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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
\(\textbf{a}\cdot (\textbf{a} \:-\:\textbf{b}) = \textbf{a}\cdot \textbf{a}\:-\:\textbf{a}\cdot \textbf{b}\)
\(\textbf{a}\cdot (\textbf{a} \:-\:\textbf{b}) = ||\textbf{a}||^2\:-\:\textbf{a}\cdot \textbf{b}\)
\(\textbf{a}\cdot (\textbf{a} \:-\:\textbf{b}) = 5^2 \:-\:2 \)
\(\textbf{a}\cdot (\textbf{a} \:-\:\textbf{b}) = 25 \:-\:2 = 23\)