Roboguru

Diberikan kubus KLMN.OPQR dengan ruas garis berarah KL, KN dan KO berturut turut adalah wakil dari vektor a, b dan c. Nyatakan vektor-vektor tersebut dalam a, b dan c! b. QK

Pertanyaan

Diberikan kubus KLMN.OPQR dengan ruas garis berarah KL, KN dan KO berturut turut adalah wakil dari vektor a, b dan c. Nyatakan vektor-vektor tersebut dalam a, b dan c!
b. QK

Pembahasan Soal:

Perhatikan gambar dibawah ini.



Kita dapat menyatakan hubungan antara vektor komponen dan komponennya masing-masing, sebagai berikut:

F subscript x equals F subscript x straight a F subscript y equals F subscript y straight b F subscript z equals F subscript z straight c 

Sehingga, vektor F dalam komponen-komponennya sebagai berikut:

F equals F subscript x straight a plus F subscript y straight b plus F subscript z straight c 

Dan untuk QK adalah arah sebaliknya vektor F. Dengan demikian dapat dituliskan sebagai berikut:

table attributes columnalign right center left columnspacing 0px end attributes row cell Q K end cell equals cell negative F end cell row cell Q K end cell equals cell negative open parentheses F subscript x straight a plus F subscript y straight b plus F subscript z straight c close parentheses end cell end table 

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

P. Tessalonika

Mahasiswa/Alumni Universitas Negeri Medan

Terakhir diupdate 07 Juni 2021

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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Pertanyaan yang serupa

Tentukan  dan  dari : c.

Pembahasan Soal:

begin mathsize 14px style open parentheses table row 2 row 3 row 6 end table close parentheses equals open parentheses table row k row cell x plus k end cell row cell x y end cell end table close parentheses end style 

Dari baris pertama

begin mathsize 14px style k equals 2 end style 

Dari baris kedua 

begin mathsize 14px style x plus k equals 3 end style 

begin mathsize 14px style x plus 2 equals 3 end style 

begin mathsize 14px style x equals 1 end style.

Dari baris ketiga

begin mathsize 14px style x y equals 6 end style 

begin mathsize 14px style y equals 6 end style.

Jadi begin mathsize 14px style x equals 1 end style dan begin mathsize 14px style y equals 6 end style.

0

Roboguru

Diketahui vektor dan  . Jika  hasil  adalah ...

Pembahasan Soal:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell c with rightwards arrow on top end cell equals cell 2 a with rightwards arrow on top minus 3 b with rightwards arrow on top end cell row blank equals cell 2 open parentheses table row 3 row cell negative 2 end cell row 5 end table close parentheses minus 3 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 2 times 3 minus 3 times 1 end cell row cell 2 times left parenthesis negative 2 right parenthesis minus 3 times 0 end cell row cell 2 times 5 minus 3 times left parenthesis negative 4 right parenthesis end cell end table close parentheses end cell row blank equals cell open parentheses table row cell 6 minus 3 end cell row cell negative 4 end cell row cell 10 plus 12 end cell end table close parentheses end cell row blank equals cell open parentheses table row 3 row cell negative 4 end cell row 22 end table close parentheses end cell end table end style 

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell c 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 3 row cell negative 4 end cell row 22 end table close parentheses end cell end table end style, maka:

begin mathsize 14px style 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 open parentheses table row 3 row cell negative 2 end cell row 5 end table close parentheses plus 3 open parentheses table row 1 row 0 row cell negative 4 end cell end table close parentheses minus 2 open parentheses table row 3 row cell negative 4 end cell row 22 end table close parentheses end cell row blank equals cell open parentheses table row cell 3 plus 3 times 1 minus 2 times 3 end cell row cell negative 2 plus 3 times 0 minus 2 times left parenthesis negative 4 right parenthesis end cell row cell 5 plus 3 times left parenthesis negative 4 right parenthesis minus 2 times 22 end cell end table close parentheses end cell row blank equals cell open parentheses table row cell 3 plus 3 minus 6 end cell row cell negative 2 plus 0 plus 8 end cell row cell 5 minus 12 minus 44 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 end style 

Jadi, jawaban yang tepat adalah B.

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 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

Diketahui  dan . Jika  maka vektor a dapat dinyatakan ....

Pembahasan Soal:

Diketahui:

a with rightwards arrow on top equals t i with rightwards arrow on top minus 8 j with rightwards arrow on top plus h k with rightwards arrow on top

b with rightwards arrow on top equals open parentheses t plus 2 close parentheses i with rightwards arrow on top plus 4 j with rightwards arrow on top plus 2 k with rightwards arrow on top

begin mathsize 14px style a with rightwards arrow on top equals negative b with rightwards arrow on top end style

Ditanya:

Vektor a = ?

Jawab:

table attributes columnalign right center left columnspacing 0px end attributes row cell a with rightwards arrow on top end cell equals cell negative b with rightwards arrow on top end cell row cell open square brackets table row t row cell negative 8 end cell row h end table close square brackets end cell equals cell negative open square brackets table row cell t plus 2 end cell row 4 row 2 end table close square brackets end cell row blank left right arrow cell t equals negative left parenthesis t plus 2 right parenthesis end cell row cell negative 8 end cell not equal to cell negative 4 end cell row straight h equals cell negative 2 end cell row blank blank blank row blank blank cell soal space di space atas space tidak space dapat space dikerjakan end cell end table 

Jika kita asumsikan a with rightwards arrow on top equals negative 2 b with rightwards arrow on top, maka soal di atas dapat diselesaikan

table attributes columnalign right center left columnspacing 0px end attributes row cell a with rightwards arrow on top end cell equals cell negative 2 b with rightwards arrow on top end cell row cell open square brackets table row t row cell negative 8 end cell row h end table close square brackets end cell equals cell negative 2 open square brackets table row cell t plus 2 end cell row 4 row 2 end table close square brackets end cell row blank left right arrow cell t equals negative 2 open parentheses t plus 2 close parentheses end cell row t equals cell negative 2 t minus 4 end cell row cell 3 t end cell equals cell negative 4 end cell row t equals cell negative 4 over 3 end cell row cell negative 8 end cell equals cell negative 2 open parentheses 4 close parentheses end cell row cell negative 8 end cell equals cell negative 8 end cell row h equals cell negative 2 times 2 end cell row h equals cell negative 4 end cell row blank blank blank end table 

Jadi, vektor a dapat dinyatakan menjadi a with rightwards arrow on top equals negative 4 over 3 i with rightwards arrow on top minus 8 j with rightwards arrow on top minus 4 k with rightwards arrow on top.

 

0

Roboguru

Diketahui  dan . Vektor yang tegak lurus terhadap kedua vektor tersebut adalah ....

Pembahasan Soal:

Diketahui: a with rightwards harpoon with barb upwards on top equals 3 i minus j plus 2 k dan b with rightwards harpoon with barb upwards on top equals i plus 2 j minus 2 k

Sehingga, 

  • a with rightwards harpoon with barb upwards on top subscript x equals 3 comma space a with rightwards harpoon with barb upwards on top subscript y equals negative 1 comma space a with rightwards harpoon with barb upwards on top subscript z equals 2 
  • b with rightwards harpoon with barb upwards on top subscript x equals 1 comma space b with rightwards harpoon with barb upwards on top subscript y equals 2 comma space b with rightwards harpoon with barb upwards on top subscript z equals negative 2 

Vektor yang tegak lurus dengan vektor a with rightwards harpoon with barb upwards on top dan vektor b with rightwards harpoon with barb upwards on top merupakan vektor hasil dari perkalian silang a with rightwards harpoon with barb upwards on top cross times b with rightwards harpoon with barb upwards on top.

table attributes columnalign right center left columnspacing 0px end attributes row cell a with rightwards harpoon with barb upwards on top cross times b with rightwards harpoon with barb upwards on top end cell equals cell open parentheses a with rightwards harpoon with barb upwards on top subscript y b with rightwards harpoon with barb upwards on top subscript z minus a with rightwards harpoon with barb upwards on top subscript z b with rightwards harpoon with barb upwards on top subscript y close parentheses i plus open parentheses a with rightwards harpoon with barb upwards on top subscript z b with rightwards harpoon with barb upwards on top subscript x minus a with rightwards harpoon with barb upwards on top subscript x b with rightwards harpoon with barb upwards on top subscript z close parentheses j end cell row blank blank cell plus open parentheses a with rightwards harpoon with barb upwards on top subscript x b with rightwards harpoon with barb upwards on top subscript y minus a with rightwards harpoon with barb upwards on top subscript y b with rightwards harpoon with barb upwards on top subscript x close parentheses k end cell row blank equals cell open parentheses open parentheses negative 1 close parentheses times open parentheses negative 2 close parentheses minus 2 times 2 close parentheses i plus open parentheses 2 times 1 minus 3 times open parentheses negative 2 close parentheses close parentheses j end cell row blank blank cell plus open parentheses 3 times 2 minus open parentheses negative 1 close parentheses times 1 close parentheses k end cell row blank equals cell open parentheses 2 minus 4 close parentheses i plus open parentheses 2 plus 6 close parentheses j plus open parentheses 6 plus 1 close parentheses k end cell row blank equals cell negative 2 i minus 8 j plus 7 k end cell end table     

Jadi, vektor yang tegak lurus terhadap kedua vektor tersebut adalah table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell a with rightwards harpoon with barb upwards on top end cell end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cross times end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell b with rightwards harpoon with barb upwards 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 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 i 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 8 end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank j 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 7 end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank k 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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