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Jika vektor  dan vektor ,tangen sudut antara vektor  dan  adalah ....

Pertanyaan

Jika vektor a with rightwards arrow on top equals square root of 3 i minus 2 j plus square root of 5 k dan vektor b with rightwards arrow on top equals negative square root of 3 i minus 2 j plus square root of 5 k,tangen sudut antara vektor a with rightwards arrow on top dan b with rightwards arrow on top adalah ....

  1. 0

  2. 1 third

  3. 1 half

  4. 1 half square root of 3

  5. 1 third square root of 3

Pembahasan Soal:

Jika vektor a with rightwards arrow on top equals square root of 3 i minus 2 j plus square root of 5 k dan vektor b with rightwards arrow on top equals negative square root of 3 i minus 2 j plus square root of 5 k, 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 square root of 3 times open parentheses negative square root of 3 close parentheses minus 2 open parentheses negative 2 close parentheses plus square root of 5 times square root of 5 end cell row blank equals cell negative 3 plus 4 plus 5 end cell row blank equals 6 end table

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 square root of open parentheses square root of 3 close parentheses squared plus open parentheses negative sign 2 close parentheses squared plus open parentheses square root of 5 close parentheses squared end root end cell row blank equals cell square root of 3 plus 4 plus 5 end root end cell row blank equals cell square root of 12 end cell end table

table attributes columnalign right center left columnspacing 0px end attributes row cell open vertical bar italic b with rightwards arrow on top close vertical bar end cell equals cell square root of open parentheses negative sign square root of 3 close parentheses squared plus open parentheses negative sign 2 close parentheses squared plus open parentheses square root of 5 close parentheses squared end root end cell row blank equals cell square root of 3 plus 4 plus 5 end root end cell row blank equals cell square root of 12 end cell end table

Cos sudut antara vektor a with rightwards arrow on top dan b with rightwards arrow on top adalah

table attributes columnalign right center left columnspacing 0px end attributes row cell cos space italic alpha end cell equals cell fraction numerator 6 over denominator square root of 12 middle dot square root of 12 end fraction end cell row blank equals cell 6 over 12 end cell row blank equals cell 1 half end cell end table

Jadi, diperoleh alpha equals 60 degree.

tangen sudut antara vektor a with rightwards arrow on top dan b with rightwards arrow on top adalah

tan space 60 degree equals square root of 3.

Jadi, tidak ada jawaban yang tepat dipilihan ganda.

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

Y. Umi

Mahasiswa/Alumni Universitas Gadjah Mada

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.

0

Roboguru

Diketahui , dan . Besar sudut antara vektor  dan  adalah ....

Pembahasan Soal:

Diketahui open vertical bar straight a with rightwards arrow on top close vertical bar equals square root of 6 comma space open parentheses straight a with rightwards arrow on top minus straight b with rightwards arrow on top close parentheses times open parentheses straight a with rightwards arrow on top plus straight b with rightwards arrow on top close parentheses equals 0, dan straight a with rightwards arrow on top times open parentheses straight a with rightwards arrow on top minus straight b with rightwards arrow on top close parentheses equals 3. Maka:

- Menentukan panjang straight b with rightwards arrow on top 

space space space space space space space space space space space space space space open parentheses straight a with rightwards arrow on top minus straight b with rightwards arrow on top close parentheses times open parentheses straight a with rightwards arrow on top plus straight b with rightwards arrow on top close parentheses equals 0 straight a with rightwards arrow on top times straight a with rightwards arrow on top plus straight a with rightwards arrow on top times straight b with rightwards arrow on top minus straight a with rightwards arrow on top times straight b with rightwards arrow on top minus straight b with rightwards arrow on top times straight b with rightwards arrow on top equals 0 space space space space space space space space space space space space space space space space space space space space space space open vertical bar straight a with rightwards arrow on top close vertical bar squared minus open vertical bar straight b with rightwards arrow on top close vertical bar squared equals 0 space space space space space space space space space space space space space space space space space space space open parentheses square root of 6 close parentheses squared minus open vertical bar straight b with rightwards arrow on top close vertical bar squared equals 0 space space space space space space space space space space space space space space space space space space space space space space space space space space space 6 minus open vertical bar straight b with rightwards arrow on top close vertical bar squared equals 0 space space space space space space space space space space space space space space space space space space space space space space space space space space space space space minus open vertical bar straight b with rightwards arrow on top close vertical bar squared equals negative 6 space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space open vertical bar straight b with rightwards arrow on top close vertical bar squared equals 6 space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space open vertical bar straight b with rightwards arrow on top close vertical bar equals plus-or-minus square root of 6 ,

- Menentukan straight a with rightwards arrow on top times straight b with rightwards arrow on top 

space space space space space space straight a with rightwards arrow on top times open parentheses straight a with rightwards arrow on top minus straight b with rightwards arrow on top close parentheses equals 3 space space space space straight a with rightwards arrow on top times straight a with rightwards arrow on top minus straight a with rightwards arrow on top times straight b with rightwards arrow on top equals 3 space space space space open vertical bar straight a with rightwards arrow on top close vertical bar squared minus straight a with rightwards arrow on top times straight b with rightwards arrow on top equals 3 open parentheses square root of 6 close parentheses squared minus straight a with rightwards arrow on top times straight b with rightwards arrow on top equals 3 space space space space space space space space space space minus straight a with rightwards arrow on top times straight b with rightwards arrow on top equals 3 minus 6 space space space space space space space space space space minus straight a with rightwards arrow on top times straight b with rightwards arrow on top equals negative 3 space space space space space space space space space space space space space straight a with rightwards arrow on top times straight b with rightwards arrow on top equals 3 

Sehingga besar sudut antara vektor straight a with rightwards arrow on top dan straight b with rightwards arrow on top adalah:

- Untuk open vertical bar straight b with rightwards arrow on top close vertical bar equals square root of 6 dan straight a with rightwards arrow on top times straight b with rightwards arrow on top equals 3 

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 cos space straight theta equals fraction numerator 3 over denominator square root of 6 times square root of 6 end fraction cos space straight theta equals 3 over 6 cos space straight theta equals 1 half cos space straight theta equals cos space 30 degree space space space space space space space straight theta equals 30 degree space space space space space space space straight theta equals 30 degree cross times fraction numerator straight pi over denominator 180 degree end fraction space space space space space space space space straight theta equals straight pi over 6 

- Untuk open vertical bar straight b with rightwards arrow on top close vertical bar equals negative square root of 6 dan straight a with rightwards arrow on top times straight b with rightwards arrow on top equals 3 

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 cos space straight theta equals fraction numerator 3 over denominator square root of 6 times negative square root of 6 end fraction cos space straight theta equals fraction numerator 3 over denominator negative 6 end fraction cos space straight theta equals negative 1 half cos space straight theta equals cos space 120 degree space space space space space space space straight theta equals 120 degree space space space space space space space straight theta equals 120 degree cross times fraction numerator straight pi over denominator 180 degree end fraction space space space space space space space space straight theta equals 2 over 3 straight pi 

Jadi, jawaban yang benar adalah A dan E.

0

Roboguru

Besar sudut antara vektor  dan  adalah ...

Pembahasan Soal:

Rumus besar sudut antara dua vektor yaitu:

cos space alpha equals fraction numerator u with rightwards harpoon with barb upwards on top times v with rightwards harpoon with barb upwards on top over denominator open vertical bar u with rightwards harpoon with barb upwards on top close vertical bar open vertical bar v with rightwards harpoon with barb upwards on top close vertical bar end fraction

Sehingga dipeorleh:

table attributes columnalign right center left columnspacing 0px end attributes row cell cos space alpha end cell equals cell fraction numerator u with rightwards harpoon with barb upwards on top times v with rightwards harpoon with barb upwards on top over denominator open vertical bar u with rightwards harpoon with barb upwards on top close vertical bar open vertical bar v with rightwards harpoon with barb upwards on top close vertical bar end fraction end cell row cell cos space alpha end cell equals cell fraction numerator open parentheses 4 cross times 3 close parentheses plus open parentheses negative 5 cross times 2 close parentheses plus open parentheses 2 cross times open parentheses negative 1 close parentheses close parentheses over denominator square root of 4 squared plus open parentheses negative 5 close parentheses squared plus 2 squared end root times square root of 3 squared plus 2 squared plus open parentheses negative 1 close parentheses squared end root end fraction end cell row cell cos space alpha end cell equals cell fraction numerator 12 plus open parentheses negative 10 close parentheses plus open parentheses negative 2 close parentheses over denominator square root of 16 plus 25 plus 4 end root times square root of 9 plus 4 plus 1 end root end fraction end cell row cell cos space alpha end cell equals cell fraction numerator 12 minus 10 minus 2 over denominator square root of 45 times square root of 14 end fraction end cell row cell cos space alpha end cell equals cell fraction numerator 0 over denominator square root of 630 end fraction end cell row cell cos space alpha end cell equals 0 row alpha equals cell arc space cos space 0 end cell row alpha equals cell 90 degree end cell end table

Maka, besar sudut antara vektor u with rightwards harpoon with barb upwards on top equals open parentheses table row 4 row cell negative 5 end cell row 2 end table close parentheses dan v with rightwards harpoon with barb upwards on top equals open parentheses table row 3 row 2 row cell negative 1 end cell end table close parentheses adalah 90 degree.

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

Pembahasan Soal:

I n g a t space r u m u s space p e r k a l i a n space t i t i k space p a d a space v e k t o r space colon space  a space ⃗. b space ⃗ equals vertical line a space ⃗ space vertical line vertical line b space ⃗ space vertical line space space cos invisible function application A    a with rightwards arrow on top equals open parentheses table row 2 row 1 row 3 end table close parentheses space dan space b with rightwards arrow on top equals open parentheses table row cell negative 1 end cell row 2 row 2 end table close parentheses  open parentheses table row 2 row 1 row 3 end table close parentheses open parentheses table row cell negative 1 end cell row 2 row 2 end table close parentheses equals square root of 2 squared plus 1 squared plus 3 squared space end root space. space square root of open parentheses negative 1 close parentheses squared plus open parentheses 2 close parentheses squared plus open parentheses 2 close parentheses squared end root space cos invisible function application theta  minus 2 plus 2 plus 6 equals square root of left parenthesis 4 plus 1 plus 9 right parenthesis end root space space. square root of left parenthesis 1 plus 4 plus 4 right parenthesis end root space space cos invisible function application theta space  6 equals square root of 14 space space.3 space cos invisible function application theta  cos invisible function application theta equals fraction numerator 6 over denominator 3 square root of 14 end fraction equals fraction numerator 2 over denominator square root of 14 end fraction    P e r h a t i k a n space s e g i t i g a

M a k a space sin invisible function application theta equals fraction numerator square root of 10 over denominator square root of 14 end fraction equals 1 over 14 square root of 140 equals 2 over 14 square root of 35 equals fraction numerator square root of 35 over denominator 7 end fraction

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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