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Diketahui vektor  dan vektor . Jika  adalah sudut antara , tentukan nilai .

Pertanyaan

Diketahui vektor begin mathsize 14px style straight a with rightwards arrow on top equals 2 straight i with rightwards arrow on top minus 3 straight j with rightwards arrow on top plus 5 straight k with rightwards arrow on top end style dan vektor begin mathsize 14px style straight b with rightwards arrow on top equals negative 3 straight i with rightwards arrow on top minus 5 straight j with rightwards arrow on top plus 2 straight k with rightwards arrow on top end style. Jika begin mathsize 14px style straight theta end style adalah sudut antara begin mathsize 14px style straight a with rightwards arrow on top space dan space straight b with rightwards arrow on top end style, tentukan nilai begin mathsize 14px style tan space straight theta end style.

Pembahasan Soal:

1. Menentukan begin mathsize 14px style open vertical bar straight a with rightwards arrow on top close vertical bar end style

begin mathsize 14px style vertical line straight a with rightwards arrow on top vertical line equals square root of 2 squared plus open parentheses negative 3 close parentheses squared plus 5 squared end root vertical line straight a with rightwards arrow on top vertical line equals square root of 4 plus 9 plus 25 end root vertical line straight a with rightwards arrow on top vertical line equals square root of 38 end style

2. Menentukan begin mathsize 14px style open vertical bar straight b with rightwards arrow on top close vertical bar end style

begin mathsize 14px style vertical line straight b with rightwards arrow on top vertical line equals square root of left parenthesis negative 3 right parenthesis squared plus open parentheses negative 5 close parentheses squared plus 2 squared end root vertical line straight b with rightwards arrow on top vertical line equals square root of 9 plus 25 plus 4 end root vertical line straight b with rightwards arrow on top vertical line equals square root of 38 end style

3. Menentukan begin mathsize 14px style straight a with rightwards arrow on top times straight b with rightwards arrow on top end style

begin mathsize 14px style straight a with rightwards arrow on top times straight b with rightwards arrow on top equals open parentheses table row 2 row cell negative 3 end cell row 5 end table close parentheses times open parentheses table row cell negative 3 end cell row cell negative 5 end cell row 2 end table close parentheses straight a with rightwards arrow on top times straight b with rightwards arrow on top equals negative 6 plus 15 plus 10 straight a with rightwards arrow on top times straight b with rightwards arrow on top equals 19 end style

4. Menentukan tangen sudut antara begin mathsize 14px style straight a with rightwards arrow on top space dan space straight b with rightwards arrow on top end style

begin mathsize 14px style cos space straight alpha equals fraction numerator straight a with rightwards arrow on top times straight b with rightwards arrow on top over denominator vertical line straight a with rightwards arrow on top vertical line vertical line straight b with rightwards arrow on top vertical line end fraction cos space straight alpha equals fraction numerator 19 over denominator vertical line square root of 38 vertical line vertical line square root of 38 vertical line end fraction cos space straight alpha equals 19 over 38 cos space straight alpha equals 1 half space space space space space space space straight alpha equals 45 degree space tan space straight alpha equals 1 end style

Jadi, tangen sudut antara begin mathsize 14px style straight a with rightwards arrow on top space dan space straight b with rightwards arrow on top end style adalah 1.

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

R. Azizatul

Mahasiswa/Alumni Universitas Negeri Yogyakarta

Terakhir diupdate 29 Maret 2021

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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Pertanyaan yang serupa

Diketahui vektor  dan  dengan  dan . Jika   maka besar sudut yang dibentuk vektor  dan

Pembahasan Soal:

Diketahui:

  • open vertical bar a with rightwards arrow on top close vertical bar equals 1
  • open vertical bar b with rightwards arrow on top close vertical bar equals square root of 2
  • a. b equals 1

Ingat!

Perkalian skalar vektor a with rightwards arrow on top dengan vektor b with rightwards arrow on top open parentheses 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. end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell b with rightwards arrow on top end cell end table close parentheses didefinisikan sebagai hasil kali panjang vektor a with rightwards arrow on top dan panjang vektor b with rightwards arrow on top dengan kosinus sudut yang diapit oleh vektor a with rightwards arrow on top dan vektor b with rightwards arrow on top atau 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. end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell b with rightwards arrow on top 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 cell stack open vertical bar a close vertical bar with rightwards arrow on top end cell end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell open vertical bar b with rightwards arrow on top close vertical bar end cell end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank space end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cos end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank space end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank alpha end table.

Sehingga:

table attributes columnalign right center left columnspacing 0px end attributes row cell a with rightwards arrow on top. b with rightwards arrow on top end cell equals cell stack open vertical bar a close vertical bar with rightwards arrow on top open vertical bar b with rightwards arrow on top close vertical bar space cos space alpha end cell row 1 equals cell 1 cross times square root of 2 cross times cos space alpha end cell row 1 equals cell square root of 2 cos space alpha end cell row cell cos space alpha end cell 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 cell cos space alpha end cell equals cell 1 half square root of 2 end cell row alpha equals cell 45 degree end cell end table 


Jadi, jawaban yang tepat adalah B.

0

Roboguru

Diketahui titik , tentukan besar sudut

Pembahasan Soal:

Diketahui:

straight P left parenthesis 2 comma 6 comma 2 right parenthesis straight Q left parenthesis 4 comma 5 comma 2 right parenthesis straight R left parenthesis 3 comma 3 comma 3 right parenthesis 

Ditanya:

angle P Q R

Perhatikan rumus berikut:

cos space angle P Q R equals fraction numerator stack P Q with rightwards arrow on top times stack Q R with rightwards arrow on top over denominator open vertical bar stack R Q with rightwards arrow on top close vertical bar open vertical bar stack R Q with rightwards arrow on top close vertical bar end fraction

Vektor stack P Q with rightwards arrow on top:

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

Vektor stack Q R with rightwards arrow on top:

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

Maka, besar sudut PQR dapat ditentukan seperti berikut:

table attributes columnalign right center left columnspacing 0px end attributes row cell cos space angle PQR end cell equals cell fraction numerator PQ with rightwards arrow on top times QR with rightwards arrow on top over denominator open vertical bar RQ with rightwards arrow on top close vertical bar open vertical bar RQ with rightwards arrow on top close vertical bar end fraction end cell row cell cos space angle PQR end cell equals cell fraction numerator open parentheses table row 2 row cell negative 1 end cell row 0 end table close parentheses times open parentheses table row cell negative 1 end cell row cell negative 2 end cell row 1 end table close parentheses over denominator open parentheses square root of 2 squared plus open parentheses negative 1 close parentheses squared plus 0 squared end root close parentheses open parentheses square root of open parentheses negative 1 close parentheses squared plus open parentheses negative 2 close parentheses squared plus 1 squared end root close parentheses end fraction end cell row cell cos space angle PQR end cell equals cell fraction numerator 2 open parentheses negative 1 close parentheses plus open parentheses negative 1 close parentheses open parentheses negative 2 close parentheses plus 0 open parentheses 1 close parentheses over denominator open parentheses square root of 4 plus 1 end root close parentheses open parentheses square root of 1 plus 4 plus 1 end root close parentheses end fraction end cell row cell cos space angle PQR end cell equals cell fraction numerator negative 2 plus 2 plus 0 over denominator open parentheses square root of 5 close parentheses open parentheses square root of 6 close parentheses end fraction end cell row cell cos space angle PQR end cell equals cell fraction numerator 0 over denominator square root of 30 end fraction end cell row cell cos space angle PQR end cell equals 0 row cell angle PQR end cell equals cell arc space cos space 0 end cell row cell angle PQR end cell equals cell 90 degree comma space 180 degree comma space 270 degree comma space... end cell end table 

Jadi, besar sudut PQR yang mungkin adalah 90 degree atau 180 degree atau 270 degree, atau seterusnya.

0

Roboguru

Jika  dan , maka besar sudut yang dibentuk vektor  dan  sama dengan...

Pembahasan Soal:

Ingat kembali beberapa aturan berikut.

Diketahui vektor top enclose a equals left parenthesis a subscript 1 comma space a subscript 2 comma space a subscript 3 right parenthesis space dan space top enclose b equals left parenthesis b subscript 1 comma space b subscript 2 comma space b subscript 3 right parenthesis, maka

  • a bullet b equals a subscript 1 b subscript 1 plus a subscript 2 b subscript 2 plus a subscript 3 b subscript 3 

 

  • vertical line top enclose a vertical line equals square root of a subscript 1 squared plus a subscript 2 squared plus a subscript 3 squared end root 

 

  • Misalkan alpha adalah sudut yang dibentuk oleh vektor top enclose a dan top enclose b, maka

 cos space alpha equals fraction numerator a bullet b over denominator vertical line top enclose a vertical line vertical line top enclose b vertical line end fraction 

  • Merasionalkan bentuk akar

fraction numerator square root of a over denominator square root of b end fraction times fraction numerator square root of b over denominator square root of b end fraction 

 

  • table attributes columnalign right center left columnspacing 0px end attributes row cell cos space 135 degree end cell equals cell cos space left parenthesis 180 degree minus 45 degree right parenthesis end cell row blank equals cell negative cos space 45 degree end cell row blank equals cell negative 1 half square root of 2 end cell end table 

 

Dari rumus di atas, diperoleh perhitungan sebagai berikut

table attributes columnalign right center left columnspacing 0px end attributes row cell cos space alpha end cell equals cell fraction numerator a bullet b over denominator vertical line top enclose a vertical line vertical line top enclose b vertical line end fraction end cell row blank equals cell fraction numerator left parenthesis negative 1 right parenthesis cross times 1 plus 1 cross times left parenthesis negative 2 right parenthesis plus 0 cross times 2 over denominator square root of left parenthesis negative 1 right parenthesis squared plus 1 squared plus 0 squared space end root space cross times square root of 1 squared plus left parenthesis negative 2 right parenthesis squared plus 2 squared end root end fraction end cell row blank equals cell fraction numerator negative 1 minus 2 plus 0 over denominator square root of 1 plus 1 plus 0 end root cross times square root of 1 plus 4 plus 4 end root end fraction end cell row blank equals cell fraction numerator negative 3 over denominator square root of 2 cross times square root of 9 end fraction end cell row blank equals cell fraction numerator negative 3 over denominator square root of 2 cross times 3 end fraction end cell row blank equals cell fraction numerator negative 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 fraction numerator negative 1 over denominator 2 end fraction square root of 2 end cell end table 

Sehingga diperoleh

table attributes columnalign right center left columnspacing 0px end attributes row alpha equals cell cos to the power of negative 1 end exponent left parenthesis negative 1 half square root of 2 right parenthesis end cell row blank equals cell 135 degree end cell end table

Jadi besar sudut yang dibentuk vektor top enclose a dan top enclose b adalah 135 degree.

Oleh karena itu, jawaban yang benar adalah E.

 

0

Roboguru

Diketahui titik . Sudut antara vektor AB dengan AC adalah ...

Pembahasan Soal:

Akan di tentukan vektor AB 

AB===BA(2,1,1)(1,0,2)(1,1,1)  

Setelah itu akan di tentukan panjang vektor AB 

AB===(12+12+121+1+13  

Akan ditentukan vektor AC 

AC===CA(2,0,3)(1,0,2)(1,0,1)  

Setelah itu akan ditentukan panjang vektor AC 

 AC===(12+02+(1)2)1+0+12 

Menentukan besar sudut antara vektor AB dengan AC 

cosθcosθθ======ABACABAC32(1,1,1)(1,0,1)61+0+(1)60090  

Dengan demikian, sudut antara vektor AB dengan AC adalah 90.

Oleh karena itu, jawaban yang benar adalah D.

0

Roboguru

Diketahui . Hitunglah besar sudut-sudut segitiga ABC!

Pembahasan Soal:

Besar sudut segitiga tersebut dapat ditentukan dengan menggunakan aturan perkalian titik (dot product) vektor untuk menentukan besar sudut antara dua vektor. Jika alpha adalah sudut yang terletak di antara dua vektor a with rightwards arrow on top space dan space b with rightwards arrow on top, maka diperoleh hubungan berikut.

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 times open vertical bar b with rightwards arrow on top close vertical bar end fraction end cell 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:

straight A open parentheses 1 comma space 4 comma space 4 close parentheses comma space straight B open parentheses 0 comma space 2 comma space 3 close parentheses comma space dan space straight C open parentheses 1 comma space 0 comma space 2 close parentheses 

Misalkan: 

table attributes columnalign right center left columnspacing 0px end attributes row alpha equals cell angle BAC equals sudut space antara space AB with rightwards arrow on top space dan space AC with rightwards arrow on top end cell row beta equals cell angle ABC equals sudut space antara space BA with rightwards arrow on top space dan space BC with rightwards arrow on top end cell row theta equals cell angle BCA equals sudut space antara space CB with rightwards arrow on top space dan space CA with rightwards arrow on top end cell end table 

Menentukan besar sudut bold italic alpha 

table attributes columnalign right center left columnspacing 0px end attributes row cell AB with rightwards arrow on top end cell equals cell O B bond OA end cell row blank equals cell open parentheses 0 comma space 2 comma space 3 close parentheses minus sign open parentheses 1 comma space 4 comma space 4 close parentheses end cell row blank equals cell open parentheses 0 minus sign 1 comma space 2 minus sign 4 comma space 3 minus sign 4 close parentheses end cell row blank equals cell open parentheses negative sign 1 comma space minus sign 2 comma space minus sign 1 close parentheses end cell row cell open vertical bar AB with rightwards arrow on top close vertical bar end cell equals cell square root of open parentheses negative sign 1 close parentheses squared plus open parentheses negative sign 2 close parentheses squared plus open parentheses negative sign 1 close parentheses squared end root end cell row blank equals cell square root of 1 plus 4 plus 1 end root end cell row blank equals cell square root of 6 end cell row cell AC with rightwards arrow on top end cell equals cell O C bond OA end cell row blank equals cell open parentheses 1 comma space 0 comma space 2 close parentheses minus sign open parentheses 1 comma space 4 comma space 4 close parentheses end cell row blank equals cell open parentheses 1 minus sign 1 comma space 0 minus sign 4 comma space 2 minus sign 4 close parentheses end cell row blank equals cell open parentheses 0 comma space minus sign 4 comma space minus sign 2 close parentheses end cell row cell open vertical bar AC with rightwards arrow on top close vertical bar end cell equals cell square root of open parentheses 0 close parentheses squared plus open parentheses negative sign 4 close parentheses squared plus open parentheses negative sign 2 close parentheses squared end root end cell row blank equals cell square root of 0 plus 16 plus 4 end root end cell row blank equals cell square root of 20 end cell row blank equals cell 2 square root of 5 end cell row cell cos space italic alpha end cell equals cell fraction numerator AB with rightwards arrow on top middle dot AC with rightwards arrow on top over denominator open vertical bar AB with rightwards arrow on top close vertical bar middle dot open vertical bar AC with rightwards arrow on top close vertical bar end fraction end cell row blank equals cell fraction numerator negative sign 1 middle dot 0 plus open parentheses negative sign 2 close parentheses middle dot open parentheses negative sign 4 close parentheses plus open parentheses negative sign 1 close parentheses middle dot open parentheses negative sign 2 close parentheses over denominator square root of 6 middle dot 2 square root of 5 end fraction end cell row blank equals cell fraction numerator 0 plus 8 plus 2 over denominator 2 square root of 30 end fraction end cell row blank equals cell fraction numerator 10 over denominator 2 square root of 30 end fraction end cell row blank equals cell fraction numerator 5 over denominator square root of 30 end fraction end cell row blank equals cell 1 over 6 square root of 30 end cell row italic alpha equals cell cos to the power of negative sign 1 end exponent space open parentheses fraction numerator square root of 30 over denominator 6 end fraction close parentheses end cell row blank almost equal to cell 24 degree end cell end table 

Menentukan besar sudut bold italic beta

table attributes columnalign right center left columnspacing 0px end attributes row cell BA with rightwards arrow on top end cell equals cell OA minus OB end cell row blank equals cell open parentheses 1 comma space 4 comma space 4 close parentheses minus open parentheses 0 comma space 2 comma space 3 close parentheses end cell row blank equals cell open parentheses 1 minus 0 comma space 4 minus 2 comma space 4 minus 3 close parentheses end cell row blank equals cell open parentheses 1 comma space 2 comma space 1 close parentheses end cell row cell open vertical bar BA with rightwards arrow on top close vertical bar end cell equals cell square root of open parentheses 1 close parentheses squared plus open parentheses 2 close parentheses squared plus open parentheses 1 close parentheses squared end root end cell row blank equals cell square root of 1 plus 4 plus 1 end root end cell row blank equals cell square root of 6 end cell row cell BC with rightwards arrow on top end cell equals cell OC minus OB end cell row blank equals cell open parentheses 1 comma space 0 comma space 2 close parentheses minus open parentheses 0 comma space 2 comma space 3 close parentheses end cell row blank equals cell open parentheses 1 minus 0 comma space 0 minus 2 comma space 2 minus 3 close parentheses end cell row blank equals cell open parentheses 1 comma space minus 2 comma space minus 1 close parentheses end cell row cell open vertical bar BC with rightwards arrow on top close vertical bar end cell equals cell square root of open parentheses 1 close parentheses squared plus open parentheses negative 2 close parentheses squared plus open parentheses negative 1 close parentheses squared end root end cell row blank equals cell square root of 1 plus 4 plus 1 end root end cell row blank equals cell square root of 6 end cell row cell cos space beta end cell equals cell fraction numerator BA with rightwards arrow on top times BC with rightwards arrow on top over denominator open vertical bar BA with rightwards arrow on top close vertical bar times open vertical bar BC with rightwards arrow on top close vertical bar end fraction end cell row blank equals cell fraction numerator 1 times 1 plus 2 times open parentheses negative 2 close parentheses plus 1 times open parentheses negative 1 close parentheses over denominator square root of 6 times square root of 6 end fraction end cell row blank equals cell fraction numerator 1 minus 4 minus 1 over denominator 6 end fraction end cell row blank equals cell fraction numerator negative 4 over denominator 6 end fraction end cell row blank equals cell negative 2 over 3 end cell row beta equals cell cos to the power of negative 1 end exponent space open parentheses negative 2 over 3 close parentheses end cell row blank almost equal to cell 132 degree end cell end table 

Menentukan besar sudut bold italic theta  

Karena jumlah besar sudut dalam segitiga adalah 180 degree maka dapat ditentukan besar sudut theta sebagai berikut.

table attributes columnalign right center left columnspacing 0px end attributes row theta equals cell 180 degree minus left parenthesis alpha plus beta right parenthesis end cell row blank equals cell 180 degree minus open parentheses 24 plus 132 close parentheses degree end cell row blank equals cell 180 degree minus 156 degree end cell row blank equals cell 24 degree end cell end table 

Dengan demikian, diperoleh besar sudut segitiga ABC adalah 24 degree comma space 132 degree comma space dan space 24 degree.

0

Roboguru

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