Roboguru

Diketahui vektor , vektor , dan vektor . Nilai vektor  adalah

Pertanyaan

Diketahui vektor a with rightwards arrow on top equals 2 i with rightwards arrow on top minus 3 j with rightwards arrow on top plus k with rightwards arrow on top, vektor b with rightwards arrow on top equals i with rightwards arrow on top minus 3 j with rightwards arrow on top minus k with rightwards arrow on top, dan vektor c with rightwards arrow on top equals 4 i with rightwards arrow on top plus j with rightwards arrow on top minus k with rightwards arrow on top. Nilai vektor table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell a 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 cell c with rightwards arrow on top end cell end table adalah

Pembahasan Soal:

Untuk menentukan operasi vektor di atas, kita dapat langsung menjumlahkan secara langsung:

table attributes columnalign right center left columnspacing 0px end attributes row cell a with rightwards arrow on top plus c with rightwards arrow on top end cell equals cell open parentheses 2 i with rightwards arrow on top minus 3 j with rightwards arrow on top plus k with rightwards arrow on top close parentheses plus open parentheses 4 i with rightwards arrow on top plus j with rightwards arrow on top minus k with rightwards arrow on top close parentheses end cell row blank equals cell 6 i with rightwards arrow on top minus 2 j with rightwards arrow on top end cell end table 

Jadi, Nilai vektor table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell a 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 cell c with rightwards arrow on top end cell end table  adalah table attributes columnalign right center left columnspacing 0px end attributes row blank blank 6 end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell i 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 2 end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell j with rightwards arrow on top end cell end table

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

M. Nasrullah

Mahasiswa/Alumni Universitas Negeri Makassar

Terakhir diupdate 06 Juni 2021

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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Pertanyaan yang serupa

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

Diketahui koordinat titik ,  dan . Jika  dan , tentukan hasil: b.

Pembahasan Soal:

Diketahui koordinat titik straight A left parenthesis negative 4 comma space 6 right parenthesis, straight B left parenthesis negative 2 comma space minus 3 right parenthesis dan straight C left parenthesis 4 comma space 3 right parenthesis. Jika u with rightwards arrow on top equals AB with rightwards arrow on top dan v with rightwards arrow on top equals AC with rightwards arrow on top, maka:

table attributes columnalign right center left columnspacing 0px end attributes row cell u with rightwards arrow on top plus v with rightwards arrow on top end cell equals cell AB with rightwards arrow on top plus AC with rightwards arrow on top end cell row blank equals cell left parenthesis straight B minus straight A right parenthesis plus left parenthesis straight C minus straight A right parenthesis end cell row blank equals cell open parentheses table row cell negative 2 minus left parenthesis negative 4 right parenthesis end cell row cell negative 3 minus 6 end cell end table close parentheses plus open parentheses table row cell 4 minus left parenthesis negative 4 right parenthesis end cell row cell 3 minus 6 end cell end table close parentheses end cell row blank equals cell open parentheses table row 2 row cell negative 9 end cell end table close parentheses plus open parentheses table row 8 row cell negative 3 end cell end table close parentheses end cell row blank equals cell open parentheses table row cell 2 plus 8 end cell row cell negative 9 plus left parenthesis negative 3 right parenthesis end cell end table close parentheses end cell row blank equals cell open parentheses table row 10 row cell negative 12 end cell end table close parentheses end cell end table 

Jadi, hasil dari u with rightwards arrow on top plus v with rightwards arrow on top adalah left parenthesis 10 comma space 12 right parenthesis 

0

Roboguru

Diketahui vektor-vektor ; ; . Vektor  adalah ....

Pembahasan Soal:

Diketahui vektor-vektor a with rightwards arrow on top equals 2 i with rightwards arrow on top plus 3 j with rightwards arrow on top plus k with rightwards arrow on topb with rightwards arrow on top equals 3 i with rightwards arrow on top minus 2 k with rightwards arrow on top; dan c with rightwards arrow on top equals 2 j with rightwards arrow on top minus 5 k with rightwards arrow on top. Penjumlahan dan pengurangan vektor-vektor tersebut dapat dihitung seperti berikut :

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

Jadi, hasil dari vektor a with rightwards arrow on top plus 2 b with rightwards arrow on top minus 3 c with rightwards arrow on top adalah 8 i with rightwards arrow on top minus 3 j with rightwards arrow on top plus 12 k with rightwards arrow on top.

0

Roboguru

Diketahui: ;  dan . Jika  dan  maka

Pembahasan Soal:

Perhatikan perhitungan berikut

table attributes columnalign right center left columnspacing 0px end attributes row cell straight P open parentheses 4 comma 5 comma negative 2 close parentheses space dan space straight Q open parentheses 5 comma 1 comma 3 close parentheses end cell rightwards arrow cell top enclose PQ equals open parentheses table row cell 5 minus 4 end cell row cell 1 minus 5 end cell row cell 3 minus open parentheses negative 2 close parentheses end cell end table close parentheses equals open parentheses table row 1 row cell negative 4 end cell row 5 end table close parentheses end cell row cell straight R open parentheses negative 2 comma 3 comma negative 6 close parentheses space dan space straight Q open parentheses 5 comma 1 comma 3 close parentheses space end cell rightwards arrow cell top enclose RQ equals open parentheses table row cell 5 minus open parentheses negative 2 close parentheses end cell row cell 1 minus 3 end cell row cell 3 minus open parentheses negative 6 close parentheses end cell end table close parentheses equals open parentheses table row 7 row cell negative 2 end cell row 9 end table close parentheses end cell row cell straight P open parentheses 4 comma 5 comma negative 2 close parentheses space dan space straight R open parentheses negative 2 comma 3 comma negative 6 close parentheses space end cell rightwards arrow cell top enclose PR equals open parentheses table row cell negative 2 minus 4 end cell row cell 3 minus 5 end cell row cell negative 6 minus open parentheses negative 2 close parentheses end cell end table close parentheses equals open parentheses table row cell negative 6 end cell row cell negative 2 end cell row cell negative 4 end cell end table close parentheses end cell row cell top enclose QR end cell equals cell negative top enclose RQ equals negative open parentheses table row 7 row cell negative 2 end cell row 3 end table close parentheses equals open parentheses table row cell negative 7 end cell row 2 row cell negative 3 end cell end table close parentheses end cell row blank blank blank end table

Maka 

table attributes columnalign right center left columnspacing 0px end attributes row cell top enclose a end cell equals cell top enclose PQ plus top enclose RQ equals open parentheses table row 1 row cell negative 4 end cell row 5 end table close parentheses plus open parentheses table row 7 row cell negative 2 end cell row 9 end table close parentheses equals open parentheses table row 8 row cell negative 6 end cell row 14 end table close parentheses end cell row cell top enclose b end cell equals cell top enclose PR plus top enclose QR equals open parentheses table row cell negative 6 end cell row cell negative 2 end cell row cell negative 4 end cell end table close parentheses plus open parentheses table row cell negative 7 end cell row 2 row cell negative 9 end cell end table close parentheses equals open parentheses table row cell negative 13 end cell row 0 row cell negative 13 end cell end table close parentheses end cell end table

Ingat! Jika diketahui vektor

table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell top enclose a 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 cell x subscript 1 end cell row cell y subscript 1 end cell row cell z subscript 1 end cell end table close parentheses end cell end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank space end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank space end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank space end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank dan end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank space end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank space end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank space end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell top enclose b 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 cell x subscript 2 end cell row cell y subscript 2 end cell row cell z subscript 2 end cell end table close parentheses end cell end table

Maka

top enclose a times top enclose b equals x subscript 1 x subscript 2 plus y subscript 1 y subscript 2 plus z subscript 1 z subscript 2

Sehingga diperoleh

table attributes columnalign right center left columnspacing 0px end attributes row cell top enclose a times top enclose b end cell equals cell x subscript 1 x subscript 2 plus y subscript 1 y subscript 2 plus z subscript 1 z subscript 2 end cell row blank equals cell 8 times open parentheses negative 13 close parentheses plus open parentheses negative 6 close parentheses times 0 plus open parentheses 14 close parentheses times open parentheses negative 13 close parentheses end cell row blank equals cell negative 104 plus open parentheses negative 182 close parentheses end cell row blank equals cell negative 286 end cell end table

Dengan demikian table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell top enclose a end cell end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank times end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell top enclose b 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 minus end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank 286 end table.

Oleh karena itu, jawaban yang benar adalah E.

0

Roboguru

Jika  dan  maka besar sudut yang dibentuk oleh vektor  dan  sama dengan...

Pembahasan Soal:

Ingat kembali  penjumlahan vektor, panjang vektor dan besar sudut pada vektor berikut.

  • Jika top enclose a equals a subscript 1 i plus a subscript 2 j plus a subscript 3 k dan top enclose b equals b subscript 1 i plus b subscript 2 j plus b subscript 3 k maka

top enclose a plus top enclose b 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

top enclose a minus top enclose b 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 

  • Jika theta adalah sudut antara vektor top enclose a space dan space top enclose b, maka cos space theta equals fraction numerator a bullet b over denominator open vertical bar top enclose a close vertical bar times open vertical bar top enclose b close vertical bar end fraction.

 

  • open vertical bar top enclose a close vertical bar equals square root of a subscript 1 squared plus a subscript 2 squared plus a subscript 3 squared end root 

 

  • a bullet b equals a subscript 1 b subscript 1 plus a subscript 2 b subscript 2 plus a subscript 3 b subscript 3

Dari rumus di atas, diperoleh perhitungan sebagai berikut.

table attributes columnalign right center left columnspacing 0px end attributes row cell left parenthesis top enclose a plus top enclose b right parenthesis end cell equals cell left parenthesis negative 1 plus 1 right parenthesis i plus left parenthesis 1 plus left parenthesis negative 1 right parenthesis right parenthesis j plus left parenthesis 2 plus 2 right parenthesis k end cell row blank equals cell 0 i plus 0 j plus 4 k end cell row cell left parenthesis top enclose a minus top enclose b right parenthesis end cell equals cell left parenthesis negative 1 minus 1 right parenthesis i plus left parenthesis 1 minus left parenthesis negative 1 right parenthesis right parenthesis j plus left parenthesis 2 minus 2 right parenthesis k end cell row blank equals cell negative 2 i plus 2 j plus 0 k end cell row blank blank blank end table 

Sehingga besar sudut  yang dibentuk oleh vektor left parenthesis top enclose a plus top enclose b right parenthesis dan left parenthesis top enclose a minus top enclose b right parenthesis adalah

table attributes columnalign right center left columnspacing 0px end attributes row cell cos space theta end cell equals cell fraction numerator left parenthesis top enclose a plus top enclose b right parenthesis bullet left parenthesis top enclose a minus top enclose b right parenthesis over denominator open vertical bar top enclose a plus top enclose b close vertical bar times open vertical bar top enclose a minus top enclose b close vertical bar end fraction end cell row cell cos space theta end cell equals cell fraction numerator left parenthesis 0 i plus 0 j plus 4 k right parenthesis bullet left parenthesis negative 2 i plus 2 j plus 0 k right parenthesis over denominator square root of 0 squared plus 0 squared plus 4 squared end root cross times square root of left parenthesis negative 2 right parenthesis squared plus 2 squared plus 0 squared end root end fraction end cell row cell cos space theta end cell equals cell fraction numerator left parenthesis 0 cross times 0 right parenthesis plus left parenthesis 0 cross times 2 right parenthesis plus left parenthesis 4 cross times 0 right parenthesis over denominator square root of 0 plus 0 plus 16 end root cross times square root of 4 plus 4 plus end root end fraction end cell row cell cos space theta end cell equals cell fraction numerator 0 over denominator square root of 16 cross times square root of 8 end fraction end cell row cell cos space theta end cell equals 0 row cell cos space theta end cell equals cell cos space 90 degree end cell row theta equals cell 90 degree end cell end table 

Jadi, besar sudut yang dibentuk oleh vektor left parenthesis top enclose a plus top enclose b right parenthesis dan left parenthesis top enclose a minus top enclose b right parenthesis  adalah 90 degree.

Oleh karena itu, jawaban yang benar adalah D.

 

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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