Roboguru

Pertanyaan

8. Jika straight p with rightwards arrow on top equals open parentheses straight t comma 2 close parentheses space dan space straight q with rightwards arrow on top equals open parentheses 5 comma 3 close parentheses serta sudut yang terbentuk antara kedua vector adalah straight pi over 4, maka tentukan nilai t yang memenuhi 

A. Rahmawati

Master Teacher

Mahasiswa/Alumni Universitas Pendidikan Indonesia

Jawaban terverifikasi

Pembahasan

Vektor

 table attributes columnalign right center left columnspacing 0px end attributes row cell straight p with rightwards arrow on top end cell equals cell open parentheses straight t comma 2 close parentheses end cell row cell straight p with rightwards arrow on top end cell equals cell ti space plus 2 straight j end cell row cell open double vertical bar straight p with rightwards arrow on top close double vertical bar squared end cell equals cell straight u subscript 1 squared plus straight u subscript 2 squared end cell row cell open double vertical bar straight p with rightwards arrow on top close double vertical bar end cell equals cell square root of straight u subscript 1 squared plus straight u subscript 2 squared end root end cell row cell open double vertical bar straight p with rightwards arrow on top close double vertical bar end cell equals cell square root of straight t squared plus 2 squared end root end cell row cell open double vertical bar straight p with rightwards arrow on top close double vertical bar end cell equals cell square root of straight t plus 4 end root end cell row cell straight q with rightwards arrow on top end cell equals cell open parentheses 5 comma 3 close parentheses end cell row cell straight q with rightwards arrow on top end cell equals cell 5 straight i space plus 3 straight j end cell row cell open double vertical bar straight q with rightwards arrow on top close double vertical bar end cell equals cell square root of 5 squared plus 3 squared end root end cell row blank equals cell square root of 25 plus 9 end root end cell row blank equals cell square root of 34 end cell end table

Akan dicari nilai 

table attributes columnalign right center left columnspacing 0px end attributes row cell straight p with rightwards arrow on top space. space straight q with rightwards arrow on top end cell equals cell straight p subscript 1 space straight q subscript 1 space plus straight p subscript 2 space straight q subscript 2 end cell row cell dimana space straight p end cell equals cell open parentheses straight t comma 2 close parentheses space dan space straight q equals open parentheses 5 comma 3 close parentheses end cell row cell straight p with rightwards arrow on top space straight q with rightwards arrow on top space end cell equals cell space open parentheses straight t space straight x space 5 close parentheses space plus open parentheses 2 space straight x 3 close parentheses end cell row blank equals cell 5 straight t space plus 6 end cell end table

Nilai p dan 1 akan disubstitusi ke 

table attributes columnalign right center left columnspacing 0px end attributes row cell cos space straight theta space end cell equals cell fraction numerator open parentheses straight p with rightwards arrow on top space. space straight q with rightwards arrow on top close parentheses over denominator open parentheses open double vertical bar straight p with rightwards arrow on top close double vertical bar open double vertical bar straight q with rightwards arrow on top close double vertical bar close parentheses end fraction end cell row cell cos space 45 to the power of 0 space end cell equals cell fraction numerator 5 straight t plus 6 over denominator open parentheses square root of straight t plus 4 end root close parentheses open parentheses square root of 34 close parentheses end fraction end cell row cell 1 half square root of 2 end cell equals cell fraction numerator 5 straight t plus 6 over denominator 34 straight t space plus 136 end fraction end cell row cell square root of 2 space open parentheses 34 straight t plus 136 close parentheses end cell equals cell 2 open parentheses 5 straight t space plus 6 close parentheses end cell row cell square root of 2 open parentheses 34 straight t space plus 136 close parentheses space end cell equals cell 10 straight t space plus 12 end cell row blank blank cell Kedua space ruas space dikuadratkan end cell row cell 2 open parentheses 34 straight t plus 136 close parentheses end cell equals cell 100 straight t squared plus 240 straight t plus 144 end cell row cell 68 straight t space plus 272 end cell equals cell 100 straight t squared plus 240 straight t plus 144 end cell row blank blank blank end table

12

0.0 (0 rating)

Pertanyaan serupa

Diketahui segitiga ABC dengan titik-titik sudut A(4,−6,4),B(−2,0,4), dan C(0,2,8). Dengan menggunakan rumus kosinus antara dua vektor, tentukan besar sudut-sudut berikut. a. sudut BAC

25

0.0

Jawaban terverifikasi

RUANGGURU HQ

Jl. Dr. Saharjo No.161, Manggarai Selatan, Tebet, Kota Jakarta Selatan, Daerah Khusus Ibukota Jakarta 12860

Coba GRATIS Aplikasi Ruangguru

Produk Ruangguru

Produk Lainnya

Hubungi Kami

Ruangguru WhatsApp

081578200000

Email info@ruangguru.com

info@ruangguru.com

Contact 02140008000

02140008000

Ikuti Kami

©2022 Ruangguru. All Rights Reserved PT. Ruang Raya Indonesia