Roboguru

Diketahui  dengan koordinat  dan .  Tentukan:

Pertanyaan

Diketahui triangle ABC dengan koordinat straight A open parentheses 2 comma space minus 1 comma space minus 1 close parentheses comma space straight B open parentheses negative 1 comma space 4 comma space minus 2 close parentheses dan straight C open parentheses 5 comma space 0 comma space minus 3 close parentheses
Tentukan: 

  • open vertical bar 2 AB with rightwards arrow on top plus AC with rightwards arrow on top close vertical bar 

Pembahasan Soal:

Diketahui triangle ABC dengan koordinat straight A open parentheses 2 comma space minus 1 comma space minus 1 close parentheses comma space straight B open parentheses negative 1 comma space 4 comma space minus 2 close parentheses dan straight C open parentheses 5 comma space 0 comma space minus 3 close parentheses

Maka

table attributes columnalign right center left columnspacing 0px end attributes row cell AB with rightwards arrow on top end cell equals cell straight B minus straight A end cell row blank equals cell open parentheses negative 1 comma space 4 comma space minus 2 close parentheses minus open parentheses 2 comma space minus 1 comma space minus 1 close parentheses end cell row blank equals cell open parentheses negative 3 comma space 5 comma space minus 1 close parentheses end cell row blank blank blank row cell AC with rightwards arrow on top end cell equals cell straight C minus straight A end cell row blank equals cell open parentheses 5 comma space 0 comma space minus 3 close parentheses minus open parentheses 2 comma space minus 1 comma space minus 1 close parentheses end cell row blank equals cell open parentheses 3 comma space 1 comma space minus 2 close parentheses end cell end table 

Sehingga 

table attributes columnalign right center left columnspacing 0px end attributes row cell 2 AB with rightwards arrow on top plus AC with rightwards arrow on top end cell equals cell 2 open parentheses negative 3 comma space 5 comma space minus 1 close parentheses plus open parentheses 3 comma space 1 comma space minus 2 space close parentheses end cell row blank equals cell open parentheses negative 6 comma space 10 comma space minus 2 close parentheses plus open parentheses 3 comma space 1 comma space minus 2 space close parentheses end cell row blank equals cell open parentheses negative 3 comma space 11 comma space minus 4 close parentheses end cell end table 

Dengan menggunakan rumus panjang vektor, diperoleh 

table attributes columnalign right center left columnspacing 0px end attributes row cell open vertical bar 2 stack A B with rightwards arrow on top plus stack A C with rightwards arrow on top close vertical bar end cell equals cell square root of open parentheses negative 3 close parentheses squared plus open parentheses 11 close parentheses squared plus open parentheses negative 4 close parentheses squared end root end cell row blank equals cell square root of 9 plus 121 plus 16 end root end cell row blank equals cell square root of 146 end cell end table 

Jadi, open vertical bar 2 AB with rightwards arrow on top plus AC with rightwards arrow on top close vertical bar equals square root of 146

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

L. Rante

Mahasiswa/Alumni Universitas Negeri Makassar

Terakhir diupdate 05 Juni 2021

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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Pertanyaan yang serupa

Diberikan titik-titik  dan . Jika vektor-vektor    berturut-turut adalah vektor posisi titik A, B, C. Maka nilai dari

Pembahasan Soal:

Karena vektor-vektor a with rightwards arrow on top comma b with rightwards arrow on top space d a n space c with rightwards arrow on top berturut-turut adalah vektor posisi titik [endif]--> dan [endif]-->, maka

Error converting from MathML to accessible text.

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Diketahui koordinat titik P(2, -3, -8), Q(-1, 4, 3), dan R(5, -1, 6). Jika segiempat PQRS berbentuk jajargenjang, koordinat titik S adalah ....

Pembahasan Soal:

P Q equals S R q minus p equals r minus s open parentheses table row cell negative 1 end cell row 4 row 3 end table close parentheses minus open parentheses table row 2 row cell negative 3 end cell row cell negative 8 end cell end table close parentheses equals open parentheses table row 5 row cell negative 1 end cell row 6 end table close parentheses minus open parentheses table row x row y row z end table close parentheses open parentheses table row cell negative 3 end cell row 7 row 11 end table close parentheses equals open parentheses table row 5 row cell negative 1 end cell row 6 end table close parentheses minus open parentheses table row x row y row z end table close parentheses open parentheses table row x row y row z end table close parentheses equals open parentheses table row 5 row cell negative 1 end cell row 6 end table close parentheses minus open parentheses table row cell negative 3 end cell row 7 row 11 end table close parentheses open parentheses table row x row y row z end table close parentheses equals open parentheses table row 8 row cell negative 8 end cell row cell negative 5 end cell end table close parentheses

Jadi, koordinat titik S (8, -8, -5).

Roboguru

Jika vektor  dan , maka panjang vektor  yang memenuhi persamaan  adalah ....

Pembahasan Soal:

Gunakan konsep penjumlahan, pengurangan vektor dan panjang vektor.

begin mathsize 12px style m with rightwards arrow on top equals m subscript 1 i with rightwards arrow on top plus m subscript 2 j with rightwards arrow on top plus m subscript 2 k with rightwards arrow on top rightwards arrow open vertical bar m with rightwards arrow on top close vertical bar equals square root of open parentheses m subscript 1 close parentheses squared plus open parentheses m subscript 2 close parentheses squared plus open parentheses m subscript 3 close parentheses squared end root end style

Penjumlahan dan pengurangan vektor secara aljabar dengan menjumlahkan atau mengurangi unsur-unsur yang seletak.

Diketahui vektor a with rightwards arrow on top equals 3 i minus 4 j plus k dan b with rightwards arrow on top equals 5 i plus j plus 4 k, dan 2 a with rightwards arrow on top plus c with rightwards arrow on top equals 3 c with rightwards arrow on top plus 3 a with rightwards arrow on top minus b with rightwards arrow on top.

Akan ditentukan panjang vektor c with rightwards arrow on top.

*Terlebih dahulu tentukan vektor c with rightwards arrow on top.

table attributes columnalign right center left columnspacing 2px end attributes row cell 2 a with rightwards arrow on top plus c with rightwards arrow on top end cell equals cell 3 c with rightwards arrow on top plus 3 a with rightwards arrow on top minus b with rightwards arrow on top end cell row cell 2 a with rightwards arrow on top minus 3 a with rightwards arrow on top plus b with rightwards arrow on top end cell equals cell 3 c with rightwards arrow on top minus c with rightwards arrow on top end cell row cell negative a with rightwards arrow on top plus b with rightwards arrow on top end cell equals cell 2 c with rightwards arrow on top end cell row cell 2 c with rightwards arrow on top end cell equals cell negative a with rightwards arrow on top plus b with rightwards arrow on top end cell row cell c with rightwards arrow on top end cell equals cell 1 half open parentheses negative a with rightwards arrow on top plus b with rightwards arrow on top close parentheses end cell row cell c with rightwards arrow on top end cell equals cell 1 half open parentheses negative open parentheses 3 i minus 4 j plus k close parentheses plus open parentheses 5 i plus j plus 4 k close parentheses close parentheses end cell row cell c with rightwards arrow on top end cell equals cell 1 half open parentheses negative 3 i plus 4 j minus k plus 5 i plus j plus 4 k close parentheses end cell row cell c with rightwards arrow on top end cell equals cell 1 half open parentheses open parentheses negative 3 plus 5 close parentheses i plus open parentheses 4 plus 1 close parentheses j plus open parentheses negative 1 plus 4 close parentheses k close parentheses end cell row cell c with rightwards arrow on top end cell equals cell 1 half open parentheses 2 i plus 5 j plus 3 k close parentheses end cell row cell c with rightwards arrow on top end cell equals cell i plus 5 over 2 j plus 3 over 2 k end cell end table

*Menentukan panjang vektor c with rightwards arrow on top.

table attributes columnalign right center left columnspacing 2px end attributes row cell open vertical bar c with rightwards arrow on top close vertical bar end cell equals cell square root of open parentheses 1 close parentheses squared plus open parentheses 5 over 2 close parentheses squared plus open parentheses 3 over 2 close parentheses squared end root end cell row blank equals cell square root of 1 plus 25 over 4 plus 9 over 4 end root end cell row blank equals cell square root of 1 plus 34 over 4 end root end cell row blank equals cell square root of fraction numerator 4 plus 34 over denominator 4 end fraction end root end cell row blank equals cell square root of 38 over 4 end root end cell row blank equals cell 1 half square root of 38 end cell end table

Diperoleh panjang vektor c with rightwards arrow on top adalah 1 half square root of 38.

Oleh karena itu, tidak ada jawaban yang tepat, karena pilihan jawaban yang tepat adalah 1 half square root of 38.

Roboguru

Diketahui vektor  dengan panjang  dan . Jika panjang proyeksi vektor  pada vektor  sama dengan , maka nilai

Pembahasan Soal:

Panjang vektor a with rightwards arrow on top equals x i with rightwards arrow on top plus y j with rightwards arrow on top plus z k with rightwards arrow on top, yaitu open vertical bar a with rightwards arrow on top close vertical bar equals square root of x squared plus y squared plus z squared end root

Jika vektor a with rightwards arrow on top equals a subscript 1 i with rightwards arrow on top plus a subscript 2 j with rightwards arrow on top plus a subscript 3 k with rightwards arrow on top dan b with rightwards arrow on top equals b subscript 1 i with rightwards arrow on top plus b subscript 2 j with rightwards arrow on top plus b subscript 3 k with rightwards arrow on top, maka

a with rightwards arrow on top times b with rightwards arrow on top equals a subscript 1 times b subscript 1 plus a subscript 2 times b subscript 2 plus a subscript 3 times b subscript 3

Panjang proyeksi vektor a with rightwards arrow on top pada vektor b with rightwards arrow on top dapat ditentukan menggunakan rumus berikut.

open vertical bar c with rightwards arrow on top close vertical bar equals fraction numerator a with rightwards arrow on top times b with rightwards arrow on top over denominator open vertical bar b with rightwards arrow on top close vertical bar end fraction

Berdasarkan konsep di atas, dapat ditentukan persamaan-persamaan berikut.

Persamaan 1:

table attributes columnalign right center left columnspacing 0px end attributes row cell open vertical bar c with rightwards arrow on top close vertical bar end cell equals cell fraction numerator a with rightwards arrow on top times b with rightwards arrow on top over denominator open vertical bar b with rightwards arrow on top close vertical bar end fraction end cell row 3 equals cell fraction numerator open parentheses table row 3 row x row y end table close parentheses open parentheses table row 2 row 3 row 6 end table close parentheses over denominator square root of 2 squared plus 3 squared plus 6 squared end root end fraction end cell row 3 equals cell fraction numerator 3 times 2 plus 3 x plus 6 y over denominator square root of 49 end fraction end cell row 3 equals cell fraction numerator 6 plus 3 x plus 6 y over denominator 7 end fraction end cell row cell 6 plus 3 x plus 6 y end cell equals 21 row cell 3 x plus 6 y end cell equals 15 row cell x plus 2 y end cell equals 5 row x equals cell negative 2 y plus 5 end cell end table

Persamaan 2:

table attributes columnalign right center left columnspacing 0px end attributes row cell square root of 3 squared plus x squared plus y squared end root end cell equals cell square root of 14 end cell row cell 9 plus x squared plus y squared end cell equals 14 row cell x squared plus y squared end cell equals 5 end table

Substitusi persamaan 1 ke persamaan 2 diperoleh:

table attributes columnalign right center left columnspacing 0px end attributes row cell x squared plus y squared end cell equals 5 row cell open parentheses negative 2 y plus 5 close parentheses squared plus y squared end cell equals 5 row cell 4 y squared minus 20 y plus 25 plus y squared end cell equals 5 row cell 5 y squared minus 20 y plus 20 end cell equals 0 row cell y squared minus 4 y plus 4 end cell equals 0 row cell open parentheses y minus 2 close parentheses squared end cell equals 0 row y equals 2 end table

Diperoleh y equals 2 sehingga nilai x equals negative 2 y plus 5 equals negative 2 open parentheses 2 close parentheses plus 5 equals 1

Nilai x plus y equals 2 plus 1 equals 3

Oleh karena itu, jawaban yang tepat adalah B.

Roboguru

Posisi titik  dan  dalam waktu  secara berturut-turut ditentukan oleh  dan . Jarak  dan  pada saat  adalah...

Pembahasan Soal:

Pada saat t equals 3 maka

table attributes columnalign right center left columnspacing 0px end attributes row cell straight P open parentheses 4 comma 1 comma negative straight t squared close parentheses end cell equals cell straight P open parentheses 4 comma 1 comma negative 3 squared close parentheses equals straight P open parentheses 4 comma 1 comma negative 9 close parentheses end cell row cell straight R open parentheses straight t squared comma negative straight t comma negative 2 straight t close parentheses end cell equals cell straight R open parentheses 3 squared comma negative 3 comma negative 2 open parentheses 3 close parentheses close parentheses equals straight R open parentheses 9 comma negative 3 comma negative 6 close parentheses end cell end table

Maka

top enclose PR equals open parentheses table row cell 9 minus 4 end cell row cell negative 3 minus 1 end cell row cell negative 6 minus open parentheses negative 9 close parentheses end cell end table close parentheses equals open parentheses table row 5 row cell negative 4 end cell row 3 end table close parentheses

Jarak antara titik straight P space dan space straight R dapat dihitung dengan menentukan panjang vektor top enclose PR.

Ingat!

Untuk sembarang vektor top enclose v equals open parentheses table row x row y row z end table close parentheses panjang vektor top enclose v adalah open vertical bar top enclose v close vertical bar dimana

open vertical bar top enclose v close vertical bar equals square root of x squared plus y squared plus z squared end root

Diketahui

top enclose PR equals open parentheses table row 5 row cell negative 4 end cell row 3 end table close parentheses

Maka

table attributes columnalign right center left columnspacing 0px end attributes row cell open vertical bar top enclose PR close vertical bar end cell equals cell square root of 5 squared plus open parentheses negative 4 close parentheses squared plus 3 squared end root end cell row blank equals cell square root of 25 plus 16 plus 9 end root end cell row blank equals cell square root of 50 end cell row blank equals cell 5 square root of 2 end cell end table

Dengan demikian jarak antara titik straight P space dan space straight R adalah 5 square root of 2.

Oleh karena itu, jawaban yang benar adalah A.

Roboguru

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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