Roboguru

Diketahui vektor  . Jika vektor  memenuhi persamaan , hasil operasi  adalah ....

Pertanyaan

Diketahui vektor  p with rightwards arrow on top equals open parentheses table row 3 row cell negative 2 end cell row cell negative 1 end cell end table close parentheses space dan space straight q with rightwards arrow on top equals open parentheses table row 2 row 5 row cell negative 3 end cell end table close parentheses. Jika vektor r with rightwards arrow on top memenuhi persamaan 2 r with rightwards arrow on top plus 5 p with rightwards arrow on top equals 3 q with rightwards arrow on top plus 4 r with rightwards arrow on top, hasil operasi p with rightwards arrow on top plus q with rightwards arrow on top plus 2 r with rightwards arrow on top adalah ....

Pembahasan Soal:

Diketahui p with rightwards arrow on top equals open parentheses table row 3 row cell negative 2 end cell row cell negative 1 end cell end table close parentheses space dan space straight q with rightwards arrow on top equals open parentheses table row 2 row 5 row cell negative 3 end cell end table close parentheses, serta 2 r with rightwards arrow on top plus 5 p with rightwards arrow on top equals 3 q with rightwards arrow on top plus 4 r with rightwards arrow on top, maka

table attributes columnalign right center left columnspacing 0px end attributes row cell 2 r with rightwards arrow on top plus 5 p with rightwards arrow on top end cell equals cell 3 q with rightwards arrow on top plus 4 r with rightwards arrow on top end cell row cell 5 open parentheses table row 3 row cell negative 2 end cell row cell negative 1 end cell end table close parentheses end cell equals cell 3 open parentheses table row 2 row 5 row cell negative 3 end cell end table close parentheses plus 2 r with rightwards arrow on top end cell row cell 2 r with rightwards arrow on top end cell equals cell open parentheses table row 15 row cell negative 10 end cell row cell negative 5 end cell end table close parentheses minus open parentheses table row 6 row 15 row cell negative 9 end cell end table close parentheses end cell row blank equals cell open parentheses table row 9 row cell negative 25 end cell row 4 end table close parentheses end cell row blank rightwards arrow cell p with rightwards arrow on top plus q with rightwards arrow on top plus 2 r with rightwards arrow on top equals open parentheses table row 3 row cell negative 2 end cell row cell negative 1 end cell end table close parentheses plus open parentheses table row 2 row 5 row cell negative 3 end cell end table close parentheses plus open parentheses table row 9 row cell negative 25 end cell row 4 end table close parentheses end cell row blank equals cell open parentheses table row 14 row cell negative 22 end cell row 0 end table close parentheses end cell end table

Jadi, hasil operasi table attributes columnalign right center left columnspacing 0px end attributes row cell p with rightwards arrow on top plus q with rightwards arrow on top plus 2 r with rightwards arrow on top end cell equals cell open parentheses table row 14 row cell negative 22 end cell row 0 end table close parentheses end cell end table.

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

N. Indah

Mahasiswa/Alumni Universitas Diponegoro

Terakhir diupdate 13 Agustus 2021

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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Pertanyaan yang serupa

Diketahui vektor  . Jika , hasil   adalah ....

Pembahasan Soal:

Diketahui a with rightwards arrow on top equals open parentheses table row 3 row cell negative 2 end cell row 5 end table close parentheses space dan space b with rightwards arrow on top equals open parentheses table row 1 row 0 row cell negative 4 end cell end table close parentheses, serta c with rightwards arrow on top equals 2 a with rightwards arrow on top minus 3 b with rightwards arrow on top, maka

table attributes columnalign right center left columnspacing 0px end attributes row cell a with rightwards arrow on top plus 3 b with rightwards arrow on top minus 2 c with rightwards arrow on top end cell equals cell a with rightwards arrow on top plus 3 b with rightwards arrow on top minus 2 open parentheses 2 a with rightwards arrow on top minus 3 b with rightwards arrow on top close parentheses end cell row blank equals cell a with rightwards arrow on top plus 3 b with rightwards arrow on top minus 4 a with rightwards arrow on top plus 6 b with rightwards arrow on top end cell row blank equals cell negative 3 a with rightwards arrow on top plus 9 b with rightwards arrow on top end cell row blank equals cell negative 3 open parentheses table row 3 row cell negative 2 end cell row 5 end table close parentheses space plus 9 open parentheses table row 1 row 0 row cell negative 4 end cell end table close parentheses end cell row blank equals cell open parentheses table row cell negative 9 end cell row 6 row cell negative 15 end cell end table close parentheses plus open parentheses table row 9 row 0 row cell negative 36 end cell end table close parentheses end cell row blank equals cell open parentheses table row 0 row 6 row cell negative 51 end cell end table close parentheses end cell end table

Jadi, jawaban yang tepat adalah B.

0

Roboguru

Diketahui koordinat titik , dan . Jika , koordinat titik N adalah ....

Pembahasan Soal:

Diketahui, K left parenthesis negative 1 comma 5 comma negative 4 right parenthesis comma space L left parenthesis 2 comma 8 comma negative 2 right parenthesis, dan M left parenthesis 5 comma 3 comma negative 3 right parenthesis, serta stack K L with rightwards arrow on top equals stack M N with rightwards arrow on top, maka

table attributes columnalign right center left columnspacing 0px end attributes row cell stack K L with rightwards arrow on top end cell equals cell stack M N with rightwards arrow on top end cell row cell open parentheses table row cell 2 plus 1 end cell row cell 8 minus 5 end cell row cell negative 2 plus 4 end cell end table close parentheses end cell equals cell open parentheses table row cell n subscript 1 minus 5 end cell row cell n subscript 2 minus 3 end cell row cell n subscript 3 plus 3 end cell end table close parentheses end cell row cell open parentheses table row 3 row 3 row 2 end table close parentheses end cell equals cell open parentheses table row cell n subscript 1 minus 5 end cell row cell n subscript 2 minus 3 end cell row cell n subscript 3 plus 3 end cell end table close parentheses end cell row blank rightwards arrow cell n subscript 1 equals 8 semicolon n subscript 2 equals 6 semicolon n subscript 3 equals negative 1 end cell end table

maka koordinat titik N adalah N left parenthesis 8 comma 6 comma negative 1 right parenthesis.

Jadi, jawaban yang tepat adalah B.

0

Roboguru

Diketahui vektor , , dan  . Hasil dari adalah ...

Pembahasan Soal:

Diketahui bahwa vektor a with rightwards arrow on top,b with rightwards arrow on top, dan c with rightwards arrow on top merupakan vektor basis (i,j,k) yang berada pada dimensi tiga yang dapat diubah ke vektor baris berada pada dimensi tiga sehingga diperoleh

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

maka

table attributes columnalign right center left columnspacing 0px end attributes row cell a with rightwards arrow on top plus 1 half b with rightwards arrow on top minus c with rightwards arrow on top end cell equals cell open parentheses 3 comma 1 half comma negative 1 half close parentheses plus 1 half open parentheses 1 , 5 comma negative 1 close parentheses minus open parentheses 3 over 2 comma 0 , 0 close parentheses end cell row blank equals cell open parentheses 3 comma 1 half comma negative 1 half close parentheses plus open parentheses 1 half comma 5 over 2 comma negative 1 half close parentheses minus open parentheses 3 over 2 comma 0 , 0 close parentheses end cell row blank equals cell open parentheses 3 1 half comma 6 over 2 comma negative 2 over 2 close parentheses minus open parentheses 3 over 2 comma 0 , 0 close parentheses end cell row blank equals cell open parentheses 7 over 2 comma 6 over 2 comma negative 2 over 2 close parentheses minus open parentheses 3 over 2 comma 0 , 0 close parentheses end cell row blank equals cell open parentheses 4 over 2 comma 6 over 2 comma negative 2 over 2 close parentheses end cell row blank equals cell open parentheses 2 , 3 comma negative 1 close parentheses end cell end table

Jadi, hasil dari a with rightwards arrow on top plus 1 half b with rightwards arrow on top minus c with rightwards arrow on top adalah open parentheses 2 , 3 comma negative 1 close parentheses.

1

Roboguru

Diketahui ; dan . Jika maka

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

Pada soal diketahui straight p with rightwards arrow on top equals 3 straight i with rightwards arrow on top plus 2 straight j with rightwards arrow on top minus 2 straight k with rightwards arrow on top equals open parentheses table row 3 row 2 row cell negative 2 end cell end table close parentheses; straight q with rightwards arrow on top equals straight i with rightwards arrow on top minus 4 straight j with rightwards arrow on top plus straight k with rightwards arrow on top equals open parentheses table row 1 row cell negative 4 end cell row 1 end table close parentheses dan straight r with rightwards arrow on top equals 13 straight i with rightwards arrow on top plus 4 straight j with rightwards arrow on top minus 7 straight k with rightwards arrow on top equals open parentheses table row 13 row 4 row cell negative 7 end cell end table close parentheses. maka:

table attributes columnalign right center left columnspacing 0px end attributes row cell straight r with rightwards arrow on top end cell equals cell straight m straight p with rightwards arrow on top plus straight n straight q with rightwards arrow on top end cell row cell open parentheses table row 13 row 4 row cell negative 7 end cell end table close parentheses end cell equals cell straight m open parentheses table row 3 row 2 row cell negative 2 end cell end table close parentheses plus straight n open parentheses table row 1 row cell negative 4 end cell row 1 end table close parentheses end cell row cell open parentheses table row 13 row 4 row cell negative 7 end cell end table close parentheses end cell equals cell open parentheses table row cell 3 straight m end cell row cell 2 straight m end cell row cell negative 2 straight m end cell end table close parentheses plus open parentheses table row straight n row cell negative 4 straight n end cell row straight n end table close parentheses end cell row cell open parentheses table row 13 row 4 row cell negative 7 end cell end table close parentheses end cell equals cell open parentheses table row cell 3 straight m plus straight n end cell row cell 2 straight m minus 4 straight n end cell row cell negative 2 straight m plus straight n end cell end table close parentheses end cell end table

Diperoleh SPLDV yaitu:

open curly brackets table attributes columnalign left end attributes row cell 3 straight m plus straight n equals 13 space... left parenthesis straight i right parenthesis end cell row cell 2 straight m minus 4 straight n equals 4 space... left parenthesis ii right parenthesis end cell row cell negative 2 straight m plus straight n equals negative 7 space... left parenthesis iii right parenthesis end cell end table close

Ambil persamaan (ii) dan (iii) sehingga didapat nilai straight n yaitu:

bottom enclose table attributes columnalign right center left columnspacing 2px end attributes row cell 2 straight m minus 4 straight n end cell equals 4 row cell negative 2 straight m plus straight n end cell equals cell negative 7 space plus end cell end table end enclose table attributes columnalign right center left columnspacing 2px end attributes row cell space space space space space space minus 3 straight n end cell equals cell negative 3 end cell row straight n equals 1 end table

Substitusi persamaan nilai straight n ke persamaan (i) sehingga didapat nilai straight m yaitu:

table attributes columnalign right center left columnspacing 0px end attributes row cell 3 straight m plus straight n end cell equals 13 row cell 3 straight m plus 1 end cell equals 13 row cell 3 straight m end cell equals cell 13 minus 1 end cell row cell 3 straight m end cell equals 12 row straight m equals 4 end table

Dengan demikian, didapat nilai straight n equals 1 dan straight m equals 4, maka pengurangan m dengan n adalah:

table attributes columnalign right center left columnspacing 0px end attributes row cell straight m minus straight n end cell equals cell 4 minus 1 end cell row cell straight m minus straight n end cell equals 3 end table

Oleh karena itu, jawaban yang benar adalah B.

0

Roboguru

Jika , hasil  adalah ....

Pembahasan Soal:

Dengan memperhatikan operasi-operasi pada vektor, maka

table attributes columnalign right center left columnspacing 0px end attributes row cell 5 a with rightwards arrow on top plus 2 b with rightwards arrow on top end cell equals cell 5 open parentheses table row 1 row cell negative 3 end cell row 2 end table close parentheses plus 2 open parentheses table row 4 row cell negative 2 end cell row cell negative 1 end cell end table close parentheses end cell row blank equals cell open parentheses table row 5 row cell negative 15 end cell row 10 end table close parentheses plus open parentheses table row 8 row cell negative 4 end cell row cell negative 2 end cell end table close parentheses end cell row blank equals cell open parentheses table row cell 5 plus 8 end cell row cell negative 5 minus 4 end cell row cell 10 minus 2 end cell end table close parentheses end cell row blank equals cell open parentheses table row 13 row cell negative 9 end cell row 8 end table close parentheses end cell end table

Jadi, hasil operasi table attributes columnalign right center left columnspacing 0px end attributes row cell 5 a with rightwards arrow on top plus 2 b with rightwards arrow on top end cell equals cell open parentheses table row 13 row cell negative 9 end cell row 8 end table close parentheses end cell end table..

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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