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Pertanyaan

Gambar OADC.BEFG adalah paralelpipedum yaitu padatan dengan enam sisi. Setiap sisi jajargenjang yang berseberangan kongruen. O adalah titik asal. A, B, dan C masing-masing memiliki vektor posisi open parentheses table row 1 row 0 row 0 end table close parenthesesopen parentheses table row 2 row 1 row 2 end table close parentheses, dan open parentheses table row cell negative 3 end cell row 2 row 1 end table close parentheses.

  1. Tunjukkan bahwa OF with rightwards arrow on top dan CE with rightwards arrow on top tegak lurus.

A. Salim

Master Teacher

Mahasiswa/Alumni Universitas Pelita Harapan

Jawaban terverifikasi

Jawaban

terbukti bahwa OF with rightwards arrow on top dan CE with rightwards arrow on top tegak lurus.

Pembahasan

Vektor posisi dari A, B, dan C dengan titik asal O dapat ditulis berturut-turut sebagai vektor OA with rightwards arrow on top equals open parentheses table row 1 row 0 row 0 end table close parenthesesOB with rightwards arrow on top equals open parentheses table row 2 row 1 row 2 end table close parentheses, dan vektor OC with rightwards arrow on top equals open parentheses table row cell negative 3 end cell row 2 row 1 end table close parentheses.

Dari gambar pada soal, diperoleh bahwa:

OC with rightwards arrow on top equals AD with rightwards arrow on top equals BG with rightwards arrow on top equals EF with rightwards arrow on top  OB with rightwards arrow on top equals AE with rightwards arrow on top equals CD with rightwards arrow on top equals DF with rightwards arrow on top  OA with rightwards arrow on top equals CD with rightwards arrow on top equals BE with rightwards arrow on top equals GF with rightwards arrow on top

Vektor OF with rightwards arrow on top dan CE with rightwards arrow on top dapat diperoleh dengan menjumlahkan vektor sebagaimana perhitungan berikut:

table attributes columnalign right center left columnspacing 0px end attributes row cell OF with rightwards arrow on top end cell equals cell OA with rightwards arrow on top plus AE with rightwards arrow on top plus EF with rightwards arrow on top end cell row blank equals cell OA with rightwards arrow on top plus OB with rightwards arrow on top plus OC with rightwards arrow on top end cell row blank equals cell open parentheses table row 1 row 0 row 0 end table close parentheses plus open parentheses table row 2 row 1 row 2 end table close parentheses plus open parentheses table row cell negative 3 end cell row 2 row 1 end table close parentheses end cell row blank equals cell open parentheses table row cell 1 plus 2 minus 3 end cell row cell 0 plus 1 plus 2 end cell row cell 0 plus 2 plus 1 end cell end table close parentheses end cell row blank equals cell open parentheses table row 0 row 3 row 3 end table close parentheses end cell end table

table attributes columnalign right center left columnspacing 0px end attributes row cell CE with rightwards arrow on top end cell equals cell CD with rightwards arrow on top plus DA with rightwards arrow on top plus AE with rightwards arrow on top end cell row blank equals cell OA with rightwards arrow on top minus AD with rightwards arrow on top plus OB with rightwards arrow on top end cell row blank equals cell OA with rightwards arrow on top minus OC with rightwards arrow on top plus OB with rightwards arrow on top end cell row blank equals cell open parentheses table row 1 row 0 row 0 end table close parentheses minus open parentheses table row cell negative 3 end cell row 2 row 1 end table close parentheses plus open parentheses table row 2 row 1 row 2 end table close parentheses end cell row blank equals cell open parentheses table row cell 1 minus open parentheses negative 3 close parentheses plus 2 end cell row cell 0 minus 2 plus 1 end cell row cell 0 minus 1 plus 2 end cell end table close parentheses end cell row blank equals cell open parentheses table row 6 row cell negative 1 end cell row 1 end table close parentheses end cell end table

Dua vektor dapat dikatakan saling tegak lurus apabila hasil kali dot product kedua vektor tersebut memberikan nilai nol.

Ingat bahwa hasil kali dot product dari vektor bold italic a equals open parentheses table row cell a subscript 1 end cell row cell a subscript 2 end cell row cell a subscript 3 end cell end table close parentheses dan bold italic b equals open parentheses table row cell b subscript 1 end cell row cell b subscript 2 end cell row cell b subscript 3 end cell end table close parentheses dapat dirumuskan sebagai berikut:

bold italic a times bold italic b equals a subscript 1 b subscript 1 plus a subscript 2 b subscript 2 plus a subscript 3 b subscript 3

Sehingga, 

table attributes columnalign right center left columnspacing 0px end attributes row cell OF with rightwards arrow on top times CE with rightwards arrow on top end cell equals cell open parentheses table row 0 row 3 row 3 end table close parentheses times open parentheses table row 6 row cell negative 1 end cell row 1 end table close parentheses end cell row blank equals cell open parentheses 0 close parentheses open parentheses 6 close parentheses plus open parentheses 3 close parentheses open parentheses negative 1 close parentheses plus open parentheses 3 close parentheses open parentheses 1 close parentheses end cell row blank equals 0 end table

Karena hasil kali dot product antara OF with rightwards arrow on top dan CE with rightwards arrow on top bernilai nol, maka OF with rightwards arrow on top dan CE with rightwards arrow on top saling tegak lurus.

Dengan demikian, terbukti bahwa OF with rightwards arrow on top dan CE with rightwards arrow on top tegak lurus.

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