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Diketahui Jika x=ku+lv+mw dan y​=(k+l)v  maka pernyataan dibawah ini yang benar adalah .... k + l + m = 2.   Cosinus sudut antara u dan v adalah −51​.   x⋅y​​=4.   , tetapi y berlawanan arah dengan u.

Pertanyaan

Diketahui begin mathsize 14px style straight u with rightwards arrow on top equals open square brackets table row 1 row 0 row 2 end table close square brackets comma blank straight v with rightwards arrow on top equals open square brackets table row cell negative 1 end cell row 2 row 0 end table close square brackets comma blank straight w with rightwards arrow on top equals open square brackets table row 3 row 1 row 1 end table close square brackets comma blank straight x with rightwards arrow on top equals open square brackets table row 6 row cell negative 1 end cell row 5 end table close square brackets. end style

Jika begin mathsize 14px style straight x with rightwards arrow on top equals straight k straight u with rightwards arrow on top plus straight l straight v with rightwards arrow on top plus straight m straight w with rightwards arrow on top end style dan begin mathsize 14px style straight y with rightwards arrow on top equals left parenthesis straight k plus straight l right parenthesis straight v with rightwards arrow on top end style  maka pernyataan dibawah ini yang benar adalah ....

  1. = 2.
     
  2. Cosinus sudut antara undefined adalah begin mathsize 14px style negative 1 fifth end style.
     
  3. begin mathsize 14px style square root of straight x with rightwards arrow on top times straight y with rightwards arrow on top end root equals 4 end style.
     
  4. begin mathsize 14px style open vertical bar straight y with rightwards arrow on top close vertical bar equals open vertical bar straight u with rightwards arrow on top close vertical bar end style, tetapi y berlawanan arah dengan begin mathsize 14px style straight u with rightwards arrow on top end style.
  1. 1, 2, dan 3undefined 

  2. 1 dan 3undefined 

  3. 2 dan 4undefined 

  4. 4 sajaundefined 

  5. Semua pernyataan benarundefined 

S. Nur

Master Teacher

Jawaban terverifikasi

Jawaban

pilihan jawaban yang tepat untuk permasalahan ini adalah A.

Pembahasan

Diketahui :

begin mathsize 14px style straight u with rightwards arrow on top equals open square brackets table row 1 row 0 row 2 end table close square brackets comma blank straight v with rightwards arrow on top equals open square brackets table row cell negative 1 end cell row 2 row 0 end table close square brackets comma blank straight w with rightwards arrow on top equals open square brackets table row 3 row 1 row 1 end table close square brackets comma blank straight x with rightwards arrow on top equals open square brackets table row 6 row cell negative 1 end cell row 5 end table close square brackets straight x with rightwards arrow on top equals straight k straight u with rightwards arrow on top plus straight l straight v with rightwards arrow on top plus straight m straight w with rightwards arrow on top straight y with rightwards arrow on top equals left parenthesis straight k plus straight l right parenthesis straight v with rightwards arrow on top end style 


Membuktikan pernyataan 1

begin mathsize 14px style          blank straight x with rightwards arrow on top equals straight k straight u with rightwards arrow on top plus straight l straight v with rightwards arrow on top plus straight m straight w with rightwards arrow on top open square brackets table row 6 row cell negative 1 end cell row 5 end table close square brackets equals straight k open square brackets table row 1 row 0 row 2 end table close square brackets plus straight l open square brackets table row cell negative 1 end cell row 2 row 0 end table close square brackets plus straight m open square brackets table row 3 row 1 row 1 end table close square brackets open square brackets table row 6 row cell negative 1 end cell row 5 end table close square brackets equals open square brackets table row straight k row 0 row cell 2 straight k end cell end table close square brackets plus open square brackets table row cell negative straight l end cell row cell 2 straight l end cell row 0 end table close square brackets plus open square brackets table row cell 3 straight m end cell row straight m row straight m end table close square brackets open square brackets table row 6 row cell negative 1 end cell row 5 end table close square brackets equals open square brackets table row cell straight k minus straight l plus 3 straight m end cell row cell 2 straight l plus straight m end cell row cell 2 straight k plus straight m end cell end table close square brackets end style 

Berdasarkan persamaan vektor di atas maka diperoleh 3 persamaan yaitu

begin mathsize 14px style straight k minus straight l plus 3 straight m equals 6 space horizontal ellipsis space left parenthesis 1 right parenthesis 2 straight l plus straight m equals negative 1 space horizontal ellipsis space left parenthesis 2 right parenthesis 2 straight k plus straight m equals 5 space horizontal ellipsis space left parenthesis 3 right parenthesis space space end style 

Eliminasikan persamaan (2) dan (3)
begin mathsize 14px style bottom enclose 2 straight l plus straight m equals negative 1 2 straight k plus straight m equals 5 minus end enclose 2 straight l minus 2 straight k equals negative 6 straight l minus straight k equals negative 3 straight k minus straight l equals 3 end style 

Subtitusi persamaan (4) ke dalam persamaan (1)
begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell straight k minus straight l plus 3 straight m end cell equals 6 row cell 3 plus 3 straight m end cell equals 6 row cell 3 straight m end cell equals 3 row bold m bold equals bold 1 end table end style 

Subtitusi nilai = 1 ke persamaan (2)
begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell 2 straight l plus straight m end cell equals cell negative 1 end cell row cell 2 straight l plus 1 end cell equals cell negative 1 end cell row cell 2 straight l end cell equals cell negative 2 end cell row bold l bold equals cell bold minus bold 1 blank end cell end table end style 

Subtitusi nilai = 1 ke persamaan (3)
begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell 2 straight k plus straight m end cell equals 5 row cell 2 straight k plus 1 end cell equals 5 row cell 2 straight k end cell equals 4 row bold k bold equals bold 2 end table end style 

Maka nilai = 2 + -+ 1 = 2 .

Dengan demikian, Pernyataan 1 benar.


Membuktikan pernyataan 2

Misalkan undefined adalah sudut yang dibentuk oleh vektor begin mathsize 14px style straight u with rightwards arrow on top end style dan begin mathsize 14px style straight v with rightwards arrow on top end style 

begin mathsize 14px style cos invisible function application straight theta equals fraction numerator straight u with rightwards arrow on top times straight v with rightwards arrow on top over denominator open vertical bar straight u with rightwards arrow on top close vertical bar times open vertical bar straight v with rightwards arrow on top close vertical bar end fraction cos invisible function application straight theta equals fraction numerator open square brackets table row 1 row 0 row 2 end table close square brackets times open square brackets table row cell negative 1 end cell row 2 row 0 end table close square brackets over denominator square root of 1 squared plus 0 squared plus 2 squared end root times square root of left parenthesis negative 1 right parenthesis squared plus 2 squared plus 0 squared end root end fraction cos invisible function application straight theta equals fraction numerator negative 1 plus 0 plus 0 over denominator square root of 1 plus 4 end root times square root of 1 plus 4 end root end fraction cos invisible function application straight theta equals fraction numerator negative 1 over denominator square root of 5 times square root of 5 end fraction cos invisible function application straight theta equals negative 1 fifth end style   

Pernyataan 2 benar.


Membuktikan pernyataan 3

begin mathsize 14px style straight y with rightwards arrow on top equals left parenthesis straight k plus straight l right parenthesis straight u with rightwards arrow on top straight y with rightwards arrow on top equals left parenthesis 2 plus left parenthesis negative 1 right parenthesis right parenthesis straight u with rightwards arrow on top straight y with rightwards arrow on top equals straight u with rightwards arrow on top equals open square brackets table row 1 row 0 row 2 end table close square brackets end style 

Dengan demikian,

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell straight x with rightwards arrow on top times straight y with rightwards arrow on top end cell equals cell open square brackets table row 6 row cell negative 1 end cell row 5 end table close square brackets times open square brackets table row 1 row 0 row 2 end table close square brackets end cell row cell straight x with rightwards arrow on top times straight y with rightwards arrow on top end cell equals cell 6 open parentheses 1 close parentheses plus open parentheses negative 1 close parentheses open parentheses 0 close parentheses plus 5 open parentheses 2 close parentheses end cell row cell straight x with rightwards arrow on top times straight y with rightwards arrow on top end cell equals cell 6 plus 10 end cell row cell straight x with rightwards arrow on top times straight y with rightwards arrow on top end cell equals 16 row cell square root of straight x with rightwards arrow on top times straight y with rightwards arrow on top end root end cell equals cell square root of 16 equals 4 end cell end table end style 

Jadi, pernyataan 3 adalah benar


Membuktikan pernyataan 4

Diketahui bahwa
begin mathsize 14px style straight y with rightwards arrow on top equals straight u with rightwards arrow on top equals open square brackets table row 1 row 0 row 2 end table close square brackets end style 

Maka vektor begin mathsize 14px style straight y with rightwards arrow on top end style searah dengan vektor begin mathsize 14px style straight u with rightwards arrow on top. blank end style 

Dengan demikian, pernyataan 4 Salah.


Berdasarkan  pembuktian di atas, maka dapat diketahui bahwa hanya pernyataan 1, 2, dan 3 SAJA yang benar.

Dengan demikian, pilihan jawaban yang tepat untuk permasalahan ini adalah A.

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Pertanyaan serupa

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