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Jika a=3i−5k dan b=−2i+j​−k. Proyeksi skalar 2a pada b adalah ....

Pertanyaan

Jika straight a with rightwards arrow on top equals 3 straight i with rightwards arrow on top minus 5 straight k with rightwards arrow on top dan straight b with rightwards arrow on top equals negative 2 straight i with rightwards arrow on top plus straight j with rightwards arrow on top minus straight k with rightwards arrow on top. Proyeksi skalar 2 straight a with rightwards arrow on top pada straight b with rightwards arrow on top adalah ....

  1. negative 1 over 6 square root of 6 

  2. negative 1 third square root of 6 

  3. 1 over 6 square root of 6 

  4. 1 third square root of 6 

  5. square root of 6 

Pembahasan Soal:

Ingat kembali proyeksi vektor straight a with rightwards arrow on top pada straight b with rightwards arrow on top adalah:

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

Diketahui:

straight a with rightwards arrow on top equals open parentheses table row 3 row 0 row cell negative 5 end cell end table close parentheses comma space straight b with rightwards arrow on top equals open parentheses table row cell negative 2 end cell row 1 row cell negative 1 end cell end table close parentheses 

Maka:

2 straight a with rightwards arrow on top subscript straight b with rightwards arrow on top end subscript equals fraction numerator 2 straight a with rightwards arrow on top times straight b with rightwards arrow on top over denominator open vertical bar straight b with rightwards arrow on top close vertical bar end fraction 2 straight a with rightwards arrow on top subscript straight b with rightwards arrow on top end subscript equals fraction numerator 2 open parentheses table row 3 row 0 row cell negative 5 end cell end table close parentheses times open parentheses table row cell negative 2 end cell row 1 row cell negative 1 end cell end table close parentheses over denominator square root of open parentheses negative 2 close parentheses squared plus 1 squared plus open parentheses negative 1 close parentheses squared end root end fraction 2 straight a with rightwards arrow on top subscript straight b with rightwards arrow on top end subscript equals fraction numerator open parentheses table row 6 row 0 row cell negative 10 end cell end table close parentheses times open parentheses table row cell negative 2 end cell row 1 row cell negative 1 end cell end table close parentheses over denominator square root of 4 plus 1 plus 1 end root end fraction 2 straight a with rightwards arrow on top subscript straight b with rightwards arrow on top end subscript equals fraction numerator negative 12 plus 0 plus 10 over denominator square root of 6 end fraction 2 straight a with rightwards arrow on top subscript straight b with rightwards arrow on top end subscript equals fraction numerator negative 2 over denominator square root of 6 end fraction cross times fraction numerator square root of 6 over denominator square root of 6 end fraction 2 straight a with rightwards arrow on top subscript straight b with rightwards arrow on top end subscript equals fraction numerator negative 2 square root of 6 over denominator 6 end fraction 2 straight a with rightwards arrow on top subscript straight b with rightwards arrow on top end subscript equals negative 1 third square root of 6 

Jadi, jawaban yang benar adalah B.

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

W. Lestari

Mahasiswa/Alumni Universitas Sriwijaya

Terakhir diupdate 06 Oktober 2021

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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Pertanyaan yang serupa

Vektor a dengan panjang 5​membentuk sudut lancip dengan vektor b=(6,8). Jika panjang proyeksi orthogonal  pada 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.

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Roboguru

Proyeksi skalar orthogonal a=(m,−18,6) pada b=(−3,0,4) adalah 6. Nilai m yang memenuhi adalah...

Pembahasan Soal:

Beberapa konsep yang dipakai :

table attributes columnalign right center left columnspacing 0px end attributes row u equals cell open parentheses u subscript 1 comma u subscript 2 comma u subscript 3 comma..... comma u subscript n close parentheses end cell row v equals cell open parentheses v subscript 1 comma v subscript 2 comma v subscript 3 comma..... comma v subscript n close parentheses end cell row cell u times v end cell equals cell u subscript 1 v subscript 1 plus u subscript 2 v subscript 2 plus u subscript 3 v subscript 3 plus... plus u subscript n v subscript n end cell row cell open vertical bar u close vertical bar end cell equals cell square root of open parentheses u subscript 1 close parentheses squared plus open parentheses u subscript 2 close parentheses squared plus open parentheses u subscript 3 close parentheses squared plus... plus open parentheses u subscript n close parentheses squared end root end cell end table

Sehingga:

table attributes columnalign right center left columnspacing 0px end attributes row cell fraction numerator a times b over denominator open vertical bar b close vertical bar end fraction end cell equals 6 row cell fraction numerator open parentheses m cross times open parentheses negative 3 close parentheses close parentheses plus open parentheses negative 18 cross times 0 close parentheses plus left parenthesis 6 cross times 4 right parenthesis over denominator square root of open parentheses negative 3 close parentheses squared plus 0 squared plus 4 squared end root end fraction end cell equals 6 row cell fraction numerator negative 3 m plus 0 plus 24 over denominator square root of 9 plus 16 end root end fraction end cell equals 6 row cell fraction numerator negative 3 m plus 24 over denominator square root of 25 end fraction end cell equals 6 row cell fraction numerator negative 3 m plus 24 over denominator 5 end fraction end cell equals 6 row cell negative 3 m plus 24 end cell equals cell 6 cross times 5 end cell row cell negative 3 m plus 24 end cell equals 30 row cell negative 3 m end cell equals cell 30 minus 24 end cell row cell negative 3 m end cell equals 6 row m equals cell 6 divided by negative 3 end cell row m equals cell negative 2 end cell row blank blank blank end table

Dengan demikian, nilai yang memenuhi adalah 2.

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Sebuah vektor x dengan panjang 22​ membuat sudut lancip dengan y​=[3,1]. Bila vektor  diproyeksikan ke vektor y​ panjang proyeksinya sama dengan 54​10​. Vektor  tersebut adalah ...

Pembahasan Soal:

Panjang vektor a with rightwards arrow on top equals open parentheses table row x row y end table close parentheses, yaitu open vertical bar a with rightwards arrow on top close vertical bar equals square root of x squared plus y squared end root

Jika vektor a with rightwards arrow on top equals open parentheses table row cell a subscript 1 end cell row cell a subscript 2 end cell end table close parentheses dan b with rightwards arrow on top equals open parentheses table row cell b subscript 1 end cell row cell b subscript 2 end cell end table close parentheses, maka

a with rightwards arrow on top times b with rightwards arrow on top equals a subscript 1 b subscript 1 plus a subscript 2 b subscript 2

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

open vertical bar c with rightwards arrow on top close vertical bar equals 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

Berdasarkan konsep di atas dapat ditentukan hubungan berikut.

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 fraction numerator x with rightwards arrow on top times y with rightwards arrow on top over denominator open vertical bar y close vertical bar end fraction end cell row cell 4 over 5 square root of 10 end cell equals cell fraction numerator x with rightwards arrow on top times y with rightwards arrow on top over denominator square root of 3 squared plus 1 squared end root end fraction end cell row cell 4 over 5 square root of 10 end cell equals cell fraction numerator x with rightwards arrow on top times y with rightwards arrow on top over denominator square root of 10 end fraction end cell row cell x with rightwards arrow on top times y with rightwards arrow on top end cell equals cell fraction numerator 4 square root of 10 times square root of 10 over denominator 5 end fraction end cell row cell x with rightwards arrow on top times y with rightwards arrow on top end cell equals 8 end table

Misal vektor x with rightwards arrow on top equals open square brackets a comma space b close square brackets dapat ditentukan persamaan-persamaan berikut.

Persamaan 1:

table attributes columnalign right center left columnspacing 0px end attributes row cell x with rightwards arrow on top times y with rightwards arrow on top end cell equals 8 row cell open parentheses table row a row b end table close parentheses open parentheses table row 3 row 1 end table close parentheses end cell equals 8 row cell 3 a plus b end cell equals 8 row b equals cell 8 minus 3 a end cell end table

Persamaan 2:

table attributes columnalign right center left columnspacing 0px end attributes row cell square root of a squared plus b squared end root end cell equals cell open vertical bar x with rightwards arrow on top close vertical bar end cell row cell square root of a squared plus b squared end root end cell equals cell 2 square root of 2 end cell row cell a squared plus b squared end cell equals 8 end table

Substitusi persamaan 1 ke persamaan 2:

table attributes columnalign right center left columnspacing 0px end attributes row cell a squared plus b squared end cell equals 8 row cell a squared plus open parentheses 8 minus 3 a close parentheses squared end cell equals 8 row cell a squared plus 64 minus 48 a plus 9 a squared end cell equals 8 row cell 10 a squared minus 48 a plus 56 end cell equals 0 row cell 5 a squared minus 24 a plus 28 end cell equals 0 row cell open parentheses a minus 2 close parentheses open parentheses 5 a minus 14 close parentheses end cell equals 0 end table

a equals 2 atau a equals 14 over 5

Untuk a equals 2 diperoleh

table attributes columnalign right center left columnspacing 0px end attributes row b equals cell 8 minus 3 a end cell row blank equals cell 8 minus 3 times 2 end cell row blank equals 2 end table

Untuk a equals 14 over 5 diperoleh

table attributes columnalign right center left columnspacing 0px end attributes row b equals cell 8 minus 3 a end cell row blank equals cell 8 minus 3 times 14 over 5 end cell row blank equals cell 40 over 5 minus 42 over 5 end cell row blank equals cell negative 2 over 5 end cell end table

Vektor x with rightwards arrow on top equals open square brackets 2 comma space 2 close square brackets space text atau end text space open square brackets 14 over 5 comma space minus 2 over 5 close square brackets

Oleh karena itu, jawaban yang tepat adalah A.

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Diketahui titik-titik A(2,1,−3), B(10,1,3) dan C(8,4,−1). Jika titik D merupakan proyeksi titik C pada garis AB maka panjang AD 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

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Jika vektor a tegak lurus b maka proyeksi skalar orthogonal   pada  sama dengan...

Pembahasan Soal:

Ingat kembali proyeksi skalar orthogonal berikut.

Jika vektor top enclose a diproyeksikan secara orthogonal pada top enclose b, maka proyeksi skalar orthogonalnya adalah 

open vertical bar top enclose c close vertical bar equals open vertical bar open vertical bar top enclose a close vertical bar space cos space alpha close vertical bar 

Dari rumus di atas, diperoleh perhitungan sebagai berikut.

Vektor top enclose a tegak lurus top enclose b artinya angle left parenthesis top enclose a comma space top enclose b right parenthesis equals 90 degree.

Sehingga proyeksi skalar orthogonal vektor top enclose a pada top enclose b adalah

table attributes columnalign right center left columnspacing 0px end attributes row cell open vertical bar top enclose c close vertical bar end cell equals cell open vertical bar open vertical bar top enclose a close vertical bar space cos space alpha close vertical bar end cell row blank equals cell vertical line open vertical bar top enclose a close vertical bar space cos space 90 degree vertical line end cell row blank equals cell vertical line open vertical bar top enclose a close vertical bar space 0 vertical line end cell row blank equals cell vertical line 0 vertical line end cell row blank equals 0 row blank blank blank row blank blank blank end table  

Jadi, proyeksi skalar orthogonal vektor top enclose a pada top enclose b adalah 0.

Oleh karena itu, jawaban yang benar adalah B.

0

Roboguru

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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