Soal Ulangan Harian Vektor bagian 01

Ayouk berlatih persiapan ulangan harian vektor bagian 01.

Bimbel WIN:
Salahsatu alternatif untuk mempersiapkan ulangan di sekolah adalah belajar dari bentuk soal yang sudah pernah diujikan sebelumnya. Cara belajar seperti ini akan menciptakan persiapan yag lebih terarah, lebih fokus dan lebih efektif.

Berikut ini adalah contoh soal ulangan harian Vektor bagian 01 dari salahsatu sekolah. Disini kami telah membuat pembahasannya yang mudah untuk difahami, semoga bisa dibuat jadi reverensi berlatih untuk persiapan ulangan harian di sekolah mu.

💦 Soal No. 01.

Jika titik A(1, -2) dan B(2, -5) maka vektor \(\overline {AB} \) dinyatakan dengan ...

(A) (1, -3)
(B) (1, 3)
(C) (-1, -3)
(D) (-3, -1)
(E) (-3, -1)

Pembahasan

\(\begin{array}{l}\overline {AB} = B - A\\ = \left( {\begin{array}{*{20}{c}}2\\{ - 5}\end{array}} \right) - \left( {\begin{array}{*{20}{c}}1\\{ - 2}\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}1\\{ - 3}\end{array}} \right)\end{array}\)

💥 Kunci Jawabn A

💦 Soal No. 02.

Jika titik P(0, 2) dan Q(1, 3, maka dinyatakan dengan ...

(A) (-1, -5)
(B) (1, 5)
(C) (1, -5)
(D) (-5, -1)
(E) (-5, 1)

Pembahasan

\(\begin{array}{l}\overline {QP} = P - Q\\ = \left( {\begin{array}{*{20}{c}}0\\{ - 2}\end{array}} \right) - \left( {\begin{array}{*{20}{c}}1\\3\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}{ - 1}\\{ - 5}\end{array}} \right)\end{array}\)

💥 Kunci Jawaban A

💦 Soal No. 03

Diketahui vektor \(\overline {PQ} = \left( {\begin{array}{*{20}{c}}3\\{ - 1}\end{array}} \right)\) dan titik Q(2, -4), makakoordiant titik P adalah ...

(A) (-1, -3)
(B) (1, 3)
(C) (-1, 3)
(D) (1, -3)
(E) (3, 1)

Pembahasan

\(\begin{array}{l}\overline {PQ} = Q - P\\\left( {\begin{array}{*{20}{c}}3\\{ - 1}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}2\\{ - 4}\end{array}} \right) - P\\P = \left( {\begin{array}{*{20}{c}}2\\{ - 4}\end{array}} \right) - \left( {\begin{array}{*{20}{c}}3\\{ - 1}\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}{ - 1}\\{ - 3}\end{array}} \right)\end{array}\)

💥 Kunci Jawaban A

💦 Soal No. 04.

Jika \(\overline {KL} = \left( {\begin{array}{*{20}{c}}0\\2\end{array}} \right),\overline {JK} = \left( {\begin{array}{*{20}{c}}2\\5\end{array}} \right)\), titik L(-2, 7), dan titik I(2, 3) maka IJ adalah ...

(A) (-6, 3)
(B) (6, 3)
(C) (-6, -3)
(D) (-3, -6)
(E) (3, -6)

Pembahasan

\(\begin{array}{l}\overline {KL} = L - K = \left( {0,2} \right)\\{\underline {\overline {JK} = K - L = \left( {2,5} \right)} _{\,\, + }}\\L - J = \left( {2,7} \right)\\J = L - \left( {2,7} \right)\\ \Rightarrow J = \left( {\begin{array}{*{20}{c}}{ - 2}\\7\end{array}} \right) - \left( {\begin{array}{*{20}{c}}2\\7\end{array}} \right) = \left( {\begin{array}{*{20}{c}}{ - 4}\\0\end{array}} \right)\\ \Rightarrow IJ = J - I\\ = \left( {\begin{array}{*{20}{c}}{ - 4}\\0\end{array}} \right) - \left( {\begin{array}{*{20}{c}}2\\3\end{array}} \right) = \left( {\begin{array}{*{20}{c}}{ - 6}\\{ - 3}\end{array}} \right)\end{array}\)

💥 Kunci Jawaban C

💦 Soal No.05.

Diketahui koordinat titik P(2, 2), Q(3, 3), jika vektor posisi R adalah \(\overline r = \overline {QP} \) maka \(\overline r \) =

(A) (-1, -1)
(B) (1, -1)
(C) (-1, 1)
(D) (-1, 0)
(E) (0, -1)

Pembahasan

\(\begin{array}{l}\overline r = \overline {QP} = P - Q\\ = \left( {\begin{array}{*{20}{c}}2\\2\end{array}} \right) - \left( {\begin{array}{*{20}{c}}3\\3\end{array}} \right) = \left( {\begin{array}{*{20}{c}}{ - 1}\\{ - 1}\end{array}} \right)\end{array}\)

💥 Kunci Jawaban A

💦 Soal No. 06.

Diketahui \(\overline s = i + 3j\,\,dan\,\,\overline t = 2i + 5j\) maka ...

(A) \(\left( {\begin{array}{*{20}{c}}4\\{ - 9}\end{array}} \right)\)
(B) \(\left( {\begin{array}{*{20}{c}}-4\\{ - 9}\end{array}} \right)\)
(C) \(\left( {\begin{array}{*{20}{c}}4\\{ 9}\end{array}} \right)\)
(D) \(\left( {\begin{array}{*{20}{c}}9\\{ - 4}\end{array}} \right)\)
(E) \(\left( {\begin{array}{*{20}{c}}-9\\{ - 4}\end{array}} \right)\)

Pembahasan

\(\begin{array}{l}\overline s = i + 3j\\\overline t = 2i + 5j\\ \Rightarrow 2\overline s - 3\overline t = 2\left( {\begin{array}{*{20}{c}}1\\3\end{array}} \right) - 3\left( {\begin{array}{*{20}{c}}2\\5\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}2\\6\end{array}} \right) - \left( {\begin{array}{*{20}{c}}6\\{15}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}{ - 4}\\{ - 9}\end{array}} \right)\end{array}\)

Kunci Jawaban B

💦 Soal No. 07.

Diketahui vektor posisi titik P dan Q masing-masing adalah \(\overline p = \left( { - 2,1} \right),\overline q = \left( {2,a} \right)\). Jika \(\left| {\overline p } \right| = \left| {\overline q } \right|\), maka nilai a adalah ...

(A) 1
(B) 2
(C) -1
(D) \( \pm 1\)
(E) \( \pm 2\)

Pembahasan

\(\begin{array}{l} \Rightarrow \left| {\overline p } \right| = \sqrt {{{\left( { - 2} \right)}^2} + {1^2}} = \sqrt 5 \\ \Rightarrow \left| {\overline q } \right| = \sqrt {{2^2} + {a^2}} = \sqrt {4 + {a^2}} \\ \Leftrightarrow \left| {\overline p } \right| = \left| {\overline q } \right|\\\sqrt 5 = \sqrt {4 + {a^2}} (dikuadratkan)\\5 = 4 + {a^2}\\{a^2} = 1\,\,\,\,\, \to \,\,\,\,\,\,\,\,a = \pm 1\end{array}\)

💥 Kunci Jawaban D

💦 Soal No. 08.

Perhatikan gambar di bawah ini, AB mewakili vektor \(\overline u \) dan BC mewakilivektor \(\overline v \), maka HF jika dinyatakan dalam vektor \(\overline u \) dan \(\overline v \) adalah ...

(A) \(\overline u + \overline v \)
(B) \(\overline u - \overline v \)
(C) \(\overline -u + \overline -v \)
(D) \(\overline -u + \overline v \)
(E) \(\overline -u + \overline 2v \)

Pembahasan

\(\begin{array}{l} \Rightarrow AB = DC = HG = \overline u \\ \Rightarrow BC = AD = EH = \overline v \\ \Leftrightarrow EH + HF = EF\\HF = EF - EH\\ = \overline u - \overline v \end{array}\)

💥 Kunci Jawaban B

💦 Soal No.09.

Pada gambar jajaran genjang di bawah, hasil dari \((- \,\overline d + \overline e - \,\overline h - \overline b = ...\)

(A) \(\overline a \)
(B) \(\overline c \)
(C) \(\overline b \)
(D) \(\overline d \)
(E) \(\overline -d \)

Pembahasan:

\(\begin{array}{l}*)\,\,\, - \,\overline d + \overline e = \overline g \\*)\,\,\,\overline g - \,\overline h = 2\,\overline g \\*)\,\,\overline b + \,\overline a = 2\,\overline g \\ \Rightarrow 2\,\overline g - \overline b = \overline a \end{array}\) jadi \( - \,\overline d + \overline e - \,\overline h - \overline b = \overline a \)

💥 Kunci Jawaban A

💦 Soal No. 10.

Jika \(\overline a = \left( {\begin{array}{*{20}{c}}3\\2\end{array}} \right),\,\,\,\overline b = \left( {\begin{array}{*{20}{c}}{ - 8}\\9\end{array}} \right),\,\,\,\overline c = \left( {\begin{array}{*{20}{c}}{12}\\{13}\end{array}} \right)\) dan \(\overline d = \overline a + \overline b + \overline c \) maka panjang \(\overline d \) adalah ...

(A) 15
(B) 17
(C) 20
(D) 23
(E) 25

Pembahasan

\(\begin{array}{l}\overline d = \overline a + \overline b + \overline c \\\overline d = \left( {\begin{array}{*{20}{c}}3\\2\end{array}} \right) + \left( {\begin{array}{*{20}{c}}{ - 8}\\9\end{array}} \right) + \left( {\begin{array}{*{20}{c}}{12}\\{13}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}7\\{24}\end{array}} \right)\\\overline d = \sqrt {{7^2} + {{24}^2}} = 25\end{array}\)

💥 Kunci Jawaban E

💦 Soal No.11.

Jika \(\overline a = i - j,\,\,\,\overline b = 2i + 4j\) maka \(\overline a + \overline b = ...\)

(A) \(3i + 2j\)
(B) \(3i - 2j\)
(C) \(2i - 3j\)
(D) \(2i + 3j\)
(E) \(2i - 2j\)

Pembahasan

\(\begin{array}{l}\overline a = i - j\\{\underline {\overline b = 2i + 3j} _{\, + }}\\\overline a + \overline b = 3i + 2j\end{array}\)

💥 Kunci Jawaban A

💦 Soal No. 12

Diketahui vektor \(\overline a = 3i - 2j,\,\,\,\overline b = - i + 4j\,\) dan \(\overline r = 7i - 8j\,\). Jika \(\overline r = k\overline a + m\overline b \) maka nilai k + m = ...

(A) -3
(B) -2
(C) -1
(D) 1
(E) 2

Pembahasan

\(\begin{array}{l}*)\,\,\,\overline a = 3i - 2j,\,\,\,\overline b = - i + 4j\,\,dan\\\,\,\overline r = k\overline a + m\overline b \\\left( {\begin{array}{*{20}{c}}7\\{ - 8}\end{array}} \right) = k\left( {\begin{array}{*{20}{c}}3\\{ - 2}\end{array}} \right) + m\left( {\begin{array}{*{20}{c}}{ - 1}\\4\end{array}} \right)\\*)\,\,\,7 = 3k - m \to (kali4)\\28 = 12k - 4m\,\,...\,\,i)\\{\underline { - 8 = - 2k + 4m} _{\, + }}\,\,...\,\,ii)\\20 = 10k\,\, \to \,\,k = 2\\i)\,\,m = 3(2) - 7 = - 1\\ \Leftrightarrow k + m = 2 - 1 = 1\end{array}\)

💥 Kunci Jawaban D

💦 Soal No.13.

Diketahui \(2\left( {\begin{array}{*{20}{c}}1\\{ - 2}\end{array}} \right) + 4\left( {\begin{array}{*{20}{c}}1\\4\end{array}} \right) + m\left( {\begin{array}{*{20}{c}}2\\3\end{array}} \right) = \left( {\begin{array}{*{20}{c}}2\\6\end{array}} \right)\) dengan m adalah bilangan real, maka nilai m adalah ...

(A) -4
(B) -2
(C) -1
(D) 1
(E) 3

Pembahasan

\(\begin{array}{l}2\left( {\begin{array}{*{20}{c}}1\\{ - 2}\end{array}} \right) + 4\left( {\begin{array}{*{20}{c}}1\\4\end{array}} \right) + m\left( {\begin{array}{*{20}{c}}2\\3\end{array}} \right) = \left( {\begin{array}{*{20}{c}}2\\6\end{array}} \right)\\2\left( 1 \right) + 4\left( 1 \right) + m\left( 2 \right) = 2\\2m = 2 - 2 - 4\\m = - \frac{4}{2} = - 2\end{array}\)

💥 Kunci Jawaban B

💦 Soal No. 14.

Vektor satuan dari \(\overline a = 3i + 4j\) adalah ...

(A) \({\widehat e_{\overline a }} = \frac{3}{5}i + \frac{4}{5}j\)
(B) \({\widehat e_{\overline a }} = \frac{3}{5}i - \frac{4}{5}j\)
(C) \({\widehat e_{\overline a }} = - \frac{3}{5}i + \frac{4}{5}j\)
(D) \({\widehat e_{\overline a }} = \frac{3}{7}i + \frac{4}{7}j\)
(E) \({\widehat e_{\overline a }} = \frac{3}{7}i - \frac{4}{7}j\)

Pembahasan

\(\begin{array}{l}{\widehat e_{\overline a }} = \frac{1}{{\left| {\overline a } \right|}} \cdot \overline a \\ \Rightarrow \left| {\overline a } \right| = \sqrt {{3^2} + {4^2}} = 5\\ \Leftrightarrow {\widehat e_{\overline a }} = \frac{3}{5}i + \frac{4}{5}j\end{array}\)

💥 Kunci Jawaban A

💦 Soal No. 15.

Pada jajaran genjang ABCD diketahui vektor posisi \(\overline a = \left( {\begin{array}{*{20}{c}}2\\{ - 2}\end{array}} \right),\,\,\overline c = \left( {\begin{array}{*{20}{c}}{ - 4}\\3\end{array}} \right)\) dan \(\overline d = \left( {\begin{array}{*{20}{c}}1\\2\end{array}} \right)\) dengan demikian koordinat titik B adalah ...

(A) (2, 3)
(B) (-1, 3)
(C) (1, 3)
(D) (-3,-1)
(E) (3, 1)

Pembahasan

\(\begin{array}{l}\overline {AB} = \overline {DC} \\B - A = C - D\\\overline b - \overline a = \overline c - \overline d \\\overline b = \overline c - \overline d + \overline a \\\overline b = \left( {\begin{array}{*{20}{c}}{ - 4}\\3\end{array}} \right) - \left( {\begin{array}{*{20}{c}}1\\2\end{array}} \right) + \left( {\begin{array}{*{20}{c}}2\\{ - 2}\end{array}} \right)\\\overline b = \left( {\begin{array}{*{20}{c}}{ - 3}\\{ - 1}\end{array}} \right)\\B = \left( { - 3, - 1} \right)\end{array}\)

💥 Kunci Jawaban B

💦 Soal No. 16.

Pada gamabar di bawah, nilai dari \(\overline a \cdot \overline b = \)

(A) \(10\sqrt 3 \)
(B) \( - 10\sqrt 3 \)
(C) 10
(D) -10
(E) 20

Pembahasan

\(\begin{array}{l}\overline a \cdot \overline b = \left| {\overline a } \right| \cdot \left| {\overline b } \right|\cos {30^o}\\ = 4 \times 5 \times \frac{1}{2}\sqrt 3 \\ = 10\sqrt 3 \end{array}\)

💥 Kunci Jawaban A

💦 Soal No. 17.

Jika \(\overline p = 4i + j + 5k\) dan \(\overline q = 2i + j - 5k\) maka hasil kali \(\overline p \cdot \overline q = \)...

(A) -18
(B) 12
(C) -16
(D) 18
(E) 3

Pembahasan

\(\begin{array}{l}\overline p \cdot \overline q = 4\left( 2 \right) + 1\left( 1 \right) + 5\left( { - 5} \right)\\ = 8 + 1 - 25\\ = - 16\end{array}\)

💥 Kunci Jawaban C

💦 Soal No. 18.

Besar sudut antara vektor \(\overline p = - 2i + 2k,dan\,\,\overline q = 2j + 2k\) adalah

(A) \({30^o}\)
(B) \({45^o}\)
(C) \({60^o}\)
(D) \({90^o}\)
(E) \({120^o}\)

Pembahasan

\(\begin{array}{l}\overline p = - 2i + 2k\,\, \to \left| {\overline p } \right| = \sqrt {{{\left( { - 2} \right)}^2} + {2^2}} = \sqrt 8 \\\,\overline q = 2j + 2k \to \left| {\overline q } \right| = \sqrt {{2^2} + {2^2}} = \sqrt 8 \\\overline p \cdot \overline q = \left( { - 2} \right)0 + 2\left( 0 \right) + \left( 2 \right)2 = 4\\\cos \theta = \frac{4}{{\sqrt 8 \cdot \sqrt 8 }} = \frac{1}{2}\\maka\,\,\theta = {60^o}\end{array}\)

💥 Kunci Jawaban B

💦 Soal No.19.

Diketahui vektor \(\overline v = 2i - j + 4k\) dan \(\overline v = 2i - j + 4k\). jika \(\overline u \) dan \(\overline v \) membentuk sudut \(\theta \) maka \(\tan \theta = \)

(A) \(\frac{5}{7}\)
(B) \(\frac{6}{7}\)
(C) \(\frac{2}{7}\sqrt 6 \)
(D) \(\frac{5}{7}\sqrt 6 \)
(E) \(\frac{6}{7}\sqrt 6 \)

Pembahasan

\(\begin{array}{l}\overline u = \left( {3, - 2,1} \right)\,\, \to \left| {\overline u } \right| = \sqrt {{{\left( 3 \right)}^2} + {{\left( { - 2} \right)}^2} + {{\left( 1 \right)}^2}} = \sqrt {14} \\\,\overline v = \left( {2, - 1,4} \right) \to \left| {\overline v } \right| = \sqrt {{{\left( 2 \right)}^2} + {{\left( { - 1} \right)}^2} + {{\left( 4 \right)}^2}} = \sqrt {21} \\\overline u \cdot \overline v = \frac{{\left( 3 \right)\left( 2 \right) + \left( { - 2} \right)\left( { - 1} \right) + \left( 1 \right)\left( 4 \right)}}{{\sqrt {14} \sqrt {21} }}\\\cos \theta = \frac{{12}}{{7\sqrt 6 }} \equiv \frac{x}{r}\end{array}\) \(\begin{array}{l}{y^2} = {r^2} - {x^2} = {\left( {7\sqrt 6 } \right)^2} - {\left( {12} \right)^2}\\y = \sqrt {294 - 144} = \sqrt {150} = 5\sqrt 6 \\maka\,\,\,\tan \theta = \frac{y}{x} = \frac{{5\sqrt 6 }}{{12}}\end{array}\)

💥 Kunci Jawaban C

💦 Soal No. 20.

Diketahui vektor \(\overline a \) dan \(\overline b \) dimana \(\left| {\overline a } \right| = 6\) dan \(\left| {\overline b } \right| = 4\) serta \(\left| {\overline a + \overline b } \right|\, = 8\) Jika \(\theta \) adalah sudut antara vektor \(\overline a \) dan \(\overline b \) maka \(\cos \theta = ...\)

(A) \(\frac{1}{3}\)
(B) \(\frac{1}{4}\)
(C) \(\frac{2}{5}\)
(D) \( - \frac{1}{3}\)
(E) \( - \frac{2}{5}\)

Pembahasan

\(\begin{array}{l}{\left| {a + b} \right|^2} = {\left| a \right|^2} + {\left| b \right|^2} + 2\left| a \right|\left| b \right|\cos \theta \\{8^2} = {6^2} + {4^2} + 2 \cdot 6 \cdot 4\cos \theta \\48\cos \theta = 64 - 52\\\cos \theta = \frac{{12}}{{48}}\\\cos \theta = \frac{1}{4}\end{array}\)

💥 Kunci Jawaban B