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Diketahui vektor p​=i+j−4k,danq​=−2i−j. Nilai sinus sudut antara vektor p​dan q​=....

Pertanyaan

Diketahui vektor p with rightwards arrow on top equals i plus j space minus 4 k comma space d a n space q with rightwards arrow on top equals negative 2 i minus j. Nilai sinus sudut antara vektor space p with rightwards arrow on top spacedan q with rightwards arrow on top=....

  1. negative 3 over 10 square root of 10

  2. negative 1 over 10 square root of 10

  3. 1 over 10 square root of 10

  4. 1 third square root of 10

  5. 3 over 10 square root of 10

Pembahasan Soal:

c o s space theta equals fraction numerator p ⃗. q with rightwards arrow on top over denominator vertical line p ⃗ vertical line. vertical line q ⃗ vertical line end fraction equals fraction numerator i plus j minus 4 k right parenthesis. left parenthesis negative 2 i minus j right parenthesis over denominator square root of 1 squared plus 1 squared plus left parenthesis negative 4 right parenthesis squared end root. square root of left parenthesis negative 2 right parenthesis squared plus left parenthesis negative 1 right parenthesis squared plus 0 squared end root end fraction  equals fraction numerator negative 2 minus 1 minus 0 over denominator square root of 18. square root of 5 end fraction equals fraction numerator negative 3 over denominator square root of 90 end fraction equals fraction numerator negative 3 over denominator 3 square root of 10 end fraction equals fraction numerator negative 1 over denominator square root of 10 end fraction equals negative 1 over 10 square root of 10  M a k a comma D e n g a n space i d e n t i t a s space s i n space theta equals  square root of left parenthesis 1 minus cos squared theta end root  equals square root of 1 minus open parentheses negative 1 over 10 square root of 10 right parenthesis squared close parentheses end root  equals 3 over 10 square root of 10

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

Y. Endah

Mahasiswa/Alumni Institut Teknologi Bandung

Terakhir diupdate 04 Oktober 2021

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Pertanyaan yang serupa

Diketahui vektor – vektor u=9i+aj​+bk dan v=ai−bj​+ak. Sudut antara vektor u dan vektor v adalah θ dengan cosθ=116​. Proyeksi  pada  adalah p​=4i−2j​+4k. Nilai b adalah ....

Pembahasan Soal:

Perhatikan bahwa

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell straight u with rightwards arrow on top. straight v with rightwards arrow on top end cell equals cell open parentheses table row 9 row straight a row straight b end table close parentheses open parentheses table row straight a row cell negative straight b end cell row straight a end table close parentheses end cell row cell straight u with rightwards arrow on top. straight v with rightwards arrow on top end cell equals cell 9 straight a minus ab plus ab end cell row cell straight u with rightwards arrow on top. straight v with rightwards arrow on top end cell equals cell 9 straight a horizontal ellipsis horizontal ellipsis horizontal ellipsis open parentheses 1 close parentheses end cell end table end style  

Kemudian

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell blank straight u with rightwards arrow on top. straight v with rightwards arrow on top end cell equals cell open vertical bar straight u with rightwards arrow on top close vertical bar open vertical bar straight v with rightwards arrow on top close vertical bar cos invisible function application straight theta end cell row cell straight u with rightwards arrow on top. straight v with rightwards arrow on top end cell equals cell square root of 9 squared plus straight a squared plus straight b squared end root square root of straight a squared plus open parentheses negative straight b close parentheses squared plus straight a squared end root open parentheses 6 over 11 close parentheses end cell row cell straight u with rightwards arrow on top. straight v with rightwards arrow on top end cell equals cell square root of 9 squared plus straight a squared plus straight b squared end root square root of straight a squared plus straight b squared plus straight a squared end root open parentheses 6 over 11 close parentheses horizontal ellipsis horizontal ellipsis horizontal ellipsis open parentheses 2 close parentheses end cell end table end style 

Proyeksi begin mathsize 14px style straight u end style pada begin mathsize 14px style straight v end style adalah begin mathsize 14px style straight p with rightwards arrow on top equals 4 straight i with rightwards arrow on top minus 2 straight j with rightwards arrow on top plus 4 straight k with rightwards arrow on top end style, berarti

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell fraction numerator straight u with rightwards arrow on top. straight v with rightwards arrow on top over denominator open vertical bar straight v with rightwards arrow on top close vertical bar end fraction end cell equals cell square root of 4 squared plus open parentheses negative 2 close parentheses squared plus 4 squared end root end cell row cell fraction numerator straight u with rightwards arrow on top. straight v with rightwards arrow on top over denominator square root of straight a squared plus straight b squared plus straight a squared end root end fraction end cell equals 6 row cell straight u with rightwards arrow on top. straight v with rightwards arrow on top end cell equals cell 6 open parentheses square root of straight a squared plus straight b squared plus straight a squared end root close parentheses horizontal ellipsis horizontal ellipsis horizontal ellipsis open parentheses 3 close parentheses end cell end table end style 

Subtitusi persamaan (1) ke persamaan (3)

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell 9 straight a end cell equals cell 6 open parentheses square root of straight a squared plus straight b squared plus straight a squared end root close parentheses end cell row cell 3 over 2 straight a end cell equals cell square root of straight a squared plus straight b squared plus straight a squared end root end cell row cell 9 over 4 straight a squared end cell equals cell straight a squared plus straight b squared plus straight a squared end cell row cell 1 fourth straight a squared end cell equals cell straight b squared end cell row cell straight a squared end cell equals cell 4 straight b squared end cell row straight a equals cell 2 straight b horizontal ellipsis horizontal ellipsis horizontal ellipsis open parentheses 4 close parentheses end cell end table end style 

Subitusi persamaan (4) dan (1) ke persamaan (2) :

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell 9 open parentheses 2 straight b close parentheses end cell equals cell square root of 9 squared plus 4 straight b squared plus straight b squared end root square root of 4 straight b squared plus straight b squared plus 4 straight b squared end root open parentheses 6 over 11 close parentheses end cell row cell 18 straight b end cell equals cell square root of 81 plus 5 straight b squared end root open parentheses 3 straight b close parentheses open parentheses 6 over 11 close parentheses end cell row cell fraction numerator 18 straight b over denominator 3 straight b end fraction open parentheses 11 over 6 close parentheses end cell equals cell square root of 81 plus 5 straight b squared end root end cell row 11 equals cell square root of 81 plus 5 straight b squared end root end cell row 121 equals cell 81 plus 5 straight b squared end cell row cell 5 straight b squared end cell equals 40 row cell straight b squared end cell equals 8 row straight b equals cell plus-or-minus 2 square root of 2 end cell end table end style 

Jadi, nilai b adalah begin mathsize 14px style plus-or-minus 2 square root of 2 end style

Jadi, jawaban yang tepat adalah D.space space  

0

Roboguru

Diketahui . Besar sudut antara adanbadalah ....

Pembahasan Soal:

Dari soal tersebut, didapat bahwa begin mathsize 14px style a with rightwards arrow on top equals open parentheses table row 0 row cell negative 1 end cell row 1 end table close parentheses space d a n space b with rightwards arrow on top equals open parentheses table row 2 row cell negative 1 end cell row 2 end table close parentheses end style.

Oleh karena itu, kita dapatkan

begin mathsize 14px style 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 0 times 2 plus open parentheses negative 1 close parentheses times open parentheses negative 1 close parentheses plus 1 times 2 end cell row blank equals cell 0 plus 1 plus 2 end cell row blank equals 3 end table end style

Kemudian, kita dapatkan pula bahwa

begin mathsize 14px style 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 0 squared plus open parentheses negative 1 close parentheses squared plus 1 squared end root end cell row blank equals cell square root of 0 plus 1 plus 1 end root end cell row blank equals cell square root of 2 end cell end table end style

dan

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell open vertical bar b with rightwards arrow on top close vertical bar end cell equals cell square root of 2 squared plus open parentheses negative 1 close parentheses squared plus open parentheses negative 2 close parentheses squared end root end cell row blank equals cell square root of 4 plus 1 plus 4 end root end cell row blank equals cell square root of 9 end cell row blank equals 3 end table end style

Misalkan besar sudut antara begin mathsize 14px style a with rightwards arrow on top space d a n space b with rightwards arrow on top end style adalah α, maka

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell cos space alpha end cell equals cell fraction numerator a with rightwards arrow on top times b with rightwards arrow on top over denominator open vertical bar a with rightwards arrow on top close vertical bar open vertical bar b with rightwards arrow on top close vertical bar end fraction end cell row cell cos space alpha end cell equals cell fraction numerator 3 over denominator square root of 2 times 3 end fraction end cell row cell cos space alpha end cell equals cell fraction numerator 1 over denominator square root of 2 end fraction end cell row cell cos space alpha end cell equals cell fraction numerator 1 over denominator square root of 2 end fraction times fraction numerator square root of 2 over denominator square root of 2 end fraction end cell row cell cos space alpha end cell equals cell 1 half square root of 2 end cell end table end style

Didapat bahwa α = 45°.

Jadi, jawaban yang tepat adalah B.

0

Roboguru

Diketahuivektora=2i+j+3kdanb=−i+2j+2k.Sudutθadalahsudutantaravektoradanb.Nilaisinθ=....

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

Diketahui vektor a dan b membentuk sudut 120∘. Jika  dan , maka hasil dari b⋅(a−b) adalah ....

Pembahasan Soal:

Karena besar sudut yang terbentuk dari vektor undefined dan undefined adalah begin mathsize 14px style 120 degree end style, maka begin mathsize 14px style alpha equals 120 degree end style.

Kemudian, perhatikan hubungan berikut.

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell b with rightwards arrow on top times open parentheses a with rightwards arrow on top minus b with rightwards arrow on top close parentheses end cell equals cell b with rightwards arrow on top times a with rightwards arrow on top minus b with rightwards arrow on top times b with rightwards arrow on top end cell row blank equals cell a with rightwards arrow on top times b with rightwards arrow on top minus b with rightwards arrow on top times b with rightwards arrow on top end cell end table end style

Ingat bahwa begin mathsize 14px style 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 cos alpha end style dan begin mathsize 14px style b with rightwards arrow on top times b with rightwards arrow on top equals open vertical bar b with rightwards arrow on top close vertical bar open vertical bar b with rightwards arrow on top close vertical bar end style.

Oleh karena itu, kita dapatkan perhitungan sebagai berikut.

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell b with rightwards arrow on top times open parentheses a with rightwards arrow on top minus b with rightwards arrow on top close parentheses end cell equals cell a with rightwards arrow on top times b with rightwards arrow on top minus b with rightwards arrow on top times b 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 cos space alpha minus open vertical bar b with rightwards arrow on top close vertical bar open vertical bar b with rightwards arrow on top close vertical bar end cell row blank equals cell 5 times 2 times cos space 120 degree minus 2 times 2 end cell row blank equals cell 5 times 2 times open parentheses negative 1 half close parentheses minus 2 times 2 end cell row blank equals cell negative 5 minus 4 end cell row blank equals cell negative 9 end cell end table end style

Jadi, jawaban yang tepat adalah D.

0

Roboguru

Diketahui vektor a=⎝⎛​34−5​⎠⎞​danb=⎝⎛​1−22​⎠⎞​ . Nilai sinus sudut antara adanb adalah .....

Pembahasan Soal:

Ingat rumus perkalian titik pada vektor :

a with rightwards arrow on top. 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 cos space A  open parentheses table row 3 row 4 row cell negative 5 end cell end table close parentheses. open parentheses table row 1 row cell negative 2 end cell row 2 end table close parentheses equals square root of 3 squared plus 4 squared plus left parenthesis negative 5 right parenthesis squared. end root square root of 1 squared plus left parenthesis negative 2 right parenthesis squared plus left parenthesis 2 right parenthesis squared end root. cos space alpha  3 minus 8 minus 10 equals square root of 9 plus 16 plus 25 end root. square root of 1 plus 4 plus 4 end root. cos space alpha  minus 15 equals square root of 50. square root of 9. cos space alpha  minus 15 equals 5 square root of 2.3 space cos space alpha  cos space alpha equals negative fraction numerator 1 over denominator square root of 2 end fraction    sin space alpha equals square root of 1 minus cos squared alpha end root equals square root of 1 minus open parentheses negative fraction numerator 1 over denominator square root of 2 end fraction close parentheses squared end root equals square root of 1 minus 1 half end root equals 1 half square root of 2    bold Catatan bold thin space bold colon  Karena space sudut space antara space dua space vektor space bernilai space 0 degree less or equal than straight alpha less or equal than 180 degree comma space maka  sin space straight alpha space tidaklah space negatif.

0

Roboguru

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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