Roboguru

Diketahui vektor-vektor , , dan . Vektor . Dalam bentuk vektor basis adalah ....

Pertanyaan

Diketahui vektor-vektor p with bar on top equals open parentheses table row 4 row cell negative 2 end cell end table close parenthesesq with bar on top equals open parentheses table row 9 row cell negative 6 end cell end table close parentheses, dan r with bar on top equals open parentheses table row 4 row 6 end table close parentheses. Vektor s with bar on top equals 1 half p with bar on top minus 1 third q with bar on top plus 1 fourth r with bar on top. Dalam bentuk vektor basis adalah ....

Pembahasan Soal:

Diketahui vektor-vektor p with bar on top equals open parentheses table row 4 row cell negative 2 end cell end table close parenthesesq with bar on top equals open parentheses table row 9 row cell negative 6 end cell end table close parentheses, dan r with bar on top equals open parentheses table row 4 row 6 end table close parentheses. Vektor s with bar on top equals 1 half p with bar on top minus 1 third q with bar on top plus 1 fourth r with bar on top dapat dinyatakan sebagai berikut

table attributes columnalign right center left columnspacing 0px end attributes row cell s with bar on top end cell equals cell 1 half open parentheses table row 4 row cell negative 2 end cell end table close parentheses minus 1 third open parentheses table row 9 row cell negative 6 end cell end table close parentheses plus 1 fourth open parentheses table row 4 row 6 end table close parentheses end cell row blank equals cell open parentheses table row 2 row cell negative 1 end cell end table close parentheses minus open parentheses table row 3 row cell negative 2 end cell end table close parentheses plus open parentheses table row 1 row cell 6 over 4 end cell end table close parentheses end cell row blank equals cell open parentheses table row cell 2 minus 3 plus 1 end cell row cell negative 1 plus 2 plus 6 over 4 end cell end table close parentheses end cell row blank equals cell open parentheses table row 0 row cell 4 over 4 plus 6 over 4 end cell end table close parentheses end cell row blank equals cell open parentheses table row 0 row cell 10 over 4 end cell end table close parentheses end cell row blank equals cell negative 10 over 4 j end cell end table

Jadi, s with bar on top equals negative 10 over 4 j.

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

S. Nur

Terakhir diupdate 13 Agustus 2021

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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Pertanyaan yang serupa

Diketahui koordinat titik ,  dan . Jika a⇀, b⇀ 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:

2a3b+c====(8i6j)(3i15j)+(2i+3j)8i6j+3i+15j2i+3j(8+32)i+(6+15+3)j9i+12j

 

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 vektor , , dan . Panjang vektor adalah  ...

Pembahasan Soal:

Diketahui vektor tersebut berapa pada dimensi tiga. Ingat operasi pengurangan vektor, vektor basis serta vektor kolom dan rumus panjang vektor yaitu w with rightwards arrow on top equals square root of x squared plus y squared plus z squared end root.

Mengubah vektor basis ke vektor kolom

table attributes columnalign right center left columnspacing 0px end attributes row cell u with rightwards arrow on top end cell equals cell 2 i with rightwards arrow on top plus j with rightwards arrow on top minus 2 k with rightwards arrow on top end cell row cell u with rightwards arrow on top end cell equals cell open parentheses table row 2 row 1 row cell negative 2 end cell end table close parentheses end cell end table

dan

table attributes columnalign right center left columnspacing 0px end attributes row cell v with rightwards arrow on top end cell equals cell i with rightwards arrow on top plus 3 j with rightwards arrow on top plus 2 k with rightwards arrow on top end cell row cell v with rightwards arrow on top end cell equals cell open parentheses table row 1 row 3 row 2 end table close parentheses end cell end table

Menghitung vektor w with rightwards arrow on top

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

Maka panjang vektor w with rightwards arrow on top adalah

table attributes columnalign right center left columnspacing 0px end attributes row cell w with rightwards arrow on top end cell equals cell square root of x squared plus y squared plus z squared end root end cell row blank equals cell square root of 3 squared plus left parenthesis negative 1 right parenthesis squared plus left parenthesis negative 6 right parenthesis squared end root end cell row blank equals cell square root of 9 plus 1 plus 36 end root end cell row blank equals cell square root of 46 end cell end table

Oleh karena itu, jawaban yang benar adalah C.

0

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 end table close parentheses, 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 end table close parentheses dan straight k 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 end cell end table close parentheses left parenthesis ii right parenthesis space straight k straight a with rightwards arrow on top equals open parentheses table row cell ka subscript 1 end cell row cell ka subscript 2 end cell end table close parentheses

Pada soal diketahui straight a with rightwards arrow on top equals 2 straight i with rightwards arrow on top plus straight j with rightwards arrow on top equals open parentheses table row 2 row 1 end table close parenthesesstraight b with rightwards arrow on top equals 4 straight i with rightwards arrow on top plus 5 straight j with rightwards arrow on top equals open parentheses table row 4 row 5 end table close parentheses dan straight c with rightwards arrow on top equals 16 straight i with rightwards arrow on top plus 17 straight j with rightwards arrow on top equals open parentheses table row 16 row 17 end table close parentheses, maka:

table attributes columnalign right center left columnspacing 0px end attributes row cell straight c with rightwards arrow on top end cell equals cell straight m straight a with rightwards arrow on top plus straight n straight b with rightwards arrow on top end cell row cell open parentheses table row 16 row 17 end table close parentheses end cell equals cell straight m open parentheses table row 2 row 1 end table close parentheses plus straight n open parentheses table row 4 row 5 end table close parentheses end cell row cell open parentheses table row 16 row 17 end table close parentheses end cell equals cell open parentheses table row cell 2 straight m end cell row straight m end table close parentheses plus open parentheses table row cell 4 straight n end cell row cell 5 straight n end cell end table close parentheses end cell row cell open parentheses table row 16 row 17 end table close parentheses end cell equals cell open parentheses table row cell 2 straight m plus 4 straight n end cell row cell straight m plus 5 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 2 straight m plus 4 straight n equals 16 space... left parenthesis straight i right parenthesis end cell row cell straight m plus 5 straight n equals 17 space... left parenthesis ii right parenthesis end cell end table close

Ambil persamaan (ii) sehingga diperoleh persamaan (iii)

table attributes columnalign right center left columnspacing 0px end attributes row cell straight m plus 5 straight n end cell equals 17 row straight m equals cell 17 minus 5 straight n space... left parenthesis iii right parenthesis end cell end table

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

table attributes columnalign right center left columnspacing 0px end attributes row cell 2 straight m plus 4 straight n end cell equals 16 row cell 2 open parentheses 17 minus 5 straight n close parentheses plus 4 straight n end cell equals 16 row cell 34 minus 10 straight n plus 4 straight n end cell equals 16 row cell negative 6 straight n end cell equals cell 16 minus 34 end cell row cell negative 6 straight n end cell equals cell negative 18 end cell row cell 6 straight n end cell equals 18 row straight n equals cell 3 space... left parenthesis iv right parenthesis end cell end table

Substitusi pers. (iv) pada pers. (iii) sehingga didapat nilai straight m yaitu:

straight m equals 17 minus 5 straight n straight m equals 17 minus 5 open parentheses 3 close parentheses straight m equals 17 minus 15 straight m equals 2

Dengan demikian, didapat nilai straight n equals 3 dan straight m equals 2, maka penjumlahan keduanya adalah:

table attributes columnalign right center left columnspacing 0px end attributes row cell straight m plus straight n end cell equals cell 2 plus 3 end cell row cell straight m plus straight n end cell equals 5 end table

Oleh karena itu, jawaban yang benar adalah A.

0

Roboguru

Jika , dan , maka vektor 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 end table close parentheses, 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 end table close parentheses dan straight k 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 end cell end table close parentheses left parenthesis ii right parenthesis space straight k straight a with rightwards arrow on top equals open parentheses table row cell ka subscript 1 end cell row cell ka subscript 2 end cell end table close parentheses

Berdasarkan sifat-sifat operasi vektor tersebut, maka:

table attributes columnalign right center left columnspacing 0px end attributes row cell straight d with rightwards arrow on top end cell equals cell straight a with rightwards arrow on top plus 3 straight b with rightwards arrow on top minus 2 straight c with rightwards arrow on top end cell row cell straight d with rightwards arrow on top end cell equals cell open parentheses table row 2 row cell negative 1 end cell end table close parentheses plus 3 open parentheses table row 4 row cell negative 2 end cell end table close parentheses minus 2 open parentheses table row cell negative 1 end cell row 3 end table close parentheses end cell row cell straight d with rightwards arrow on top end cell equals cell open parentheses table row 2 row cell negative 1 end cell end table close parentheses plus open parentheses table row 12 row cell negative 6 end cell end table close parentheses minus open parentheses table row cell negative 2 end cell row 6 end table close parentheses end cell row cell straight d with rightwards arrow on top end cell equals cell open parentheses table row cell 2 plus 12 plus 2 end cell row cell negative 1 minus 6 minus 6 end cell end table close parentheses end cell row cell straight d with rightwards arrow on top end cell equals cell open parentheses table row 16 row cell negative 13 end cell end table close parentheses 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 cell straight d with rightwards arrow on top end cell end table table attributes columnalign right center left columnspacing 0px end attributes row blank equals blank end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell open parentheses table row 16 row cell negative 13 end cell end table close parentheses end cell end table.

Oleh karena itu, tidak ada jawaban yang benar.

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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