PAS MATEMATIKA BLOG: Vektor pada Ruang

$A. Vektor Di Ruang$

Perhatikanlah ilustrasi gambar berikut

\begin{array}{cc} Nama & R^{3} \\ Vektor Satuan & Ruang (Bidang XYZ) \\ {\hat{e}}_{\bar{a}} = \frac{\bar{a}}{| \bar{a} |} & {\begin{cases} i = & vektor satuan \\ yang searah sumbu X \\ j = & Vektor satuan \\ yang searah sumbu Y \\ k = & Vektor satuan \\ searah sumbu Z \end{cases} \\ Vektor nol & \vec{O} = (\begin{matrix} 0 \\ 0 \\ 0 \end{matrix}) \\ Vektor posisi & \vec{O P} = \vec{p} = (\begin{matrix} p_{1} \\ p_{2} \\ p_{3} \end{matrix}) = p_{1} \bar{i} + p_{2} \bar{j} + p_{3} \bar{j} \\ Besar Vektor & \vec{O P} = \sqrt{p_{1}^{2} + p_{2}^{2} + p_{3}^{2}} \end{array}

B. Operasi Vektor

1. Sifat-Sifat Aljabar Vektor

\begin{array}{lll} 1. & Komutatif penjumlahan & \vec{a} + \vec{b} = \vec{b} + \vec{a} \\ 2. & Asosiatif penjumlahan & (\vec{a} + \vec{b}) + \vec{c} = \vec{a} + (\vec{b} + \vec{c}) \\ 3. & Elemen Identitas & \vec{a} + \vec{0} = \vec{0} + \vec{a} = \vec{a} \\ 4. & Invers Penjumlahan & \vec{a} + (- \vec{a}) = (- \vec{a}) + \vec{a} = \vec{0} \\ 5. & Perkalian dengan skalar & k (l \vec{a}) = (k l) \vec{a} \\ k (\vec{a} + \vec{b}) = k \vec{a} + k \vec{b} \\ k (\vec{a} - \vec{b}) = k \vec{a} - k \vec{b} \\ 6. & \begin{aligned} Jika A, B, dan C segaris \\ (Kolinear) \end{aligned} & {\begin{cases} \vec{A B} = k \vec{B C} \\ \vec{A C} = k \vec{A B} \\ d l l \end{cases} \end{array}

\begin{array}{cc} Vektor & Contoh \\ \vec{z} = a \vec{i} + b \vec{j} + c \vec{k} & \begin{aligned} diketahui \vec{p} = \vec{i} - 2 \vec{j} + 2 \vec{k} \\ maka pangjang vektor \vec{p} adalah \\ | \vec{p} | = \sqrt{1^{2} + (- 2)^{2} + 2^{2}} \\ = \sqrt{1 + 4 + 4} = \sqrt{9} = 3 \end{aligned} \\ \begin{aligned} Vektor satuan dari \vec{p} adalah \\ {\vec{e}}_{\vec{p}} = \frac{\vec{p}}{| \vec{p} |} = \frac{(\begin{array}{c} 1 \\ - 2 \\ 2 \end{array})}{3} \\ = \frac{1}{3} (\begin{array}{c} 1 \\ - 2 \\ 2 \end{array}) = (\begin{array}{c} \frac{1}{3} \\ - \frac{2}{3} \\ \frac{2}{3} \end{array}) \end{aligned} \end{array}

CONTOH SOAL

\begin{array}{ll} 1. & Diketahui vektor-vektor \vec{a} = (\begin{array}{c} 2 \\ 1 \\ - 4 \end{array}) \\ \vec{b} = (\begin{array}{c} - 3 \\ - 5 \\ 2 \end{array}), dan \vec{c} = (\begin{array}{c} 7 \\ 0 \\ 4 \end{array}), \\ tentukanlah hasil dari \\ a . \vec{a} + \vec{b} \\ b . 6 \vec{a} + 2 \vec{b} \\ c . 2 \vec{a} - \vec{b} + \vec{c} \\ d . \frac{1}{2} \vec{c} - \vec{a} + \frac{3}{4} \vec{b} \\ Jawab : \\ \begin{aligned} a & \vec{a} + \vec{b} = (\begin{array}{c} 2 \\ 1 \\ - 4 \end{array}) + (\begin{array}{c} - 3 \\ - 5 \\ 2 \end{array}) \\ = (\begin{array}{c} 2 + (- 3) \\ 1 + (- 5) \\ (- 4) + 2 \end{array}) \\ = (\begin{array}{c} 2 - 3 \\ 1 - 5 \\ - 4 + 2 \end{array}) = (\begin{array}{c} - 1 \\ - 4 \\ - 2 \end{array}) \\ b . & 6 \vec{a} + 2 \vec{b} = 6 (\begin{array}{c} 2 \\ 1 \\ - 4 \end{array}) + 2 (\begin{array}{c} - 3 \\ - 5 \\ 2 \end{array}) \\ = (\begin{array}{c} 18 - 6 \\ 6 - 10 \\ - 24 + 4 \end{array}) = (\begin{array}{c} 12 \\ - 4 \\ - 20 \end{array}) \\ c . & 2 \vec{a} - \vec{b} + \vec{c} \\ 2 (\begin{array}{c} 2 \\ 1 \\ - 4 \end{array}) - (\begin{array}{c} - 3 \\ - 5 \\ 2 \end{array}) + (\begin{array}{c} 7 \\ 0 \\ 4 \end{array}) \\ = (\begin{array}{c} 4 + 3 + 7 \\ 2 + 5 + 0 \\ - 8 - 2 + 4 \end{array}) = (\begin{array}{c} 17 \\ 7 \\ - 6 \end{array}) \\ d . & \frac{1}{2} \vec{c} - \vec{a} + \frac{3}{4} \vec{b} = \dots \end{aligned} \end{array}

\begin{array}{ll} 2. & Diketahui vektor-vektor \vec{a} = (\begin{array}{c} 2 \\ 1 \\ - 4 \end{array}) \\ tentukanlah | \vec{a} | \\ Jawab : \\ \begin{aligned} | \vec{a} | & = \sqrt{2^{2} + 1^{2} + (- 4)^{2}} \\ = \sqrt{4 + 1 + 16} \\ = \sqrt{21} \end{aligned} \end{array}

2. Perkalian Skalar Dua Vektor

Konsep perkalian skalar dua buah vektor di ruang sama persis dengan konsep di bidang, yaitu:

\vec{a} \cdot \vec{b} = | \vec{a} | | \vec{b} | \cos θ

Misalkan diketahui

\begin{aligned} \vec{a} = (\begin{array}{c} a_{1} \\ a_{2} \\ a_{3} \end{array}), \vec{b} = (\begin{array}{c} b_{1} \\ b_{2} \\ b_{3} \end{array}), maka \\ \vec{a} \cdot \vec{b} = (\begin{array}{c} a_{1} \\ a_{2} \\ a_{3} \end{array}) \cdot (\begin{array}{c} b_{1} \\ b_{2} \\ b_{3} \end{array}) \\ = a_{1} b_{1} + a_{2} b_{2} + a_{3} b_{3} \end{aligned}

CONTOH SOAL

\begin{array}{ll} 1. & Diketahui vektor-vektor \vec{a} = (\begin{array}{c} 2 \\ 1 \\ - 4 \end{array}) \\ \vec{b} = (\begin{array}{c} - 3 \\ - 5 \\ 2 \end{array}), dan \vec{c} = (\begin{array}{c} 7 \\ 0 \\ 4 \end{array}), \\ tentukanlah hasil dari \\ a . \vec{a} \cdot \vec{b} \\ b . \vec{a} \cdot \vec{c} \\ c . \vec{b} \cdot \vec{c} \\ Jawab : \\ \begin{aligned} a & \vec{a} \cdot \vec{b} = (\begin{array}{c} 2 \\ 1 \\ - 4 \end{array}) \cdot (\begin{array}{c} - 3 \\ - 5 \\ 2 \end{array}) \\ = (2) (- 3) + (1) (- 5) + (- 4) (2) \\ = - 6 - 5 - 8 = - 19 \\ b & \vec{a} \cdot \vec{c} = (\begin{array}{c} 2 \\ 1 \\ - 4 \end{array}) \cdot (\begin{array}{c} 7 \\ 0 \\ 4 \end{array}) \\ = (2) (7) + (1) (0) + (- 4) (4) \\ = 14 + 0 - 16 = - 2 \\ c & \vec{b} \cdot \vec{c} = (\begin{array}{c} - 3 \\ - 5 \\ 2 \end{array}) \cdot (\begin{array}{c} 7 \\ 0 \\ 4 \end{array}) \\ = (- 3) (7) + (- 5) (0) + (2) (4) \\ = - 21 + 0 + 8 = - 13 \end{aligned} \end{array}

\begin{array}{ll} 2. & Tentukanlah nilai t jika \\ \vec{p} = 3 \bar{i} + t \bar{j} + \bar{k} dan \vec{p} \cdot \vec{p} = 13 \\ Jawab : \\ \begin{aligned} \vec{p} \cdot \vec{p} = 13 \\ \vec{p} \cdot \vec{p} = | \vec{p} | | \vec{p} | \cos 0^{\circ} = 13, \\ ingat bahwa ∠ (\vec{p}, \vec{p}) = 0^{\circ} \\ dan nilai \cos 0^{\circ} = 1, \\ maka \\ \vec{p} \cdot \vec{p} = {| \vec{p} |}^{2} .1 = 13 \Leftrightarrow {| \vec{p} |}^{2} = 13 \\ \Leftrightarrow {(\sqrt{3^{2} + t^{2} + 1^{2}})}^{2} = 13 \\ \Leftrightarrow 3^{2} + t^{2} + 1^{2} = 13 \\ \Leftrightarrow 9 + t^{2} + 1 = 13 \\ \Leftrightarrow t^{2} = 13 - 9 - 1 = 10 \\ \Leftrightarrow t^{2} = 3 \\ \Leftrightarrow t = \pm \sqrt{3} \end{aligned} \end{array}

\begin{array}{ll} 3. & Diketahui \vec{p} = (\begin{array}{c} - 2 \\ 1 \\ 3 \end{array}) dan \vec{q} = (\begin{array}{c} 4 \\ - 1 \\ t \end{array}) \\ Jika \vec{p} tegak lurus \vec{q} maka \\ tentukanlah nilai t \\ Jawab : \\ \begin{aligned} Diketahui bahwa \\ \vec{p} = (\begin{array}{c} - 2 \\ 1 \\ 3 \end{array}) dan \vec{q} = (\begin{array}{c} 4 \\ - 1 \\ t \end{array}) \\ dengan \vec{p} dan \vec{q} tegak lurus \\ artinya ∠ (\vec{p}, \vec{q}) = 90^{\circ} . Sehingga \\ nilai \cos 90^{\circ} = 0 \\ maka \\ \vec{p} \cdot \vec{q} = | \vec{p} | | \vec{q} | \cos θ \\ \vec{p} \cdot \vec{q} = | \vec{p} | | \vec{q} | \cdot 0 = 0 \\ \Leftrightarrow (\begin{array}{c} - 2 \\ 1 \\ 3 \end{array}) \cdot (\begin{array}{c} 4 \\ - 1 \\ t \end{array}) = 0 \\ \Leftrightarrow (- 2) (4) + (1) (- 1) + (3) (t) = 0 \\ \Leftrightarrow - 8 - 1 + 3 t = 0 \\ \Leftrightarrow 3 t = 9 \\ \Leftrightarrow t = 3 \end{aligned} \end{array}

$\begin{array}{ll} 4. & Jika | \overset{―}{u} | = 6, | \overset{―}{v} | = 4 \sqrt{3}, dan | \overset{―}{u} - \overset{―}{v} | = 8 \\ tentukanlah nilai dari \\ a . \overset{―}{u} ∙ \overset{―}{v} \\ b . | \overset{―}{u} + \overset{―}{v} | \\ c . cosinus sudut antara \overset{―}{u} dan \overset{―}{v} \\ Jawab : \\ \begin{aligned} a . & \overset{―}{u} ∙ \overset{―}{v} = \dots \\ 2. \overset{―}{u} ∙ \overset{―}{v} = {| \overset{―}{u} |}^{2} + {| \overset{―}{v} |}^{2} - {| \overset{―}{u} - \overset{―}{v} |}^{2} \\ 2. \overset{―}{u} ∙ \overset{―}{v} = 6^{2} + (4 \sqrt{3})^{2} - 8^{2} \\ 2. \overset{―}{u} ∙ \overset{―}{v} = 36 + 48 - 64 = 84 - 64 = 20 \\ \overset{―}{u} ∙ \overset{―}{v} = \frac{20}{2} = 10 \\ b . & {| \overset{―}{u} + \overset{―}{v} |}^{2} = {| \overset{―}{u} |}^{2} + {| \overset{―}{v} |}^{2} + 2. \overset{―}{u} ∙ \overset{―}{v} \\ {| \overset{―}{u} + \overset{―}{v} |}^{2} = 6^{2} + (4 \sqrt{3})^{2} + 20 \\ = 84 + 20 = 104 \\ | \overset{―}{u} + \overset{―}{v} | = \sqrt{104} \\ c . & \cos ∠ (\overset{―}{u}, \overset{―}{v}) = \frac{\overset{―}{u} ∙ \overset{―}{v}}{| \overset{―}{u} | . | \overset{―}{v} |} = \frac{10}{6. (4 \sqrt{3})} \times \frac{\sqrt{3}}{\sqrt{3}} \\ = \frac{10 \sqrt{3}}{72} = \frac{5}{36} \sqrt{3} \\ Berikut ilustrasi gambarnya \end{aligned} \end{array}$ .

\begin{array}{ll} 5. & Diketahui | \vec{a} | = \sqrt{3}, | \vec{b} | = 1, dan | \vec{a} - \vec{b} | = 1 \\ maka panjang vektor \vec{a} + \vec{b} adalah . . . . \\ \begin{array}{llllllll} a . & \sqrt{3} & d . & 2 \sqrt{2} \\ b . & \sqrt{5} & c . & \sqrt{7} & e . & 3 \end{array} \\ Jawab \\ \begin{aligned} Diketahui & sebagaimana pada soal \\ {| \vec{a} - \vec{b} |}^{2} & = {| \vec{a} |}^{2} + {| \vec{b} |}^{2} - 2 | \vec{a} | | \vec{b} | \cos θ \\ 1^{2} & = {(\sqrt{3})}^{2} + 1^{2} - 2. \sqrt{3} .1 . \cos θ \\ 2 \sqrt{3} \cos θ & = 3 \\ maka pan & jang vektor \vec{a} + \vec{b} adalah \\ | \vec{a} + \vec{b} | & = \sqrt{{| \vec{a} |}^{2} + {| \vec{b} |}^{2} + 2 | \vec{a} | | \vec{b} | \cos θ} \\ = \sqrt{{(\sqrt{3})}^{2} + 1^{2} + 3} \\ = \sqrt{3 + 1 + 3} \\ = \sqrt{7} \end{aligned} \end{array}

RANGKUMAN

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DAFTAR PUSTAKA

Johanes, Kastolan, Sulasim. 2006. Kompetensi Matematika Program IPA 3A SMA Kelas XII Semester Pertama. Jakarta: YUDHISTIRA.
Miyanto, Aksin, N., Suparno. 2021. Buku Interaktif Matematika untuk SMA/MA Peminatan Matematika dan Ilmu-Ilmu Alam Kelas X Semester 2. Yogyakarta: INTAN PARIWARA.
Yuana, R.A., Indriyastuti. 2017. Persektif Matematika untuk Kelas X SMA dan MA Kelompok Peminatan Matematika dan Ilmu Alam. Solo: PT TIGA SERANGKAI MANDIRI.

PAS MATEMATIKA BLOG

Vektor pada Ruang

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