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Vektor  mempunyai panjang . Jika . Tentukan .

Pertanyaan

Vektor bold italic a mempunyai panjang 2 square root of 3 space. Jika bold italic a bold times bold left parenthesis bold a bold plus bold b bold right parenthesis equals 15 space dan space angle open parentheses bold italic a bold comma bold italic b close parentheses bold equals straight pi over 6. Tentukan open vertical bar bold italic b close vertical bar.

Pembahasan Soal:

Ingat : a times b equals open vertical bar a close vertical bar open vertical bar b close vertical bar cos space theta

Maka,

table attributes columnalign right center left columnspacing 0px end attributes row cell a times left parenthesis straight a plus straight b right parenthesis end cell equals 15 row cell a times a plus a times b end cell equals 15 row cell open vertical bar a close vertical bar open vertical bar a close vertical bar c o s space 0 plus open vertical bar a close vertical bar open vertical bar b close vertical bar c o s space straight pi over 6 end cell equals 15 row cell open parentheses 2 square root of 3 close parentheses squared plus 2 square root of 3 open vertical bar b close vertical bar open parentheses 1 half square root of 3 close parentheses end cell equals 15 row cell 12 plus 3 open vertical bar b close vertical bar end cell equals 15 row cell open vertical bar b close vertical bar end cell equals 1 end table

Jadi, diperoleh open vertical bar b close vertical bar equals 1.

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

H. Janatu

Mahasiswa/Alumni Universitas Riau

Terakhir diupdate 03 Mei 2021

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Pertanyaan yang serupa

Jika  dan  maka proyeksi skalar orthogonal vektor  pada  = ...

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 fraction numerator top enclose a bullet top enclose b over denominator open vertical bar top enclose b close vertical bar end fraction close vertical bar 

dengan

top enclose a bullet top enclose b equals a subscript 1 b subscript 1 plus a subscript 2 b subscript 2 plus a subscript 3 b subscript 3 space dan space open vertical bar top enclose b close vertical bar equals square root of b subscript 1 squared plus b subscript 2 squared plus b subscript 3 squared end root   

Dari rumus di atas, diperoleh perhitungan sebagai berikut.

Menentukan top enclose a bullet top enclose b 

table attributes columnalign right center left columnspacing 0px end attributes row cell top enclose a bullet top enclose b end cell equals cell 3 left parenthesis 6 right parenthesis plus 1 left parenthesis 2 right parenthesis plus left parenthesis negative 2 right parenthesis left parenthesis 3 right parenthesis end cell row blank equals cell 18 plus 2 minus 6 end cell row blank equals 14 end table 

Menentukan open vertical bar top enclose b close vertical bar 

table attributes columnalign right center left columnspacing 0px end attributes row cell open vertical bar top enclose b close vertical bar end cell equals cell square root of 6 squared plus 2 squared plus 3 squared end root end cell row blank equals cell square root of 36 plus 4 plus 9 end root end cell row blank equals cell square root of 49 end cell row blank equals 7 end table 

Maka 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 fraction numerator top enclose a bullet top enclose b over denominator open vertical bar top enclose b close vertical bar end fraction close vertical bar end cell row blank equals cell vertical line 14 over 7 vertical line end cell row blank equals cell vertical line 2 vertical line end cell row blank equals 2 end table  

Jadi, proyeksi skalar orthogonal vektor top enclose a pada top enclose b adalah 2.

Oleh karena itu, jawaban yang benar adalah B.

Roboguru

Diketahui vektor  . Nilai sinus sudut antara  adalah .....

Pembahasan Soal:

Ingat rumus perkalian titik pada vektor :

a with rightwards arrow on top. b with rightwards arrow on top equals open vertical bar a with rightwards arrow on top close vertical bar open vertical bar b with rightwards arrow on top close vertical bar cos space A  open parentheses table row 3 row 4 row cell negative 5 end cell end table close parentheses. open parentheses table row 1 row cell negative 2 end cell row 2 end table close parentheses equals square root of 3 squared plus 4 squared plus left parenthesis negative 5 right parenthesis squared. end root square root of 1 squared plus left parenthesis negative 2 right parenthesis squared plus left parenthesis 2 right parenthesis squared end root. cos space alpha  3 minus 8 minus 10 equals square root of 9 plus 16 plus 25 end root. square root of 1 plus 4 plus 4 end root. cos space alpha  minus 15 equals square root of 50. square root of 9. cos space alpha  minus 15 equals 5 square root of 2.3 space cos space alpha  cos space alpha equals negative fraction numerator 1 over denominator square root of 2 end fraction    sin space alpha equals square root of 1 minus cos squared alpha end root equals square root of 1 minus open parentheses negative fraction numerator 1 over denominator square root of 2 end fraction close parentheses squared end root equals square root of 1 minus 1 half end root equals 1 half square root of 2    bold Catatan bold thin space bold colon  Karena space sudut space antara space dua space vektor space bernilai space 0 degree less or equal than straight alpha less or equal than 180 degree comma space maka  sin space straight alpha space tidaklah space negatif.

Roboguru

Deketahui segitiga ABCD dengan koordinat ,  dan . Proyeksi vektor orthognal  pada  adalah ...

Pembahasan Soal:

Ingat konsep proyeksi vektor orthognal AB with rightwards arrow on top pada AC with rightwards arrow on top adalah 

Proyeksi space vektor space AC with rightwards arrow on top equals fraction numerator stack AB times with rightwards arrow on top AC with rightwards arrow on top over denominator open vertical bar AC with rightwards arrow on top close vertical bar squared end fraction stack times AC with rightwards arrow on top  

Diketahui straight A equals open parentheses 2 comma space minus 1 comma space minus 1 close parenthesesstraight B open parentheses negative 1 comma space 4 comma space minus 2 close parentheses dan straight C open parentheses 5 comma space 0 comma space minus 3 close parentheses maka:

table attributes columnalign right center left columnspacing 0px end attributes row cell AB with rightwards arrow on top end cell equals cell straight B with rightwards arrow on top minus straight A with rightwards arrow on top end cell row blank equals cell open parentheses negative 1 comma space 4 comma space minus 2 close parentheses minus open parentheses 2 comma space minus 1 comma space minus 1 close parentheses end cell row blank equals cell open parentheses negative 3 comma space 5 comma space minus 1 close parentheses end cell row cell AC with rightwards arrow on top end cell equals cell straight C with rightwards arrow on top minus straight A with rightwards arrow on top end cell row blank equals cell open parentheses 5 comma space 0 comma space minus 3 close parentheses minus open parentheses 2 comma space minus 1 comma space minus 1 close parentheses end cell row blank equals cell open parentheses 3 comma space 1 comma space minus 2 close parentheses end cell end table

maka 

table attributes columnalign right center left columnspacing 0px end attributes row cell stack AB times with rightwards arrow on top AC with rightwards arrow on top end cell equals cell open parentheses negative 3 comma space 5 comma space minus 1 close parentheses times open parentheses 3 comma space 1 comma space minus 2 close parentheses end cell row blank equals cell negative 9 plus 5 plus 2 end cell row blank equals cell negative 2 end cell row cell open vertical bar AC with rightwards arrow on top close vertical bar end cell equals cell square root of 3 squared plus 1 squared plus open parentheses negative 2 close parentheses squared end root end cell row blank equals cell square root of 9 plus 1 plus 4 end root end cell row blank equals cell square root of 14 end cell end table 

sehingga

table attributes columnalign right center left columnspacing 0px end attributes row cell Proyeksi space vektor space AC with rightwards arrow on top end cell equals cell fraction numerator AB with rightwards arrow on top times AC with rightwards arrow on top over denominator open vertical bar AC with rightwards arrow on top close vertical bar squared end fraction times AC with rightwards arrow on top end cell row blank equals cell fraction numerator negative 2 over denominator open parentheses square root of 14 close parentheses squared end fraction open parentheses 3 comma space 1 comma space minus 2 close parentheses end cell row blank equals cell fraction numerator negative 2 over denominator 14 end fraction open parentheses 3 comma space 1 comma space minus 2 close parentheses end cell row blank equals cell negative 1 over 7 open parentheses 3 comma space 1 comma space minus 2 close parentheses end cell row blank equals cell open parentheses negative 3 over 7 comma space minus 1 over 7 comma space 2 over 7 close parentheses end cell end table  

Dengan demikian proyeksi vektor orthognal AB with rightwards arrow on top pada AC with rightwards arrow on top adalah table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell open parentheses negative 3 over 7 comma space minus 1 over 7 comma space 2 over 7 close parentheses end cell end table.

Roboguru

Diketahui vektor  dan . Jika panjang proyeksi vektor  pada vektor  adalah , maka nilai  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 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, nilai x dapat ditentukan sebagai 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 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 end cell row 2 equals cell fraction numerator open parentheses table row 2 row cell negative 2 end cell row x end table close parentheses open parentheses table row cell negative 3 end cell row cell negative 4 end cell row 12 end table close parentheses over denominator square root of open parentheses negative 3 close parentheses squared plus open parentheses negative 4 close parentheses squared plus 12 squared end root end fraction end cell row 2 equals cell fraction numerator 2 open parentheses negative 3 close parentheses plus open parentheses negative 2 close parentheses open parentheses negative 4 close parentheses plus 12 x over denominator square root of 9 plus 16 plus 144 end root end fraction end cell row 2 equals cell fraction numerator negative 6 plus 8 plus 12 x over denominator square root of 169 end fraction end cell row 2 equals cell fraction numerator 2 plus 12 x over denominator 13 end fraction end cell row 26 equals cell 2 plus 12 x end cell row cell 12 x end cell equals 24 row x equals 2 end table

Oleh karena itu, jawaban yang tepat adalah E.

Roboguru

Misal vektor  dan vektor . Jika panjang proyeksi vektor  pada  sama dengan 3, maka nilai z yang tidak bulat sama dengan...

Pembahasan Soal:

Ingat kembali panjang proyeksi berikut.

  • Jika vektor top enclose a diproyeksikan secara orthogonal pada top enclose b, maka panjang proyeksi vektornya adalah 

open vertical bar top enclose c close vertical bar equals open vertical bar fraction numerator top enclose a bullet top enclose b over denominator open vertical bar top enclose b close vertical bar end fraction close vertical bar 

             dengan

top enclose a bullet top enclose b equals a subscript 1 b subscript 1 plus a subscript 2 b subscript 2 plus a subscript 3 b subscript 3 space dan space open vertical bar top enclose b close vertical bar equals square root of b subscript 1 squared plus b subscript 2 squared plus b subscript 3 squared end root   

Dari rumus di atas, diperoleh perhitungan sebagai berikut.

Karena  panjang proyeksi vektor top enclose a pada top enclose b sama dengan 3, maka

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 fraction numerator top enclose a bullet top enclose b over denominator open vertical bar top enclose b close vertical bar end fraction close vertical bar end cell row 3 equals cell open vertical bar fraction numerator 1 left parenthesis 3 right parenthesis plus left parenthesis negative 3 right parenthesis left parenthesis negative 2 right parenthesis plus 2 left parenthesis z right parenthesis over denominator square root of 3 squared plus left parenthesis negative 2 right parenthesis squared plus z squared end root end fraction close vertical bar end cell row 3 equals cell open vertical bar fraction numerator 3 plus 6 plus 2 z over denominator square root of 9 plus 4 plus z squared end root end fraction close vertical bar end cell row 3 equals cell open vertical bar fraction numerator 9 plus 2 z over denominator square root of 13 plus z squared end root end fraction close vertical bar space space end cell row 9 equals cell fraction numerator 81 plus 36 z plus 4 z squared over denominator 13 plus z squared end fraction end cell row cell 9 left parenthesis 13 plus z squared right parenthesis end cell equals cell 81 plus 36 z plus 4 z squared end cell row cell 117 plus 9 z squared end cell equals cell 81 plus 36 z plus 4 z squared end cell row cell 9 z squared minus 4 z squared minus 36 z plus 117 minus 81 end cell equals 0 row cell 5 z squared minus 36 z plus 36 end cell equals 0 row cell left parenthesis 5 z minus 6 right parenthesis left parenthesis z minus 6 right parenthesis end cell equals 0 row z equals cell 6 over 5 space atau space z equals 6 end cell end table    

Karena yang diminta adalah nilai z yang tidak bulat, maka

z equals 6 over 5 equals 1 1 fifth  

Oleh karena itu, jawaban yang benar adalah D.

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