Roboguru

Diketahui koordinat titik A(4,−3),B(−1,−5),danC(−2,3). Jika a,b,danc berturut-turut vektor posisi titik A, B, dan C, hasil 2a−3b+c adalah ...

Pertanyaan

Diketahui koordinat titik straight A open parentheses 4 comma negative 3 close parentheses comma space straight B open parentheses negative 1 comma negative 5 close parentheses comma space dan space straight C open parentheses negative 2 comma 3 close parentheses. Jika straight a with rightwards arrow on top comma space straight b with rightwards arrow on top comma space dan space straight c with rightwards arrow on top berturut-turut vektor posisi titik A, B, dan C, hasil 2 straight a with rightwards arrow on top minus 3 straight b with rightwards arrow on top plus straight c with rightwards arrow on top adalah ...

  1. open parentheses table row 9 row 12 end table close parentheses

  2. open parentheses table row 9 row 10 end table close parentheses

  3. open parentheses table row 9 row 6 end table close parentheses

  4. open parentheses table row 8 row 12 end table close parentheses

  5. open parentheses table row 8 row 6 end table close parentheses

Pembahasan Soal:

Karena  straight a with rightwards arrow on top comma space straight b with rightwards arrow on top comma space dan space straight c with rightwards arrow on top berturut-turut vektor posisi titik A, B, dan C, maka

table attributes columnalign right center left columnspacing 0px end attributes row cell 2 straight a with rightwards arrow on top minus 3 straight b with rightwards arrow on top plus straight c end cell equals cell 2 open parentheses table row 4 row cell negative 3 end cell end table close parentheses minus 3 open parentheses table row cell negative 1 end cell row cell negative 5 end cell end table close parentheses plus open parentheses table row cell negative 2 end cell row 3 end table close parentheses end cell row blank equals cell open parentheses table row 8 row cell negative 6 end cell end table close parentheses minus open parentheses table row cell negative 3 end cell row cell negative 15 end cell end table close parentheses plus open parentheses table row cell negative 2 end cell row 3 end table close parentheses end cell row blank equals cell open parentheses table row 9 row 12 end table close parentheses end cell end table

Sehingga, 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 a 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 3 end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell straight b 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 straight c 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 9 row 12 end table close parentheses end cell end table

Jadi, pilihan jawaban yang tepat adalah A.

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

R. Azizatul

Mahasiswa/Alumni Universitas Negeri Yogyakarta

Terakhir diupdate 06 Oktober 2021

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Pertanyaan yang serupa

Vektor-vektor dari posisi A, B, dan C adalah a=(4i,3j), b=(−2i,3j), c=(1i,5j). Hitunglah vektor 2a+b−3c

Pembahasan Soal:

Jika diketahui vektor a with rightwards arrow on top equals open parentheses table row cell a subscript 1 end cell row cell a subscript 2 end cell end table close parentheses dan b with rightwards arrow on top equals open parentheses table row cell b subscript 1 end cell row cell b subscript 2 end cell end table close parentheses, maka berlaku

a with rightwards arrow on top plus b with rightwards arrow on top equals open parentheses table row cell a subscript 1 end cell row cell a subscript 2 end cell end table close parentheses plus open parentheses table row cell b subscript 1 end cell row cell b subscript 2 end cell end table close parentheses equals open parentheses table row cell a subscript 1 plus b subscript 1 end cell row cell a subscript 2 plus b subscript 2 end cell end table close parentheses

a with rightwards arrow on top minus b with rightwards arrow on top equals open parentheses table row cell a subscript 1 end cell row cell a subscript 2 end cell end table close parentheses minus open parentheses table row cell b subscript 1 end cell row cell b subscript 2 end cell end table close parentheses equals open parentheses table row cell a subscript 1 minus b subscript 1 end cell row cell a subscript 2 minus b subscript 2 end cell end table close parentheses

k times a with rightwards arrow on top equals k open parentheses table row cell a subscript 1 end cell row cell a subscript 2 end cell end table close parentheses equals open parentheses table row cell k times a subscript 1 end cell row cell k times a subscript 2 end cell end table close parentheses

 

Hasil dari operasi hitung vektor tersebut adalah sebagai berikut.

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

Dengan demikian, hasil dari 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 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 b 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 3 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 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 3 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 6 end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank j end table 

0

Roboguru

Diketahui vektor a=5i−12j​ dan b=(3,4), vektor c=−2i−6j​. Tentukan: a) vektor a−2b−c b) vektor 2a+b−21​c

Pembahasan Soal:

Ingat kemabli konsep operasi hitung vektor.

Untuk a dan b vektor-vektor pada dua dimensi, berlaku:

a±b==(a1,a2)+(b1,b2)(a1+b1,a2+b2)

Perkalian skalar dengan vektor.

Jika k skalar tak nol dan vektor a=(a1,a2,...an), maka ka=(ka1,ka2,...kan).

Diketahui:

vektor a with rightwards arrow on top equals 5 i with hat on top minus 12 j with hat on top dan b with rightwards arrow on top equals left parenthesis 3 comma space 4 right parenthesis equals 3 i with hat on top plus 4 j with hat on top, vektor c with rightwards arrow on top equals negative 2 i with hat on top minus 6 j with hat on top.

a) menentukan vektor a with rightwards arrow on top minus 2 b with rightwards arrow on top minus c with rightwards arrow on top

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

b) menentukan vektor 2 a with rightwards arrow on top plus b with rightwards arrow on top minus 1 half c with rightwards arrow on top

table attributes columnalign right center left columnspacing 2px end attributes row cell 2 a with rightwards arrow on top plus b with rightwards arrow on top minus 1 half c with rightwards arrow on top end cell equals cell 2 open parentheses 5 i with hat on top minus 12 j with hat on top close parentheses plus open parentheses 3 i with hat on top minus 4 j with hat on top close parentheses minus 1 half open parentheses negative 2 i with hat on top minus 6 j with hat on top close parentheses end cell row blank equals cell 10 i with hat on top minus 24 j with hat on top plus 3 i with hat on top minus 4 j with hat on top plus i with hat on top plus 3 j with hat on top end cell row cell 2 a with rightwards arrow on top plus b with rightwards arrow on top minus 1 half c with rightwards arrow on top end cell equals cell 14 i with hat on top minus 25 j with hat on top end cell end table

Jadi, diperoleh a with rightwards arrow on top minus 2 b with rightwards arrow on top minus c with rightwards arrow on top equals i with hat on top plus 2 j with hat on top dan 2 a with rightwards arrow on top plus b with rightwards arrow on top minus 1 half c with rightwards arrow on top equals 14 i with hat on top minus 25 j with hat on top.

0

Roboguru

Diketahui a⇀=4i+j+5k dan b⇀=2i+j−5k  a. a⇀+b⇀  b. a⇀−b⇀  c. a⇀−2b⇀  d. 2a⇀⋅3b⇀

Pembahasan Soal:

Diketahui a with rightwards harpoon with barb upwards on top equals 4 i plus j plus 5 k dan b with rightwards harpoon with barb upwards on top equals 2 i plus j minus 5 k 

a. Ingat kembali jika a with rightwards harpoon with barb upwards on top equals a subscript 1 i plus a subscript 2 j plus a subscript 3 k dan b with rightwards harpoon with barb upwards on top equals b subscript 1 i plus b subscript 2 j plus b subscript 3 k maka a with rightwards harpoon with barb upwards on top plus b with rightwards harpoon with barb upwards on top 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 

table attributes columnalign right center left columnspacing 0px end attributes row cell a with rightwards harpoon with barb upwards on top plus b with rightwards harpoon with barb upwards on top end cell equals cell left parenthesis 4 plus 2 right parenthesis i plus left parenthesis 1 plus 1 right parenthesis j plus left parenthesis 5 plus left parenthesis negative 5 right parenthesis right parenthesis k end cell row blank equals cell 6 i plus 2 j end cell end table 

Jadi, a with rightwards harpoon with barb upwards on top plus b with rightwards harpoon with barb upwards on top equals 6 i plus 2 j.

b. Ingat kembali bahwa jika a with rightwards harpoon with barb upwards on top equals a subscript 1 i plus a subscript 2 j plus a subscript 3 k dan b with rightwards harpoon with barb upwards on top equals b subscript 1 i plus b subscript 2 j plus b subscript 3 k maka a with rightwards harpoon with barb upwards on top minus b with rightwards harpoon with barb upwards on top 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 

table attributes columnalign right center left columnspacing 0px end attributes row cell a with rightwards harpoon with barb upwards on top minus b with rightwards harpoon with barb upwards on top end cell equals cell left parenthesis 4 minus 2 right parenthesis i plus left parenthesis 1 minus 1 right parenthesis j plus left parenthesis 5 minus left parenthesis negative 5 right parenthesis k end cell row blank equals cell 2 i plus 10 k end cell end table 

Jadi, a with rightwards harpoon with barb upwards on top minus b with rightwards harpoon with barb upwards on top equals 2 i plus 10 k.

c. Ingat kembali bahwa misalkan k adalah bilangan skalar dan a with rightwards harpoon with barb upwards on top equals a subscript 1 i plus a subscript 2 j plus a subscript 3 k maka  k a with rightwards harpoon with barb upwards on top equals k a subscript 1 i plus k a subscript 2 k plus k a subscript 3 k 

table attributes columnalign right center left columnspacing 0px end attributes row cell a with rightwards harpoon with barb upwards on top minus 2 b with rightwards harpoon with barb upwards on top end cell equals cell left parenthesis 4 i plus j plus 5 k right parenthesis minus left parenthesis 2 left parenthesis 2 i plus j minus 5 k right parenthesis right parenthesis end cell row blank equals cell left parenthesis 4 i plus j plus 5 k right parenthesis minus left parenthesis 4 i plus 2 j minus 10 k right parenthesis end cell row blank equals cell left parenthesis 4 minus 4 right parenthesis i plus left parenthesis 1 minus 2 right parenthesis j plus left parenthesis 5 minus left parenthesis negative 10 right parenthesis right parenthesis k end cell row blank equals cell negative j plus 15 k end cell end table 

Jadi, a with rightwards harpoon with barb upwards on top minus 2 b with rightwards harpoon with barb upwards on top equals negative j plus 15 k.

d. Ingat kembali bahwa jika a with rightwards harpoon with barb upwards on top equals a subscript 1 i plus a subscript 2 j plus a subscript 3 k dan b with rightwards harpoon with barb upwards on top equals b subscript 1 i plus b subscript 2 j plus b subscript 3 k maka a with rightwards harpoon with barb upwards on top times b with rightwards harpoon with barb upwards on top equals a subscript 1 times b subscript 1 plus a subscript 2 times b subscript 2 plus a subscript 3 times b subscript 3 

table attributes columnalign right center left columnspacing 0px end attributes row cell 2 a with rightwards harpoon with barb upwards on top times 3 b with rightwards harpoon with barb upwards on top end cell equals cell 2 left parenthesis 4 comma space 1 comma space 5 right parenthesis times 3 left parenthesis 2 comma space 1 comma space minus 5 right parenthesis end cell row blank equals cell left parenthesis 8 comma space 2 comma space 10 right parenthesis times left parenthesis 6 comma space 3 comma space minus 15 right parenthesis end cell row blank equals cell 8 times 6 plus 2 times 3 plus 10 times negative 15 end cell row blank equals cell 48 plus 6 minus 150 end cell row blank equals cell 54 minus 150 end cell row blank equals cell negative 96 end cell end table 

Jadi, 2 a with rightwards harpoon with barb upwards on top times 3 b with rightwards harpoon with barb upwards on top equals negative 96.

Dengan demikian, hasil yang diperoleh adalah a with rightwards harpoon with barb upwards on top plus b with rightwards harpoon with barb upwards on top equals 6 i plus 2 ja with rightwards harpoon with barb upwards on top minus b with rightwards harpoon with barb upwards on top equals 2 i plus 10 k dan 2 a with rightwards harpoon with barb upwards on top times 3 b with rightwards harpoon with barb upwards on top equals negative 96.

0

Roboguru

Diketahui A(2,6,−5), B(−1,3,7), dan C(4,−1,8). AB dan BC berturut-turut mewakili vektor u dan v. Tentukan: d. (2u−3v)2

Pembahasan Soal:

Jika a with rightwards arrow on top equals open parentheses table row cell a subscript 1 end cell row cell a subscript 2 end cell end table close parentheses dan b with rightwards arrow on top equals open parentheses table row cell b subscript 1 end cell row cell b subscript 2 end cell end table close parentheses, maka a with rightwards arrow on top times b with rightwards arrow on top adalah:

table attributes columnalign right center left columnspacing 0px end attributes row cell a with rightwards arrow on top times b with rightwards arrow on top end cell equals cell open parentheses table row cell a subscript 1 end cell row cell a subscript 2 end cell end table close parentheses times open parentheses table row cell b subscript 1 end cell row cell b subscript 2 end cell end table close parentheses end cell row blank equals cell a subscript 1 b subscript 1 plus a subscript 2 b subscript 2 end cell end table 

Vektor Posisi adalah vektor yang berpangkal di pusat koordinat open parentheses 0 comma space 0 close parentheses dan berujung di suatu titik open parentheses x comma space y close parentheses.

Diketahui straight A open parentheses 2 comma space 6 comma space minus 5 close parentheses, straight B open parentheses negative 1 comma space 3 comma space 7 close parentheses, dan straight C open parentheses 4 comma space minus 1 comma space 8 close parentheses.

Nilai vektor posisi akan sama dengan koordinat titik ujungnya, maka tentukan AB with rightwards arrow on top dan BC with rightwards arrow on top.

table attributes columnalign right center left columnspacing 0px end attributes row cell AB with rightwards arrow on top end cell equals cell b with rightwards arrow on top minus a with rightwards arrow on top end cell row cell u with rightwards arrow on top end cell equals cell open parentheses table row cell negative 1 end cell row 3 row 7 end table close parentheses minus open parentheses table row 2 row 6 row cell negative 5 end cell end table close parentheses end cell row cell u with rightwards arrow on top end cell equals cell open parentheses table row cell negative 1 minus 2 end cell row cell 3 minus 6 end cell row cell 7 plus 5 end cell end table close parentheses end cell row cell u with rightwards arrow on top end cell equals cell open parentheses table row cell negative 3 end cell row cell negative 3 end cell row 12 end table close parentheses end cell row cell BC with rightwards arrow on top end cell equals cell c with rightwards arrow on top minus b with rightwards arrow on top end cell row cell v with rightwards arrow on top end cell equals cell open parentheses table row 4 row cell negative 1 end cell row 8 end table close parentheses minus open parentheses table row cell negative 1 end cell row 3 row 7 end table close parentheses end cell row cell v with rightwards arrow on top end cell equals cell open parentheses table row cell 4 plus 1 end cell row cell negative 1 minus 3 end cell row cell 8 minus 7 end cell end table close parentheses end cell row cell v with rightwards arrow on top end cell equals cell open parentheses table row 5 row cell negative 4 end cell row 1 end table close parentheses end cell end table   

d. open parentheses 2 u with rightwards arrow on top minus 3 v with rightwards arrow on top close parentheses squared

begin mathsize 12px style table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell open parentheses 2 u with rightwards arrow on top minus 3 v with rightwards arrow on top close parentheses squared end cell row blank equals cell 4 open parentheses u with rightwards arrow on top close parentheses squared minus 12 open parentheses u with rightwards arrow on top times v with rightwards arrow on top close parentheses plus 9 open parentheses v with rightwards arrow on top close parentheses squared end cell row blank equals cell 4 open parentheses table row cell negative 3 end cell row cell negative 3 end cell row 12 end table close parentheses squared minus 12 open parentheses open parentheses table row cell negative 3 end cell row cell negative 3 end cell row cell negative 3 end cell end table close parentheses times open parentheses table row 5 row cell negative 4 end cell row 1 end table close parentheses close parentheses plus 9 open parentheses table row 5 row cell negative 4 end cell row 1 end table close parentheses squared end cell row blank equals cell 4 open parentheses open parentheses negative 3 close parentheses squared plus open parentheses negative 3 close parentheses squared plus 12 squared close parentheses minus 12 open parentheses negative 15 plus 12 plus 12 close parentheses plus 9 open parentheses 5 squared plus open parentheses negative 4 close parentheses squared plus 1 squared close parentheses end cell row blank equals cell 4 times 162 minus 12 times 9 plus 9 times 42 end cell row blank equals cell 648 minus 108 plus 378 end cell row blank equals 918 end table end style 

Jadi, Error converting from MathML to accessible text..

0

Roboguru

Diketahui vektor a=(24​), vektor b=(−5−7​), jika c=2a+b maka hasil dari a−3b+2c adalah ....

Pembahasan Soal:

Diketahui vektor begin mathsize 14px style straight a with rightwards arrow on top equals open parentheses table row 2 row 4 end table close parentheses end style, vektor begin mathsize 14px style straight b with rightwards arrow on top equals open parentheses table row cell negative 5 end cell row cell negative 7 end cell end table close parentheses end style, dan begin mathsize 14px style straight c with rightwards arrow on top equals 2 straight a with rightwards arrow on top plus straight b with rightwards arrow on top end style. Maka:

begin mathsize 14px style straight c with rightwards arrow on top equals 2 straight a with rightwards arrow on top plus straight b with rightwards arrow on top straight c with rightwards arrow on top equals 2 open parentheses table row 2 row 4 end table close parentheses plus open parentheses table row cell negative 5 end cell row cell negative 7 end cell end table close parentheses straight c with rightwards arrow on top equals open parentheses table row 4 row 8 end table close parentheses plus open parentheses table row cell negative 5 end cell row cell negative 7 end cell end table close parentheses straight c with rightwards arrow on top equals open parentheses table row cell negative 1 end cell row 1 end table close parentheses end style 

Sehingga:

 begin mathsize 14px style straight a with rightwards arrow on top minus 3 straight b with rightwards arrow on top plus 2 straight c with rightwards arrow on top equals open parentheses table row 2 row 4 end table close parentheses minus 3 open parentheses table row cell negative 5 end cell row cell negative 7 end cell end table close parentheses plus 2 open parentheses table row cell negative 1 end cell row 1 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 2 row 4 end table close parentheses minus open parentheses table row cell negative 15 end cell row cell negative 21 end cell end table close parentheses plus open parentheses table row cell negative 2 end cell row 2 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 15 row 27 end table close parentheses end style 

Jadi, 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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