Roboguru

Jika diketahui titik , dan , tentukan: a. Vektor b. Vektor

Pertanyaan

Jika diketahui titik A left parenthesis 3 comma space 1 comma space 1 right parenthesis, dan B left parenthesis 5 comma space 3 comma space 4 right parenthesis, tentukan:
a. Vektor stack A B with rightwards arrow on top
b. Vektor stack B A with rightwards arrow on top

Pembahasan Soal:

a. Vektor stack A B with rightwards arrow on top 

equals left parenthesis 5 minus 3 comma space 3 minus 1 comma space 4 minus 1 right parenthesis equals left parenthesis 2 comma 2 comma 3 right parenthesis


b. Vektor stack B A with rightwards arrow on top

equals left parenthesis 3 minus 5 comma space 1 minus 3 comma space 1 minus 4 right parenthesis equals left parenthesis 2 comma space minus 2 comma space minus 3 right parenthesis

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

G. Albiah

Mahasiswa/Alumni Universitas Galuh Ciamis

Terakhir diupdate 01 Mei 2021

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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Pertanyaan yang serupa

Jika  dan titik  maka koordinat titik  adalah...

Pembahasan Soal:

Diketahui:

 top enclose PQ equals open parentheses table row 3 row 5 row 1 end table close parentheses space dan space top enclose PR equals open parentheses table row 4 row 1 row 2 end table close parentheses  

straight Q left parenthesis 1 comma space 3 comma space minus 2 right parenthesis

Sehingga,

table attributes columnalign right center left columnspacing 0px end attributes row cell top enclose PQ end cell equals cell straight q with rightwards arrow on top minus straight p with rightwards arrow on top end cell row cell open parentheses table row 3 row 5 row 1 end table close parentheses end cell equals cell open parentheses table row 1 row 3 row cell negative 2 end cell end table close parentheses minus straight p with rightwards arrow on top end cell row cell straight p with rightwards arrow on top end cell equals cell open parentheses table row 1 row 3 row cell negative 2 end cell end table close parentheses minus open parentheses table row 3 row 5 row 1 end table close parentheses end cell row blank equals cell open parentheses table row cell negative 2 end cell row cell negative 2 end cell row cell negative 3 end cell end table close parentheses end cell end table

Jadi, koordinat titik straight R adalah

table attributes columnalign right center left columnspacing 0px end attributes row cell top enclose PR end cell equals cell straight r with rightwards arrow on top minus straight p with rightwards arrow on top end cell row cell open parentheses table row 4 row 1 row 2 end table close parentheses end cell equals cell straight r with rightwards arrow on top minus open parentheses table row cell negative 2 end cell row cell negative 2 end cell row cell negative 3 end cell end table close parentheses end cell row cell straight r with rightwards arrow on top end cell equals cell open parentheses table row 4 row 1 row 2 end table close parentheses plus open parentheses table row cell negative 2 end cell row cell negative 2 end cell row cell negative 3 end cell end table close parentheses end cell row blank equals cell open parentheses table row 2 row cell negative 1 end cell row cell negative 1 end cell end table close parentheses end cell end table


Oleh karena itu, jawaban yang benar adalah A.

0

Roboguru

Diketahui vektor dan . Jika vektor  tegak lurus terhadap , nilai sama dengan...

Pembahasan Soal:

Diketahui:

  • straight u with rightwards arrow on top equals 3 straight i with rightwards arrow on top plus 2 straight j with rightwards arrow on top minus straight k with rightwards arrow on top
  • straight v with rightwards arrow on top equals 3 straight i with rightwards arrow on top plus 9 straight j with rightwards arrow on top minus 12 straight k with rightwards arrow on top
  • vektor 2 straight u with rightwards arrow on top minus straight a straight v with rightwards arrow on top tegak lurus terhadap straight v with rightwards arrow on top

Sehingga,

table attributes columnalign right center left columnspacing 0px end attributes row cell 2 straight u with rightwards arrow on top minus straight a straight v with rightwards arrow on top end cell equals cell 2 open parentheses table row 3 row 2 row cell negative 1 end cell end table close parentheses minus straight a open parentheses table row 3 row 9 row cell negative 12 end cell end table close parentheses end cell row blank equals cell open parentheses table row 6 row 4 row cell negative 2 end cell end table close parentheses minus open parentheses table row cell 3 straight a end cell row cell 9 straight a end cell row cell negative 12 straight a end cell end table close parentheses end cell row blank equals cell open parentheses table row cell 6 minus 3 straight a end cell row cell 4 minus 9 straight a end cell row cell negative 2 plus 12 straight a end cell end table close parentheses end cell end table

Karena vektor 2 straight u with rightwards arrow on top minus straight a straight v with rightwards arrow on top tegak lurus terhadap straight v with rightwards arrow on top, maka:

table attributes columnalign right center left columnspacing 0px end attributes row cell left parenthesis 2 straight u with rightwards arrow on top minus straight a straight v with rightwards arrow on top right parenthesis times straight v with rightwards arrow on top end cell equals 0 row cell open parentheses table row cell 6 minus 3 straight a end cell row cell 4 minus 9 straight a end cell row cell negative 2 plus 12 straight a end cell end table close parentheses times open parentheses table row 3 row 9 row cell negative 12 end cell end table close parentheses end cell equals 0 row cell open parentheses 6 minus 3 straight a close parentheses left parenthesis 3 right parenthesis plus left parenthesis 4 minus 9 straight a right parenthesis left parenthesis 9 right parenthesis plus left parenthesis negative 2 plus 12 straight a right parenthesis left parenthesis negative 12 right parenthesis end cell equals 0 row cell 18 – 9 straight a plus 36 – 81 straight a plus 24 – 144 straight a end cell equals 0 row cell 78 – 234 straight a end cell equals cell 0 space end cell row cell – 234 straight a end cell equals cell – 78 end cell row straight a equals cell 1 third end cell end table


Oleh karena itu, jawaban yang benar adalah C.

 

0

Roboguru

Koordinat titik berat segitiga dengan titik , dan adalah...

Pembahasan Soal:

Misalkan koordinat titik berat adalah S dan koordinat titik merupakan koordinat titik berat segitiga, maka berlaku

table attributes columnalign right center left columnspacing 0px end attributes row S equals cell 1 third open square brackets P plus Q plus R close square brackets end cell row S equals cell 1 third open square brackets open parentheses negative 1 comma 2 comma 4 close parentheses plus open parentheses 4 comma negative 1 comma 7 close parentheses plus open parentheses 3 comma 5 comma 1 close parentheses close square brackets end cell row blank equals cell 1 third open parentheses open square brackets table row cell negative 1 end cell row 2 row 4 end table close square brackets plus open square brackets table row 4 row cell negative 1 end cell row 7 end table close square brackets plus open square brackets table row 3 row 5 row 1 end table close square brackets close parentheses end cell row blank equals cell 1 third open square brackets table row cell negative 1 plus 4 plus 3 end cell row cell 2 minus 1 plus 5 end cell row cell 4 plus 7 plus 1 end cell end table close square brackets end cell row blank equals cell 1 third open square brackets table row 6 row 6 row 12 end table close square brackets end cell row blank equals cell open square brackets table row cell 6 over 3 end cell row cell 6 over 3 end cell row cell 12 over 3 end cell end table close square brackets end cell row blank equals cell open square brackets table row 2 row 2 row 4 end table close square brackets end cell row S equals cell open parentheses 2 comma 2 comma 4 close parentheses end cell end table

Jadi, koordinat titik berat segitiga P Q R adalah left parenthesis 2 comma 2 comma 4 right parenthesis.

0

Roboguru

Diketahui titik , , dan titik . Nyatakan vektor-vektor berikut dalam bentuk vektor basis kemudian tentukan panjang masing-masing vektor tersebut. a.

Pembahasan Soal:

Vektor basis dan panjang vektor.

table attributes columnalign right center left columnspacing 2px end attributes row cell AB with rightwards arrow on top end cell equals cell straight B minus straight A end cell row cell Vektor space basis space AB with rightwards arrow on top end cell rightwards arrow cell AB with rightwards arrow on top equals x i with rightwards arrow on top plus y j with rightwards arrow on top plus z k with rightwards arrow on top end cell row cell Panjang space vektor space AB with rightwards arrow on top end cell rightwards arrow cell open vertical bar AB with rightwards arrow on top close vertical bar equals square root of x squared plus y squared plus z squared end root end cell end table

Diketahui:

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

Akan ditentukan vektor basis dan panjang vekto AB with rightwards arrow on top.

Terlebih dahulu tentukan AB with rightwards arrow on top.

AB with rightwards arrow on top equals straight B minus straight A equals open parentheses 1 minus 5 comma space 5 minus open parentheses negative 3 close parentheses comma space minus 7 minus 4 close parentheses equals open parentheses negative 4 comma space 8 comma space minus 11 close parentheses

Sehingga diperoleh vektor basisnya adalah AB with rightwards arrow on top equals negative 4 i with rightwards arrow on top plus 8 j with rightwards arrow on top minus 11 k with rightwards arrow on top.

Menghitung panjang vektor AB with rightwards arrow on top.

table attributes columnalign right center left columnspacing 2px end attributes row cell open vertical bar AB with rightwards arrow on top close vertical bar end cell equals cell square root of open parentheses negative 4 close parentheses squared plus open parentheses 8 close parentheses squared plus open parentheses negative 11 close parentheses squared end root end cell row blank equals cell square root of 16 plus 64 plus 121 end root end cell row cell open vertical bar AB with rightwards arrow on top close vertical bar end cell equals cell square root of 201 end cell end table

Diperoleh panjang vektornya adalah square root of 201.

0

Roboguru

Diketahui vektor ; dan . Hasil dari adalah ...

Pembahasan Soal:

Misalkan terdapat vektor straight a with rightwards arrow on top equals open parentheses table row cell straight a subscript 1 end cell row cell straight a subscript 2 end cell row cell straight a subscript 3 end cell end table close parentheses equals straight a subscript 1 straight i with rightwards arrow on top plus straight a subscript 2 straight j with rightwards arrow on top plus straight a subscript 3 straight k with rightwards arrow on top, straight b with rightwards arrow on top equals open parentheses table row cell straight b subscript 1 end cell row cell straight b subscript 2 end cell row cell straight b subscript 3 end cell end table close parentheses equals straight b subscript 1 straight i with rightwards arrow on top plus straight b subscript 2 straight j with rightwards arrow on top plus straight b subscript 3 straight k with rightwards arrow on top dan straight z adalah skalar. maka berlaku:

open parentheses straight i close parentheses space straight a with rightwards arrow on top plus-or-minus straight b with rightwards arrow on top equals open parentheses table row cell straight a subscript 1 plus-or-minus straight b subscript 1 end cell row cell straight a subscript 2 plus-or-minus straight b subscript 2 straight a subscript 3 plus-or-minus straight b subscript 3 end cell end table close parentheses equals open parentheses straight a subscript 1 plus-or-minus straight b subscript 1 close parentheses straight i with rightwards arrow on top plus open parentheses straight a subscript 2 plus-or-minus straight b subscript 2 close parentheses straight j with rightwards arrow on top plus open parentheses straight a subscript 3 plus-or-minus straight b subscript 3 close parentheses straight k with rightwards arrow on top left parenthesis ii right parenthesis space straight z straight a with rightwards arrow on top equals open parentheses table row cell za subscript 1 end cell row cell za subscript 2 end cell row cell za subscript 3 end cell end table close parentheses equals za subscript 1 straight i with rightwards arrow on top plus za subscript 2 straight j with rightwards arrow on top plus za subscript 3 straight k with rightwards arrow on top

Berdasarkan sifat-sifat operasi vektor tersebut, maka:

table attributes columnalign right center left columnspacing 0px end attributes row cell straight a with rightwards arrow on top plus 1 half straight b with rightwards arrow on top minus straight c with rightwards arrow on top end cell equals cell open parentheses table row 3 row cell begin inline style 1 half end style end cell row cell negative begin inline style 1 half end style end cell end table close parentheses plus 1 half open parentheses table row 1 row 5 row cell negative 1 end cell end table close parentheses minus open parentheses table row cell begin inline style 3 over 2 end style end cell row 0 row 0 end table close parentheses end cell row blank equals cell open parentheses table row 3 row cell begin inline style 1 half end style end cell row cell negative begin inline style 1 half end style end cell end table close parentheses plus open parentheses table row cell begin inline style 1 half end style end cell row cell begin inline style 5 over 2 end style end cell row cell negative begin inline style 1 half end style end cell end table close parentheses minus open parentheses table row cell begin inline style 3 over 2 end style end cell row 0 row 0 end table close parentheses end cell row blank equals cell open parentheses table row cell 3 plus begin inline style 1 half end style minus begin inline style 3 over 2 end style end cell row cell begin inline style 1 half end style plus begin inline style 5 over 2 end style end cell row cell negative begin inline style 1 half end style minus begin inline style 1 half end style end cell end table close parentheses end cell row blank equals cell open parentheses table row 2 row 3 row cell negative 1 end cell end table close parentheses end cell row blank equals cell 2 straight i with rightwards arrow on top plus 3 straight j with rightwards arrow on top minus straight k with rightwards arrow on top end cell end table

Dengan demikian, hasil operasi vektor pada soal adalah table attributes columnalign right center left columnspacing 0px end attributes row blank blank 2 end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell straight i with rightwards arrow on top end cell end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank plus end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank 3 end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell straight j with rightwards arrow on top end cell end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank minus end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell straight k with rightwards arrow on top end cell end table.

Oleh karena itu, jawaban yang benar adalah E.

0

Roboguru

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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