Kategori: Operasi Vektor

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!

Anda telah menyelesaikan kuis sebelumnya. Oleh karena itu, Anda tidak dapat memulainya lagi.

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Ragu-Ragu
Pertanyaan 1 dari 10

1. Pertanyaan
1 points

Diketahui vektor \(\textbf{a} = 2\textbf{i}\:-\:\textbf{j} + 3\textbf{k}\) dan \(\textbf{b} =5\textbf{j} \:-\:\textbf{k}\). Hasil dari \(\textbf{a}\cdot \textbf{b} = \dotso\)

\(-8\)

\(-9\)

\(8\)

\(-5\textbf{j}\:-\:3\textbf{k}\)

\(5\textbf{j}\:-\:3\textbf{k}\)

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

\(\textbf{a} = \left(\begin{array}{c}2\\ -1\\3\end{array}\right)\)

\(\textbf{b} = \left(\begin{array}{c}0\\ 5\\-1\end{array}\right)\)

\(\textbf{a}\cdot \textbf{b} = \left(\begin{array}{c}2\\ -1\\3\end{array}\right) \cdot \left(\begin{array}{c}0\\ 5\\-1\end{array}\right)\)

\(\textbf{a}\cdot \textbf{b} = 2(0) + (-1)(5) + 3(-1)\)

\(\textbf{a}\cdot \textbf{b} = 0\:-\:5\:-3\)

\(\textbf{a}\cdot \textbf{b} = -8\)

Pertanyaan 2 dari 10

2. Pertanyaan
1 points

Diketahui \(\textbf{a} = \left(\begin{array}{c}-4\\ 2\\-1\end{array}\right)\) dan \(\textbf{b} = \left(\begin{array}{c}1\\ 3\\6\end{array}\right)\), Hasil dari \((3\textbf{a} + \textbf{b})(2\textbf{a} \:-\:\textbf{b}) = \dotso\)

84

85

82

80

88

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

\(3\textbf{a} + \textbf{b} = 3\left(\begin{array}{c}-4\\ 2\\-1\end{array}\right) + \left(\begin{array}{c}1\\ 3\\6\end{array}\right)\)

\(3\textbf{a} + \textbf{b} = \left(\begin{array}{c}-12\\ 6\\-3\end{array}\right) + \left(\begin{array}{c}1\\ 3\\6\end{array}\right)\)

\(3\textbf{a} + \textbf{b} = \left(\begin{array}{c}-11\\ 9\\3\end{array}\right)\)

\(2\textbf{a} \:-\:\textbf{b} = 2\left(\begin{array}{c}-4\\ 2\\-1\end{array}\right)\:-\:\left(\begin{array}{c}1\\ 3\\6\end{array}\right)\)

\(2\textbf{a} \:-\:\textbf{b} = \left(\begin{array}{c}-8\\ 4\\-2\end{array}\right)\:-\:\left(\begin{array}{c}1\\ 3\\6\end{array}\right)\)

\(2\textbf{a} \:-\:\textbf{b} = \left(\begin{array}{c}-9\\1\\-8\end{array}\right)\)

\((3\textbf{a} + \textbf{b})(2\textbf{a} \:-\:\textbf{b}) =\left(\begin{array}{c}-11\\ 9\\3\end{array}\right)\cdot \left(\begin{array}{c}-9\\1\\-8\end{array}\right)\)

\((3\textbf{a} + \textbf{b})(2\textbf{a} \:-\:\textbf{b}) = -11(-9) + 9(1) + 3(-8)\)

\((3\textbf{a} + \textbf{b})(2\textbf{a} \:-\:\textbf{b}) = 99 + 9 \:-\:24 = 84\)

Pertanyaan 3 dari 10

3. Pertanyaan
1 points

Diketahui 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 \(m = \dotso\)

4

3

2

1

5

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

\(\textbf{u}\cdot \textbf{v} = 5\)

\(\left(\begin{array}{c}m\\ -1\\-2\end{array}\right) \cdot \left(\begin{array}{c}4\\ 3\\m\end{array}\right) = 5\)

\(4m + (-1)(3) + (-2)(m) = 5\)

\(4m\:-\:3\:-\:2m = 5\)

\(2m = 5 + 3\)

\(2m = 8\)

\(m = 4\)

Pertanyaan 4 dari 10

4. Pertanyaan
1 points

Jika 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 \(m = \dotso\)

\(-3\) atau \(1\)

\(3\) atau \(-1\)

\(3\) atau \(1\)

\(-3\) atau \(3\)

\(-1\) atau \(1\)

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 \(1\)

Pertanyaan 5 dari 10

5. Pertanyaan
1 points

Diketahui 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 \(\cos \alpha = \dotso\)

\(-\dfrac{1}{9}\)

\(-\dfrac{1}{11}\)

\(\dfrac{1}{9}\)

\(\dfrac{1}{12}\)

\(\dfrac{1}{13}\)

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

\(\cos \alpha = \dfrac{\textbf{u}\cdot \textbf{v}}{||\textbf{u}||\cdot ||\textbf{v}||}\)

\(\cos \alpha = \dfrac{\left(\begin{array}{c}2\\ -2\\1\end{array}\right)\cdot \left(\begin{array}{c}4\\ 4\\-2\end{array}\right)}{\sqrt{2^2 + (-2)^2 + 1^2}\cdot \sqrt{4^2 + 4^2 + (-2)^2}}\)

\(\cos \alpha = \dfrac{2(4) + (-2)(4) + 1(-2)}{\sqrt{4 + 4 + 1}\cdot \sqrt{16 + 16 +4}}\)

\(\cos \alpha = \dfrac{8\:-\:8 \:-\:2}{\sqrt{9}\cdot \sqrt{36}}\)

\(\cos \alpha = \dfrac{-2}{3\times 6}\)

\(\cos \alpha = \dfrac{-2}{18}\)

\(\cos \alpha = -\dfrac{1}{9}\)

Pertanyaan 6 dari 10

6. Pertanyaan
1 points

Diketahui \(||\textbf{a}|| = 2\) dan \(||\textbf{b}|| = 3\). Jika besar sudut antara vektor \(\textbf{a}\) dan \(\textbf{b}\) adalah \(120^{\circ}\), maka \(||\textbf{a} + \textbf{b}|| = \dotso\)

\(\sqrt{7}\)

\(\sqrt{6}\)

\(\sqrt{8}\)

\(\sqrt{10}\)

\(\sqrt{15}\)

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

\(||\textbf{a} + \textbf{b}|| = \sqrt{||\textbf{a}||^2 +||\textbf{b}||^2 + 2\cdot ||\textbf{a}||\cdot ||\textbf{b}||\cdot \cos \alpha}\)

\(||\textbf{a} + \textbf{b}|| = \sqrt{2^2 + 3^2 + 2\cdot 2\cdot 3\cdot \cos 120^{\circ}}\)

\(||\textbf{a} + \textbf{b}|| = \sqrt{4 + 9 + 2\cdot 2\cdot 3\cdot (-\frac{1}{2})}\)

\(||\textbf{a} + \textbf{b}|| = \sqrt{13 \:-\: 6}\)

\(||\textbf{a} + \textbf{b}|| = \sqrt{7}\)

Pertanyaan 7 dari 10

7. Pertanyaan
1 points

Diketahui \(||\textbf{a}|| = 4\) dan \(||\textbf{b}|| = 5\). Jika besar sudut antara vektor \(\textbf{a}\) dan \(\textbf{b}\) adalah \(60^{\circ}\), maka \(||\textbf{a} \:-\: \textbf{b}|| = \dotso\)

\(\sqrt{21}\)

\(\sqrt{22}\)

\(\sqrt{20}\)

\(\sqrt{19}\)

\(\sqrt{23}\)

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

\(||\textbf{a} \:-\: \textbf{b}|| = \sqrt{||\textbf{a}||^2 +||\textbf{b}||^2 \:-\:2\cdot ||\textbf{a}||\cdot ||\textbf{b}||\cdot \cos \alpha}\)

\(||\textbf{a} \:-\: \textbf{b}|| = \sqrt{4^2 + 5^2 \:-\:2\cdot 4\cdot 5\cdot \cos 60^{\circ}}\)

\(||\textbf{a} \:-\: \textbf{b}|| = \sqrt{16 + 25 \:-\:2\cdot 4\cdot 5\cdot \dfrac{1}{2}}\)

\(||\textbf{a} \:-\: \textbf{b}|| = \sqrt{41 \:-\:20}\)

\(||\textbf{a} \:-\: \textbf{b}|| = \sqrt{21}\)

Pertanyaan 8 dari 10

8. Pertanyaan
1 points

Jika \(||\textbf{a}|| = 5\), \(||\textbf{b}|| = 2\), dan \(||\textbf{a} + \textbf{b}|| = \sqrt{31}\) maka nilai dari \(\textbf{a}\cdot \textbf{b} = \dotso\)

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2

3

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5

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

\(||\textbf{a} + \textbf{b}|| = \sqrt{||\textbf{a}||^2 +||\textbf{b}||^2 + 2\cdot ||\textbf{a}||\cdot ||\textbf{b}||\cdot \cos \alpha}\)

\(||\textbf{a} + \textbf{b}|| = \sqrt{||\textbf{a}||^2 +||\textbf{b}||^2 + 2\cdot \textbf{a}\cdot \textbf{b}}\)

\(\sqrt{31} = \sqrt{5^2 + 2^2 + 2\cdot \textbf{a}\cdot \textbf{b}}\)

\(31 = 25 + 4 + 2\cdot \textbf{a}\cdot \textbf{b}\)

\(31 \:-\:29 = 2\cdot \textbf{a}\cdot \textbf{b}\)

\(2 = 2\cdot \textbf{a}\cdot \textbf{b}\)

\(\textbf{a}\cdot \textbf{b} = 1\)

Pertanyaan 9 dari 10

9. Pertanyaan
1 points

Diketahui vektor \(||\textbf{a}|| = 3\) dan \(||\textbf{b}|| = 5\). Jika besar sudut antara vektor \(||\textbf{a}||\) dan \(||\textbf{b}||\) adalah \(60^{\circ}\), maka nilai dari \(||2\textbf{a} + \textbf{b}|| = \dotso\)

\(\sqrt{91}\)

\(\sqrt{90}\)

\(\sqrt{92}\)

\(\sqrt{93}\)

\(\sqrt{94}\)

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

\(||\textbf{a} + \textbf{b}|| = \sqrt{||\textbf{a}||^2 +||\textbf{b}||^2 + 2\cdot ||\textbf{a}||\cdot ||\textbf{b}||\cdot \cos \alpha}\)

\(||\textbf{2a} + \textbf{b}|| = \sqrt{||2\textbf{a}||^2 +||\textbf{b}||^2 + 2\cdot ||2\textbf{a}||\cdot ||\textbf{b}||\cdot \cos \alpha}\)

\(||\textbf{2a} + \textbf{b}|| = \sqrt{4||\textbf{a}||^2 +||\textbf{b}||^2 + 4\cdot ||\textbf{a}||\cdot ||\textbf{b}||\cdot \cos \alpha}\)

\(||\textbf{2a} + \textbf{b}|| = \sqrt{4(3)^2 + (5)^2 + 4\cdot 3\cdot 5\cdot \cos 60^{\circ}}\)

\(||\textbf{2a} + \textbf{b}|| = \sqrt{36 + 25 + 4\cdot 3\cdot 5\cdot \dfrac{1}{2}}\)

\(||\textbf{2a} + \textbf{b}|| = \sqrt{36 + 25 + 30}\)

\(||\textbf{2a} + \textbf{b}|| = \sqrt{91}\)

Pertanyaan 10 dari 10

10. Pertanyaan
1 points

Jika \(||\textbf{a}|| = 5\) dan \(\textbf{a}\cdot \textbf{b} = 2\), maka \(\textbf{a}\cdot (\textbf{a} \:-\:\textbf{b}) = \dotso\)

23

22

21

24

25

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

\(\textbf{a}\cdot (\textbf{a} \:-\:\textbf{b}) = \textbf{a}\cdot \textbf{a}\:-\:\textbf{a}\cdot \textbf{b}\)

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

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

\(\textbf{a}\cdot (\textbf{a} \:-\:\textbf{b}) = 25 \:-\:2 = 23\)
19/01/2022
Kuis Vektor 01
Batas Waktu: 0
Quiz-summary

0 of 13 questions completed

Pertanyaan:

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6

7

8

9

10

11

12

13
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!

Anda telah menyelesaikan kuis sebelumnya. Oleh karena itu, Anda tidak dapat memulainya lagi.

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Dijawab

Ragu-Ragu
Pertanyaan 1 dari 13

1. Pertanyaan
1 points

Diketahui koordinat titik A\((-1, 2, 0)\) dan titik B\((3, -4, 2)\). Vektor satuan dari \(\overrightarrow{\text{BA}}\) adalah …

\(\dfrac{1}{\sqrt{14}}\left(\begin{array}{c}-2\\ 3\\-1\end{array}\right)\)

\(\dfrac{1}{\sqrt{14}}\left(\begin{array}{c}-4\\ 3\\-2\end{array}\right)\)

\(\dfrac{1}{14}\left(\begin{array}{c}-2\\ 3\\-1\end{array}\right)\)

\(\dfrac{2}{\sqrt{14}}\left(\begin{array}{c}-2\\ 3\\-1\end{array}\right)\)

\(\dfrac{1}{\sqrt{14}}\left(\begin{array}{c}2\\ 3\\-1\end{array}\right)\)

\(\dfrac{1}{\sqrt{14}}\left(\begin{array}{c}-2\\ 3\\2\end{array}\right)\)

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

\(\widehat{\text{BA}}=\dfrac{\textbf{BA}}{||\textbf{BA}||}\)

\(\overrightarrow{\text{BA}} = \textbf{a}\:-\:\textbf{b}\)

\(\overrightarrow{\text{BA}} = \left(\begin{array}{c}-1\\ 2\\0\end{array}\right)\:-\:\left(\begin{array}{c}3\\-4\\2\end{array}\right)\)

\(\overrightarrow{\text{BA}} = \left(\begin{array}{c}-4\\ 6\\-2\end{array}\right)\)

\(\widehat{\text{BA}}=\dfrac{\left(\begin{array}{c}-4\\ 6\\-2\end{array}\right)}{\sqrt{(-4)^2 + 6^2 + (-2)^2}}\)

\(\widehat{\text{BA}}=\dfrac{\left(\begin{array}{c}-4\\ 6\\-2\end{array}\right)}{\sqrt{16 + 36 + 4}}\)

\(\widehat{\text{BA}}=\dfrac{\left(\begin{array}{c}-4\\ 6\\-2\end{array}\right)}{\sqrt{56}}\)

\(\widehat{\text{BA}}=\dfrac{\left(\begin{array}{c}-4\\ 6\\-2\end{array}\right)}{2\sqrt{14}}\)

\(\widehat{\text{BA}}=\dfrac{1}{\sqrt{14}}\left(\begin{array}{c}-2\\ 3\\-1\end{array}\right)\)

Pertanyaan 2 dari 13

2. Pertanyaan
1 points

Diketahui \(\textbf{u} = 2\textbf{i} \:-\:5\textbf{j} + \textbf{k}\) dan \(\textbf{v} = \textbf{j} + 3\textbf{k}\). Pernyataan di bawah ini yang benar adalah …

\(||\textbf{u}|| \cdot ||\textbf{v}|| =\sqrt{30}\cdot \sqrt{10} = 3\sqrt{10}\)

\(||\textbf{u}||\:-\: ||\textbf{v}|| = \sqrt{20}\)

\(||\textbf{u}\:-\:2\textbf{v} || = \sqrt{78}\)

\(||\textbf{u}\:-\:3\textbf{v} || = \sqrt{87}\)

\(2||\textbf{u}\:-\:3\textbf{v} || = \sqrt{89}\)

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

\(\textbf{u} = 2\textbf{i} \:-\:5\textbf{j} + \textbf{k}\)

\(||\textbf{u}|| = \sqrt{2^2 + (-5)^2 + 1^2}\)

\(||\textbf{u}|| = \sqrt{4 + 25 + 1}\)

\(||\textbf{u}|| = \sqrt{30}\)

\(\textbf{v} = \textbf{j} + 3\textbf{k}\)

\(||\textbf{v}|| = \sqrt{1^2 + 3^2}\)

\(||\textbf{v}|| = \sqrt{1 + 9}\)

\(||\textbf{v}|| = \sqrt{10}\)

\(||\textbf{u}|| \cdot ||\textbf{v}|| =\sqrt{30}\cdot \sqrt{10} = 10\sqrt{3}\)

\(||\textbf{u}||\:-\: ||\textbf{v}|| =\sqrt{30}\:-\: \sqrt{10}\)

\(\textbf{u}\:-\:2\textbf{v} = 2\textbf{i} \:-\:5\textbf{j} + \textbf{k}\:-\:(\textbf{j} + 3\textbf{k})\)

\(\textbf{u}\:-\:2\textbf{v} = 2\textbf{i} \:-\:7\textbf{j} \:-\:5\textbf{k})\)

\(||\textbf{u}\:-\:2\textbf{v} || = \sqrt{2^2 + (-7)^2 + (-5)^2}\)

\(||\textbf{u}\:-\:2\textbf{v} || = \sqrt{4 + 49 + 25}\)

\(||\textbf{u}\:-\:2\textbf{v} || = \sqrt{78}\)

Pertanyaan 3 dari 13

3. Pertanyaan
1 points

Jika \(\textbf{p} = \left(\begin{array}{c}1\\ 3\\0\end{array}\right)\), \(\textbf{q} = \left(\begin{array}{c}0\\ -1\\2\end{array}\right)\) dan \(\textbf{r} = \left(\begin{array}{c}1\\ 3\\-3\end{array}\right)\), maka nilai \(||(2\textbf{p} + \textbf{q})\:-\:(\textbf{p}\:-\:3\textbf{r})|| = \dotso\)

\(\sqrt{186}\)

\(\sqrt{185}\)

\(\sqrt{184}\)

\(\sqrt{188}\)

\(\sqrt{187}\)

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

\(2\textbf{p} + \textbf{q}\:-\:(\textbf{p}\:-\:3\textbf{r})\)

\(2\textbf{p} + \textbf{q}\:-\:\textbf{p}+ 3\textbf{r}\)

\(\textbf{p} + \textbf{q}+ 3\textbf{r}\)

\(\left(\begin{array}{c}1\\ 3\\0\end{array}\right) + \left(\begin{array}{c}0\\ -1\\2\end{array}\right) + 3\left(\begin{array}{c}1\\ 3\\-3\end{array}\right)\)

\(\textbf{p} + \textbf{q}+ 3\textbf{r} = \left(\begin{array}{c}1 + 0 + 3\\3\:-\:1 + 9\\0 + 2 \:-\:9\end{array}\right)\)

\(\textbf{p} + \textbf{q}+ 3\textbf{r} = \left(\begin{array}{c}4\\11\\-7\end{array}\right)\)

\(||\textbf{p} + \textbf{q}+ 3\textbf{r} ||= \sqrt{4^2 + 11^2 + (-7)^2}\)

\(\sqrt{16 + 121 +49}\)

\(\sqrt{186}\)

Pertanyaan 4 dari 13

4. Pertanyaan
1 points

Diketahui titik A\((2, 0, -6)\) dan titik B\((-4, 2, 4)\). Titik P terletak di tengah-tengah AB. Vektor posisi dari P adalah …

\(\textbf{p} = \left(\begin{array}{c}-1\\1\\-1\end{array}\right)\)

\(\textbf{p} = \left(\begin{array}{c}-2\\1\\-1\end{array}\right)\)

\(\textbf{p} = \left(\begin{array}{c}-1\\0\\2\end{array}\right)\)

\(\textbf{p} = \left(\begin{array}{c}1\\2\\-1\end{array}\right)\)

\(\textbf{p} = \left(\begin{array}{c}2\\2\\2\end{array}\right)\)

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

\(\textbf{p} = \dfrac{\textbf{a} + \textbf{b}}{2}\)

\(\textbf{p} = \dfrac{\left(\begin{array}{c}2\\ 0\\-6\end{array}\right) + \left(\begin{array}{c}-4\\ 2\\4\end{array}\right)}{2}\)

\(\textbf{p} = \dfrac{\left(\begin{array}{c}2\:-\:4\\ 0 + 2\\-6 + 4\end{array}\right) }{2}\)

\(\textbf{p} = \dfrac{\left(\begin{array}{c}-2\\2\\-2\end{array}\right) }{2}\)

\(\textbf{p} = \left(\begin{array}{c}-1\\1\\-1\end{array}\right)\)

Pertanyaan 5 dari 13

5. Pertanyaan
1 points

Diketahui 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 …

\(\textbf{p} =3\textbf{i} + 6\textbf{j}\)

\(\textbf{p} =3\textbf{i} + 5\textbf{j}\)

\(\textbf{p} =2\textbf{i} + 6\textbf{j}\)

\(\textbf{p} =3\textbf{i} \:-\: 6\textbf{j}\)

\(\textbf{p} =4\textbf{i} + 3\textbf{j}\)

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

\(\textbf{p} = \dfrac{1\cdot \textbf{b} + 3\cdot \textbf{a}}{1 + 3}\)

\(\textbf{p} = \dfrac{\left(\begin{array}{c}3\\3\end{array}\right)+ 3\left(\begin{array}{c}3\\7\end{array}\right)}{4}\)

\(\textbf{p} = \dfrac{\left(\begin{array}{c}3 + 9\\3 + 21\end{array}\right)}{4}\)

\(\textbf{p} = \dfrac{\left(\begin{array}{c}12\\24\end{array}\right)}{4}\)

\(\textbf{p} = \left(\begin{array}{c}3\\6\end{array}\right)\)

\(\textbf{p} =3\textbf{i} + 6\textbf{j}\)

Pertanyaan 6 dari 13

6. Pertanyaan
1 points

Titik 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 \(\textbf{p} = \dotso\)

\(\textbf{p} = \left(\begin{array}{c}3\\8\\-10\end{array}\right)\)

\(\textbf{p} = \left(\begin{array}{c}-3\\8\\-10\end{array}\right)\)

\(\textbf{p} = \left(\begin{array}{c}3\\4\\-10\end{array}\right)\)

\(\textbf{p} = \left(\begin{array}{c}4\\-8\\10\end{array}\right)\)

\(\textbf{p} = \left(\begin{array}{c}3\\8\\4\end{array}\right)\)

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

\(\dfrac{\overrightarrow{\text{AP}}}{\overrightarrow{\text{PB}}} = \dfrac{- 2}{5}\)

\(\dfrac{\textbf{p}\:-\:\textbf{a}}{\textbf{b}\:-\:\textbf{p}} = -2 : 5\)

\(5\textbf{p}\:-\:5\textbf{a} = -2\textbf{b} + 2\textbf{p}\)

\(5\textbf{p}\:-\:2\textbf{p} = 5\textbf{a}\:-\:2\textbf{b}\)

\(3\textbf{p} = 5\textbf{a}\:-\:2\textbf{b}\)

\(\textbf{p} = \dfrac{5}{3}\textbf{a}\:-\:\dfrac{2}{3}\textbf{b}\)

\(\textbf{p} = \dfrac{5}{3}\left(\begin{array}{c}3\\0\\-6\end{array}\right)\:-\:\dfrac{2}{3}\left(\begin{array}{c}3\\-12\\0\end{array}\right)\)

\(\textbf{p} = \left(\begin{array}{c}5\\0\\-10\end{array}\right)\:-\:\left(\begin{array}{c}2\\-8\\0\end{array}\right)\)

\(\textbf{p} = \left(\begin{array}{c}3\\8\\-10\end{array}\right)\)

Pertanyaan 7 dari 13

7. Pertanyaan
1 points

Besar sudut antara vektor \(\textbf{a}\) dan \(\textbf{b}\) di atas adalah …

\(110^{\circ}\)

\(130^{\circ}\)

\(150^{\circ}\)

\(-50^{\circ}\)

\(50^{\circ}\)

Benar

Salah

Pertanyaan 8 dari 13

8. Pertanyaan
1 points

Misalkan 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 \(\overrightarrow{\text{PC}} = \dotso\)

\(\frac{1}{2}\textbf{a} + \frac{1}{2}\textbf{b}\)

\(\frac{1}{4}\textbf{a} + \frac{1}{4}\textbf{b}\)

\(\frac{1}{2}\textbf{a} \:-\: \frac{1}{2}\textbf{b}\)

\(\textbf{a} + \frac{1}{2}\textbf{b}\)

\(\textbf{a} \:-\: \frac{1}{2}\textbf{b}\)

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 \(\overrightarrow{\text{PC}}\)

\(\overrightarrow{\text{PC}} = \dfrac{1}{2}\overrightarrow{\text{AP}}\)

\(\overrightarrow{\text{PC}} = \dfrac{1}{2}\left[\dfrac{1}{2}\textbf{a} + \dfrac{1}{2}\textbf{b}\right]\)

\(\overrightarrow{\text{PC}} = \dfrac{1}{4}\textbf{a} + \dfrac{1}{4}\textbf{b}\)

Pertanyaan 9 dari 13

9. Pertanyaan
1 points

Perhatikan 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 \(\overrightarrow{\text{AE}} +\overrightarrow{\text{OD}} = \dotso\)

\(\frac{3}{2}\textbf{u} + \frac{3}{2}\textbf{v}\)

\(\frac{1}{3}\textbf{u} + \frac{3}{2}\textbf{v}\)

\(\frac{3}{2}\textbf{u} \:-\: \frac{3}{2}\textbf{v}\)

\(\frac{1}{2}\textbf{u} + \frac{1}{2}\textbf{v}\)

\(\frac{1}{3}\textbf{u} \:-\: \frac{3}{2}\textbf{v}\)

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 \(\overrightarrow{\text{AE}} + \overrightarrow{\text{OD}}\)

\(\overrightarrow{\text{AE}} + \overrightarrow{\text{OD}} = 2\textbf{u} + \textbf{v} + -\dfrac{1}{2}\textbf{u} + \dfrac{1}{2}\textbf{v}\)

\(\overrightarrow{\text{AE}} + \overrightarrow{\text{OD}} = \dfrac{3}{2}\textbf{u} + \dfrac{3}{2}\textbf{v}\)

Pertanyaan 10 dari 13

10. Pertanyaan
1 points

Diketahui \(\textbf{p} = \left(\begin{array}{c}9\\4\\5\end{array}\right)\), \(\textbf{q} = \left(\begin{array}{c}3\\-2\\3\end{array}\right)\) 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 \(m + n = \dotso\)

1

2

3

4

5

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, \(m + n = 3 + 2 = 5\)

Pertanyaan 11 dari 13

11. Pertanyaan
1 points

Perhatikan jajargenjang ABCD di bawah ini:

Jika koordinat titik A\((-2, 0, 3)\), B\((-2, 1, 2)\) dan D\((-3, 1, 1)\), maka \(\overrightarrow{\text{AC}}=\dotso\)

\(\textbf{i} + 2\textbf{j}\:-\:3\textbf{k}\)

\(\textbf{i} + 3\textbf{j}\:-\:3\textbf{k}\)

\(-\textbf{i} + 2\textbf{j} + 3\textbf{k}\)

\(-\textbf{i} + 2\textbf{j} \:-\: 3\textbf{k}\)

\(-2\textbf{i} + 3\textbf{j} \:-\: \textbf{k}\)

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 \(\overrightarrow{\text{AC}}\)

\(\overrightarrow{\text{AC}}= \overrightarrow{\text{AB}} + \overrightarrow{\text{BC}}\)

\(\overrightarrow{\text{AC}}= \left(\begin{array}{c}0\\1\\-1\end{array}\right) + \left(\begin{array}{c}-1\\1\\-2\end{array}\right)\)

\(\overrightarrow{\text{AC}}= \left(\begin{array}{c}-1\\2\\-3\end{array}\right) \)

\(\overrightarrow{\text{AC}}= -\textbf{i} + 2\textbf{j} \:-\: 3\textbf{k}\)

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 …

\(\frac{1}{3}\textbf{u}\:-\:\frac{1}{2}\textbf{v}\)

\(\textbf{u}\:-\:\frac{1}{2}\textbf{v}\)

\(-\frac{1}{4}\textbf{u}\:-\:\frac{1}{2}\textbf{v}\)

\(-\frac{1}{4}\textbf{u} + \frac{1}{2}\textbf{v}\)

\(\frac{3}{4}\textbf{u}\:-\:\frac{1}{2}\textbf{v}\)

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 \(\overrightarrow{\text{PR}}\)

\(\overrightarrow{\text{PR}} = \overrightarrow{\text{PO}} + \overrightarrow{\text{OR}}\)

\(\overrightarrow{\text{PR}} = -\dfrac{3}{8}\textbf{u} + \dfrac{3}{8}\textbf{v} + \dfrac{1}{8}\textbf{u} + \dfrac{1}{8}\textbf{v}\)

\(\overrightarrow{\text{PR}} = -\dfrac{2}{8}\textbf{u} + \dfrac{4}{8}\textbf{v}\)

\(\overrightarrow{\text{PR}} = -\dfrac{1}{4}\textbf{u} + \dfrac{1}{2}\textbf{v}\)

Pertanyaan 13 dari 13

13. Pertanyaan
1 points

Perhatikan 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 \(\overrightarrow{\text{OP}} = \dotso\)

\(\frac{2}{3}\textbf{u} + \textbf{v} + \textbf{w}\)

\(\frac{2}{3}\textbf{u} \:-\: \textbf{v} + \textbf{w}\)

\(\frac{1}{3}\textbf{u} + \textbf{v} + \textbf{w}\)

\(\frac{2}{3}\textbf{u} + \textbf{v} + \frac{1}{3}\textbf{w}\)

\(\frac{2}{3}\textbf{u} + \textbf{v} \:-\:\frac{1}{3} \textbf{w}\)

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

\(\overrightarrow{\text{OP}} = \overrightarrow{\text{OA}} + \overrightarrow{\text{AB}} + \overrightarrow{\text{BQ}} + \overrightarrow{\text{QP}}\)

\(\overrightarrow{\text{OP}} = \textbf{u} + \textbf{v} \:-\: \dfrac{1}{3}\textbf{u} + \textbf{w}\)

\(\overrightarrow{\text{OP}} = \dfrac{2}{3}\textbf{u} + \textbf{v} + \textbf{w}\)
19/01/2022
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.

Pembahasan Antiruwet

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.

Pembahasan Antiruwet

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 \((0, 10, 2)\)

\(\overrightarrow{\text{PQ}} = \textbf{q}\:-\:\textbf{p}\)

\(\overrightarrow{\text{AP}} = \left(\begin{array}{c}0\\10\\2\end{array}\right)\:-\:\left(\begin{array}{c}10\\ 4\\10\end{array}\right) = \left(\begin{array}{c}-10\\6\\-8\end{array}\right)\)

\(||\overrightarrow{\text{PQ}}|| =\sqrt{(-10)^2 + 6^2 + (-8)^2}\)

\(||\overrightarrow{\text{PQ}}|| =\sqrt{100 + 36 + 64}\)

\(||\overrightarrow{\text{PQ}}|| =\sqrt{200}\)

\(||\overrightarrow{\text{PQ}}|| = 10\sqrt{2}\text { satuan}\)
19/01/2022
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 \(\textbf{b} = 4\textbf{i} \:-\:\textbf{j} + 5\textbf{k}\)

Pembahasan Antiruwet

\(\textbf{a} \times \textbf{b} = \begin{vmatrix}\textbf{i}&\textbf{j}&\textbf{k}\\-1 & 2 & 3\\4 & -1 & 5\end{vmatrix}\)

\(\textbf{a} \times \textbf{b} =\begin{vmatrix}2 & 3\\ -1 & 5\end{vmatrix}\textbf{i}\:-\:\begin{vmatrix}-1 & 3\\ 4 & 5\end{vmatrix}\textbf{j} + \begin{vmatrix}-1 & 2\\ 4 & -1\end{vmatrix}\textbf{k}\)

\(\textbf{a} \times \textbf{b} =(2\cdot 5\:-\:3\cdot (-1))\textbf{i}\:-\:(-1\cdot 5\:-\:3\cdot 4)\textbf{j} + (-1\cdot (-1)\:-\:2\cdot 4)\textbf{k}\)

\(\textbf{a} \times \textbf{b} =13\textbf{i}\:-\:(-17)\textbf{j} + (-7)\textbf{k}\)

\(\textbf{a} \times \textbf{b} =13\textbf{i} + 17\textbf{j} \:-\:7\textbf{k}\)

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)

Pembahasan Antiruwet

\(\overrightarrow{\text{AB}}=\textbf{b}\:-\:\textbf{a}\)

\(\overrightarrow{\text{AB}}=\left(\begin{array}{c}3\\ -2\\0\end{array}\right) \:-\: \left(\begin{array}{c}2\\0\\-1\end{array}\right) = \left(\begin{array}{c}1\\-2\\1\end{array}\right)\)

\(\overrightarrow{\text{AC}}=\textbf{c}\:-\:\textbf{a}\)

\(\overrightarrow{\text{AC}}=\left(\begin{array}{c}4\\ 4\\1\end{array}\right) \:-\: \left(\begin{array}{c}2\\0\\-1\end{array}\right) = \left(\begin{array}{c}2\\4\\2\end{array}\right)\)

\(\color{blue} \text{Area of triangle } = \dfrac{1}{2} \left|\overrightarrow{\text{AB}} \times \overrightarrow{\text{AC}}\right|\)

\(\overrightarrow{\text{AB}} \times \overrightarrow{\text{AC}} = \left(\begin{array}{c}1\\-2\\1\end{array}\right)\times \left(\begin{array}{c}2\\4\\2\end{array}\right)\)

\(\overrightarrow{\text{AB}} \times \overrightarrow{\text{AC}} =\begin{vmatrix}\textbf{i}&\textbf{j}&\textbf{k}\\1 & -2 & 1\\2 & 4 & 2\end{vmatrix}\)

\(\overrightarrow{\text{AB}}\times \overrightarrow{\text{AC}} =\begin{vmatrix}-2 & 1\\ 4& 2\end{vmatrix}\textbf{i}\:-\:\begin{vmatrix}1&1\\ 2& 2\end{vmatrix}\textbf{j} + \begin{vmatrix}1 & -2\\ 2 & 4\end{vmatrix}\textbf{k}\)

\(\overrightarrow{\text{AB}}\times \overrightarrow{\text{AC}} =(-2\cdot 2\:-\:1\cdot 4)\textbf{i}\:-\:(1\cdot 2\:-\:1\cdot 2)\textbf{j} + (1\cdot 4\:-\:(-2)\cdot 2)\textbf{k}\)

\(\overrightarrow{\text{AB}} \times \overrightarrow{\text{AC}} =-8\textbf{i}\:-\:0\textbf{j} +8\textbf{k}\)

\(\overrightarrow{\text{AB}} \times \overrightarrow{\text{AC}} = -8\textbf{i} + 8\textbf{k}\)

\(\text{Area of triangle } = \dfrac{1}{2}\sqrt{(-8)^2 + 8^2}\)

\(\text{Area of triangle } = \dfrac{1}{2}\sqrt{64 + 64}\)

\(\text{Area of triangle } = \dfrac{1}{2}\sqrt{2\times 64}\)

\(\text{Area of triangle } = \dfrac{8}{2}\sqrt{2}\)

\(\text{Area of triangle } = 4\sqrt{2} \text{ square units}\)

19/01/2022
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 \(\textbf{v} = 3\textbf{i} \:-\: \textbf{j}\).

Pembahasan Antiruwet

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 \(\textbf{u} = -2\textbf{i} \:-\: 3\textbf{j}\).

Pembahasan Antiruwet

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

19/01/2022
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:

Pembahasan Anti Ruwet

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.

Pembahasan Anti Ruwet

\(\cos \theta = \dfrac{\textbf{a} \cdot \textbf{b}}{||\textbf{a}|| \cdot ||\textbf{b}||}\)

\(\cos \theta = \dfrac{\left(\begin{array}{c}2\\ -1\\2\end{array}\right) \cdot \left(\begin{array}{c}0\\ 3\\-4\end{array}\right)}{\sqrt{2^2 + (-1)^2 + 2^2}\cdot \sqrt{0^2 + 3^2 + (-4)^2}}\)

\(\cos \theta = \dfrac{0\:-\:3\:-\:8}{\sqrt{9}\cdot \sqrt{25}}\)

\(\cos \theta = \dfrac{-11}{3\cdot 5}\)

\(\cos \theta = \dfrac{-11}{15}\)

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.

Pembahasan Anti Ruwet

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.

Pembahasan Anti Ruwet

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 \(\theta\).

Pembahasan Anti Ruwet

\(\cos \theta = \dfrac{\textbf{a} \cdot \textbf{b}}{|\textbf{a}| \cdot |\textbf{b}|}\)

\(\cos \theta = \dfrac{-\frac{1}{2}\sqrt{2}\cancel{||\textbf{a}||\cdot ||\textbf{b}||}}{\cancel{||\textbf{a}|| \cdot ||\textbf{b}||}}\)

\(\cos \theta = -\dfrac{1}{2}\sqrt{2}\)

\(\theta = 180^{\circ}\:-\:45^{\circ}\)

\(\theta = 135^{\circ}\)

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.

Pembahasan Anti Ruwet

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 \(\frac{1}{6}\sqrt{30}\)

19/01/2022
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 \(\textbf{v} = \left(\begin{array}{c}-5\\ 1\end{array}\right)\)

Pembahasan Antiruwet

\(\left(\begin{array}{c}-3\\2\end{array}\right)\cdot \left(\begin{array}{c}-5\\1\end{array}\right) = (-3)(-5) + 2(1)\)

\(\left(\begin{array}{c}-3\\2\end{array}\right)\cdot \left(\begin{array}{c}-5\\1\end{array}\right) = 15 + 2 = 17\)

Soal 2

Tentukan hasil perkalian skalar antara vektor \(\textbf{u} = \left(\begin{array}{c}-1\\ 0\\2\end{array}\right)\) dan \(\textbf{v} = \left(\begin{array}{c}3\\8\\-5\end{array}\right)\)

Pembahasan Antiruwet

\(\left(\begin{array}{c}-1\\ 0\\2\end{array}\right) \cdot \left(\begin{array}{c}3\\8\\-5\end{array}\right) = (-1)(3) + 0(8) + 2(-5)\)

\(\left(\begin{array}{c}-1\\ 0\\2\end{array}\right) \cdot \left(\begin{array}{c}3\\8\\-5\end{array}\right) = -3 + 0 \:-\:10 = -13\)

19/01/2022
Perbandingan Vektor

Diketahui titik A, B, dan C terletak pada ruas garis yang sama.

Vektor posisi titik A, B, dan C berturut-turut adalah \(\textbf{a}\), \(\textbf{b}\), dan \(\textbf{c}\).

Titik C berada diantara titik A dan B dalam perbandingan AC : CB = m : n.

Vektor posisi \(\textbf{c}\) dapat dicari sebagai berikut:

\(\textbf{c} = \dfrac{m\textbf{b} + n\textbf{a} }{m + n}\)

Contoh Soal

Soal 1

Diketahui koordinat titik A(4, 3) dan B(-3, 4). Titik C berada pada ruas garis AB, dengan perbandingan \(\text{AC} : \text{CB} = 1 : 3\). Tentukan koordinat titik C tersebut.

Pembahasan Antiruwet

Diketahui:

\(\textbf{a} = \left(\begin{array}{c}4\\ 3\end{array}\right)\)

\(\textbf{b} = \left(\begin{array}{c}-3\\ 4\end{array}\right)\)

\(\text{AC} : \text{CB} = m : n = 1 : 3\)

Vektor posisi \(\textbf{c}\) dapat dicari sebagai berikut:

\(\textbf{c} = \dfrac{m\textbf{b} + n\textbf{a} }{m + n}\)

\(\textbf{c} = \dfrac{1\cdot \left(\begin{array}{c}-3\\ 4\end{array}\right) + 3\cdot\left(\begin{array}{c}4\\ 3\end{array}\right)}{1 + 3}\)

\(\textbf{c} = \dfrac{\left(\begin{array}{c}-3\\ 4\end{array}\right) + \left(\begin{array}{c}12\\ 9\end{array}\right)}{4}\)

\(\textbf{c} = \dfrac{\left(\begin{array}{c}-3 + 12\\ 4 + 9\end{array}\right)}{4}\)

\(\textbf{c} = \dfrac{\left(\begin{array}{c}9\\ 13\end{array}\right)}{4}\)

\(\textbf{c} = \left(\begin{array}{c}\frac{9}{4}\\ \frac{13}{4}\end{array}\right)\)

Jadi koordinat titik C adalah \((\frac{9}{4},\:\frac{13}{4})\)

Soal 2

Perhatikan gambar di bawah ini:

Diketahui OC : CB = 5 : 1 dan BD : DA = 4 : 2. Garis OD berpotongan dengan garis AC di titik P sehingga OP : PD = x : y dan AP : PC = m : n. Tentukan:

(A) x : y

(B) m : n

Pembahasan Antiruwet

Langkah 1: Nyatakan vektor posisi \(\overrightarrow{\text{OD}}\)

\(\overrightarrow{\text{OD}}= \dfrac{2\cdot \overrightarrow{\text{OB}} + 4\cdot \overrightarrow{\text{OA}}}{2 + 4}\)

\(\overrightarrow{\text{OD}}= \dfrac{2\cdot \overrightarrow{\text{OB}} + 4\cdot \overrightarrow{\text{OA}}}{6}\)

Langkah 2: Nyatakan vektor \(\overrightarrow{\text{AC}}\)

\(\overrightarrow{\text{AC}}= \dfrac{5\cdot \overrightarrow{\text{AB}} + 1\cdot \overrightarrow{\text{AO}}}{5+1}\)

\(\overrightarrow{\text{AC}}= \dfrac{5\cdot \overrightarrow{\text{AB}} + 1\cdot \overrightarrow{\text{AO}}}{6}\)

\(\text{Note:}\)

\(\color{blue} \overrightarrow{\text{OA}} + \overrightarrow{\text{AB}} =\color{blue} \overrightarrow{\text{OB}}\)

\(\color{blue} \overrightarrow{\text{AB}} =\color{blue} \overrightarrow{\text{OB}}\:-\:\overrightarrow{\text{OA}}\)

\(\overrightarrow{\text{AC}}= \dfrac{5\cdot (\overrightarrow{\text{OB}}\:-\:\overrightarrow{\text{OA}})+ 1\cdot \overrightarrow{\text{AO}}}{6}\)

\(\overrightarrow{\text{AC}}= \dfrac{5\cdot (\overrightarrow{\text{OB}}\:-\:\overrightarrow{\text{OA}})\:-\: \overrightarrow{\text{OA}}}{6}\)

\(\overrightarrow{\text{AC}}= \dfrac{5\cdot\overrightarrow{\text{OB}}\:-\:6\cdot\overrightarrow{\text{OA}}}{6}\)

Langkah 3: \(\overrightarrow{\text{OA}} = \overrightarrow{\text{OP}} + \overrightarrow{\text{PA}}\)

\(\overrightarrow{\text{OP}} = \dfrac{x}{x + y}\cdot \overrightarrow{\text{OD}}\)

\(\overrightarrow{\text{PA}} = \dfrac{m}{m + n}\cdot \overrightarrow{\text{CA}}\)

\(\overrightarrow{\text{PA}} = -\dfrac{m}{m + n}\cdot \overrightarrow{\text{AC}}\)

\(\overrightarrow{\text{OA}} = \dfrac{x}{x + y}\cdot \overrightarrow{\text{OD}} \:-\: \dfrac{m}{m + n}\cdot \overrightarrow{\text{AC}}\)

\(\overrightarrow{\text{OA}} = \dfrac{x}{x + y}\left(\dfrac{2\cdot \overrightarrow{\text{OB}} + 4\cdot \overrightarrow{\text{OA}}}{6}\right) \:-\: \dfrac{m}{m + n}\left(\dfrac{5\cdot\overrightarrow{\text{OB}}\:-\:6\cdot\overrightarrow{\text{OA}}}{6}\right)\)

\(\textbf{a} = \dfrac{x}{x + y}\left(\dfrac{\textbf{b} + 2\textbf{a}}{3}\right) \:-\: \dfrac{m}{m + n}\left(\dfrac{5\textbf{b}\:-\:6\textbf{a}}{6}\right)\)

\(\textbf{a} =\dfrac{x}{3(x+y)}\textbf{b} + \dfrac{2x}{3(x + y)}\textbf{a}\:-\:\dfrac{5m}{6(m +n)}\textbf{b} + \dfrac{m}{m+n}\textbf{a}\)

\(0\textbf{b} + 1\textbf{a} =\dfrac{x}{3(x+y)}\textbf{b} \:-\:\dfrac{5m}{6(m +n)}\textbf{b} + \dfrac{2x}{3(x + y)}\textbf{a}+ \dfrac{m}{m+n}\textbf{a}\)

Samakan koefisien dari masing-masing vektor

\(0 = \dfrac{x}{3(x+y)}\:-\:\dfrac{5m}{6(m +n)}\dotso\dotso\color{blue}(1)\)

\(1 = \dfrac{2x}{3(x + y)} + \dfrac{m}{m+n}\dotso\dotso\color{blue}(2)\)

Selanjutnya eliminasi kedua persamaan tersebut

Persamaan pertama dikali 2, persamaan kedua dikali 1

\(0 = \dfrac{2x}{3(x+y)}\:-\:\dfrac{10m}{6(m +n)}\dotso\dotso\color{blue}(1)\)

\(1 = \dfrac{2x}{3(x + y)} + \dfrac{m}{m+n}\dotso\dotso\color{blue}(2)\)

Kurangkan persamaan (1) dengan persamaan (2)

\(-1 = -\dfrac{8}{3}\cdot \dfrac{m}{m+n}\)

\(1 = \dfrac{8m}{3(m+n)}\)

\(3(m + n) = 8m\)

\(3m + 3n = 8m\)

\(3n = 5m\)

\(\dfrac{3}{5} = \dfrac{m}{n}\)

Jadi m : n = 3 : 5

Dengan menggunakan cara eliminasi lagi

Persamaan pertama dikali 1, persamaan kedua dikali \(\frac{5}{6}\)

\(0 = \dfrac{x}{3(x+y)} \:-\:\dfrac{5m}{6(m + n)}\)

\(\dfrac{5}{6} = \dfrac{10x}{18(x + y)} + \dfrac{5m}{6(m + n)}\)

Jumlahkan persamaan pertama dengan persamaan kedua

\(\dfrac{5}{6} = \dfrac{8x}{9(x + y)}\)

\(45(x + y) = 48x\)

\(45x + 45y = 48x\)

\(45y = 3x\)

\(15y = x\)

\(\dfrac{x}{y} = \dfrac{15}{1}\)

Jadi x : y = 15 : 1

19/01/2022
Penjumlahan Vektor

Penjumlahan Vektor

Jika \(\textbf{u} = \left(\begin{array}{c}a\\ b\end{array}\right)\) dan \(\textbf{v} = \left(\begin{array}{c}c\\ d\end{array}\right)\) maka:

\(\textbf{u} + \textbf{v} = \left(\begin{array}{c}a\\ b\end{array}\right) + \left(\begin{array}{c}c\\ d\end{array}\right)\)

\(\textbf{u} +\textbf{v} = \left(\begin{array}{c}a + c\\ b + d\end{array}\right)\)

Pengurangan Vektor

Jika \(\textbf{u} = \left(\begin{array}{c}a\\ b\end{array}\right)\) dan \(\textbf{v} = \left(\begin{array}{c}c\\ d\end{array}\right)\) maka:

\(\textbf{u} \:-\: \textbf{v} = \left(\begin{array}{c}a\\ b\end{array}\right) \:-\: \left(\begin{array}{c}c\\ d\end{array}\right)\)

\(\textbf{u} \:-\: \textbf{v} = \left(\begin{array}{c}a \:-\: c\\ b \:-\: d\end{array}\right)\)

Contoh Soal

Soal 1

Jika \(\textbf{u} = \left(\begin{array}{c}-12\\ 5\end{array}\right)\) dan \(\textbf{v} = \left(\begin{array}{c}4\\ -3\end{array}\right)\), tentukan nilai \(\textbf{u} +\textbf{v} \)

Pembahasan Antiruwet

\(\textbf{u} +\textbf{v} = \left(\begin{array}{c}-12\\ 5\end{array}\right) + \left(\begin{array}{c}4\\ -3\end{array}\right)\)

\(\textbf{u} +\textbf{v} = \left(\begin{array}{c}-12 + 4\\ 5 + (-3)\end{array}\right)\)

\(\textbf{u} +\textbf{v} = \left(\begin{array}{c}-8\\ 2\end{array}\right)\)

Soal 2

Jika \(\textbf{u} = \left(\begin{array}{c}12\\ 7\end{array}\right)\) dan \(\textbf{v} = \left(\begin{array}{c}4\\ 1\end{array}\right)\), tentukan \(||\textbf{u} \:-\:\textbf{v}||\)

Pembahasan Antiruwet

\(\textbf{u} \:-\:\textbf{v} = \left(\begin{array}{c}12\\ 7\end{array}\right) \:-\: \left(\begin{array}{c}4\\ 1\end{array}\right)\)

\(\textbf{u} \:-\:\textbf{v} = \left(\begin{array}{c}12\:-\: 4\\ 7\:-\:1\end{array}\right)\)

\(\textbf{u} \:-\:\textbf{v} = \left(\begin{array}{c}8\\ 6\end{array}\right)\)

\(||\textbf{u} \:-\:\textbf{v}|| = \sqrt{8^2 + 6^2}\)

\(||\textbf{u} \:-\:\textbf{v}|| = \sqrt{64 + 36}\)

\(||\textbf{u} \:-\:\textbf{v}|| = \sqrt{100} = 10 \text{ satuan}\)

Soal 3

Tentukan nilai \(m\) yang memenuhi persamaan \(-2\left(\begin{array}{c}3\\ 1\end{array}\right) + \left(\begin{array}{c}-4\\ 2\end{array}\right) = \left(\begin{array}{c}1+m\\ 0\end{array}\right)\)

Pembahasan Antiruwet

\(-2\left(\begin{array}{c}3\\ 1\end{array}\right) + \left(\begin{array}{c}-4\\ 2\end{array}\right) = \left(\begin{array}{c}1+m\\ 0\end{array}\right)\)

\(\left(\begin{array}{c}-6\\ -2\end{array}\right) + \left(\begin{array}{c}-4\\ 2\end{array}\right) = \left(\begin{array}{c}1+m\\ 0\end{array}\right)\)

\(\left(\begin{array}{c}-6 + (-4)\\ -2 + 2\end{array}\right) = \left(\begin{array}{c}1+m\\ 0\end{array}\right)\)

\(\left(\begin{array}{c}-10\\ 0\end{array}\right) = \left(\begin{array}{c}1+m\\ 0\end{array}\right)\)

\(-10 = 1 + m\)

\(-10\:-\:1 = m\)

\(m = -11\)

19/01/2022

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