Roboguru

Diketahui titik  dan .  wakil vektor  dan  wakil vektor , tentukan nilai cosinus sudut antara .

Pertanyaan

Diketahui titik begin mathsize 14px style straight P open parentheses 5 comma 7 comma negative 5 close parentheses comma space straight Q open parentheses 4 comma 7 comma negative 3 close parentheses end style dan begin mathsize 14px style straight R open parentheses 2 comma 7 comma negative 4 close parentheses end style. begin mathsize 14px style PQ with rightwards arrow on top end style wakil vektor undefined dan size 14px PR with size 14px rightwards arrow on top wakil vektor undefined, tentukan nilai cosinus 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.

Pembahasan Soal:

1. Menentukan vektor undefined

begin mathsize 14px style straight a with rightwards arrow on top equals PQ with rightwards arrow on top straight a with rightwards arrow on top equals open parentheses table row 4 row 7 row cell negative 3 end cell end table close parentheses minus open parentheses table row 5 row 7 row cell negative 5 end cell end table close parentheses straight a with rightwards arrow on top equals open parentheses table row cell negative 1 end cell row 0 row 2 end table close parentheses end style

2. Menentukan vektor undefined

begin mathsize 14px style straight b with rightwards arrow on top equals PR with rightwards arrow on top straight b with rightwards arrow on top equals open parentheses table row 2 row 7 row cell negative 4 end cell end table close parentheses minus open parentheses table row 5 row 7 row cell negative 5 end cell end table close parentheses straight b with rightwards arrow on top equals open parentheses table row cell negative 3 end cell row 0 row 1 end table close parentheses 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 cell negative 1 end cell row 0 row 2 end table close parentheses times open parentheses table row cell negative 3 end cell row 0 row 1 end table close parentheses straight a with rightwards arrow on top times straight b with rightwards arrow on top equals 3 plus 0 plus 2 straight a with rightwards arrow on top times straight b with rightwards arrow on top equals 5 end style

4. Menentukan undefined

begin mathsize 14px style open vertical bar straight a with rightwards arrow on top close vertical bar equals square root of open parentheses negative 1 close parentheses squared plus 0 squared plus 2 squared end root open vertical bar straight a with rightwards arrow on top close vertical bar equals square root of 1 plus 0 plus 4 end root open vertical bar straight a with rightwards arrow on top close vertical bar equals square root of 5 end style

5. Menentukan begin mathsize 14px style vertical line straight b with rightwards arrow on top vertical line end style

begin mathsize 14px style open vertical bar straight b with rightwards arrow on top close vertical bar equals square root of open parentheses negative 3 close parentheses squared plus 0 squared plus 1 squared end root open vertical bar straight b with rightwards arrow on top close vertical bar equals square root of 9 plus 0 plus 1 end root open vertical bar straight b with rightwards arrow on top close vertical bar equals square root of 10 end style

6. Menentukan sinus 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 open vertical bar straight a with rightwards arrow on top close vertical bar open vertical bar straight b with rightwards arrow on top close vertical bar end fraction cos space straight alpha equals fraction numerator 5 over denominator open vertical bar square root of 5 close vertical bar open vertical bar square root of 10 close vertical bar end fraction cos space straight alpha equals fraction numerator 5 over denominator square root of 50 end fraction cos space straight alpha equals 1 over 10 square root of 50 cos space straight alpha equals 1 over 10 open parentheses 5 square root of 2 close parentheses cos space straight alpha equals 1 half square root of 2 space space space space space space space straight alpha equals 45 degree space sin space straight alpha equals 1 half square root of 2 end style

Jadi, sinus 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 begin mathsize 14px style 1 half square root of 2 end style.

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 . Nilai sinus sudut antara vektor  dan  pada kuadran II adalah ....

Pembahasan Soal:

Dengan menggunakan sifat dot product, maka :

table attributes columnalign right center left columnspacing 0px end attributes row cell cos space alpha end cell equals cell fraction numerator p times q over denominator open vertical bar p close vertical bar times open vertical bar q close vertical bar end fraction end cell row blank equals cell fraction numerator open parentheses 1 cross times open parentheses negative 2 close parentheses plus 1 cross times left parenthesis negative 1 right parenthesis plus left parenthesis negative 4 right parenthesis cross times 0 close parentheses over denominator square root of 1 squared plus 1 squared plus left parenthesis negative 4 right parenthesis squared end root cross times square root of left parenthesis negative 2 right parenthesis squared plus left parenthesis negative 1 right parenthesis squared end root end fraction end cell row blank equals cell fraction numerator negative 2 minus 1 over denominator square root of 18 cross times square root of 5 end fraction end cell row blank equals cell fraction numerator negative 3 over denominator square root of 90 end fraction end cell row blank equals cell fraction numerator negative 3 over denominator 3 square root of 10 end fraction end cell row cell cos space alpha end cell equals cell fraction numerator negative square root of 10 over denominator 10 end fraction end cell end table 

Sisi pada depan sudut segitiga :

table attributes columnalign right center left columnspacing 0px end attributes row cell square root of 10 squared minus open parentheses negative square root of 10 close parentheses squared end root end cell equals cell square root of 100 minus 10 end root end cell row blank equals cell square root of 90 end cell row blank equals cell 3 square root of 10 end cell end table 

Sinus sudut kedua vektor pada kuadran II :

sin space alpha equals negative fraction numerator 3 square root of 10 over denominator 10 end fraction 

Maka nilai sinus sudut antara vektor p with rightwards harpoon with barb upwards on top dan q with rightwards harpoon with barb upwards on top pada kuadran II adalah negative 3 over 10 square root of 10 

Oleh karena itu, jawaban yang benar adalah A

0

Roboguru

Diketahui vektor  dan , maka besar sudut antara vektor  dan vektor  adalah ....

Pembahasan Soal:

Jika begin mathsize 14px style straight alpha end style adalah sudut antara vektor vektor begin mathsize 14px style a with rightwards harpoon with barb upwards on top end style dan begin mathsize 14px style b with rightwards harpoon with barb upwards on top end style, maka nilai begin mathsize 14px style straight alpha end style dapat ditentukan dengan: begin mathsize 14px style cos space straight alpha equals fraction numerator a with rightwards harpoon with barb upwards on top times b with rightwards harpoon with barb upwards on top over denominator open vertical bar a with rightwards harpoon with barb upwards on top close vertical bar open vertical bar b with rightwards harpoon with barb upwards on top close vertical bar end fraction end style

begin mathsize 14px style cos space straight alpha equals fraction numerator open parentheses table row 1 row 3 row 2 end table close parentheses times open parentheses table row 3 row 2 row cell negative 1 end cell end table close parentheses over denominator square root of 1 squared plus 3 squared plus 2 squared end root square root of 3 squared plus 2 squared plus open parentheses negative 1 close parentheses squared end root end fraction cos space straight alpha equals fraction numerator 1 times 3 plus 3 times 2 plus 2 times open parentheses negative 1 close parentheses over denominator square root of 1 plus 9 plus 4 end root square root of 9 plus 4 plus 1 end root end fraction cos space straight alpha equals fraction numerator 3 plus 6 minus 2 over denominator square root of 14 square root of 14 end fraction cos space straight alpha equals 7 over 14 cos space straight alpha equals 1 half straight alpha equals 60 degree end style 

Jadi, besar sudut antara vektor begin mathsize 14px style straight a with rightwards harpoon with barb upwards on top end style dan vektor begin mathsize 14px style straight b with rightwards harpoon with barb upwards on top end style adalah begin mathsize 14px style 60 degree end style.

 

 

 

 

0

Roboguru

Diketahui titik-titik ,  dan . wakil dari , wakil dari . Kosinus sudut antara vektor dan vektor  adalah ...

Pembahasan Soal:

Misalkan terdapat dua buah titik straight A open parentheses x subscript 1 comma space y subscript 1 comma space z subscript 1 close parentheses dan straight B open parentheses x subscript 2 comma space y subscript 2 comma space z subscript 2 close parentheses. Vektor AB dapat didefinisikan sebagai:

AB with bar on top equals straight B minus straight A equals open parentheses table row cell straight x subscript 2 minus straight x subscript 1 end cell row cell straight y subscript 2 minus straight y subscript 1 end cell row cell straight z subscript 2 minus straight z subscript 1 end cell end table close parentheses

Sedangkan panjang vektor AB with rightwards arrow on top adalah:

open vertical bar AB with rightwards arrow on top close vertical bar equals square root of open parentheses x subscript 2 minus x subscript 1 close parentheses squared plus open parentheses y subscript 2 minus y subscript 1 close parentheses squared plus open parentheses z subscript 2 minus z subscript 1 close parentheses squared end root.

Lalu ingat juga rumus dot product antara dua vektor yaitu:

table attributes columnalign right center left columnspacing 0px end attributes row cell cos space straight theta end cell equals cell 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 open vertical bar straight b with rightwards arrow on top close vertical bar end fraction end cell end table.

Pada soal diketahui straight A left parenthesis 2 comma space 4 comma space 1 right parenthesisstraight B left parenthesis 4 comma space 6 comma space 1 right parenthesis dan straight C left parenthesis 3 comma space 5 comma space 5 right parenthesis. AB with rightwards arrow on top wakil dari straight u with rightwards arrow on top, 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 cell straight u with rightwards arrow on top end cell equals cell open parentheses table row 4 row 6 row 1 end table close parentheses minus open parentheses table row 2 row 4 row 1 end table close parentheses end cell row cell straight u with rightwards arrow on top end cell equals cell open parentheses table row 2 row 2 row 0 end table close parentheses end cell end table

Panjang vektor straight u with rightwards arrow on top adalah:

table attributes columnalign right center left columnspacing 0px end attributes row cell open vertical bar straight u with rightwards arrow on top close vertical bar end cell equals cell square root of 2 squared plus 2 squared plus 0 squared end root end cell row blank equals cell square root of 4 plus 4 plus 0 end root end cell row blank equals cell square root of 8 end cell row blank equals cell 2 square root of 2 end cell end table

Sedangkan jika AC with rightwards arrow on top wakil dari straight v with rightwards arrow on top, maka:

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

Panjang vektor straight v with rightwards arrow on top adalah:

table attributes columnalign right center left columnspacing 0px end attributes row cell open vertical bar straight v with rightwards arrow on top close vertical bar end cell equals cell square root of 1 squared plus 1 squared plus 4 squared end root end cell row blank equals cell square root of 1 plus 1 plus 16 end root end cell row blank equals cell square root of 18 end cell row blank equals cell 3 square root of 2 end cell end table

Selanjutnya, kosinus sudut antara vektor straight u with rightwards arrow on top dan vektor straight v with rightwards arrow on top dapat dicari sebagai berikut.

table attributes columnalign right center left columnspacing 0px end attributes row cell cos space straight theta end cell equals cell fraction numerator straight u with rightwards arrow on top times straight v with rightwards arrow on top over denominator 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 end fraction end cell row blank equals cell fraction numerator open parentheses 2 comma space 2 comma space 0 close parentheses times open parentheses 1 comma space 1 comma space 4 close parentheses over denominator 2 square root of 2 open parentheses 3 square root of 2 close parentheses end fraction end cell row blank equals cell fraction numerator 2 open parentheses 1 close parentheses plus 2 open parentheses 1 close parentheses plus 0 open parentheses 4 close parentheses over denominator 6 open parentheses 2 close parentheses end fraction end cell row blank equals cell 4 over 12 end cell row blank equals cell 1 third end cell end table

Dengan demikian, jawaban yang benar adalah B.

0

Roboguru

Diketahui  dan . Hitunglah besar sudut  antara vektor  dan .

Pembahasan Soal:

Ingat kembali:

begin mathsize 14px style cos space straight theta equals fraction numerator straight a with rightwards arrow on top times straight b with rightwards arrow on top over denominator stack open vertical bar straight a close vertical bar with rightwards arrow on top times open vertical bar straight b with rightwards arrow on top close vertical bar end fraction end style 

Diketahui:

begin mathsize 14px style straight a with rightwards arrow on top equals open parentheses table row 0 row cell negative 4 end cell row cell negative 4 end cell end table close parentheses end style dan begin mathsize 14px style b with rightwards arrow on top equals open parentheses table row cell negative 4 end cell row 0 row cell negative 4 end cell end table close parentheses end style.

Maka:

 begin mathsize 14px style cos space straight theta equals fraction numerator straight a with rightwards arrow on top times straight b with rightwards arrow on top over denominator stack vertical line straight a vertical line with rightwards arrow on top times vertical line straight b with rightwards arrow on top vertical line end fraction cos space straight theta equals fraction numerator open parentheses table row 0 row cell negative 4 end cell row cell negative 4 end cell end table close parentheses times open parentheses table row cell negative 4 end cell row 0 row cell negative 4 end cell end table close parentheses over denominator square root of 0 squared plus open parentheses negative 4 squared close parentheses plus open parentheses negative 4 close parentheses squared end root times square root of open parentheses negative 4 close parentheses squared plus 0 squared plus open parentheses negative 4 close parentheses squared end root end fraction cos space straight theta equals fraction numerator 0 plus 0 plus 16 over denominator square root of 0 plus 16 plus 16 end root times square root of 16 plus 0 plus 16 end root end fraction cos space straight theta equals fraction numerator 16 over denominator square root of 32 times square root of 32 end fraction cos space straight theta equals 16 over 32 cos space straight theta equals 1 half cos space straight theta equals cos space 60 degree space space space space space space straight theta equals 60 degree end style 

Jadi, besar sudut begin mathsize 14px style straight theta end style antara vektor begin mathsize 14px style straight a with rightwards arrow on top end style dan begin mathsize 14px style straight b with rightwards arrow on top end style adalah begin mathsize 14px style 60 degree end style.

0

Roboguru

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

Pembahasan Soal:

Berdasarkan konsep sudut antara dua vektor maka:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell cos space theta end cell equals cell fraction numerator p with bar on top times q with bar on top over denominator open vertical bar p with bar on top close vertical bar times open vertical bar q with bar on top close vertical bar end fraction end cell row blank equals cell fraction numerator open parentheses table row 1 row 2 end table close parentheses times open parentheses table row 4 row 2 end table close parentheses over denominator square root of 1 squared plus 2 squared end root times square root of 4 squared plus 2 squared end root end fraction end cell row blank equals cell fraction numerator 4 plus 4 over denominator square root of 5 times square root of 20 end fraction end cell row blank equals cell fraction numerator 8 over denominator square root of 100 end fraction end cell row blank equals cell 8 over 10 end cell row blank equals cell 4 over 5 end cell end table end style 

Oleh karena definisi cos adalah 

begin mathsize 14px style cos space straight theta equals samping over miring equals 4 over 5 end style  

maka berdasarkan teorema Pythagoras diperoleh sisi depan:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell sisi space depan end cell equals cell square root of 5 squared minus 4 squared end root end cell row blank equals cell square root of 25 minus 16 end root end cell row blank equals cell square root of 9 end cell row blank equals 3 end table end style 

sehingga diperoleh nilai:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell sin space straight theta end cell equals cell depan over miring end cell row blank equals cell 3 over 5 end cell end table end style 

Oleh karena itu, jawaban yang benar adalah A.

1

Roboguru

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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