Roboguru

Vektor  dengan panjang membentuk sudut lancip dengan vektor . Jika panjang proyeksi orthogonal  pada  adalah 2, tentukan vektor a.

Pertanyaan

Vektor bold italic a dengan panjang square root of 5 spacemembentuk sudut lancip dengan vektor bold italic b equals open parentheses 6 comma space 8 close parentheses. Jika panjang proyeksi orthogonal bold italic a pada bold italic b adalah 2, tentukan vektor a.

Pembahasan Soal:

Misalkan vekor bold italic a equals open parentheses x comma space y close parentheses, dengan rumus panjang proyeksi orthogonal  bold italic a pada bold italic b, maka

table attributes columnalign right center left columnspacing 0px end attributes row cell open vertical bar c close vertical bar end cell equals cell fraction numerator a times b over denominator open vertical bar b close vertical bar end fraction end cell row 2 equals cell fraction numerator open parentheses table row x row y end table close parentheses open parentheses table row 6 row 8 end table close parentheses over denominator square root of 6 squared plus 8 squared end root end fraction end cell row 2 equals cell fraction numerator 6 x plus 8 y over denominator 10 end fraction end cell row 20 equals cell 6 x plus 8 y end cell row 10 equals cell 3 x plus 4 y... left parenthesis 1 right parenthesis end cell end table

Selanjutnya diketahui vektor bold italic a dengan panjang square root of 5 space, maka

open vertical bar a close vertical bar equals square root of 5 square root of x squared plus y squared end root equals square root of 5 x squared plus y squared equals 5... left parenthesis 2 right parenthesis

Substitusi persamaan (1) ke persamaan (2),

3 x plus 4 y equals 10 rightwards arrow x equals fraction numerator 10 minus 4 y over denominator 3 end fraction x squared plus y squared equals 5 open parentheses fraction numerator 10 minus 4 y over denominator 3 end fraction close parentheses squared plus y squared equals 5 fraction numerator 100 minus 80 y plus 16 y squared over denominator 9 end fraction plus y squared equals 5 space left parenthesis k a l i space 9 right parenthesis 16 y squared minus 80 y plus 100 plus 9 y squared equals 45 25 y squared minus 80 y plus 55 equals 0 5 y squared minus 16 y plus 11 equals 0 left parenthesis 5 y minus 11 right parenthesis left parenthesis y minus 1 right parenthesis equals 0 y subscript 1 equals 11 over 5 space atau space y subscript 2 equals 1 x subscript 1 equals fraction numerator 10 minus 4 open parentheses begin display style 11 over 5 end style close parentheses over denominator 3 end fraction equals 2 over 5 x subscript 2 equals fraction numerator 10 minus 4 left parenthesis 1 right parenthesis over denominator 3 end fraction equals 2

Jadi, diperoleh vektor bold italic a equals left parenthesis 2 comma space 1 right parenthesis space atau space bold italic a equals open parentheses 2 over 5 comma space 11 over 5 close parentheses.

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

H. Janatu

Mahasiswa/Alumni Universitas Riau

Terakhir diupdate 03 Mei 2021

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

Tapi Roboguru masih mau belajar. Menurut kamu pembahasan kali ini sudah membantu, belum?

Membantu

Kurang Membantu

Apakah pembahasan ini membantu?

Belum menemukan yang kamu cari?

Post pertanyaanmu ke Tanya Jawab, yuk

Mau Bertanya

Pertanyaan yang serupa

Diketahui vektor . Jika panjang proyeksi vektor a pada b adalah , maka salah satu nilai x adalah ....

Pembahasan Soal:

V e k t o r space a equals open parentheses table row cell negative 2 end cell row 3 row 4 end table close parentheses space d a n space b equals space open parentheses table row x row 0 row 3 end table close parentheses space P a n j a n g space p r o y e k s i space v e k t o r stack space a with rightwards arrow on top space p a d a space b with rightwards arrow on top space a d a l a h space 4 over 5  fraction numerator a with rightwards arrow on top. b with rightwards arrow on top over denominator stack open vertical bar b close vertical bar with rightwards arrow on top end fraction equals 4 over 5  fraction numerator negative 2 x plus 10 plus 12 over denominator square root of x squared plus 0 plus 32 end root end fraction equals 4 over 5  fraction numerator negative 2 x plus 12 over denominator square root of x squared plus 9 end root end fraction equals 4 over 5  5 left parenthesis negative 2 x space plus space 12 right parenthesis space equals space 4 square root of x squared plus 9 end root  left parenthesis negative 10 x plus 60 right parenthesis space equals space 4 square root of x squared plus 9 end root  left parenthesis negative 10 x plus 60 right parenthesis squared space equals space left parenthesis 42 space right parenthesis left parenthesis x 2 space plus 9 right parenthesis squared  100 x squared space minus 1200 x space plus space 3600 space equals space 16 space left parenthesis x squared space plus space 9 right parenthesis  100 x squared space minus 1200 x space plus space 3600 space equals space 16 x squared space plus space 144  84 x squared space minus 1200 x space plus space 3456 space equals space 0  7 space x squared space minus space 100 x space plus space 288 space equals space 0  left parenthesis 7 x minus 72 right parenthesis left parenthesis x minus 4 right parenthesis equals 0  x space equals space 72 over 7 space a t a u space x space equals space 4

0

Roboguru

Diketahui titik-titik ,  dan . Jika titik  merupakan proyeksi titik  pada garis  maka panjang  sama dengan ...

Pembahasan Soal:

Panjang vektor a with rightwards arrow on top equals x i with rightwards arrow on top plus y j with rightwards arrow on top plus z k with rightwards arrow on top, yaitu open vertical bar a with rightwards arrow on top close vertical bar equals square root of x squared plus y squared plus z squared end root

Jika vektor a with rightwards arrow on top equals a subscript 1 i with rightwards arrow on top plus a subscript 2 j with rightwards arrow on top plus a subscript 3 k with rightwards arrow on top dan b with rightwards arrow on top equals b subscript 1 i with rightwards arrow on top plus b subscript 2 j with rightwards arrow on top plus b subscript 3 k with rightwards arrow on top, maka

a with rightwards arrow on top times b with rightwards arrow 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

Panjang proyeksi vektor stack text a end text with rightwards arrow on top pada vektor stack text b end text with rightwards arrow on top dapat ditentukan menggunakan rumus berikut.

open vertical bar c with rightwards arrow on top close vertical bar equals open vertical bar fraction numerator a with rightwards arrow on top times b with rightwards arrow on top over denominator open vertical bar b with rightwards arrow on top close vertical bar end fraction close vertical bar

Komponen vektor stack A B with rightwards arrow on top dan stack A C with rightwards arrow on top pada soal di atas dapat ditentukan sebagai berikut.

table attributes columnalign right center left columnspacing 0px end attributes row cell stack A B with rightwards arrow on top end cell equals cell text B-A end text end cell row blank equals cell open parentheses table row 10 row 1 row 3 end table close parentheses minus open parentheses table row 2 row 1 row cell negative 3 end cell end table close parentheses end cell row blank equals cell open parentheses table row 8 row 0 row 6 end table close parentheses end cell end table

table attributes columnalign right center left columnspacing 0px end attributes row cell stack A C with rightwards arrow on top end cell equals cell text C-A end text end cell row blank equals cell open parentheses table row 8 row 4 row cell negative 1 end cell end table close parentheses minus open parentheses table row 2 row 1 row cell negative 3 end cell end table close parentheses end cell row blank equals cell open parentheses table row 6 row 3 row 2 end table close parentheses end cell end table

Berdasarkan konsep di atas, panjang proyeksi vektor stack A C with rightwards arrow on top pada vektor stack A B with rightwards arrow on top adalah sebagai berikut.

table attributes columnalign right center left columnspacing 0px end attributes row cell open vertical bar text AD end text close vertical bar end cell equals cell open vertical bar fraction numerator stack A C with rightwards arrow on top times stack A B with rightwards arrow on top over denominator open vertical bar stack A B with rightwards arrow on top close vertical bar end fraction close vertical bar end cell row blank equals cell open vertical bar fraction numerator open parentheses table row 6 row 3 row 2 end table close parentheses open parentheses table row 8 row 0 row 6 end table close parentheses over denominator square root of 8 squared plus 0 squared plus 6 squared end root end fraction close vertical bar end cell row blank equals cell open vertical bar fraction numerator 6 times 8 plus 3 times 0 plus 2 times 6 over denominator square root of 100 end fraction close vertical bar end cell row blank equals cell open vertical bar 60 over 10 close vertical bar end cell row blank equals 6 end table

Diperoleh panjang text AD=6 end text 

Oleh karena itu, jawaban yang tepat adalah C.

0

Roboguru

Hitunglah proyeksi skalar ortogonal  pada  dan juga  pada  dari vektor-vektor berikut. b.  dan

Pembahasan Soal:

Proyeksi skalar ortogonal begin mathsize 14px style a with rightwards arrow on top end style pada begin mathsize 14px style b with rightwards arrow on top end style adalah sebagai berikut:

begin mathsize 14px style vertical line stack O C with rightwards arrow on top vertical line equals vertical line c with rightwards arrow on top vertical line equals fraction numerator a with rightwards arrow on top times b with rightwards arrow on top over denominator vertical line b with rightwards arrow on top vertical line end fraction end style

Maka,

begin mathsize 14px style table attributes columnalign right center left end attributes row cell vertical line c with rightwards arrow on top vertical line end cell equals cell fraction numerator a with rightwards arrow on top times b with rightwards arrow on top over denominator vertical line b with rightwards arrow on top vertical line end fraction end cell row space equals cell fraction numerator left parenthesis 3 cross times 2 right parenthesis plus left parenthesis 2 cross times 3 right parenthesis over denominator square root of 2 squared plus 3 squared end root end fraction end cell row space equals cell fraction numerator 6 plus 6 over denominator square root of 4 plus 9 end root end fraction end cell row space equals cell fraction numerator 12 over denominator square root of 13 end fraction end cell row space equals cell fraction numerator 12 over denominator square root of 13 end fraction cross times fraction numerator square root of 13 over denominator square root of 13 end fraction end cell row space equals cell fraction numerator 12 square root of 13 over denominator 13 end fraction end cell row space space space end table end style 

Maka, proyeksi skalar ortogonal begin mathsize 14px style a with rightwards arrow on top end style pada begin mathsize 14px style b with rightwards arrow on top end style adalah begin mathsize 14px style fraction numerator 12 square root of 13 over denominator 13 end fraction end style.

Proyeksi skalar ortogonal begin mathsize 14px style b with rightwards arrow on top end style pada begin mathsize 14px style a with rightwards arrow on top end style adalah sebagai berikut:

begin mathsize 14px style vertical line stack O D with rightwards arrow on top vertical line equals vertical line d with rightwards arrow on top vertical line equals fraction numerator b with rightwards arrow on top times a with rightwards arrow on top over denominator vertical line a with rightwards arrow on top vertical line end fraction end style

Maka,

begin mathsize 14px style table attributes columnalign center center left end attributes row cell vertical line d with rightwards arrow on top vertical line end cell equals cell fraction numerator b with rightwards arrow on top times a with rightwards arrow on top over denominator vertical line a with rightwards arrow on top vertical line end fraction end cell row space equals cell fraction numerator left parenthesis 2 cross times 3 right parenthesis plus left parenthesis 3 cross times 2 right parenthesis over denominator square root of 3 squared plus 2 squared end root end fraction end cell row space equals cell fraction numerator 6 plus 6 over denominator square root of 9 plus 4 end root end fraction end cell row space equals cell fraction numerator 12 over denominator square root of 13 end fraction end cell row space equals cell fraction numerator 12 over denominator square root of 13 end fraction cross times fraction numerator square root of 13 over denominator square root of 13 end fraction end cell row space equals cell fraction numerator 12 square root of 13 over denominator 13 end fraction end cell end table end style 

Maka, proyeksi skalar ortogonal begin mathsize 14px style b with rightwards arrow on top end style pada begin mathsize 14px style a with rightwards arrow on top end style adalah begin mathsize 14px style fraction numerator 12 square root of 13 over denominator 13 end fraction end style.

0

Roboguru

Diketahui vektor dan . Jika panjang proyeksi vektor  pada  adalah , nilai p = ....

Pembahasan Soal:

begin mathsize 14px style open vertical bar straight c with rightwards arrow on top close vertical bar equals fraction numerator straight a with rightwards arrow on top straight space. straight space straight b with rightwards arrow on top over denominator open vertical bar straight b with rightwards arrow on top close vertical bar end fraction  fraction numerator 6 over denominator square root of 26 end fraction equals fraction numerator open parentheses table row 2 row cell negative 2 straight p end cell row 4 end table close parentheses. open parentheses table row 1 row cell negative 3 end cell row 4 end table close parentheses over denominator square root of 1 squared plus open parentheses negative 3 close parentheses squared plus 4 squared end root end fraction  fraction numerator 6 over denominator square root of 26 end fraction equals fraction numerator 2 plus 6 straight p plus 16 over denominator square root of 26 end fraction    Maka  6 equals 2 plus 6 straight p plus 16  6 straight p equals 6 minus 2 minus 16  6 straight p equals negative 12  straight p equals negative 2 end style

0

Roboguru

Jika  dan , proyeksi skalar  pada  adalah ....

Pembahasan Soal:

Diketahui straight a with rightwards arrow on top equals open parentheses table row 3 row cell negative 1 end cell end table close parentheses dan straight b with rightwards arrow on top equals open parentheses table row cell negative 2 end cell row 4 end table close parentheses, maka:

straight a with rightwards arrow on top minus straight b with rightwards arrow on top equals open parentheses table row 3 row cell negative 1 end cell end table close parentheses minus open parentheses table row cell negative 2 end cell row 4 end table close parentheses equals open parentheses table row 5 row cell negative 5 end cell end table close parentheses 2 straight a with rightwards arrow on top plus straight b with rightwards arrow on top equals 2 open parentheses table row 3 row cell negative 1 end cell end table close parentheses plus open parentheses table row cell negative 2 end cell row 4 end table close parentheses equals open parentheses table row 6 row cell negative 2 end cell end table close parentheses plus open parentheses table row cell negative 2 end cell row 4 end table close parentheses equals open parentheses table row 4 row 2 end table close parentheses open vertical bar 2 straight a with rightwards arrow on top plus straight b with rightwards arrow on top close vertical bar equals square root of 4 squared plus 2 squared end root equals square root of 16 plus 4 end root equals square root of 20 equals 2 square root of 5 

Sehingga:

open parentheses straight a with rightwards arrow on top minus straight b with rightwards arrow on top close parentheses subscript open parentheses 2 straight a with rightwards arrow on top plus straight b with rightwards arrow on top close parentheses end subscript equals fraction numerator open parentheses straight a with rightwards arrow on top minus straight b with rightwards arrow on top close parentheses times open parentheses 2 straight a with rightwards arrow on top plus straight b with rightwards arrow on top close parentheses over denominator open vertical bar 2 straight a with rightwards arrow on top plus straight b with rightwards arrow on top close vertical bar end fraction open parentheses straight a with rightwards arrow on top minus straight b with rightwards arrow on top close parentheses subscript open parentheses 2 straight a with rightwards arrow on top plus straight b with rightwards arrow on top close parentheses end subscript equals fraction numerator open parentheses table row 5 row cell negative 5 end cell end table close parentheses times open parentheses table row 4 row 2 end table close parentheses over denominator 2 square root of 5 end fraction open parentheses straight a with rightwards arrow on top minus straight b with rightwards arrow on top close parentheses subscript open parentheses 2 straight a with rightwards arrow on top plus straight b with rightwards arrow on top close parentheses end subscript equals fraction numerator 20 minus 10 over denominator 2 square root of 5 end fraction open parentheses straight a with rightwards arrow on top minus straight b with rightwards arrow on top close parentheses subscript open parentheses 2 straight a with rightwards arrow on top plus straight b with rightwards arrow on top close parentheses end subscript equals fraction numerator 10 over denominator 2 square root of 5 end fraction open parentheses straight a with rightwards arrow on top minus straight b with rightwards arrow on top close parentheses subscript open parentheses 2 straight a with rightwards arrow on top plus straight b with rightwards arrow on top close parentheses end subscript equals fraction numerator 5 over denominator square root of 5 end fraction cross times fraction numerator square root of 5 over denominator square root of 5 end fraction open parentheses straight a with rightwards arrow on top minus straight b with rightwards arrow on top close parentheses subscript open parentheses 2 straight a with rightwards arrow on top plus straight b with rightwards arrow on top close parentheses end subscript equals fraction numerator 5 square root of 5 over denominator 5 end fraction open parentheses straight a with rightwards arrow on top minus straight b with rightwards arrow on top close parentheses subscript open parentheses 2 straight a with rightwards arrow on top plus straight b with rightwards arrow on top close parentheses end subscript equals square root of 5 

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?

Membantu

Kurang Membantu

Apakah pembahasan ini membantu?

Belum menemukan yang kamu cari?

Post pertanyaanmu ke Tanya Jawab, yuk

Mau Bertanya

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

Ikuti Kami

©2021 Ruangguru. All Rights Reserved