{"id":248,"date":"2023-07-10T12:56:19","date_gmt":"2023-07-10T12:56:19","guid":{"rendered":"https:\/\/mathority.org\/id\/posisi-relatif-dua-garis-dalam-ruang-r3-contoh-soal-yang-diselesaikan\/"},"modified":"2023-07-10T12:56:19","modified_gmt":"2023-07-10T12:56:19","slug":"posisi-relatif-dua-garis-dalam-ruang-r3-contoh-soal-yang-diselesaikan","status":"publish","type":"post","link":"https:\/\/mathority.org\/id\/posisi-relatif-dua-garis-dalam-ruang-r3-contoh-soal-yang-diselesaikan\/","title":{"rendered":"Posisi relatif dua garis dalam ruang"},"content":{"rendered":"<p>Di sini Anda akan menemukan semua posisi relatif dua garis dalam ruang (di R3). Selain itu, dijelaskan cara mencari posisi relatif antara dua garis menggunakan 2 metode yang mungkin: berdasarkan rentang atau dari suatu titik dan vektor setiap garis. Anda bahkan akan dapat melihat contoh dan latihan yang diselesaikan langkah demi langkah. <\/p>\n<div class=\"adsb30\" style=\" margin:12px; text-align:center\">\n<div id=\"ezoic-pub-ad-placeholder-104\"><\/div>\n<\/div>\n<h2 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"%c2%bfcuales-son-las-posiciones-relativas-de-dos-rectas-en-el-espacio\"><\/span> Berapakah kedudukan relatif dua garis dalam ruang? <span class=\"ez-toc-section-end\"><\/span><\/h2>\n<div class=\"adsb30\" style=\" margin:12px; text-align:center\">\n<div id=\"ezoic-pub-ad-placeholder-105\"><\/div>\n<\/div>\n<p> Dalam geometri analitik, ketika bekerja dalam ruang tiga dimensi (dalam R3) ada 4 kemungkinan posisi relatif antara dua garis: dua garis dapat berupa <strong>garis penggabungan<\/strong> , <strong>garis sejajar<\/strong> , <strong>garis potong<\/strong> , atau <strong>garis potong<\/strong> . <\/p>\n<div class=\"wp-block-columns is-layout-flex wp-container-79\">\n<div class=\"wp-block-column is-layout-flow\">\n<p class=\"has-text-align-center has-text-color has-medium-font-size\" style=\"color:#ff6f00\"> <strong>Garis sejajar<\/strong> <\/p>\n<div class=\"wp-block-image\">\n<figure class=\"aligncenter size-large is-resized\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/uploads\/2023\/07\/droites-paralleles-a-langle.webp\" alt=\"kedudukan relatif dua garis sejajar\" class=\"wp-image-1643\" width=\"222\" height=\"200\" srcset=\"\" sizes=\"auto, \"><\/figure>\n<\/div>\n<p class=\"has-text-align-left\"> Dua garis dikatakan sejajar jika arahnya sama tetapi tidak mempunyai titik persekutuan. Selain itu, garis sejajar selalu memiliki jarak yang sama satu sama lain. <\/p>\n<\/div>\n<div class=\"wp-block-column is-layout-flow\">\n<p class=\"has-text-align-center has-text-color has-medium-font-size\" style=\"color:#ff6f00\"> <strong>garis yang bertepatan<\/strong> <\/p>\n<div class=\"wp-block-image\">\n<figure class=\"aligncenter size-large is-resized\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/uploads\/2023\/07\/angle-coincident-lignes.webp\" alt=\"posisi relatif dari dua garis yang berhimpitan\" class=\"wp-image-1646\" width=\"202\" height=\"179\" srcset=\"\" sizes=\"auto, \"><\/figure>\n<\/div>\n<p> Dua garis berhimpitan jika arahnya sama dan terlebih lagi jika semua titiknya bersekutu. <\/p>\n<\/div>\n<\/div>\n<div class=\"adsb30\" style=\" margin:12px; text-align:center\">\n<div id=\"ezoic-pub-ad-placeholder-106\"><\/div>\n<\/div>\n<div class=\"wp-block-columns is-layout-flex wp-container-82\">\n<div class=\"wp-block-column is-layout-flow\">\n<p class=\"has-text-align-center has-text-color has-medium-font-size\" style=\"color:#ff6f00\"> <strong>garis-garis yang berpotongan<\/strong> <\/p>\n<div class=\"wp-block-image\">\n<figure class=\"aligncenter size-large is-resized\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/uploads\/2023\/07\/angles-droits-secants.webp\" alt=\"kedudukan relatif dua garis yang berpotongan\" class=\"wp-image-1644\" width=\"222\" height=\"208\" srcset=\"\" sizes=\"auto, \"><\/figure>\n<\/div>\n<p> Dua garis berpotongan mempunyai arah yang berbeda namun bersentuhan pada satu titik. <\/p>\n<\/div>\n<div class=\"wp-block-column is-layout-flow\">\n<p class=\"has-text-align-center has-text-color has-medium-font-size\" style=\"color:#ff6f00\"> <strong>Garis persimpangan<\/strong> <\/p>\n<div class=\"wp-block-image\">\n<figure class=\"aligncenter size-large is-resized\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/uploads\/2023\/07\/lignes-dintersection-1.webp\" alt=\"\" class=\"wp-image-2692\" width=\"228\" height=\"221\" srcset=\"\" sizes=\"auto, \"><\/figure>\n<\/div>\n<p> Dua garis yang berpotongan mempunyai arah yang berbeda dan tidak berpotongan di titik manapun. Oleh karena itu, dua garis yang bersilangan tidak berada pada bidang yang sama. Misalnya pada representasi grafis di atas garis<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-ae1901659f469e6be883797bfd30f4f8_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"s\" title=\"Rendered by QuickLaTeX.com\" height=\"8\" width=\"8\" style=\"vertical-align: 0px;\"><\/p>\n<p> selalu terdepan<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-c409433a9e2dfcdb83360a974d243f18_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"r\" title=\"Rendered by QuickLaTeX.com\" height=\"8\" width=\"8\" style=\"vertical-align: 0px;\"><\/p>\n<p> , jadi mereka tidak akan pernah saling bersentuhan.<\/p>\n<\/div>\n<\/div>\n<p> Ada 2 cara untuk mengetahui posisi relatif antara dua garis, karena keduanya bergantung pada cara persamaan kedua garis dinyatakan:<\/p>\n<ul>\n<li> Jika garis-garisnya berbentuk vektor, parametrik, atau persamaan kontinu, yang terbaik <strong>adalah menghitung posisi relatif dari suatu titik dan vektor setiap garis<\/strong> (penjelasan metode ini diberikan di bawah).<\/li>\n<li> Sebaliknya, jika suatu garis didefinisikan dalam bentuk persamaan implisit (atau umum), maka akan lebih mudah <strong>untuk mengetahui posisi relatif antara dua garis tersebut dengan menghitung pangkat dua matriks<\/strong> (lihat penjelasan di bawah). <\/li>\n<\/ul>\n<h2 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"determinar-la-posicion-relativa-de-dos-rectas-a-partir-de-un-punto-y-un-vector\"><\/span> Menentukan kedudukan relatif dua garis dari suatu titik dan suatu vektor <span class=\"ez-toc-section-end\"><\/span><\/h2>\n<div class=\"adsb30\" style=\" margin:12px; text-align:center\">\n<div id=\"ezoic-pub-ad-placeholder-109\"><\/div>\n<\/div>\n<p> Anda dapat mengetahui posisi relatif antara dua garis dengan titik dan vektor pada setiap garis. Metode ini cocok digunakan ketika garis didefinisikan dalam bentuk persamaan vektor, persamaan parametrik, atau persamaan kontinu.<\/p>\n<p> Jadi, misalkan vektor arah dan titik mana pun pada kedua garis tersebut adalah:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-1bdc3a31a3a5a8aa3da312bb2badb356_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle r: \\ \\begin{cases} \\vv{\\text{v}} = (\\text{v}}_x, \\text{v}}_y,\\text{v}}_z})\\\\[2ex] P(P_x,P_y,P_z)\\end{cases} \\qquad\\qquad s: \\ \\begin{cases} \\vv{\\text{v}}' = (\\text{v}}_x', \\text{v}}_y',\\text{v}}_z'})\\\\[2ex] P'(P_x',P_y',P_z')\\end{cases}\" title=\"Rendered by QuickLaTeX.com\" height=\"66\" width=\"417\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p> Jadi, untuk mencari posisi relatif dua garis, kita perlu mengikuti prosedur berikut:<\/p>\n<p> <strong><span style=\"color:#ff6f00;\">\u2023<\/span><\/strong> Hal pertama yang perlu kita lakukan adalah melihat apakah vektor kedua garis sebanding atau tidak dan, bergantung pada kasusnya, kita melakukan hal berikut:<\/p>\n<ul style=\"color:#ff6f00; font-weight: bold;list-style-type:disc\">\n<li> <span style=\"color:#000000;font-weight: normal;\">Jika kedua vektornya sebanding, maka garis-garisnya bisa sejajar atau berimpit. Oleh karena itu kita harus memeriksa apakah titik pada satu garis memenuhi persamaan garis lainnya:<\/span>\n<ul style=\"list-style-type:circle\">\n<li> <span style=\"color:#000000;font-weight: normal;\">Jika titik suatu garis memenuhi persamaan garis yang lain, berarti kedua garis tersebut berimpit.<\/span><\/li>\n<li> <span style=\"color:#000000;font-weight: normal;\">Jika tidak, berarti kedua garis tersebut sejajar.<\/span><\/li>\n<\/ul>\n<\/li>\n<li> <span style=\"color:#000000;font-weight: normal;\">Jika kedua vektor tidak sebanding, maka garis tersebut dapat berpotongan atau berpotongan. Dalam hal ini kita harus menyelesaikan determinan 3\u00d73 berikut:<\/span>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-84a440053c71b2d4287cf246ff1d2f4b_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\begin{vmatrix} \\text{v}}_x &amp; \\text{v}}_x' &amp; P_x-P_x' \\\\[1.1ex] \\text{v}}_y &amp; \\text{v}}_y' &amp; P_y-P_y' \\\\[1.1ex]\\text{v}}_z &amp; \\text{v}}_z' &amp; P_z-P_z'  \\end{vmatrix}\" title=\"Rendered by QuickLaTeX.com\" height=\"86\" width=\"138\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<ul style=\"list-style-type:circle\">\n<li> <span style=\"color:#000000;font-weight: normal;\">Jika determinan sebelumnya sama dengan nol, maka kedua garis tersebut berpotongan di suatu titik (berpotongan).<\/span><\/li>\n<li> <span style=\"color:#000000;font-weight: normal;\">Jika determinan sebelumnya berbeda dengan nol, maka kedua garis tersebut berpotongan.<\/span><\/li>\n<\/ul>\n<\/li>\n<\/ul>\n<p> Grafik berikut merangkum seluruh prosedur: <\/p>\n<div class=\"wp-block-image\">\n<figure class=\"aligncenter size-large is-resized\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/uploads\/2023\/07\/positions-relatives-de-deux-lignes.webp\" alt=\"posisi relatif dua garis\" class=\"wp-image-2731\" width=\"827\" height=\"544\" srcset=\"\" sizes=\"auto, \"><\/figure>\n<\/div>\n<h3 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"ejemplo-de-como-determinar-la-posicion-relativa-entre-dos-rectas\"><\/span> Contoh penentuan posisi relatif antara dua garis<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p> Prosedur sebelumnya mungkin tampak sedikit rumit, namun agar Anda dapat melihat bahwa justru sebaliknya, kami akan menyelesaikan masalah sebagai contoh:<\/p>\n<ul>\n<li> Tentukan posisi relatif antara dua garis berikut:<\/li>\n<\/ul>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-16d9c8120b44ee56adb7cd6358b34f44_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"r: \\ (x,y,z)=(2,0,1)+t(4,-1,1)\" title=\"Rendered by QuickLaTeX.com\" height=\"19\" width=\"264\" style=\"vertical-align: -5px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-0d49353d91f4eb7753b4cb7229859f81_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"s: \\ (x,y,z)=(1,-3,1)+t(1,2,0)\" title=\"Rendered by QuickLaTeX.com\" height=\"19\" width=\"264\" style=\"vertical-align: -5px;\"><\/p>\n<\/p>\n<p> Kedua garis tersebut dinyatakan sebagai persamaan vektor, dengan vektor arah setiap garis adalah:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-c5e74a2aac28c26437e62531f2bb7c88_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\vv{r} =(4,-1,1) \\qquad \\qquad \\vv{s}=(1,2,0)\" title=\"Rendered by QuickLaTeX.com\" height=\"19\" width=\"260\" style=\"vertical-align: -5px;\"><\/p>\n<\/p>\n<p> Dan titik yang dilalui setiap garis adalah:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-e9386dff14919cdeaf57a882471de177_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"R(2,0,1)\\qquad \\qquad S(1,-3,1)\" title=\"Rendered by QuickLaTeX.com\" height=\"19\" width=\"222\" style=\"vertical-align: -5px;\"><\/p>\n<\/p>\n<p> Setelah kita mengetahui suatu titik dan vektor arah setiap garis, kita menerapkan metode yang terlihat di atas. Pertama-tama, kita harus memeriksa apakah koordinat vektor-vektornya proporsional:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-fe839d3477334c9358279b3d97bcf4d3_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\cfrac{4}{1} \\neq \\cfrac{-1}{2} \\neq \\cfrac{1}{0} \\quad \\longrightarrow \\quad \\text{No proporcionales}\" title=\"Rendered by QuickLaTeX.com\" height=\"38\" width=\"314\" style=\"vertical-align: -12px;\"><\/p>\n<\/p>\n<p> Karena kedua vektor tersebut tidak sebanding satu sama lain, maka garis-garis tersebut hanya dapat bersinggungan atau bersilangan. Oleh karena itu, sekarang kita perlu menyelesaikan determinan berikut yang dibentuk oleh vektor arah dan sebuah titik pada setiap garis:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-225a68c152f54a250471b7c4c2254b89_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\begin{vmatrix} \\text{v}}_x &amp; \\text{v}}_x' &amp; P_x-P_x' \\\\[1.1ex] \\text{v}}_y &amp; \\text{v}}_y' &amp; P_y-P_y' \\\\[1.1ex]\\text{v}}_z &amp; \\text{v}}_z' &amp; P_z-P_z' \\end{vmatrix}\" title=\"Rendered by QuickLaTeX.com\" height=\"86\" width=\"138\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p> Kami mengganti nilainya ke dalam rumus:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-fac3bc2228451f94261e296aeecb5de6_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\begin{vmatrix} 4 &amp; 1 &amp; 2-1 \\\\[1.1ex] -1 &amp; 2 &amp; 0-(-3) \\\\[1.1ex]1&amp; 0 &amp; 1-1 \\end{vmatrix}\" title=\"Rendered by QuickLaTeX.com\" height=\"86\" width=\"139\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p> Dan kami menghitung determinannya, untuk ini Anda dapat menggunakan metode apa pun (aturan Sarrus, metode komplemen atau kofaktor, dll.):<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-538af230a8105fceefc5a30f41237ea3_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\begin{vmatrix} 4 &amp; 1 &amp; 1 \\\\[1.1ex] -1 &amp; 2 &amp; 3 \\\\[1.1ex]1&amp; 0 &amp; 0 \\end{vmatrix} = 1\" title=\"Rendered by QuickLaTeX.com\" height=\"86\" width=\"116\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p> Jika hasil penentuannya nol, berarti garis tersebut bersilangan (bersentuhan). Namun determinannya berbeda dengan 0 sehingga <strong>garisnya berpotongan<\/strong> . <\/p>\n<h2 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"hallar-la-posicion-relativa-de-dos-rectas-por-rangos\"><\/span> Temukan posisi relatif dua baris demi baris<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p> Cara lain untuk mencari posisi relatif dua baris adalah dengan menghitung pangkat dua matriks beton, seperti yang akan kita lihat selanjutnya. Cara ini sangat berguna bila kedua garis berada dalam bentuk persamaan implisit (atau umum).<\/p>\n<p> Jadi, jika kita mempunyai dua garis yang dinyatakan dengan persamaan implisit (atau umum) dalam ruang tiga dimensi (dalam R3):<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-500405383e97627c17d01023fd9dd198_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle r: \\ \\begin{cases}A_1x+B_1y+C_1z+D_1=0 \\\\[2ex] A_2x+B_2y+C_2z+D_2=0 \\end{cases}\" title=\"Rendered by QuickLaTeX.com\" height=\"65\" width=\"256\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-1c96b6990dae5ce476ee55689cf4f4fb_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle s: \\ \\begin{cases}A_3x+B_3y+C_3z+D_3=0 \\\\[2ex] A_4x+B_4y+C_4z+D_4=0 \\end{cases}\" title=\"Rendered by QuickLaTeX.com\" height=\"65\" width=\"256\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p> Misalkan A adalah matriks yang terdiri dari koefisien dua garis:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-9199790c5f157691d9307604f25fc873_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle A=\\begin{pmatrix}A_1 &amp; B_1 &amp; C_1\\\\[1.1ex]A_2 &amp; B_2 &amp; C_2\\\\[1.1ex]A_3 &amp; B_3 &amp; C_3\\\\[1.1ex]A_4 &amp; B_4 &amp; C_4 \\end{pmatrix}\" title=\"Rendered by QuickLaTeX.com\" height=\"108\" width=\"158\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p> Dan diketahui matriks A&#8217; yang diperluas, yaitu matriks yang dibentuk oleh semua parameter kedua garis:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-f087aea2d9209341c2acf240eab2bc77_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle A'=\\begin{pmatrix}A_1 &amp; B_1 &amp; C_1&amp;D_1\\\\[1.1ex]A_2 &amp; B_2 &amp; C_2&amp;D_2\\\\[1.1ex]A_3 &amp; B_3 &amp; C_3&amp;D_3\\\\[1.1ex]A_4 &amp; B_4 &amp; C_4&amp;D_4 \\end{pmatrix}\" title=\"Rendered by QuickLaTeX.com\" height=\"108\" width=\"201\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p> Kemudian posisi relatif kedua baris tersebut dapat ditentukan oleh range kedua matriks sebelumnya sesuai tabel berikut: <\/p>\n<div class=\"wp-block-image\">\n<figure class=\"aligncenter size-large is-resized\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/uploads\/2023\/07\/positions-relatives-de-deux-lignes-par-plages.webp\" alt=\"posisi relatif dua baris demi baris\" class=\"wp-image-2752\" width=\"494\" height=\"223\" srcset=\"\" sizes=\"auto, \"><\/figure>\n<\/div>\n<p> <strong>Oleh karena itu, untuk mencari posisi relatif antara dua baris kita harus menghitung rangking kedua matriks dan bergantung pada rangking masing-masing matriks, hal tersebut akan tergantung pada satu atau lain kasus.<\/strong><\/p>\n<p> Teorema ini dapat dibuktikan dengan menggunakan teorema Rouch\u00e9-Frobenius (metode yang digunakan untuk menyelesaikan sistem persamaan linear), namun pada halaman ini kami tidak akan melakukan pembuktiannya karena cukup rumit dan tidak menambah banyak. <\/p>\n<h3 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"ejemplo-de-como-hallar-la-posicion-relativa-de-dos-rectas-por-rangos\"><\/span> Contoh cara mencari posisi relatif dua garis berdasarkan rentang<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p> Setelah kita melihat teori tentang posisi relatif antara dua garis demi baris, mari kita lihat bagaimana teori tersebut dipraktikkan melalui sebuah contoh:<\/p>\n<ul>\n<li> Temukan posisi relatif dari dua garis berikut:<\/li>\n<\/ul>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-0d930886e4afd4cd3b14f1bd788c6da5_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle r: \\ \\begin{cases}3x+2y+z+4=0 \\\\[2ex] 4x+2z+2=0 \\end{cases}\" title=\"Rendered by QuickLaTeX.com\" height=\"65\" width=\"198\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-7704e8cf4de26fa0c249eaabdefa4150_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle s: \\ \\begin{cases}3x+4z-1=0 \\\\[2ex] x-5y-2z-2=0 \\end{cases}\" title=\"Rendered by QuickLaTeX.com\" height=\"65\" width=\"197\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p> Kedua garis tersebut berbentuk persamaan umum (atau implisit), sehingga kita akan menggunakan metode rangking untuk mencari posisi relatif antara kedua garis tersebut. Oleh karena itu, kita membuat matriks A dan matriks perluasan A&#8217; dengan koefisien garis:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-3726bfaa82678d1fffdbae281882572a_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle A=\\begin{pmatrix}3 &amp; 2 &amp; 1\\\\[1.1ex]4 &amp; 0 &amp; 2\\\\[1.1ex]3 &amp; 0 &amp; 4\\\\[1.1ex]1 &amp; -5 &amp; -2 \\end{pmatrix} \\qquad \\qquad A'=\\begin{pmatrix}3 &amp; 2 &amp; 1&amp;4\\\\[1.1ex]4 &amp; 0 &amp; 2&amp;2\\\\[1.1ex]3 &amp; 0 &amp; 4&amp;-1\\\\[1.1ex]1 &amp; -5 &amp; -2 &amp;-2  \\end{pmatrix}\" title=\"Rendered by QuickLaTeX.com\" height=\"107\" width=\"422\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p> Setelah kita memiliki kedua matriks, kita perlu menghitung rank masing-masing matriks. Pertama kita hitung rank matriks A berdasarkan determinannya: <\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-18eb85dd3870c1e7c865705200cf6414_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"rg(A)= \\ ?\" title=\"Rendered by QuickLaTeX.com\" height=\"19\" width=\"77\" style=\"vertical-align: -5px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-1e643845acc44a30e16f7628e85955d0_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle \\begin{vmatrix}3 &amp; 2 &amp; 1\\\\[1.1ex]4 &amp; 0 &amp; 2\\\\[1.1ex]3 &amp; 0 &amp; 4 \\end{vmatrix}  = -20 \\bm{\\neq 0}\" title=\"Rendered by QuickLaTeX.com\" height=\"86\" width=\"152\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-466d66279a42e8aeb422d568e1547ef9_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"rg(A)=3\" title=\"Rendered by QuickLaTeX.com\" height=\"19\" width=\"77\" style=\"vertical-align: -5px;\"><\/p>\n<\/p>\n<p> Matriks A memuat determinan submatriks 3\u00d73 bukan nol, sehingga <strong>matriks A mempunyai rangking 3<\/strong> .<\/p>\n<p> Dan sekarang kita menghitung ruang lingkup matriks A&#8217; yang diperluas. Matriks A&#8217; setidak-tidaknya akan selalu berada pada rangking matriks A, yang dalam hal ini bernilai 3, jadi cukup diperiksa apakah matriks tersebut berpangkat 4 atau berpangkat 3. Untuk melakukannya, kita selesaikan determinannya dari matriks 4\u00d7 4 dengan penjumlahan (atau kofaktor): <\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-d97c98cb8c416b0ce39b18628124923c_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"rg(A')= \\ ?\" title=\"Rendered by QuickLaTeX.com\" height=\"19\" width=\"82\" style=\"vertical-align: -5px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-dd21ed058cb7405e6aee811315086225_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle det(A')=\\begin{vmatrix}3 &amp; 2 &amp; 1&amp;4\\\\[1.1ex]4 &amp; 0 &amp; 2&amp;2\\\\[1.1ex]3 &amp; 0 &amp; 4&amp;-1\\\\[1.1ex]1 &amp; -5 &amp; -2 &amp;-2 \\end{vmatrix} =\" title=\"Rendered by QuickLaTeX.com\" height=\"108\" width=\"234\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-0f99f68649b9ca6274e4531a1d172315_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle =3 \\cdot\\begin{vmatrix} 0 &amp; 2&amp;2\\\\[1.1ex] 0 &amp; 4&amp;-1\\\\[1.1ex] -5 &amp; -2 &amp;-2 \\end{vmatrix}-2\\cdot\\begin{vmatrix}4 &amp; 2&amp;2\\\\[1.1ex]3 &amp; 4&amp;-1\\\\[1.1ex]1 &amp; -2 &amp;-2 \\end{vmatrix}+1\\cdot\\begin{vmatrix}4 &amp; 0 &amp;2\\\\[1.1ex]3 &amp; 0 &amp; -1\\\\[1.1ex]1 &amp; -5 &amp; -2 \\end{vmatrix}-4\\cdot \\begin{vmatrix}4 &amp; 0 &amp; 2\\\\[1.1ex]3 &amp; 0 &amp; 4\\\\[1.1ex]1 &amp; -5 &amp; -2 \\end{vmatrix}\" title=\"Rendered by QuickLaTeX.com\" height=\"86\" width=\"553\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-39e083e5164c9719ef34fae3bcc2fe29_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle =3 \\cdot 50-2\\cdot(-50)+1\\cdot(-50)-4\\cdot 50\" title=\"Rendered by QuickLaTeX.com\" height=\"19\" width=\"297\" style=\"vertical-align: -5px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-3de941669508660aa0a9173155a34685_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle =150+100-50-200\" title=\"Rendered by QuickLaTeX.com\" height=\"15\" width=\"182\" style=\"vertical-align: -2px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-df99174f84ad1be8de3746d1ed3e245c_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle \\bm{=0}\" title=\"Rendered by QuickLaTeX.com\" height=\"12\" width=\"28\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-e8c9889f409b449fd7809f4e02394de2_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"rg(A')=3\" title=\"Rendered by QuickLaTeX.com\" height=\"19\" width=\"82\" style=\"vertical-align: -5px;\"><\/p>\n<\/p>\n<p> Penentu seluruh matriks yang diperluas adalah nol, sehingga <strong>matriks A&#8217; juga berpangkat 3<\/strong> .<\/p>\n<p> Jadi matriks A dan matriks A&#8217; mempunyai rangking 3 sehingga <strong>kedua garis tersebut berpotongan<\/strong> . Artinya, hanya ada satu titik potong di antara keduanya.<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-48635c0a23e1d45677b03099c38205e7_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\color{orange} \\boxed{\\color{black} \\quad rg(A) = rg(A')= 3 \\quad \\longrightarrow \\quad \\text{Rectas Secantes} \\quad \\vphantom{\\Bigl(}}\" title=\"Rendered by QuickLaTeX.com\" height=\"43\" width=\"488\" style=\"vertical-align: -17px;\"><\/p>\n<\/p>\n<p> Ingatlah bahwa di atas terdapat tabel yang merangkum semua kemungkinan kasus posisi relatif antara dua garis menurut rentang matriks A dan A&#8217;. <\/p>\n<h2 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"ejercicios-resueltos-de-la-posicion-relativa-entre-dos-rectas-en-el-espacio\"><\/span> Menyelesaikan masalah kedudukan relatif antara dua garis dalam ruang<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<h3 class=\"wp-block-heading\"> Latihan 1<\/h3>\n<p> Temukan posisi relatif antara dua garis berikut: <\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-0ee51f157ce008a022df46fc1930955e_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"r: \\ (x,y,z)=(3,4,0)+t(2,1,-3)\" title=\"Rendered by QuickLaTeX.com\" height=\"19\" width=\"264\" style=\"vertical-align: -5px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-b3980ce534f631ae1f4781492158bcd7_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"s: \\ (x,y,z)=(1,-2,2)+t(0,5,1)\" title=\"Rendered by QuickLaTeX.com\" height=\"19\" width=\"264\" style=\"vertical-align: -5px;\"><\/p>\n<\/p>\n<div class=\"wp-block-otfm-box-spoiler-start otfm-sp__wrapper otfm-sp__box js-otfm-sp-box__closed otfm-sp__E4F0FE\" role=\"button\" tabindex=\"0\" aria-expanded=\"false\" data-otfm-spc=\"#E4F0FE\" style=\"text-align:center\">\n<div class=\"otfm-sp__title\"> <strong>lihat solusi<\/strong><\/div>\n<\/div>\n<p class=\"has-text-align-left\"> Karena kedua garis dinyatakan sebagai persamaan vektor, kita akan mencari posisi relatif antara dua garis dengan metode satu titik dan satu vektor pada setiap garis.<\/p>\n<p class=\"has-text-align-left\"> Vektor arah setiap garis adalah:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-0112cb63d5449bc1868594300ae69609_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\vv{r} =(2,1,-3) \\qquad \\qquad \\vv{s}=(0,5,1)\" title=\"Rendered by QuickLaTeX.com\" height=\"19\" width=\"260\" style=\"vertical-align: -5px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Dan titik yang dimiliki setiap baris adalah:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-62532678cedab3090235b715a54792d8_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"R(3,4,0)\\qquad \\qquad S(1,-2,2)\" title=\"Rendered by QuickLaTeX.com\" height=\"19\" width=\"222\" style=\"vertical-align: -5px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Jadi, untuk menerapkan prosedur ini, pertama-tama perlu diperiksa apakah komponen-komponen vektor arah sebanding:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-c9d7a07d74505c2794d50f3ecb878ed7_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\cfrac{2}{0} \\neq \\cfrac{1}{5} \\neq \\cfrac{-3}{1} \\quad \\longrightarrow \\quad \\text{No proporcionales}\" title=\"Rendered by QuickLaTeX.com\" height=\"38\" width=\"314\" style=\"vertical-align: -12px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Karena kedua vektor tidak sebanding satu sama lain, maka garis hanya dapat berpotongan atau berpotongan. Oleh karena itu, sekarang kita perlu menyelesaikan determinan berikut yang terdiri dari vektor arah dan sebuah titik pada setiap garis:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-225a68c152f54a250471b7c4c2254b89_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\begin{vmatrix} \\text{v}}_x &amp; \\text{v}}_x' &amp; P_x-P_x' \\\\[1.1ex] \\text{v}}_y &amp; \\text{v}}_y' &amp; P_y-P_y' \\\\[1.1ex]\\text{v}}_z &amp; \\text{v}}_z' &amp; P_z-P_z' \\end{vmatrix}\" title=\"Rendered by QuickLaTeX.com\" height=\"86\" width=\"138\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Kami mengganti nilainya ke dalam rumus:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-b5b88268ae0a4248a6289d0f789250a9_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\begin{vmatrix} 2 &amp; 0 &amp; 3-1 \\\\[1.1ex] 1 &amp; 5 &amp; 4-(-2) \\\\[1.1ex]-3&amp; 1 &amp; 0-2 \\end{vmatrix}\" title=\"Rendered by QuickLaTeX.com\" height=\"86\" width=\"139\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Dan kami menghitung determinannya:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-99043262ed64a3fce660dc55e943a93a_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\begin{vmatrix} 2 &amp; 0 &amp; 2 \\\\[1.1ex] 1 &amp; 5 &amp; 6 \\\\[1.1ex]-3&amp; 1 &amp; -2 \\end{vmatrix}= 0\" title=\"Rendered by QuickLaTeX.com\" height=\"86\" width=\"125\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Hasil determinannya setara dengan 0, sehingga <strong>garis-garisnya berpotongan<\/strong> .<\/p>\n<div class=\"wp-block-otfm-box-spoiler-end otfm-sp_end\"><\/div>\n<h3 class=\"wp-block-heading\"> Latihan 2<\/h3>\n<p> Hitung posisi relatif dua garis berikut: <\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-1848a24fa2901265ca9b0d141c7e2d4b_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle  r: \\ \\begin{cases} x=3+2t \\\\[1.7ex] y=1+3t \\\\[1.7ex] z=2-t \\end{cases} \\qquad \\qquad s: \\ \\cfrac{x+1}{-4}=\\cfrac{y+5}{-6} = \\cfrac{z-4}{2}\" title=\"Rendered by QuickLaTeX.com\" height=\"107\" width=\"421\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<div class=\"wp-block-otfm-box-spoiler-start otfm-sp__wrapper otfm-sp__box js-otfm-sp-box__closed otfm-sp__E4F0FE\" role=\"button\" tabindex=\"0\" aria-expanded=\"false\" data-otfm-spc=\"#E4F0FE\" style=\"text-align:center\">\n<div class=\"otfm-sp__title\"> <strong>lihat solusi<\/strong><\/div>\n<\/div>\n<p class=\"has-text-align-left\"> Baris pertama berbentuk persamaan parametrik dan baris kedua berbentuk persamaan kontinu, yang dengannya kita akan menentukan posisi relatif antara dua garis dari metode vektor satu titik setiap garis.<\/p>\n<p class=\"has-text-align-left\"> Koordinat vektor arah ke kanan<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-c409433a9e2dfcdb83360a974d243f18_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"r\" title=\"Rendered by QuickLaTeX.com\" height=\"8\" width=\"8\" style=\"vertical-align: 0px;\"><\/p>\n<p> adalah koefisien di depan parameter<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-3bdeeaff7b1444be22ec3b07e3219ac5_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"t,\" title=\"Rendered by QuickLaTeX.com\" height=\"16\" width=\"10\" style=\"vertical-align: -4px;\"><\/p>\n<p> dan koordinat vektor arah garis<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-ae1901659f469e6be883797bfd30f4f8_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"s\" title=\"Rendered by QuickLaTeX.com\" height=\"8\" width=\"8\" style=\"vertical-align: 0px;\"><\/p>\n<p> adalah bilangan penyebutnya:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-a5b857b3af4daf70ac0819c2c43dccce_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\vv{r} =(2,3,-1) \\qquad \\qquad \\vv{s}=(-4,-6,2)\" title=\"Rendered by QuickLaTeX.com\" height=\"19\" width=\"288\" style=\"vertical-align: -5px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Dan titik yang dimiliki setiap baris adalah:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-5e174175d9f030d32e11a291f9478360_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"R(3,1,2)\\qquad \\qquad S(-1,-5,4)\" title=\"Rendered by QuickLaTeX.com\" height=\"19\" width=\"235\" style=\"vertical-align: -5px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Jadi, untuk menerapkan prosedur ini, pertama-tama perlu diperiksa apakah komponen-komponen vektor arah sebanding:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-53e7c51a55384b233a7d54ad4a710600_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\cfrac{2}{-4} = \\cfrac{3}{-6}= \\cfrac{-1}{2} \\quad \\longrightarrow \\quad \\text{Proporcionales}\" title=\"Rendered by QuickLaTeX.com\" height=\"38\" width=\"316\" style=\"vertical-align: -12px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Kedua vektor tersebut sebanding satu sama lain, sehingga garis-garisnya hanya boleh sejajar atau berhimpitan. Untuk menghilangkan keraguan ini, perlu dilakukan substitusi titik pada garis<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-c409433a9e2dfcdb83360a974d243f18_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"r\" title=\"Rendered by QuickLaTeX.com\" height=\"8\" width=\"8\" style=\"vertical-align: 0px;\"><\/p>\n<p> dalam persamaan garis<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-ae1901659f469e6be883797bfd30f4f8_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"s\" title=\"Rendered by QuickLaTeX.com\" height=\"8\" width=\"8\" style=\"vertical-align: 0px;\"><\/p>\n<p> (atau sebaliknya) untuk melihat apakah memenuhi persamaan tersebut: <\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-539533362856566b59b5d14d058dda23_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"R(3,1,2) \\ \\longrightarrow \\ \\cfrac{x+1}{-4}=\\cfrac{y+5}{-6} = \\cfrac{z-4}{2}\" title=\"Rendered by QuickLaTeX.com\" height=\"39\" width=\"294\" style=\"vertical-align: -12px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-ec6e9d0de4528399fec953e4db21063e_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\cfrac{3+1}{-4}=\\cfrac{1+5}{-6} = \\cfrac{2-4}{2}\" title=\"Rendered by QuickLaTeX.com\" height=\"39\" width=\"170\" style=\"vertical-align: -12px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-4cfe9c6942d02fe58a00c5ee85055405_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\cfrac{4}{-4}=\\cfrac{6}{-6} = \\cfrac{-2}{2}\" title=\"Rendered by QuickLaTeX.com\" height=\"38\" width=\"127\" style=\"vertical-align: -12px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-e236b00fec23266ff19f015885c2dbd6_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"-1=-1 = -1\" title=\"Rendered by QuickLaTeX.com\" height=\"12\" width=\"114\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Dengan mensubstitusi titik pada garis kita memperoleh persamaan, sehingga titik pada satu garis memenuhi persamaan garis lainnya dan, terlebih lagi, vektor arahnya sebanding. Oleh karena itu, <strong>kedua garis tersebut berimpit.<\/strong><\/p>\n<div class=\"wp-block-otfm-box-spoiler-end otfm-sp_end\"><\/div>\n<h3 class=\"wp-block-heading\"> Latihan 3<\/h3>\n<p> Temukan posisi relatif dari dua garis berikut: <\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-33fbac4dc2933022ff39a0ed9d457200_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle r: \\ \\begin{cases}2x-2y-2z+5=0 \\\\[2ex] 2x-y-1=0 \\end{cases}\" title=\"Rendered by QuickLaTeX.com\" height=\"65\" width=\"206\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-8afae699c3b46dc2153236605e254fdc_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle s: \\ \\begin{cases}4x-y+2z+3=0 \\\\[2ex] x-2y-3z+6=0 \\end{cases}\" title=\"Rendered by QuickLaTeX.com\" height=\"65\" width=\"197\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<div class=\"wp-block-otfm-box-spoiler-start otfm-sp__wrapper otfm-sp__box js-otfm-sp-box__closed otfm-sp__E4F0FE\" role=\"button\" tabindex=\"0\" aria-expanded=\"false\" data-otfm-spc=\"#E4F0FE\" style=\"text-align:center\">\n<div class=\"otfm-sp__title\"> <strong>lihat solusi<\/strong><\/div>\n<\/div>\n<p class=\"has-text-align-left\"> Kedua garis tersebut merupakan bentuk persamaan umum (atau tersirat), jadi kita akan menggunakan metode rank untuk mencari posisi relatif antara kedua garis tersebut. Oleh karena itu kita buat matriks A dan matriks diperluas A&#8217; dengan koefisien garis-garisnya:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-6133960b0b951ec6c455384a1b2274c5_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle A=\\begin{pmatrix}2 &amp; -2 &amp; -2\\\\[1.1ex]2 &amp; -1 &amp; 0\\\\[1.1ex]4 &amp; -1 &amp; 2\\\\[1.1ex]1 &amp; -2 &amp; -3\\end{pmatrix} \\qquad \\qquad A'=\\begin{pmatrix}2 &amp; -2 &amp; -2&amp;5\\\\[1.1ex]2 &amp; -1 &amp; 0&amp;-1\\\\[1.1ex]4 &amp; -1 &amp; 2&amp;3\\\\[1.1ex]1 &amp; -2 &amp; -3&amp;6  \\end{pmatrix}\" title=\"Rendered by QuickLaTeX.com\" height=\"108\" width=\"422\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Setelah kita memiliki kedua matriks, kita perlu menghitung rank masing-masing matriks. Pertama kita hitung rank matriks A berdasarkan determinannya: <\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-18eb85dd3870c1e7c865705200cf6414_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"rg(A)= \\ ?\" title=\"Rendered by QuickLaTeX.com\" height=\"19\" width=\"77\" style=\"vertical-align: -5px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-85353e6e57de74332ffdb3d5cd44caaa_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle \\begin{vmatrix}2 &amp; -2 &amp; -2\\\\[1.1ex]2 &amp; -1 &amp; 0\\\\[1.1ex]4 &amp; -1 &amp; 2 \\end{vmatrix} =0 \\qquad \\begin{vmatrix}2 &amp; -2 &amp; -2\\\\[1.1ex]2 &amp; -1 &amp; 0\\\\[1.1ex]1 &amp; -2 &amp; -3 \\end{vmatrix}=0\" title=\"Rendered by QuickLaTeX.com\" height=\"86\" width=\"298\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-d33b143e636093057c08fbcd9d91ab54_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle \\begin{vmatrix}2 &amp; -2 &amp; -2\\\\[1.1ex]4 &amp; -1 &amp; 2\\\\[1.1ex]1 &amp; -2 &amp; -3 \\end{vmatrix} =0 \\qquad \\begin{vmatrix}2 &amp; -1 &amp; 0\\\\[1.1ex]4 &amp; -1 &amp; 2\\\\[1.1ex]1 &amp; -2 &amp; -3 \\end{vmatrix}=0\" title=\"Rendered by QuickLaTeX.com\" height=\"86\" width=\"298\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-5b4b570b6d9d92b7974b11afbe0939bd_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle \\begin{vmatrix}2 &amp; -2 \\\\[1.1ex]2 &amp; -1 \\end{vmatrix}=2 \\neq 0\" title=\"Rendered by QuickLaTeX.com\" height=\"54\" width=\"123\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-6865ac4ae80c8b5a6eff791a9da5a937_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"rg(A)=2\" title=\"Rendered by QuickLaTeX.com\" height=\"19\" width=\"76\" style=\"vertical-align: -5px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Semua determinan 3&#215;3 pada matriks A adalah nol, tetapi terdapat determinan 2&#215;2 yang bukan nol di dalam matriks tersebut, sehingga <strong>matriks A mempunyai rank 2<\/strong> .<\/p>\n<p class=\"has-text-align-left\"> Dan sekarang kita menghitung ruang lingkup matriks A&#8217; yang diperluas. Matriks A&#8217; paling sedikit selalu merupakan range dari matriks A yang dalam hal ini adalah 2, sehingga perlu dicek apakah matriks tersebut mempunyai determinan 3\u00d73 yang tidak menghilangkan dan juga berapa determinan dari matriks tersebut. seluruh matriks: <\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-d97c98cb8c416b0ce39b18628124923c_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"rg(A')= \\ ?\" title=\"Rendered by QuickLaTeX.com\" height=\"19\" width=\"82\" style=\"vertical-align: -5px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-d253c404b2ac9caf3119795e80acfbfb_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle \\displaystyle \\begin{vmatrix}2 &amp; -2 &amp; 5\\\\[1.1ex]2 &amp; -1 &amp;-1\\\\[1.1ex]4 &amp; -1 &amp;3\\end{vmatrix}=22 \\neq 0\" title=\"Rendered by QuickLaTeX.com\" height=\"86\" width=\"170\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-8744aa6a413d43832344c6eba827c7e6_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle det(A')=\\begin{vmatrix}2 &amp; -2 &amp; -2&amp;5\\\\[1.1ex]2 &amp; -1 &amp; 0&amp;-1\\\\[1.1ex]4 &amp; -1 &amp; 2&amp;3\\\\[1.1ex]1 &amp; -2 &amp; -3&amp;6 \\end{vmatrix}=0\" title=\"Rendered by QuickLaTeX.com\" height=\"108\" width=\"249\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-e8c9889f409b449fd7809f4e02394de2_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"rg(A')=3\" title=\"Rendered by QuickLaTeX.com\" height=\"19\" width=\"82\" style=\"vertical-align: -5px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Matriks A&#8217; yang diperluas memang mengandung subdeterminan bukan nol 3\u00d73 dan terlebih lagi determinan seluruh matriks yang diperluas sama dengan 0, sehingga <strong>matriks A&#8217; mempunyai rangking 3<\/strong> .<\/p>\n<p class=\"has-text-align-left\"> Jadi matriks A mempunyai rangking 2 dan matriks A&#8217; mempunyai rangking 3, maka <strong>kedua garis tersebut sejajar<\/strong> . Artinya, mereka tidak mempunyai kesamaan.<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-042e97ffdc2f223e6a3ba51d9f942b5d_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\color{orange} \\boxed{\\color{black} \\quad rg(A) = 2 \\neq rg(A')= 3 \\quad \\longrightarrow \\quad \\text{Rectas Paralelas} \\quad \\vphantom{\\Bigl(}}\" title=\"Rendered by QuickLaTeX.com\" height=\"43\" width=\"525\" style=\"vertical-align: -17px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Ingatlah bahwa dalam penjelasan metode (di atas) Anda mempunyai tabel yang merangkum semua kemungkinan kasus posisi relatif antara dua garis menurut pangkat matriks A dan A&#8217;.<\/p>\n<div class=\"wp-block-otfm-box-spoiler-end otfm-sp_end\"><\/div>\n<h3 class=\"wp-block-heading\"> Latihan 4<\/h3>\n<p> Temukan posisi relatif dari dua garis berikut: <\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-b39d8ac101e2b4db7f5a17a3f27066b5_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle r: \\ \\begin{cases}4x-y+2z=0 \\\\[2ex] x+y+3z-1=0 \\end{cases}\" title=\"Rendered by QuickLaTeX.com\" height=\"65\" width=\"189\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-ddbbfa691034e2f7acd17d6391e9719e_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle s: \\ \\begin{cases}2x+5y-z-2=0 \\\\[2ex] 2x+3z+1=0 \\end{cases}\" title=\"Rendered by QuickLaTeX.com\" height=\"65\" width=\"197\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<div class=\"wp-block-otfm-box-spoiler-start otfm-sp__wrapper otfm-sp__box js-otfm-sp-box__closed otfm-sp__E4F0FE\" role=\"button\" tabindex=\"0\" aria-expanded=\"false\" data-otfm-spc=\"#E4F0FE\" style=\"text-align:center\">\n<div class=\"otfm-sp__title\"> <strong>lihat solusi<\/strong><\/div>\n<\/div>\n<p class=\"has-text-align-left\"> Dalam hal ini, kedua garis tersebut berbentuk persamaan Kartesius (atau implisit), jadi kita akan menggunakan metode pengurutan untuk mencari posisi relatif antara kedua garis tersebut. Oleh karena itu, kita membuat matriks A dan matriks perluasan A&#8217; dengan koefisien garis:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-cfc6a66630bf6a5cc24f5a006db629af_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle A=\\begin{pmatrix}4 &amp; -1 &amp; 2\\\\[1.1ex]1 &amp; 1 &amp; 3\\\\[1.1ex]2 &amp; 5 &amp; -1\\\\[1.1ex]2 &amp; 0 &amp; 3 \\end{pmatrix} \\qquad \\qquad A'=\\begin{pmatrix}4 &amp; -1 &amp; 2&amp;0\\\\[1.1ex]1 &amp; 1 &amp; 3&amp;-1\\\\[1.1ex]2 &amp; 5 &amp; -1&amp;-2\\\\[1.1ex]2 &amp; 0 &amp; 3 &amp;1  \\end{pmatrix}\" title=\"Rendered by QuickLaTeX.com\" height=\"107\" width=\"422\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Setelah kita mengetahui kedua matriks tersebut, kita perlu menghitung rank masing-masing matriks. Pertama-tama kita akan menghitung rank matriks A berdasarkan determinannya: <\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-18eb85dd3870c1e7c865705200cf6414_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"rg(A)= \\ ?\" title=\"Rendered by QuickLaTeX.com\" height=\"19\" width=\"77\" style=\"vertical-align: -5px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-7a682d947fc58cb847cb2dd60f8772d8_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle \\begin{vmatrix}4 &amp; -1 &amp; 2\\\\[1.1ex]1 &amp; 1 &amp; 3\\\\[1.1ex]2 &amp; 5 &amp; -1 \\end{vmatrix}  = -65 \\bm{\\neq 0}\" title=\"Rendered by QuickLaTeX.com\" height=\"86\" width=\"179\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-466d66279a42e8aeb422d568e1547ef9_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"rg(A)=3\" title=\"Rendered by QuickLaTeX.com\" height=\"19\" width=\"77\" style=\"vertical-align: -5px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Matriks A memuat submatriks berukuran 3\u00d73 yang determinannya bukan nol, sehingga <strong>matriks A mempunyai rangking 3<\/strong> .<\/p>\n<p class=\"has-text-align-left\"> Dan sekarang kita menghitung ruang lingkup matriks A&#8217; yang diperluas. Matriks A&#8217; paling sedikit akan selalu mempunyai rangking dari matriks A, yang dalam hal ini bernilai 3, jadi cukup diperiksa apakah matriks tersebut mempunyai rangking 4 atau rangking 3. Untuk melakukannya, kita selesaikan determinan dari himpunan matriks 4\u00d74 dengan penjumlahan (atau kofaktor): <\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-d97c98cb8c416b0ce39b18628124923c_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"rg(A')= \\ ?\" title=\"Rendered by QuickLaTeX.com\" height=\"19\" width=\"82\" style=\"vertical-align: -5px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-fc223e99519a7cc9359cb6e08591e039_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle det(A')=\\begin{vmatrix}4 &amp; -1 &amp; 2&amp;0\\\\[1.1ex]1 &amp; 1 &amp; 3&amp;-1\\\\[1.1ex]2 &amp; 5 &amp; -1&amp;-2\\\\[1.1ex]2 &amp; 0 &amp; 3 &amp;1 \\end{vmatrix} =\" title=\"Rendered by QuickLaTeX.com\" height=\"108\" width=\"234\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-ed6f0611953cbd025d6e3bed38994fe1_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle =4 \\cdot\\begin{vmatrix} 1 &amp; 3&amp;-1\\\\[1.1ex] 5 &amp; -1&amp;-2\\\\[1.1ex] 0 &amp; 3 &amp;1 \\end{vmatrix}-(-1)\\cdot\\begin{vmatrix}1  &amp; 3&amp;-1\\\\[1.1ex]2 &amp; -1 &amp; -2\\\\[1.1ex]2 &amp; 3 &amp;1 \\end{vmatrix}+2\\cdot\\begin{vmatrix}1 &amp; 1 &amp;-1\\\\[1.1ex]2 &amp; 5 &amp; -2\\\\[1.1ex]2 &amp; 0 &amp;1 \\end{vmatrix}-0\\cdot \\begin{vmatrix}1 &amp; 1 &amp; 3\\\\[1.1ex]2 &amp; 5 &amp; -1\\\\[1.1ex]2 &amp; 0 &amp; 3 \\end{vmatrix}\" title=\"Rendered by QuickLaTeX.com\" height=\"86\" width=\"553\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-a7873f68c97807b541c64943ddddfa4a_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle =4 \\cdot(-25)+1\\cdot(-21)+2\\cdot 9\" title=\"Rendered by QuickLaTeX.com\" height=\"19\" width=\"227\" style=\"vertical-align: -5px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-b1575ce994cf8d65216e53d682fa4403_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle =-100-21+18\" title=\"Rendered by QuickLaTeX.com\" height=\"14\" width=\"138\" style=\"vertical-align: -2px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-6184cc67c626c38b308144a4114a4de4_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle =-103\\bm{\\neq0}\" title=\"Rendered by QuickLaTeX.com\" height=\"17\" width=\"87\" style=\"vertical-align: -4px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-bd9638d9df3f5ccb112e984363d10da6_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"rg(A')=4\" title=\"Rendered by QuickLaTeX.com\" height=\"19\" width=\"82\" style=\"vertical-align: -5px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Penentu seluruh matriks yang diperluas adalah bukan nol, sehingga <strong>matriks A&#8217; mempunyai rangking 4<\/strong> .<\/p>\n<p class=\"has-text-align-left\"> Sehingga matriks A mempunyai rangking 3 dan sebaliknya matriks A&#8217; mempunyai rangking 4, maka <strong>kedua garis tersebut berpotongan<\/strong> di satu titik.<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-449445957cebd657da5bda15f232ee56_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\color{orange} \\boxed{\\color{black} \\quad rg(A) = 3 \\neq rg(A')= 4 \\quad \\longrightarrow \\quad \\text{Rectas se cruzan} \\quad \\vphantom{\\Bigl(}}\" title=\"Rendered by QuickLaTeX.com\" height=\"43\" width=\"527\" style=\"vertical-align: -17px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Ingatlah bahwa dalam penjelasan prosedur (di atas) Anda mempunyai tabel yang berisi semua kemungkinan kasus posisi relatif antara dua garis menurut pangkat matriks A dan A&#8217;.<\/p>\n<div class=\"wp-block-otfm-box-spoiler-end otfm-sp_end\"><\/div>\n","protected":false},"excerpt":{"rendered":"<p>Di sini Anda akan menemukan semua posisi relatif dua garis dalam ruang (di R3). Selain itu, dijelaskan cara mencari posisi relatif antara dua garis menggunakan 2 metode yang mungkin: berdasarkan rentang atau dari suatu titik dan vektor setiap garis. Anda bahkan akan dapat melihat contoh dan latihan yang diselesaikan langkah demi langkah. Berapakah kedudukan relatif &hellip;<\/p>\n<p class=\"read-more\"> <a class=\"\" href=\"https:\/\/mathority.org\/id\/posisi-relatif-dua-garis-dalam-ruang-r3-contoh-soal-yang-diselesaikan\/\"> <span class=\"screen-reader-text\">Posisi relatif dua garis dalam ruang<\/span> Selengkapnya &raquo;<\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"","ast-site-content-layout":"","site-content-style":"default","site-sidebar-style":"default","ast-global-header-display":"","ast-banner-title-visibility":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","astra-migrate-meta-layouts":"","footnotes":""},"categories":[47],"tags":[],"class_list":["post-248","post","type-post","status-publish","format-standard","hentry","category-titik-garis-dan-bidang"],"yoast_head":"<!-- This site is optimized with the Yoast SEO plugin v21.2 - https:\/\/yoast.com\/wordpress\/plugins\/seo\/ -->\n<title>Posisi relatif dua garis dalam ruang - Mathority<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/mathority.org\/id\/posisi-relatif-dua-garis-dalam-ruang-r3-contoh-soal-yang-diselesaikan\/\" \/>\n<meta property=\"og:locale\" content=\"id_ID\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"Posisi relatif dua garis dalam ruang - Mathority\" \/>\n<meta property=\"og:description\" content=\"Di sini Anda akan menemukan semua posisi relatif dua garis dalam ruang (di R3). 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