Roboguru

Diketahui titik , , dan . Titik  membagi  sehingga , maka vektor yang diwakili  adalah ...

Pertanyaan

Diketahui titik text A end text open parentheses 3 comma space 4 comma space minus 12 close parenthesestext B end text open parentheses 6 comma space 4 comma space 3 close parentheses, dan text C end text open parentheses 2 comma space 1 comma space minus 1 close parentheses. Titik text P end text membagi text AB end text sehingga text AP:PB=2:1 end text, maka vektor yang diwakili stack text PC end text with rightwards arrow on top adalah ...

  1. 0 with rightwards arrow on top 

  2. 3 stack text i end text with rightwards arrow on top minus 3 stack text j end text with rightwards arrow on top plus stack text k end text with rightwards arrow on top

  3. negative 3 stack text i end text with rightwards arrow on top minus 3 stack text j end text with rightwards arrow on top plus 1 stack text k end text with rightwards arrow on top 

  4. 3 stack text i end text with rightwards arrow on top minus 3 stack text j end text with rightwards arrow on top minus stack text k end text with rightwards arrow on top 

  5. negative 3 stack text i end text with rightwards arrow on top minus 3 stack text k end text with rightwards arrow on top 

Pembahasan Soal:

Jika text P end text berada di antara titik text A end text dan text B end text dengan text AP:PB=m:n end text dan a with rightwards arrow on topb with rightwards arrow on top, dan p with rightwards arrow on top berturut-turut menyatakan vektor posisi titik text A end texttext B end text, dan text P end text, maka:

p with rightwards arrow on top equals fraction numerator m times b with rightwards arrow on top plus n times a with rightwards arrow on top over denominator m plus n end fraction

Komponen vektor stack A B with rightwards arrow on top dapat ditentukan, yaitu stack A B with rightwards arrow on top equals b with rightwards arrow on top minus a with rightwards arrow on top

Pada soal di atas, vektor posisi p with rightwards arrow on top adalah sebagai berikut.

table attributes columnalign right center left columnspacing 0px end attributes row cell p with rightwards arrow on top end cell equals cell fraction numerator m times b with rightwards arrow on top plus n stack times a with rightwards arrow on top over denominator m plus n end fraction end cell row blank equals cell fraction numerator 2 times open parentheses table row 6 row 4 row 3 end table close parentheses plus 1 times open parentheses table row 3 row 4 row cell negative 12 end cell end table close parentheses over denominator 2 plus 1 end fraction end cell row blank equals cell fraction numerator open parentheses table row 12 row 8 row 6 end table close parentheses plus open parentheses table row 3 row 4 row cell negative 12 end cell end table close parentheses over denominator 3 end fraction end cell row blank equals cell fraction numerator open parentheses table row 15 row 12 row cell negative 6 end cell end table close parentheses over denominator 3 end fraction end cell row blank equals cell open parentheses table row 5 row 4 row cell negative 2 end cell end table close parentheses end cell end table

Vektor stack P C with rightwards arrow on top dapat ditentukan sebagai berikut.

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

Diperoleh stack P C with rightwards arrow on top equals negative 3 i with rightwards arrow on top minus 3 j with rightwards arrow on top plus k with rightwards arrow on top 

Oleh karena itu, jawaban yang tepat adalah C.

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

H. Eka

Mahasiswa/Alumni Universitas Pendidikan Indonesia

Terakhir diupdate 11 Juli 2021

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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Pertanyaan yang serupa

Diketahui , , dan .  dan  berturut-turut mewakili vektor  dan . Tentukan: d.

Pembahasan Soal:

Jika a with rightwards arrow on top equals open parentheses table row cell a subscript 1 end cell row cell a subscript 2 end cell end table close parentheses dan b with rightwards arrow on top equals open parentheses table row cell b subscript 1 end cell row cell b subscript 2 end cell end table close parentheses, maka a with rightwards arrow on top times b with rightwards arrow on top adalah:

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 cell a subscript 1 end cell row cell a subscript 2 end cell end table close parentheses times open parentheses table row cell b subscript 1 end cell row cell b subscript 2 end cell end table close parentheses end cell row blank equals cell a subscript 1 b subscript 1 plus a subscript 2 b subscript 2 end cell end table 

Vektor Posisi adalah vektor yang berpangkal di pusat koordinat open parentheses 0 comma space 0 close parentheses dan berujung di suatu titik open parentheses x comma space y close parentheses.

Diketahui straight A open parentheses 2 comma space 6 comma space minus 5 close parentheses, straight B open parentheses negative 1 comma space 3 comma space 7 close parentheses, dan straight C open parentheses 4 comma space minus 1 comma space 8 close parentheses.

Nilai vektor posisi akan sama dengan koordinat titik ujungnya, maka tentukan AB with rightwards arrow on top dan BC with rightwards arrow on top.

table attributes columnalign right center left columnspacing 0px end attributes row cell AB with rightwards arrow on top end cell equals cell b with rightwards arrow on top minus a with rightwards arrow on top end cell row cell u with rightwards arrow on top end cell equals cell open parentheses table row cell negative 1 end cell row 3 row 7 end table close parentheses minus open parentheses table row 2 row 6 row cell negative 5 end cell end table close parentheses end cell row cell u with rightwards arrow on top end cell equals cell open parentheses table row cell negative 1 minus 2 end cell row cell 3 minus 6 end cell row cell 7 plus 5 end cell end table close parentheses end cell row cell u with rightwards arrow on top end cell equals cell open parentheses table row cell negative 3 end cell row cell negative 3 end cell row 12 end table close parentheses end cell row cell BC with rightwards arrow on top end cell equals cell c with rightwards arrow on top minus b with rightwards arrow on top end cell row cell v with rightwards arrow on top end cell equals cell open parentheses table row 4 row cell negative 1 end cell row 8 end table close parentheses minus open parentheses table row cell negative 1 end cell row 3 row 7 end table close parentheses end cell row cell v with rightwards arrow on top end cell equals cell open parentheses table row cell 4 plus 1 end cell row cell negative 1 minus 3 end cell row cell 8 minus 7 end cell end table close parentheses end cell row cell v with rightwards arrow on top end cell equals cell open parentheses table row 5 row cell negative 4 end cell row 1 end table close parentheses end cell end table   

d. open parentheses 2 u with rightwards arrow on top minus 3 v with rightwards arrow on top close parentheses squared

begin mathsize 12px style table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell open parentheses 2 u with rightwards arrow on top minus 3 v with rightwards arrow on top close parentheses squared end cell row blank equals cell 4 open parentheses u with rightwards arrow on top close parentheses squared minus 12 open parentheses u with rightwards arrow on top times v with rightwards arrow on top close parentheses plus 9 open parentheses v with rightwards arrow on top close parentheses squared end cell row blank equals cell 4 open parentheses table row cell negative 3 end cell row cell negative 3 end cell row 12 end table close parentheses squared minus 12 open parentheses open parentheses table row cell negative 3 end cell row cell negative 3 end cell row cell negative 3 end cell end table close parentheses times open parentheses table row 5 row cell negative 4 end cell row 1 end table close parentheses close parentheses plus 9 open parentheses table row 5 row cell negative 4 end cell row 1 end table close parentheses squared end cell row blank equals cell 4 open parentheses open parentheses negative 3 close parentheses squared plus open parentheses negative 3 close parentheses squared plus 12 squared close parentheses minus 12 open parentheses negative 15 plus 12 plus 12 close parentheses plus 9 open parentheses 5 squared plus open parentheses negative 4 close parentheses squared plus 1 squared close parentheses end cell row blank equals cell 4 times 162 minus 12 times 9 plus 9 times 42 end cell row blank equals cell 648 minus 108 plus 378 end cell row blank equals 918 end table end style 

Jadi, Error converting from MathML to accessible text..

0

Roboguru

Diketahui titik , , dan . Titik  membagi  sehingga , maka vektor yang diwakili  adalah ...

Pembahasan Soal:

Jika text P end text berada di antara titik text A end text dan text B end text dengan text AP:PB=m:n end text dan a with rightwards arrow on topb with rightwards arrow on top, dan p with rightwards arrow on top berturut-turut menyatakan vektor posisi titik text A end texttext B end text, dan text P end text, maka:

p with rightwards arrow on top equals fraction numerator m times b with rightwards arrow on top plus n times a with rightwards arrow on top over denominator m plus n end fraction

Komponen vektor stack A B with rightwards arrow on top dapat ditentukan, yaitu stack A B with rightwards arrow on top equals b with rightwards arrow on top minus a with rightwards arrow on top

Pada soal di atas, vektor posisi p with rightwards arrow on top adalah sebagai berikut.

table attributes columnalign right center left columnspacing 0px end attributes row cell p with rightwards arrow on top end cell equals cell fraction numerator m times b with rightwards arrow on top plus n stack times a with rightwards arrow on top over denominator m plus n end fraction end cell row blank equals cell fraction numerator 1 times open parentheses table row cell negative 8 end cell row cell negative 6 end cell row 4 end table close parentheses plus 3 times open parentheses table row 8 row 18 row cell negative 12 end cell end table close parentheses over denominator 1 plus 3 end fraction end cell row blank equals cell fraction numerator open parentheses table row cell negative 8 end cell row cell negative 6 end cell row 4 end table close parentheses plus open parentheses table row 24 row 54 row cell negative 36 end cell end table close parentheses over denominator 4 end fraction end cell row blank equals cell fraction numerator open parentheses table row 16 row 48 row cell negative 32 end cell end table close parentheses over denominator 4 end fraction end cell row blank equals cell open parentheses table row 4 row 12 row cell negative 8 end cell end table close parentheses end cell end table

Vektor stack P C with rightwards arrow on top dapat ditentukan sebagai berikut.

table attributes columnalign right center left columnspacing 0px end attributes row cell stack P C with rightwards arrow on top end cell equals cell c with rightwards arrow on top minus p with rightwards arrow on top end cell row blank equals cell open parentheses table row 7 row 9 row cell negative 3 end cell end table close parentheses minus open parentheses table row 4 row 12 row cell negative 8 end cell end table close parentheses end cell row blank equals cell open parentheses table row 3 row cell negative 3 end cell row 5 end table close parentheses end cell end table

Diperoleh stack P C with rightwards arrow on top equals 3 i with rightwards arrow on top minus 3 j with rightwards arrow on top plus 5 k with rightwards arrow on top

Oleh karena itu, jawaban yang tepat adalah B.

0

Roboguru

Jika , ,  dan  adalah vektor posisi dari ; ;  dan , maka nilai  yang memenuhi  adalah....

Pembahasan Soal:

Diketahui

table attributes columnalign right center left columnspacing 0px end attributes row cell straight P open parentheses negative 1 comma 2 comma 1 half close parentheses end cell rightwards arrow cell top enclose p equals open parentheses table row cell negative 1 end cell row 2 row cell 1 half end cell end table close parentheses end cell row cell straight Q open parentheses 3 comma 0 comma 2 close parentheses end cell rightwards arrow cell top enclose q equals open parentheses table row 3 row 0 row 2 end table close parentheses end cell row cell straight R open parentheses 2 comma 3 comma 1 close parentheses end cell rightwards arrow cell top enclose r equals open parentheses table row 2 row 3 row 1 end table close parentheses end cell row cell straight T open parentheses 9 comma negative 14 comma 2 close parentheses end cell rightwards arrow cell top enclose t equals open parentheses table row 9 row cell negative 14 end cell row 2 end table close parentheses end cell row blank blank blank end table

Perhatikan perhitungan berikut ini

table attributes columnalign right center left columnspacing 0px end attributes row cell k top enclose p plus 3 top enclose q minus 2 top enclose r end cell equals cell top enclose t end cell row cell k open parentheses table row cell negative 1 end cell row 2 row cell 1 half end cell end table close parentheses plus 3 open parentheses table row 3 row 0 row 2 end table close parentheses minus 2 open parentheses table row 2 row 3 row 1 end table close parentheses end cell equals cell open parentheses table row 9 row cell negative 14 end cell row 2 end table close parentheses end cell row cell open parentheses table row cell negative k end cell row cell 2 k end cell row cell 1 half k end cell end table close parentheses plus open parentheses table row 9 row 0 row 6 end table close parentheses minus open parentheses table row 4 row 6 row 2 end table close parentheses end cell equals cell open parentheses table row 9 row cell negative 14 end cell row 2 end table close parentheses end cell row cell open parentheses table row cell negative k plus 9 minus 4 end cell row cell 2 k plus 0 minus 6 end cell row cell k plus 6 minus 2 end cell end table close parentheses end cell equals cell open parentheses table row 9 row cell negative 14 end cell row 2 end table close parentheses end cell row cell open parentheses table row cell negative k plus 5 end cell row cell 2 k minus 6 end cell row cell 1 half k plus 4 end cell end table close parentheses end cell equals cell open parentheses table row 9 row cell negative 14 end cell row 2 end table close parentheses end cell end table

Dengan menggunakan kesamaan vektor diperoleh

table attributes columnalign right center left columnspacing 0px end attributes row cell negative k plus 5 end cell equals 9 row cell negative k end cell equals cell 9 minus 5 end cell row cell negative k end cell equals 4 row k equals cell negative 4 end cell end table

Dengan demikian nilai k yang memenuhi adalah negative 4.

Oleh karena itu, jawaban yang benar adalah E.

0

Roboguru

Diketahui koordinat titik ,  dan . Jika , ,dan  berturut-turut vektor posisi titik A, B, dan C, hasil  adalah

Pembahasan Soal:

Vektor posisi titik A, B,dan C yaitu:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell a with rightwards harpoon with barb upwards on top end cell equals cell 4 i with hat on top minus 3 j with hat on top end cell row cell b with rightwards harpoon with barb upwards on top end cell equals cell negative i with hat on top minus 5 j with hat on top end cell row cell c with rightwards harpoon with barb upwards on top end cell equals cell negative 2 i with hat on top plus 3 j with hat on top end cell end table end style 

Dengan konsep perkalian vektor dengan skalar diperoleh:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell 2 a with rightwards harpoon with barb upwards on top end cell equals cell 2 open parentheses 4 i with hat on top minus 3 j with hat on top close parentheses end cell row blank equals cell 8 i with hat on top minus 6 j with hat on top end cell row cell 3 b with rightwards harpoon with barb upwards on top end cell equals cell 3 open parentheses negative i with hat on top minus 5 j with hat on top close parentheses end cell row blank equals cell negative 3 i with hat on top minus 15 j with hat on top end cell end table end style

Operasi penjumlahan dan pengurangan vektor secara aljabar dari begin mathsize 14px style 2 a with rightwards harpoon with barb upwards on top minus 3 b with rightwards harpoon with barb upwards on top plus c with rightwards harpoon with barb upwards on top end style yaitu:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell 2 a with rightwards harpoon with barb upwards on top minus 3 b with rightwards harpoon with barb upwards on top plus c with rightwards harpoon with barb upwards on top end cell equals cell open parentheses 8 i with hat on top minus 6 j with hat on top close parentheses minus open parentheses negative 3 i with hat on top minus 15 j with hat on top close parentheses plus open parentheses negative 2 i with hat on top plus 3 j with hat on top close parentheses end cell row blank equals cell open parentheses 8 plus 3 minus 2 close parentheses i with hat on top plus open parentheses negative 6 plus 15 plus 3 close parentheses j with hat on top end cell row blank equals cell 9 i with hat on top plus 12 j with hat on top end cell end table end style 

Jadi, hasil begin mathsize 14px style 2 a with rightwards harpoon with barb upwards on top minus 3 b with rightwards harpoon with barb upwards on top plus c with rightwards harpoon with barb upwards on top end style adalah begin mathsize 14px style 9 i with hat on top plus 12 j with hat on top end style.

0

Roboguru

Diketahui , , dan . Jika , , dan  merupakan vektor posisi dari A, B, dan C, nyatakan vektor berikut ke dalam bentuk vektor satuan . c.

Pembahasan Soal:

Diketahui A left parenthesis minus sign 2 comma space 1 comma space 3 right parenthesisB left parenthesis minus sign 3 comma space 1 comma space 2 right parenthesis, dan C left parenthesis 3 comma space 2 comma minus sign 1 right parenthesis.

DIberikan a with rightwards arrow on top, b with rightwards arrow on top, dan c with rightwards arrow on top merupakan vektor posisi dari A, B, dan C, maka a with rightwards arrow on top equals negative 2 i with rightwards arrow on top plus j with rightwards arrow on top plus 3 k with rightwards arrow on top, b with rightwards arrow on top equals negative 3 i with rightwards arrow on top plus j with rightwards arrow on top plus 2 k with rightwards arrow on top, dan c with rightwards arrow on top equals 3 i with rightwards arrow on top plus 2 j with rightwards arrow on top minus k with rightwards arrow on top. Dengan demikian,

table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell 3 c with rightwards arrow on top plus 4 a with rightwards arrow on top minus 5 b with rightwards arrow on top end cell row blank equals cell 3 open parentheses table row 3 row 2 row cell negative 1 end cell end table close parentheses plus 4 open parentheses table row cell negative 2 end cell row 1 row 3 end table close parentheses minus 5 open parentheses table row cell negative 3 end cell row 1 row 2 end table close parentheses end cell row blank equals cell open square brackets open parentheses table row 9 row 6 row cell negative 3 end cell end table close parentheses plus open parentheses table row cell negative 8 end cell row 4 row 12 end table close parentheses close square brackets minus open parentheses table row cell negative 15 end cell row 5 row 10 end table close parentheses end cell row blank equals cell open parentheses table row 1 row 10 row 9 end table close parentheses minus open parentheses table row cell negative 15 end cell row 5 row 10 end table close parentheses end cell row blank equals cell open parentheses table row 16 row 5 row cell negative 1 end cell end table close parentheses end cell end table 

Jadi, hasil dari 3 c with rightwards arrow on top plus 4 a with rightwards arrow on top minus 5 b with rightwards arrow on top dalam vektor satuan adalah 16 i with rightwards arrow on top plus 5 j with rightwards arrow on top minus k with rightwards arrow on top.

0

Roboguru

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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

Membantu

Kurang Membantu

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Belum menemukan yang kamu cari?

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