Roboguru

Hitunglah  jika diketahui , dan sudut antara  dan  adalah  !

Pertanyaan

Hitunglah straight a with rightwards arrow on top times straight b with rightwards arrow on top jika diketahui open vertical bar straight a with rightwards arrow on top close vertical bar equals 3 comma space open vertical bar straight b with rightwards arrow on top close vertical bar equals 4, dan sudut antara straight a with rightwards arrow on top dan straight b with rightwards arrow on top adalah 60 degree !

Pembahasan Soal:

Ingat kembali:

cos space straight theta equals fraction numerator straight a with rightwards arrow on top times straight b with rightwards arrow on top over denominator open vertical bar straight a with rightwards arrow on top close vertical bar times open vertical bar straight b with rightwards arrow on top close vertical bar end fraction space rightwards double arrow straight a with rightwards arrow on top times straight b with rightwards arrow on top equals open vertical bar straight a with rightwards arrow on top close vertical bar times open vertical bar straight b with rightwards arrow on top close vertical bar times cos space straight theta 

Diketahui:

open vertical bar straight a with rightwards arrow on top close vertical bar equals 3 open vertical bar straight b with rightwards arrow on top close vertical bar equals 4 straight theta equals 60 degree 

maka:

straight a with rightwards arrow on top times straight b with rightwards arrow on top equals open vertical bar straight a with rightwards arrow on top close vertical bar times open vertical bar straight b with rightwards arrow on top close vertical bar times cos space straight theta straight a with rightwards arrow on top times straight b with rightwards arrow on top equals 3 times 4 times cos space 60 degree straight a with rightwards arrow on top times straight b with rightwards arrow on top equals 3 times 4 times 1 half straight a with rightwards arrow on top times straight b with rightwards arrow on top equals 6 

Jadi, hasil dari straight a with rightwards arrow on top times straight b with rightwards arrow on top adalah 6.

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

W. Lestari

Mahasiswa/Alumni Universitas Sriwijaya

Terakhir diupdate 06 Juni 2021

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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Pertanyaan yang serupa

Jika vektor  dan vektor  membentuk sudut , panjang vektor  dan panjang vektor , maka  sama dengan ...

Pembahasan Soal:

Perhatikan gambar berikut.



 

Perkalian titik (dot product) antara a with rightwards arrow on top dan b with rightwards arrow on top dapat dirumuskan sebagai berikut:

a with rightwards arrow on top times b with rightwards arrow on top equals open vertical bar a with rightwards arrow on top close vertical bar open vertical bar b with rightwards arrow on top close vertical bar space cos space theta 

dimana

table attributes columnalign right center left columnspacing 0px end attributes row cell open vertical bar a with rightwards arrow on top close vertical bar end cell equals cell panjang space vektor space a with rightwards arrow on top end cell row cell open vertical bar b with rightwards arrow on top close vertical bar end cell equals cell panjang space vektor space b with rightwards arrow on top end cell end table  

Diketahui:

table attributes columnalign right center left columnspacing 0px end attributes row cell open vertical bar a with rightwards arrow on top close vertical bar end cell equals 4 row cell open vertical bar b with rightwards arrow on top close vertical bar end cell equals 6 row theta equals cell 120 degree end cell end table 

Karena a with rightwards arrow on top times a with rightwards arrow on top equals open vertical bar a with rightwards arrow on top close vertical bar squared, maka nilai a with rightwards arrow on top times open parentheses b with rightwards arrow on top plus a with rightwards arrow on top close parentheses adalah:

table attributes columnalign right center left columnspacing 0px end attributes row cell a with rightwards arrow on top times open parentheses b with rightwards arrow on top plus a with rightwards arrow on top close parentheses end cell equals cell a with rightwards arrow on top times b with rightwards arrow on top plus a with rightwards arrow on top times a with rightwards arrow on top end cell row blank equals cell open vertical bar a with rightwards arrow on top close vertical bar open vertical bar b with rightwards arrow on top close vertical bar space cos space theta plus open vertical bar a with rightwards arrow on top close vertical bar squared end cell row blank equals cell 4 times 6 times cos space 120 degree plus 4 squared end cell row blank equals cell 24 times cos space left parenthesis 180 degree minus 120 degree right parenthesis plus 16 end cell row blank equals cell 24 times open parentheses negative cos space 60 degree close parentheses plus 16 end cell row blank equals cell 24 times open parentheses negative 1 half close parentheses plus 16 end cell row blank equals cell negative 12 plus 16 end cell row blank equals 4 end table 

Sehingga, table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell a with rightwards arrow on top end cell end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank times end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell open parentheses b with rightwards arrow on top plus a with rightwards arrow on top close parentheses end cell end table table attributes columnalign right center left columnspacing 0px end attributes row blank equals blank end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank 4 end table.

Jadi, jawaban yang tepat adalah A.

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Diketahui titik ,  dan . Agar vektor posisi dari  tegak lurus pada vektor posisi dari  dan vektor posisi dari , maka  adalah ...

Pembahasan Soal:

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

Jika vektor a with rightwards arrow on top tegak lurus dengan vektor b with rightwards arrow on top, maka 

table attributes columnalign right center left columnspacing 0px end attributes row cell a with rightwards arrow on top times b with rightwards arrow on top end cell equals cell open vertical bar a with rightwards arrow on top close vertical bar open vertical bar b with rightwards arrow on top close vertical bar space cos space 90 degree end cell row cell a with rightwards arrow on top times b with rightwards arrow on top end cell equals 0 end table

Misal vektor posisi dari text C end text open parentheses x comma space y comma space z close parentheses.

Vektor posisi c with rightwards arrow on top tegak lurus a with rightwards arrow on top sehingga diperoleh persamaan (1) berikut.

table attributes columnalign right center left columnspacing 0px end attributes row cell c with rightwards arrow on top times a with rightwards arrow on top end cell equals 0 row cell open parentheses table row x row y row z end table close parentheses open parentheses table row cell negative 1 end cell row 2 row 1 end table close parentheses end cell equals 0 row cell negative x plus 2 y plus z end cell equals 0 end table

Vektor posisi c with rightwards arrow on top tegak lurus b with rightwards arrow on top sehingga diperoleh persamaan (2) berikut.

table attributes columnalign right center left columnspacing 0px end attributes row cell c with rightwards arrow on top times b with rightwards arrow on top end cell equals 0 row cell open parentheses table row x row y row z end table close parentheses open parentheses table row 2 row cell negative 2 end cell row 2 end table close parentheses end cell equals 0 row cell 2 x minus 2 y plus 2 z end cell equals 0 end table

Dari persamaan (1) dan (2) diperoleh persamaan (3) berikut.

table row cell negative x plus 2 y plus z end cell equals cell 0 space space space space end cell row cell 2 x minus 2 y plus 2 z end cell equals cell 0 space plus end cell row cell x plus 3 z end cell equals cell 0 space space space space end cell end table

Diperoleh x equals negative 3 z sehingga dengan metode sbstitusi dapat ditentukan nilai y berikut.

table attributes columnalign right center left columnspacing 0px end attributes row cell negative x plus 2 y plus z end cell equals 0 row cell negative open parentheses negative 3 z close parentheses plus 2 y plus z end cell equals 0 row cell 4 z plus 2 y end cell equals 0 row cell 2 y end cell equals cell negative 4 z end cell row y equals cell negative 2 z end cell end table

Dapat ditentukan perbandingan berikut.

table attributes columnalign right center left columnspacing 0px end attributes row cell x colon y colon z end cell equals cell negative 3 z colon negative 2 z colon z end cell row blank equals cell 3 colon 2 colon negative 1 end cell end table

Diperoleh text C end text open parentheses 3 comma space 2 comma space 1 close parentheses 

Oleh karena itu, tidak ada jawaban yang tepat.

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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

Jika vektor a with rightwards arrow on top dan b with rightwards arrow on top membentuk sudut alpha, maka 

a with rightwards arrow on top times b with rightwards arrow on top equals open vertical bar a with rightwards arrow on top close vertical bar open vertical bar b with rightwards arrow on top close vertical bar space cos space alpha

Komponen dari vektor stack A B with rightwards arrow on top dan stack A C with rightwards arrow on top adalah sebagai berikut.

table attributes columnalign right center left columnspacing 0px end attributes row cell stack A B with rightwards arrow on top end cell equals cell text B end text minus text A end text end cell row blank equals cell open parentheses table row 7 row 6 row 5 end table close parentheses minus open parentheses table row 3 row 1 row cell negative 2 end cell end table close parentheses end cell row blank equals cell open parentheses table row 4 row 5 row 7 end table close parentheses end cell end table

table attributes columnalign right center left columnspacing 0px end attributes row cell stack A C with rightwards arrow on top end cell equals cell text C-A end text end cell row blank equals cell open parentheses table row 1 row 6 row 2 end table close parentheses minus open parentheses table row 3 row 1 row cell negative 2 end cell end table close parentheses end cell row blank equals cell open parentheses table row cell negative 2 end cell row 5 row 4 end table close parentheses end cell end table

Berdasarkan konsep di atas, dapat ditentukan besar sudut text BAC end text sebagai berikut.

Misal: alpha equals angle text BAC end text

table attributes columnalign right center left columnspacing 0px end attributes row cell stack A B with rightwards arrow on top times stack A C with rightwards arrow on top end cell equals cell open vertical bar stack A B with rightwards arrow on top close vertical bar open vertical bar stack A C with rightwards arrow on top close vertical bar space cos space alpha end cell row cell cos space alpha end cell equals cell fraction numerator stack A B with rightwards arrow on top times stack A C with rightwards arrow on top over denominator open vertical bar stack A B with rightwards arrow on top close vertical bar times open vertical bar stack A C with rightwards arrow on top close vertical bar end fraction end cell row blank equals cell fraction numerator open parentheses table row 4 row 5 row 7 end table close parentheses open parentheses table row cell negative 2 end cell row 5 row 4 end table close parentheses over denominator square root of 4 squared plus 5 squared plus 7 squared end root times square root of open parentheses negative 2 close parentheses squared plus 5 squared plus 4 squared end root end fraction end cell row blank equals cell fraction numerator 4 times open parentheses negative 2 close parentheses plus 5 times 5 plus 7 times 4 over denominator square root of 16 plus 25 plus 49 end root times square root of 4 plus 25 plus 16 end root end fraction end cell row blank equals cell fraction numerator negative 8 plus 25 plus 28 over denominator square root of 90 times square root of 45 end fraction end cell row blank equals cell fraction numerator 45 over denominator 3 square root of 10 times 3 square root of 5 end fraction end cell row blank equals cell fraction numerator 45 over denominator 45 square root of 2 end fraction end cell row blank equals cell fraction numerator 1 over denominator square root of 2 end fraction cross times fraction numerator square root of 2 over denominator square root of 2 end fraction end cell row blank equals cell 1 half square root of 2 end cell end table

Diperoleh nilai cos space alpha equals 1 half square root of 2 sehingga alpha equals 45 degree 

Oleh karena itu, jawaban yang tepat adalah B.

0

Roboguru

Jika  dan  dan , maka konstanta positif  adalah ...

Pembahasan Soal:

Perkalian titik dari vektor a with rightwards arrow on top dan b with rightwards arrow on top didefinisikan

a with rightwards arrow on top times b with rightwards arrow on top equals open vertical bar a with rightwards arrow on top close vertical bar times open vertical bar b with rightwards arrow on top close vertical bar times cos space theta

Jika a with rightwards arrow on top equals open parentheses 2 comma space k close parentheses, maka open vertical bar a with rightwards arrow on top close vertical bar equals square root of 4 plus k squared end root

Jika b with rightwards arrow on top equals open parentheses 3 comma space 5 close parentheses, maka open vertical bar b with rightwards arrow on top close vertical bar equals square root of 3 squared plus 5 squared end root equals square root of 9 plus 25 end root equals square root of 34

Nilai a with rightwards arrow on top times b with rightwards arrow on top dapat ditentukan sebagai berikut.

table attributes columnalign right center left columnspacing 0px end attributes row cell a with rightwards arrow on top times b with rightwards arrow on top end cell equals cell open parentheses table row 2 row k end table close parentheses open parentheses table row 3 row 5 end table close parentheses end cell row blank equals cell 6 plus 5 k end cell end table

Nilai konstanta positif k adalah sebagai berikut.

table attributes columnalign right center left columnspacing 0px end attributes row cell a with rightwards arrow on top times b with rightwards arrow on top end cell equals cell open vertical bar a with rightwards arrow on top close vertical bar times open vertical bar b with rightwards arrow on top close vertical bar times cos space theta end cell row cell 6 plus 5 k end cell equals cell open parentheses square root of 4 plus k squared end root close parentheses open parentheses square root of 34 close parentheses times cos space straight pi over 4 end cell row cell 6 plus 5 k end cell equals cell open parentheses square root of 34 times open parentheses 4 plus k squared close parentheses end root close parentheses times 1 half square root of 2 end cell row cell 6 plus 5 k end cell equals cell open parentheses square root of 136 plus 34 k squared end root close parentheses times 1 half square root of 2 end cell row cell 12 plus 10 k end cell equals cell open parentheses square root of 136 plus 34 k squared end root close parentheses open parentheses square root of 2 close parentheses end cell row cell 12 plus 10 k end cell equals cell square root of 272 plus 68 k squared end root end cell row cell open parentheses 12 plus 10 k close parentheses squared end cell equals cell open parentheses square root of 272 plus 68 k squared end root close parentheses squared end cell row cell 144 plus 240 k plus 100 k squared end cell equals cell 272 plus 68 k squared end cell row cell 32 k squared plus 240 k minus 128 end cell equals 0 row cell 2 k squared plus 15 k minus 8 end cell equals 0 row cell open parentheses 2 k minus 1 close parentheses open parentheses k plus 8 close parentheses end cell equals 0 end table

k equals 1 half space text atau end text space k equals negative 8

Konstanta positif k adalah 1 half

Oleh karena itu, jawaban yang tepat adalah B.

0

Roboguru

Diketahui vektor  dan . Tentukan sudut antara  dan .

Pembahasan Soal:

Jika vektor begin mathsize 14px style u with rightwards arrow on top equals open parentheses table row cell x subscript 1 end cell row cell y subscript 1 end cell row cell z subscript 1 end cell end table close parentheses end style dan begin mathsize 14px style v with rightwards arrow on top equals open parentheses table row cell x subscript 2 end cell row cell y subscript 2 end cell row cell z subscript 2 end cell end table close parentheses end style, maka hasil kali titik (dot product) dari dua vektor adalah

begin mathsize 14px style u with rightwards arrow on top times v with rightwards arrow on top equals x subscript 1 x subscript 2 plus y subscript 1 y subscript 2 plus z subscript 1 z subscript 2 end style

dan

begin mathsize 14px style u with rightwards arrow on top times v with rightwards arrow on top equals open vertical bar u with rightwards arrow on top close vertical bar open vertical bar v with rightwards arrow on top close vertical bar cos theta end style

dengan begin mathsize 14px style theta end style adalah sudut di antara kedua vektor.

Perhatikan bahwa

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell open vertical bar u with rightwards arrow on top close vertical bar end cell equals cell square root of x subscript 1 squared plus y subscript 1 squared plus z subscript 1 squared end root end cell row blank equals cell square root of 2 squared plus open parentheses negative 1 close parentheses squared plus 1 squared end root end cell row blank equals cell square root of 4 plus 1 plus 1 end root end cell row blank equals cell square root of 6 end cell end table end style

dan

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell vertical line v with rightwards arrow on top vertical line end cell equals cell square root of x subscript 2 squared plus y subscript 2 squared plus z subscript 2 squared end root end cell row blank equals cell square root of 7 squared plus 0 squared plus 1 squared end root end cell row blank equals cell square root of 49 plus 0 plus 1 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 end style

Akibatnya

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell u with rightwards arrow on top times v with rightwards arrow on top end cell equals cell vertical line u with rightwards arrow on top vertical line vertical line v with rightwards arrow on top vertical line space cos space theta end cell row cell open parentheses table row 2 row cell negative 1 end cell row 1 end table close parentheses times open parentheses table row 7 row 0 row 1 end table close parentheses end cell equals cell open parentheses square root of 6 close parentheses open parentheses 5 square root of 2 close parentheses space cos space theta end cell row cell 14 plus 0 plus 1 end cell equals cell 5 square root of 12 space cos space theta end cell row 15 equals cell 10 square root of 3 space cos space theta end cell row cell fraction numerator 15 over denominator 10 square root of 3 end fraction end cell equals cell cos space theta end cell row cell fraction numerator 15 over denominator 10 square root of 3 end fraction times fraction numerator square root of 3 over denominator square root of 3 end fraction end cell equals cell cos space theta end cell row cell 15 over 30 square root of 3 end cell equals cell cos space theta end cell row cell 1 half square root of 3 end cell equals cell cos space theta end cell row cell 30 degree end cell equals theta end table end style

Jadi, besar sudut antara begin mathsize 14px style u with rightwards arrow on top end style dan begin mathsize 14px style v with rightwards arrow on top end style adalah begin mathsize 14px style 30 degree end style.

0

Roboguru

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

Tapi Roboguru masih mau belajar. Menurut kamu pembahasan kali ini sudah membantu, belum?

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