Roboguru

Pada balok  diketahui vektor  mewakili  mewakili , dan  mewakili . Nyatakan hasil operasi vektor berikut dalam  dan . a.

Pertanyaan

Pada balok A B C D. E F G H diketahui vektor D C mewakili u with rightwards arrow on top comma space top enclose E H end enclose mewakili v with rightwards arrow on top, dan top enclose G C end enclose mewakili w with rightwards arrow on top. Nyatakan hasil operasi vektor berikut dalam u with rightwards arrow on top comma space v with rightwards arrow on top comma dan w with rightwards arrow on top.

a. top enclose A H end enclose plus top enclose D F end enclose plus top enclose B C end enclose 

Pembahasan Soal:

Diketahui:

top enclose A B end enclose equals top enclose E F end enclose equals top enclose H G end enclose equals top enclose D C end enclose equals u with rightwards arrow on top top enclose B A end enclose equals top enclose F E end enclose equals top enclose G H end enclose equals top enclose C D end enclose equals negative u with rightwards arrow on top top enclose A D end enclose equals top enclose B C end enclose equals top enclose F G end enclose equals top enclose E H end enclose equals v with rightwards arrow on top top enclose D A end enclose equals top enclose C B end enclose equals top enclose G F end enclose equals top enclose H E end enclose equals negative v with rightwards arrow on top top enclose E A end enclose equals top enclose F B end enclose equals top enclose H D end enclose equals top enclose G C end enclose equals w with rightwards arrow on top top enclose A E end enclose equals top enclose B F end enclose equals top enclose D H end enclose equals top enclose C G end enclose equals negative w with rightwards arrow on top  

Sehingga hasil operasi vektor top enclose A H end enclose plus top enclose D F end enclose plus top enclose B C end enclose:

table attributes columnalign right center left columnspacing 0px end attributes row cell top enclose A H end enclose plus top enclose D F end enclose plus top enclose B C end enclose end cell equals cell open parentheses top enclose A D end enclose plus top enclose D H end enclose close parentheses plus open parentheses top enclose D B end enclose plus top enclose B F end enclose close parentheses plus v with rightwards arrow on top end cell row blank equals cell open parentheses v with rightwards arrow on top minus w with rightwards arrow on top close parentheses plus open parentheses open parentheses top enclose B A end enclose plus top enclose A D end enclose close parentheses minus w with rightwards arrow on top close parentheses plus v with rightwards arrow on top end cell row blank equals cell 2 v with rightwards arrow on top minus w with rightwards arrow on top plus open parentheses open parentheses negative u with rightwards arrow on top plus v with rightwards arrow on top close parentheses minus w with rightwards arrow on top close parentheses end cell row blank equals cell negative u with rightwards arrow on top plus 3 v with rightwards arrow on top minus 2 w with rightwards arrow on top end cell end table 

Jadi, hasil operasi vektor top enclose A H end enclose plus top enclose D F end enclose plus top enclose B C end enclose dalam u with rightwards arrow on top comma space v with rightwards arrow on top comma dan w with rightwards arrow on top adalah 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 u 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 v 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 w with rightwards arrow on top end cell end table.

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

E. Dwi

Mahasiswa/Alumni Universitas Sriwijaya

Terakhir diupdate 07 Juni 2021

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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Pertanyaan yang serupa

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

Pembahasan Soal:

Kita ingat operasi pada vektor.

3 p with rightwards arrow on top minus q with rightwards arrow on top plus 2 r with rightwards arrow on top equals 3 open parentheses table row 10 row cell negative 3 end cell row 0 end table close parentheses minus open parentheses table row 3 row 2 row cell negative 1 end cell end table close parentheses plus 2 open parentheses table row 0 row 4 row 7 end table close parentheses space space space space space space space space space space space space space space space space space space space equals open parentheses table row 30 row cell negative 9 end cell row 0 end table close parentheses minus open parentheses table row 3 row 2 row cell negative 1 end cell end table close parentheses plus open parentheses table row 0 row 8 row 14 end table close parentheses space space space space space space space space space space space space space space space space space space space equals open parentheses table row 27 row cell negative 3 end cell row 15 end table close parentheses 

Jadi, jawaban yang tepat adalah C.

0

Roboguru

Perhatikan gambar berikut. Pernyataan yang benar adalah ....

Pembahasan Soal:

Perhatikan gambar dibawah ini.



Berdasarkan cara tersebut, maka diperoleh:

begin mathsize 14px style u with rightwards arrow on top equals v with rightwards arrow on top plus w with rightwards arrow on top v with rightwards arrow on top plus w with rightwards arrow on top minus u with rightwards arrow on top equals 0 end style 

Tidak ada pilihan jawaban yang benar. Jjawaban yang benar adalah begin mathsize 14px style v with rightwards arrow on top plus w with rightwards arrow on top minus u with rightwards arrow on top equals 0 end style.

0

Roboguru

Hasil operasi vektor  adalah ...

Pembahasan Soal:

Operasi tersebut adalah penjumlahan dan pengurangan vektor.

open parentheses table row cell negative 4 end cell row 7 end table close parentheses minus open parentheses table row cell negative 3 end cell row cell negative 5 end cell end table close parentheses plus open parentheses table row cell negative 6 end cell row 2 end table close parentheses equals open parentheses table row cell negative 7 end cell row 14 end table close parentheses

Sehingga, open parentheses table row cell negative 4 end cell row 7 end table close parentheses minus open parentheses table row cell negative 3 end cell row cell negative 5 end cell end table close parentheses plus open parentheses table row cell negative 6 end cell row 2 end table close parentheses equals open parentheses table row cell negative 7 end cell row 14 end table close parentheses.

Jadi, pilihan jawaban yang tepat adalah B.

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

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

Pembahasan Soal:

Diberikan vektor-vektor : a with rightwards arrow on top equals open parentheses table row 3 row cell 1 half end cell row cell negative 1 half end cell end table close parenthesesb with rightwards arrow on top equals open parentheses table row 1 row 5 row cell negative 1 end cell end table close parenthesesc with rightwards arrow on top equals open parentheses table row cell 3 over 2 end cell row 0 row 0 end table close parentheses. Penjumlahan dari ketiga vektor tersebut adalah

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 square brackets open parentheses table row 3 row cell 1 half end cell row cell negative 1 half end cell end table close parentheses plus 1 half times open parentheses table row 1 row 5 row cell negative 1 end cell end table close parentheses close square brackets minus open parentheses table row cell 3 over 2 end cell row 0 row 0 end table close parentheses end cell row blank equals cell open square brackets open parentheses table row 3 row cell 1 half end cell row cell negative 1 half end cell end table close parentheses plus open parentheses table row cell 1 half end cell row cell 5 over 2 end cell row cell negative 1 half end cell end table close parentheses close square brackets minus open parentheses table row cell 3 over 2 end cell row 0 row 0 end table close parentheses end cell row blank equals cell open parentheses table row cell 3 plus 1 half end cell row cell 1 half plus 5 over 2 end cell row cell negative 1 half minus 1 half end cell end table close parentheses minus open parentheses table row cell 3 over 2 end cell row 0 row 0 end table close parentheses end cell row blank equals cell open parentheses table row cell 7 over 2 end cell row 3 row cell negative 1 end cell end table close parentheses minus open parentheses table row cell 3 over 2 end cell row 0 row 0 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 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 equals 2 i with rightwards arrow on top plus 3 j with rightwards arrow on top minus k with rightwards arrow on top.

0

Roboguru

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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