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Pertanyaan

Diketahui titik A(3,-6,5) dan B(-1,2,1). Titik C terletak pada perpanjangan AB sehingga CB : BA = 3 : 4. Jika undefined masing-masing menyatakan vektor posisi dari titik A, B, dan C, maka besar sudut yang dibentuk oleh vektor posisi begin mathsize 14px style b with rightwards arrow on top space d a n space c with rightwards arrow on top end style adalah ....

  1. begin mathsize 14px style arccos space open parentheses 3 over 14 square root of 3 close parentheses end style

  2. begin mathsize 14px style arccos space open parentheses 3 over 14 square root of 14 close parentheses end style

  3. begin mathsize 14px style arccos space open parentheses 3 over 7 square root of 3 close parentheses end style

  4. begin mathsize 14px style arccos space open parentheses 3 over 7 square root of 7 close parentheses end style

  5. begin mathsize 14px style arccos space open parentheses 3 over 7 square root of 14 close parentheses end style

A. Rizky

Master Teacher

Mahasiswa/Alumni Universitas Negeri Malang

Jawaban terverifikasi

Jawaban

jawaban yang tepat adalah B.

Pembahasan

Karena titik C terletak pada perpanjangan AB dan CB : BA = 3 : 4, maka didapatkan ilustrasi sebagai berikut

 

Misalkan koordinat titik C adalah (x, y, z).

Jika undefined  masing-masing menyatakan vektor posisi dari titik A, B, dan C, maka didapat bahwa

begin mathsize 14px style a with rightwards arrow on top equals open parentheses table row 3 row cell negative 6 end cell row 5 end table close parentheses b with rightwards arrow on top equals open parentheses table row cell negative 1 end cell row 2 row 1 end table close parentheses c with rightwards arrow on top equals open parentheses table row x row y row z end table close parentheses end style

Perhatikan bahwa
CB : BA = m : n = 3 : 4
Maka dapat diperoleh m = 3 dan n = 4.

Oleh karena itu, didapat bahwa


begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell b with rightwards arrow on top end cell equals cell fraction numerator m a with rightwards arrow on top plus n c with rightwards arrow on top over denominator m plus n end fraction end cell row cell open parentheses table row cell negative 1 end cell row 2 row 1 end table close parentheses end cell equals cell fraction numerator 3 open parentheses table row 3 row cell negative 6 end cell row 5 end table close parentheses plus 4 open parentheses table row x row y row z end table close parentheses over denominator 3 plus 4 end fraction end cell row cell open parentheses table row cell negative 1 end cell row 2 row 1 end table close parentheses end cell equals cell fraction numerator open parentheses table row 9 row cell negative 18 end cell row 15 end table close parentheses plus 4 open parentheses table row x row y row z end table close parentheses over denominator 7 end fraction end cell row cell 7 open parentheses table row cell negative 1 end cell row 2 row 1 end table close parentheses end cell equals cell open parentheses table row 9 row cell negative 18 end cell row 15 end table close parentheses plus 4 open parentheses table row x row y row z end table close parentheses end cell row cell open parentheses table row cell negative 7 end cell row 14 row 7 end table close parentheses end cell equals cell open parentheses table row 9 row cell negative 18 end cell row 15 end table close parentheses plus 4 open parentheses table row x row y row z end table close parentheses end cell row cell open parentheses table row cell negative 7 end cell row 14 row 7 end table close parentheses minus open parentheses table row 9 row cell negative 18 end cell row 15 end table close parentheses end cell equals cell 4 open parentheses table row x row y row z end table close parentheses end cell row cell open parentheses table row cell negative 16 end cell row 32 row cell negative 8 end cell end table close parentheses end cell equals cell 4 open parentheses table row x row y row z end table close parentheses end cell row cell 1 fourth open parentheses table row cell negative 16 end cell row 32 row cell negative 8 end cell end table close parentheses end cell equals cell open parentheses table row x row y row z end table close parentheses end cell row cell open parentheses table row x row y row z end table close parentheses end cell equals cell open parentheses table row cell negative 4 end cell row 8 row cell negative 2 end cell end table close parentheses end cell end table end style

Sehingga didapat x=-4, y=8, dan z=-2.
Maka koordinat dari titik C adalah (-4,8,-2).

Kemudian, didapat bahwa begin mathsize 14px style b with rightwards arrow on top equals open parentheses table row cell negative 1 end cell row 2 row 1 end table close parentheses end style  dan begin mathsize 14px style c with rightwards arrow on top equals open parentheses table row cell negative 4 end cell row 8 row cell negative 2 end cell end table close parentheses end style. Misalkan α adalah sudut yang dibentuk oleh vektor posisi begin mathsize 14px style b with rightwards arrow on top space d a n space c with rightwards arrow on top end style. Oleh karena itu, didapatkan begin mathsize 14px style cos space alpha equals fraction numerator b with rightwards arrow on top times c with rightwards arrow on top over denominator open vertical bar b with rightwards arrow on top close vertical bar open vertical bar c with rightwards arrow on top close vertical bar end fraction end style

Karena  begin mathsize 14px style b with rightwards arrow on top equals open parentheses table row cell negative 1 end cell row 2 row 1 end table close parentheses end style dan begin mathsize 14px style c with rightwards arrow on top equals open parentheses table row cell negative 4 end cell row 8 row cell negative 2 end cell end table close parentheses end style, maka

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell b with rightwards arrow on top times c with rightwards arrow on top end cell equals cell open parentheses table row cell negative 1 end cell row 2 row 1 end table close parentheses times open parentheses table row cell negative 4 end cell row 8 row cell negative 2 end cell end table close parentheses end cell row blank equals cell open parentheses negative 1 close parentheses times open parentheses negative 4 close parentheses plus 2 times 8 plus 1 times open parentheses negative 2 close parentheses end cell row blank equals cell 4 plus 16 minus 2 end cell row blank equals 18 end table end style

Kemudian, didapatkan pula bahwa

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell open vertical bar b with rightwards arrow on top close vertical bar end cell equals cell square root of open parentheses negative 1 close parentheses squared plus 2 squared plus 1 squared end root end cell row blank equals cell square root of 1 plus 4 plus 1 end root end cell row blank equals cell square root of 6 end cell end table end style

dan

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell open vertical bar c with rightwards arrow on top close vertical bar end cell equals cell square root of open parentheses negative 4 close parentheses squared plus 8 squared plus open parentheses negative 2 close parentheses squared end root end cell row blank equals cell square root of 16 plus 64 plus 4 end root end cell row blank equals cell square root of 84 end cell row blank equals cell 2 square root of 21 end cell end table end style

Oleh karena itu, didapatkan bahwa

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell cos space alpha end cell equals cell fraction numerator b with rightwards arrow on top times c with rightwards arrow on top over denominator open vertical bar b with rightwards arrow on top close vertical bar open vertical bar c with rightwards arrow on top close vertical bar end fraction end cell row cell cos space alpha end cell equals cell fraction numerator 18 over denominator square root of 6 times 2 square root of 21 end fraction end cell row cell cos space alpha end cell equals cell fraction numerator 18 over denominator 2 square root of 126 end fraction end cell row cell cos space alpha end cell equals cell fraction numerator 9 over denominator square root of 126 end fraction end cell row cell cos space alpha end cell equals cell fraction numerator 9 over denominator 3 square root of 14 end fraction end cell row cell cos space alpha end cell equals cell fraction numerator 3 over denominator square root of 14 end fraction end cell row cell cos space alpha end cell equals cell fraction numerator 3 over denominator square root of 14 end fraction times fraction numerator square root of 14 over denominator square root of 14 end fraction end cell row cell cos space alpha end cell equals cell 3 over 14 square root of 14 end cell row alpha equals cell arccos space open parentheses 3 over 14 square root of 14 close parentheses end cell end table end style

Jadi, jawaban yang tepat adalah B.

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