Roboguru

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

Pertanyaan

Diketahui titik text A end text open parentheses 8 comma space 18 comma space minus 12 close parenthesestext B end text open parentheses negative 8 comma space minus 6 comma space 4 close parentheses, dan text C end text open parentheses 7 comma space 9 comma space minus 3 close parentheses. Titik text P end text membagi text AB end text sehingga text AP:PB=1:3 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 5 stack text k end text with rightwards arrow on top

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

  5. negative 5 stack text i end text with rightwards arrow on top minus 4 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 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.

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 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 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.

0

Roboguru

Misal titik  berada di antara titik  dan  dengan , maka koordinat titik  adalah...

Pembahasan Soal:

Ingat!

Jika straight C berada di antara titik straight A dan straight B dengan AC colon CB equals m colon n dan   top enclose a comma space top enclose b comma space top enclose c   menyatakan vektor posisi dari titik straight A comma space straight B comma space straight C  maka

top enclose c equals fraction numerator n top enclose a plus m top enclose b over denominator m plus n end fraction

Diketahui:

table attributes columnalign right center left columnspacing 0px end attributes row cell straight A open parentheses 5 comma negative 1 comma 5 close parentheses space end cell rightwards arrow cell top enclose straight a equals open parentheses table row 3 row 5 row 7 end table close parentheses end cell row cell straight B open parentheses 6 comma negative 4 comma negative 11 close parentheses end cell rightwards arrow cell top enclose straight b equals open parentheses table row 6 row cell negative 4 end cell row cell negative 11 end cell end table close parentheses end cell row cell AC colon CB end cell equals cell 2 colon 1 rightwards arrow m equals 2 comma space n equals 1 end cell row blank blank blank end table

Sehingga diperoleh

table attributes columnalign right center left columnspacing 0px end attributes row cell top enclose c end cell equals cell fraction numerator n top enclose a plus m top enclose b over denominator m plus n end fraction end cell row blank equals cell fraction numerator 1 open parentheses table row 3 row 5 row 7 end table close parentheses plus 2 open parentheses table row 6 row cell negative 4 end cell row cell negative 11 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 3 row 5 row 7 end table close parentheses plus open parentheses table row 12 row cell negative 8 end cell row cell negative 22 end cell end table close parentheses over denominator 3 end fraction end cell row blank equals cell 1 third open parentheses table row cell 3 plus 12 end cell row cell 5 minus 8 end cell row cell 7 minus 22 end cell end table close parentheses end cell row blank equals cell 1 third open parentheses table row 15 row cell negative 3 end cell row cell negative 15 end cell end table close parentheses end cell row blank equals cell open parentheses table row 5 row cell negative 1 end cell row cell negative 5 end cell end table close parentheses end cell end table

Dengan demikian koosrdinat titik straight C adalah open parentheses 5 comma negative 1 comma negative 5 close parentheses.

Oleh karena itu, jawaban yang benar B.

0

Roboguru

Diketahui titik  dan titik . Jika  berada di antara titik  dan  dengan , maka koordinat titik  adalah....

Pembahasan Soal:

Ingat!

Jika straight P berada di antara titik straight A dan straight B dengan AP colon PB equals m colon n dan top enclose a comma space top enclose b comma space top enclose p  menyatakan vektor posisi dari titik straight A comma space straight B comma space straight P  maka

top enclose p equals fraction numerator n top enclose a plus m top enclose b over denominator m plus n end fraction

Diketahui:

straight A open parentheses 1 comma 2 comma 1 close parentheses,  straight B open parentheses 1 comma 5. negative 5 close parentheses, dan AP colon PB equals 2 colon 1

Maka

table attributes columnalign right center left columnspacing 0px end attributes row cell top enclose a end cell equals cell open parentheses table row 1 row 2 row 1 end table close parentheses end cell row cell top enclose b end cell equals cell open parentheses table row 1 row 5 row cell negative 5 end cell end table close parentheses end cell row m equals 2 row n equals 1 end table

Sehingga

table attributes columnalign right center left columnspacing 0px end attributes row cell top enclose p end cell equals cell fraction numerator n top enclose a plus m top enclose b over denominator m plus n end fraction end cell row blank equals cell fraction numerator 1 times open parentheses table row 1 row 2 row 1 end table close parentheses plus 2 times open parentheses table row 1 row 5 row cell negative 5 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 1 row 2 row 1 end table close parentheses plus open parentheses table row 2 row 10 row cell negative 10 end cell end table close parentheses over denominator 3 end fraction end cell row blank equals cell 1 third open parentheses table row 3 row 12 row cell negative 9 end cell end table close parentheses end cell row blank equals cell open parentheses table row 1 row 4 row cell negative 3 end cell end table close parentheses end cell end table

Dengan demikian koordinat titik straight P adalah open parentheses 1 comma 4 comma negative 3 close parentheses.

Oleh karena itu, jawaban yang benar adalah A.

0

Roboguru

Diketahui  dan

Pembahasan Soal:

Diketahui a with rightwards harpoon with barb upwards on top equals 4 i plus j plus 5 k dan b with rightwards harpoon with barb upwards on top equals 2 i plus j minus 5 k 

a. Ingat kembali jika a with rightwards harpoon with barb upwards on top equals a subscript 1 i plus a subscript 2 j plus a subscript 3 k dan b with rightwards harpoon with barb upwards on top equals b subscript 1 i plus b subscript 2 j plus b subscript 3 k maka a with rightwards harpoon with barb upwards on top plus b with rightwards harpoon with barb upwards on top equals left parenthesis a subscript 1 plus b subscript 1 right parenthesis i plus left parenthesis a subscript 2 plus b subscript 2 right parenthesis j plus left parenthesis a subscript 3 plus b subscript 3 right parenthesis k 

table attributes columnalign right center left columnspacing 0px end attributes row cell a with rightwards harpoon with barb upwards on top plus b with rightwards harpoon with barb upwards on top end cell equals cell left parenthesis 4 plus 2 right parenthesis i plus left parenthesis 1 plus 1 right parenthesis j plus left parenthesis 5 plus left parenthesis negative 5 right parenthesis right parenthesis k end cell row blank equals cell 6 i plus 2 j end cell end table 

Jadi, a with rightwards harpoon with barb upwards on top plus b with rightwards harpoon with barb upwards on top equals 6 i plus 2 j.

b. Ingat kembali bahwa jika a with rightwards harpoon with barb upwards on top equals a subscript 1 i plus a subscript 2 j plus a subscript 3 k dan b with rightwards harpoon with barb upwards on top equals b subscript 1 i plus b subscript 2 j plus b subscript 3 k maka a with rightwards harpoon with barb upwards on top minus b with rightwards harpoon with barb upwards on top equals left parenthesis a subscript 1 minus b subscript 1 right parenthesis i plus left parenthesis a subscript 2 minus b subscript 2 right parenthesis j plus left parenthesis a subscript 3 minus b subscript 3 right parenthesis k 

table attributes columnalign right center left columnspacing 0px end attributes row cell a with rightwards harpoon with barb upwards on top minus b with rightwards harpoon with barb upwards on top end cell equals cell left parenthesis 4 minus 2 right parenthesis i plus left parenthesis 1 minus 1 right parenthesis j plus left parenthesis 5 minus left parenthesis negative 5 right parenthesis k end cell row blank equals cell 2 i plus 10 k end cell end table 

Jadi, a with rightwards harpoon with barb upwards on top minus b with rightwards harpoon with barb upwards on top equals 2 i plus 10 k.

c. Ingat kembali bahwa misalkan k adalah bilangan skalar dan a with rightwards harpoon with barb upwards on top equals a subscript 1 i plus a subscript 2 j plus a subscript 3 k maka  k a with rightwards harpoon with barb upwards on top equals k a subscript 1 i plus k a subscript 2 k plus k a subscript 3 k 

table attributes columnalign right center left columnspacing 0px end attributes row cell a with rightwards harpoon with barb upwards on top minus 2 b with rightwards harpoon with barb upwards on top end cell equals cell left parenthesis 4 i plus j plus 5 k right parenthesis minus left parenthesis 2 left parenthesis 2 i plus j minus 5 k right parenthesis right parenthesis end cell row blank equals cell left parenthesis 4 i plus j plus 5 k right parenthesis minus left parenthesis 4 i plus 2 j minus 10 k right parenthesis end cell row blank equals cell left parenthesis 4 minus 4 right parenthesis i plus left parenthesis 1 minus 2 right parenthesis j plus left parenthesis 5 minus left parenthesis negative 10 right parenthesis right parenthesis k end cell row blank equals cell negative j plus 15 k end cell end table 

Jadi, a with rightwards harpoon with barb upwards on top minus 2 b with rightwards harpoon with barb upwards on top equals negative j plus 15 k.

d. Ingat kembali bahwa jika a with rightwards harpoon with barb upwards on top equals a subscript 1 i plus a subscript 2 j plus a subscript 3 k dan b with rightwards harpoon with barb upwards on top equals b subscript 1 i plus b subscript 2 j plus b subscript 3 k maka a with rightwards harpoon with barb upwards on top times b with rightwards harpoon with barb upwards on top equals a subscript 1 times b subscript 1 plus a subscript 2 times b subscript 2 plus a subscript 3 times b subscript 3 

table attributes columnalign right center left columnspacing 0px end attributes row cell 2 a with rightwards harpoon with barb upwards on top times 3 b with rightwards harpoon with barb upwards on top end cell equals cell 2 left parenthesis 4 comma space 1 comma space 5 right parenthesis times 3 left parenthesis 2 comma space 1 comma space minus 5 right parenthesis end cell row blank equals cell left parenthesis 8 comma space 2 comma space 10 right parenthesis times left parenthesis 6 comma space 3 comma space minus 15 right parenthesis end cell row blank equals cell 8 times 6 plus 2 times 3 plus 10 times negative 15 end cell row blank equals cell 48 plus 6 minus 150 end cell row blank equals cell 54 minus 150 end cell row blank equals cell negative 96 end cell end table 

Jadi, 2 a with rightwards harpoon with barb upwards on top times 3 b with rightwards harpoon with barb upwards on top equals negative 96.

Dengan demikian, hasil yang diperoleh adalah a with rightwards harpoon with barb upwards on top plus b with rightwards harpoon with barb upwards on top equals 6 i plus 2 ja with rightwards harpoon with barb upwards on top minus b with rightwards harpoon with barb upwards on top equals 2 i plus 10 k dan 2 a with rightwards harpoon with barb upwards on top times 3 b with rightwards harpoon with barb upwards on top equals negative 96.

0

Roboguru

Jika , hasil  adalah ...

Pembahasan Soal:

Operasi tersebut adalah perkalian vektor dengan skalar, penjumlahan vektor, dan pengurangan vektor.

table attributes columnalign right center left columnspacing 0px end attributes row cell 2 over 3 straight p with rightwards arrow on top plus 2 straight q with rightwards arrow on top minus 1 half straight r with rightwards arrow on top end cell equals cell 2 over 3 open parentheses table row cell negative 3 end cell row 6 row 12 end table close parentheses plus 2 open parentheses table row 2 row 3 row cell negative 1 end cell end table close parentheses minus 1 half open parentheses table row 4 row cell negative 2 end cell row cell negative 6 end cell end table close parentheses end cell row blank equals cell open parentheses table row cell negative 2 end cell row 4 row 8 end table close parentheses plus open parentheses table row 4 row 6 row cell negative 2 end cell end table close parentheses minus open parentheses table row 2 row cell negative 1 end cell row cell negative 3 end cell end table close parentheses end cell row blank equals cell open parentheses table row 0 row 11 row 9 end table close parentheses end cell end table

Sehingga, 2 over 3 straight p with rightwards arrow on top plus 2 straight q with rightwards arrow on top minus 1 half straight r with rightwards arrow on top equals open parentheses table row 0 row 11 row 9 end table close parentheses.

Jadi, pilihan jawaban yang tepat adalah A.

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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