Dua Vektor
Sudut terkecil yang dibentuk oleh vektor \(\textbf{u}\) dan \(\textbf{v}\), dihitung sebagai berikut:
\(\cos \theta = \dfrac{\textbf{u} \cdot \textbf{v}}{||\textbf{u}|| \cdot ||\textbf{v}||}\)
Dua vektor yang saling tegak lurus (membentuk sudut 90°)
\(\cos 90^{\circ} = \dfrac{\textbf{u} \cdot \textbf{v}}{||\textbf{u}|| \cdot ||\textbf{v}||}\)
\(0 = \dfrac{\textbf{u} \cdot \textbf{v}}{||\textbf{u}|| \cdot ||\textbf{v}||}\)
\(\color{blue} \textbf{u} \cdot \textbf{v} = 0\)
Dua vektor yang berhimpit (membentuk sudut 0°)
\(\cos 0^{\circ} = \dfrac{\textbf{u} \cdot \textbf{v}}{||\textbf{u}|| \cdot ||\textbf{v}||}\)
\(1 = \dfrac{\textbf{u} \cdot \textbf{v}}{||\textbf{u}|| \cdot ||\textbf{v}||}\)
\(\color{blue} \textbf{u} \cdot \textbf{v} = ||\textbf{u}|| \cdot ||\textbf{v}||\)
Contoh Soal
Soal 1
Tentukan besar sudut yang dibentuk oleh vektor \(\textbf{u}\) dan \(\textbf{v}\) di bawah ini:
Sudut yang dibentuk antara vektor \(\textbf{u}\) dan \(\textbf{v}\) adalah \(180^{\circ}\:-\:60^{\circ} = 120^{\circ}\)
Soal 2
Diketahui vektor \(\textbf{a} = 2\textbf{i}\:-\:\textbf{j} + 2\textbf{k}\) dan vektor \(\textbf{b} = 3\textbf{j}\:-\:4\textbf{k}\). Tentukan cosinus sudut antara kedua vektor tersebut, dan tentukan jenis sudutnya.
\(\cos \theta = \dfrac{\textbf{a} \cdot \textbf{b}}{||\textbf{a}|| \cdot ||\textbf{b}||}\)
\(\cos \theta = \dfrac{\left(\begin{array}{c}2\\ -1\\2\end{array}\right) \cdot \left(\begin{array}{c}0\\ 3\\-4\end{array}\right)}{\sqrt{2^2 + (-1)^2 + 2^2}\cdot \sqrt{0^2 + 3^2 + (-4)^2}}\)
\(\cos \theta = \dfrac{0\:-\:3\:-\:8}{\sqrt{9}\cdot \sqrt{25}}\)
\(\cos \theta = \dfrac{-11}{3\cdot 5}\)
\(\cos \theta = \dfrac{-11}{15}\)
Karena hasil \(\cos \theta\) negatif, maka jenis sudut \(\theta\) adalah sudut tumpul.
Soal 3
Jika sudut yang dibentuk oleh vektor \(\textbf{u} = -2\textbf{i} + 6\textbf{j}\) dan \(\textbf{v} = m\textbf{i} + 3\textbf{j}\) adalah \(90^{\circ}\), maka tentukan nilai \(m\) yang memenuhi.
Vektor \(\textbf{u}\) dan \(\textbf{v}\) saling tegak lurus (membentuk sudut 90°), maka berlaku \(\textbf{u}\cdot \textbf{v} = 0\)
\(\left(\begin{array}{c}-2\\ 6\end{array}\right) \cdot \left(\begin{array}{c}m\\ 3\end{array}\right) = 0\)
\(-2m + 6(3) = 0\)
\(-2m + 18 = 0\)
\(-2m = -18\)
\(m = \frac{-18}{-2} = 9\)
Soal 4
Tentukan nilai \(m\) yang memenuhi agar vektor \(\textbf{a} = (m + 2)\textbf{i} \:-\:\textbf{j} + 2\textbf{k}\) dan vektor \(\textbf{b} = \textbf{i} \:-\:m\textbf{j} + \textbf{k}\) saling tegak lurus.
Vektor \(\textbf{a}\) dan \(\textbf{b}\) saling tegak lurus, maka \(\textbf{a}\cdot \textbf{b} = 0\)
\(\left(\begin{array}{c}m + 2\\ -1\\2\end{array}\right) \cdot \left(\begin{array}{c}1\\ -m\\1\end{array}\right) = 0\)
\(m + 2 + m + 2 = 0\)
\(2m + 4 = 0\)
\(2m = -4\)
\(m = -2\)
Soal 5
Jika \(\theta\) adalah sudut antara vektor \(\textbf{a}\) dan \(\textbf{b}\) dan hasil dari \(\textbf{a}\cdot \textbf{b} = -\dfrac{1}{2}\sqrt{2}||\textbf{a}||\cdot ||\textbf{b}||\), maka tentukan nilai \(\theta\).
\(\cos \theta = \dfrac{\textbf{a} \cdot \textbf{b}}{|\textbf{a}| \cdot |\textbf{b}|}\)
\(\cos \theta = \dfrac{-\frac{1}{2}\sqrt{2}\cancel{||\textbf{a}||\cdot ||\textbf{b}||}}{\cancel{||\textbf{a}|| \cdot ||\textbf{b}||}}\)
\(\cos \theta = -\dfrac{1}{2}\sqrt{2}\)
\(\theta = 180^{\circ}\:-\:45^{\circ}\)
\(\theta = 135^{\circ}\)
Soal 6
Suatu segitiga ABC, memiliki koordinat titik A(0, 2, 3), B(1, 4, 4), dan C(1, 0 , 2). Tentukan besar cosinus sudut ABC.
Sudut ABC dibentuk oleh vektor \(\overrightarrow{\text{BC}}\) dan \(\overrightarrow{\text{BA}}\) (titik pangkal kedua vektor berada di titik B)
\(\overrightarrow{\text{BC}} = \textbf{c}\:-\:\textbf{b}\)
\(\overrightarrow{\text{BC}} = \left(\begin{array}{c}1\\ 0\\2\end{array}\right)\:-\:\left(\begin{array}{c}1\\ 4\\4\end{array}\right) =\left(\begin{array}{c}0\\ -4\\-2\end{array}\right) \)
\(\overrightarrow{\text{BA}} = \textbf{a}\:-\:\textbf{b}\)
\(\overrightarrow{\text{BA}} = \left(\begin{array}{c}0\\ 2\\3\end{array}\right)\:-\:\left(\begin{array}{c}1\\ 4\\4\end{array}\right) = \left(\begin{array}{c}-1\\ -2\\-1\end{array}\right)\)
\(\cos \theta = \dfrac{\overrightarrow{\text{BC}} \cdot \overrightarrow{\text{BA}}}{|\overrightarrow{\text{BC}}|\cdot |\overrightarrow{\text{BA}}|}\)
\(\cos \theta = \dfrac{\left(\begin{array}{c}0\\ -4\\-2\end{array}\right)\cdot \left(\begin{array}{c}-1\\ -2\\-1\end{array}\right)}{\sqrt{0^2 + (-4)^2 + (-2)^2} \cdot \sqrt{(-1)^2 + (-2)^2 + (-1)^2}}\)
\(\cos \theta = \dfrac{(0)(-1) + (-4)(-2) + (-2)(-1)}{\sqrt{20} \cdot \sqrt{6}}\)
\(\cos \theta = \dfrac{10}{\sqrt{120}}\)
\(\cos \theta = \dfrac{10}{2\sqrt{30}}\)
\(\cos \theta = \dfrac{5}{\sqrt{30}} \times \color{red}\dfrac{\sqrt{30}}{\sqrt{30}} \)
\(\cos \theta = \dfrac{5\sqrt{30}}{30}\)
\(\cos \theta = \frac{1}{6}\sqrt{30}\)
Jadi, besar cosinus sudut ABC adalah \(\frac{1}{6}\sqrt{30}\)
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Perkalian Skalar (Dot Product)
Diketahui vektor \(\textbf{u} = \left(\begin{array}{c}a\\ b\end{array}\right)\) dan \(\textbf{v} = \left(\begin{array}{c}c\\ d\end{array}\right)\).
Perkalian skalar antara vektor \(\textbf{u}\) dan \(\textbf{v}\) dihitung sebagai berikut:
\(\left(\begin{array}{c}a\\ b\end{array}\right)\cdot \left(\begin{array}{c}c\\ d\end{array}\right) = ac + bd\)
Berapakah hasil \(\textbf{u}\cdot \textbf{u}\)?
\(\textbf{u}\cdot \textbf{u} = ||\textbf{u}||^2\)
Contoh Soal
Soal 1
Tentukan hasil perkalian skalar antara vektor \(\textbf{u} = \left(\begin{array}{c}-3\\ 2\end{array}\right)\) dan \(\textbf{v} = \left(\begin{array}{c}-5\\ 1\end{array}\right)\)
\(\left(\begin{array}{c}-3\\2\end{array}\right)\cdot \left(\begin{array}{c}-5\\1\end{array}\right) = (-3)(-5) + 2(1)\)
\(\left(\begin{array}{c}-3\\2\end{array}\right)\cdot \left(\begin{array}{c}-5\\1\end{array}\right) = 15 + 2 = 17\)
Soal 2
Tentukan hasil perkalian skalar antara vektor \(\textbf{u} = \left(\begin{array}{c}-1\\ 0\\2\end{array}\right)\) dan \(\textbf{v} = \left(\begin{array}{c}3\\8\\-5\end{array}\right)\)
\(\left(\begin{array}{c}-1\\ 0\\2\end{array}\right) \cdot \left(\begin{array}{c}3\\8\\-5\end{array}\right) = (-1)(3) + 0(8) + 2(-5)\)
\(\left(\begin{array}{c}-1\\ 0\\2\end{array}\right) \cdot \left(\begin{array}{c}3\\8\\-5\end{array}\right) = -3 + 0 \:-\:10 = -13\)