Roboguru

Diberikan matriks dan vektor-vektor sebagai berikut: , ;  dan  menyatakan transpose dari . Jika vekor  tegak lurus vektor  maka nilai  sama dengan ...

Pertanyaan

Diberikan matriks dan vektor-vektor sebagai berikut: text A= end text open parentheses table row 1 3 row 1 cell negative 2 end cell row 2 1 end table close parenthesesstack text a end text with rightwards arrow on top equals open parentheses table row 1 row cell negative 1 end cell row cell negative 3 end cell end table close parenthesesstack text b end text with rightwards arrow on top equals open parentheses table row cell negative text p end text end cell row cell text q end text end cell end table close parentheses dan text A end text to the power of text T end text end exponent menyatakan transpose dari text A end text. Jika vekor text A end text to the power of text T end text end exponent stack text a end text with rightwards arrow on top tegak lurus vektor stack text b end text with rightwards arrow on top maka nilai text q end text sama dengan ... 

  1. text p end text 

  2. text -p end text 

  3. text 2p end text 

  4. text -2p end text 

  5. text 3p end text 

Pembahasan Soal:

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

Jika vektor a with rightwards arrow on top tegak lurus vektor b with rightwards arrow on top, maka 

table attributes columnalign right center left columnspacing 0px end attributes row cell a with rightwards arrow on top times b with rightwards arrow on top end cell equals cell open vertical bar a with rightwards arrow on top close vertical bar open vertical bar b with rightwards arrow on top close vertical bar space cos space alpha end cell row cell a with rightwards arrow on top times b with rightwards arrow on top end cell equals 0 end table

Transpose matriks A equals open parentheses table row a b row c d end table close parentheses adalah A to the power of text T end text end exponent equals open parentheses table row a c row b d end table close parentheses

Hasil perkalian transpose matriks A dengan vektor a with rightwards arrow on top adalah sebagai berikut.

table attributes columnalign right center left columnspacing 0px end attributes row cell A to the power of text T end text end exponent times a with rightwards arrow on top end cell equals cell open parentheses table row 1 1 2 row 3 cell negative 2 end cell 1 end table close parentheses open parentheses table row 1 row cell negative 1 end cell row cell negative 3 end cell end table close parentheses end cell row blank equals cell open parentheses table row cell 1 times 1 plus 1 open parentheses negative 1 close parentheses plus 2 open parentheses negative 3 close parentheses end cell row cell 3 times 1 plus negative 2 open parentheses negative 1 close parentheses plus 1 open parentheses negative 3 close parentheses end cell end table close parentheses end cell row blank equals cell open parentheses table row cell 1 minus 1 minus 6 end cell row cell 3 plus 2 minus 3 end cell end table close parentheses end cell row blank equals cell open parentheses table row cell negative 6 end cell row 2 end table close parentheses end cell end table

Jika vektor A to the power of text T end text end exponent a with rightwards arrow on top tegak lurus vektor b with rightwards arrow on top, maka diperoleh:

table attributes columnalign right center left columnspacing 0px end attributes row cell A to the power of text T end text end exponent a with rightwards arrow on top times b with rightwards arrow on top end cell equals 0 row cell open parentheses table row cell negative 6 end cell row 2 end table close parentheses open parentheses table row cell negative p end cell row q end table close parentheses end cell equals 0 row cell negative 6 open parentheses negative p close parentheses plus 2 times q end cell equals 0 row cell 6 p plus 2 q end cell equals 0 row cell 2 q end cell equals cell negative 6 p end cell row q equals cell negative 3 p end cell end table

Oleh karena itu, tidak terdapat pilihan jawaban yang tepat.

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

H. Eka

Mahasiswa/Alumni Universitas Pendidikan Indonesia

Terakhir diupdate 11 Juli 2021

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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Pertanyaan yang serupa

Diketahui vektor  tegak lurus dengan vektor . Jika  maka

Pembahasan Soal:

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

Jika vektor a with rightwards arrow on top tegak lurus vektor b with rightwards arrow on top, maka 

table attributes columnalign right center left columnspacing 0px end attributes row cell a with rightwards arrow on top times b with rightwards arrow on top end cell equals cell open vertical bar a with rightwards arrow on top close vertical bar open vertical bar b with rightwards arrow on top close vertical bar space cos space alpha end cell row cell a with rightwards arrow on top times b with rightwards arrow on top end cell equals 0 end table

Berdasarkan konsep di atas, nilai x dapat ditentukan sebagai berikut.

table attributes columnalign right center left columnspacing 0px end attributes row cell a with rightwards arrow on top times b with rightwards arrow on top end cell equals 0 row cell open parentheses table row 2 row 3 row x end table close parentheses open parentheses table row 3 row cell negative 6 end cell row 2 end table close parentheses end cell equals 0 row cell 2 times 3 plus 3 open parentheses negative 6 close parentheses plus 2 x end cell equals 0 row cell 6 minus 18 plus 2 x end cell equals 0 row cell negative 12 plus 2 x end cell equals 0 row cell 2 x end cell equals 12 row x equals 6 end table

Diperoleh vektor x equals 6 sehingga a with rightwards arrow on top equals 2 i with rightwards arrow on top plus 3 j with rightwards arrow on top plus 6 k with rightwards arrow on top dan c with rightwards arrow on top equals 6 i with rightwards arrow on top minus 5 j with rightwards arrow on top plus k with rightwards arrow on top. Dapat ditentukan operasi penjumlahan dan pengurangan vektor berikut.

table attributes columnalign right center left columnspacing 0px end attributes row cell a with rightwards arrow on top minus b with rightwards arrow on top plus c with rightwards arrow on top end cell equals cell open parentheses table row 2 row 3 row 6 end table close parentheses minus open parentheses table row 3 row cell negative 6 end cell row 2 end table close parentheses plus open parentheses table row 6 row cell negative 5 end cell row 1 end table close parentheses end cell row blank equals cell open parentheses table row cell 2 minus 3 plus 6 end cell row cell 3 minus open parentheses negative 6 close parentheses plus open parentheses negative 5 close parentheses end cell row cell 6 minus 2 plus 1 end cell end table close parentheses end cell row blank equals cell open parentheses table row 5 row 4 row 5 end table close parentheses end cell end table

Diperoleh table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell a with rightwards arrow on top end cell end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank minus end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell b with rightwards arrow on top end cell end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank plus end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell c with rightwards arrow on top end cell end table table attributes columnalign right center left columnspacing 0px end attributes row blank equals blank end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank 5 end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell i with rightwards arrow on top end cell end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank plus end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank 4 end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell j with rightwards arrow on top end cell end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank plus end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank 5 end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell k with rightwards arrow on top end cell end table

Oleh karena itu, jawaban yang tepat adalah B.

Roboguru

Diketahui vektor  dan . Tentukan sudut antara  dan .

Pembahasan Soal:

Jika vektor begin mathsize 14px style u with rightwards arrow on top equals open parentheses table row cell x subscript 1 end cell row cell y subscript 1 end cell row cell z subscript 1 end cell end table close parentheses end style dan begin mathsize 14px style v with rightwards arrow on top equals open parentheses table row cell x subscript 2 end cell row cell y subscript 2 end cell row cell z subscript 2 end cell end table close parentheses end style, maka hasil kali titik (dot product) dari dua vektor adalah

begin mathsize 14px style u with rightwards arrow on top times v with rightwards arrow on top equals x subscript 1 x subscript 2 plus y subscript 1 y subscript 2 plus z subscript 1 z subscript 2 end style

dan

begin mathsize 14px style u with rightwards arrow on top times v with rightwards arrow on top equals open vertical bar u with rightwards arrow on top close vertical bar open vertical bar v with rightwards arrow on top close vertical bar cos theta end style

dengan begin mathsize 14px style theta end style adalah sudut di antara kedua vektor.

Perhatikan bahwa

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell open vertical bar u with rightwards arrow on top close vertical bar end cell equals cell square root of x subscript 1 squared plus y subscript 1 squared plus z subscript 1 squared end root end cell row blank equals cell square root of 2 squared plus open parentheses negative 1 close parentheses squared plus 1 squared end root end cell row blank equals cell square root of 4 plus 1 plus 1 end root end cell row blank equals cell square root of 6 end cell end table end style

dan

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell vertical line v with rightwards arrow on top vertical line end cell equals cell square root of x subscript 2 squared plus y subscript 2 squared plus z subscript 2 squared end root end cell row blank equals cell square root of 7 squared plus 0 squared plus 1 squared end root end cell row blank equals cell square root of 49 plus 0 plus 1 end root end cell row blank equals cell square root of 50 end cell row blank equals cell 5 square root of 2 end cell end table end style

Akibatnya

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell u with rightwards arrow on top times v with rightwards arrow on top end cell equals cell vertical line u with rightwards arrow on top vertical line vertical line v with rightwards arrow on top vertical line space cos space theta end cell row cell open parentheses table row 2 row cell negative 1 end cell row 1 end table close parentheses times open parentheses table row 7 row 0 row 1 end table close parentheses end cell equals cell open parentheses square root of 6 close parentheses open parentheses 5 square root of 2 close parentheses space cos space theta end cell row cell 14 plus 0 plus 1 end cell equals cell 5 square root of 12 space cos space theta end cell row 15 equals cell 10 square root of 3 space cos space theta end cell row cell fraction numerator 15 over denominator 10 square root of 3 end fraction end cell equals cell cos space theta end cell row cell fraction numerator 15 over denominator 10 square root of 3 end fraction times fraction numerator square root of 3 over denominator square root of 3 end fraction end cell equals cell cos space theta end cell row cell 15 over 30 square root of 3 end cell equals cell cos space theta end cell row cell 1 half square root of 3 end cell equals cell cos space theta end cell row cell 30 degree end cell equals theta end table end style

Jadi, besar sudut antara begin mathsize 14px style u with rightwards arrow on top end style dan begin mathsize 14px style v with rightwards arrow on top end style adalah begin mathsize 14px style 30 degree end style.

Roboguru

Diketahui titik ,  dan . Agar vektor posisi dari  tegak lurus pada vektor posisi dari  dan vektor posisi dari , maka  adalah ...

Pembahasan Soal:

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

Jika vektor a with rightwards arrow on top tegak lurus dengan vektor b with rightwards arrow on top, maka 

table attributes columnalign right center left columnspacing 0px end attributes row cell a with rightwards arrow on top times b with rightwards arrow on top end cell equals cell open vertical bar a with rightwards arrow on top close vertical bar open vertical bar b with rightwards arrow on top close vertical bar space cos space 90 degree end cell row cell a with rightwards arrow on top times b with rightwards arrow on top end cell equals 0 end table

Misal vektor posisi dari text C end text open parentheses x comma space y comma space z close parentheses.

Vektor posisi c with rightwards arrow on top tegak lurus a with rightwards arrow on top sehingga diperoleh persamaan (1) berikut.

table attributes columnalign right center left columnspacing 0px end attributes row cell c with rightwards arrow on top times a with rightwards arrow on top end cell equals 0 row cell open parentheses table row x row y row z end table close parentheses open parentheses table row cell negative 1 end cell row 2 row 1 end table close parentheses end cell equals 0 row cell negative x plus 2 y plus z end cell equals 0 end table

Vektor posisi c with rightwards arrow on top tegak lurus b with rightwards arrow on top sehingga diperoleh persamaan (2) berikut.

table attributes columnalign right center left columnspacing 0px end attributes row cell c with rightwards arrow on top times b with rightwards arrow on top end cell equals 0 row cell open parentheses table row x row y row z end table close parentheses open parentheses table row 2 row cell negative 2 end cell row 2 end table close parentheses end cell equals 0 row cell 2 x minus 2 y plus 2 z end cell equals 0 end table

Dari persamaan (1) dan (2) diperoleh persamaan (3) berikut.

table row cell negative x plus 2 y plus z end cell equals cell 0 space space space space end cell row cell 2 x minus 2 y plus 2 z end cell equals cell 0 space plus end cell row cell x plus 3 z end cell equals cell 0 space space space space end cell end table

Diperoleh x equals negative 3 z sehingga dengan metode sbstitusi dapat ditentukan nilai y berikut.

table attributes columnalign right center left columnspacing 0px end attributes row cell negative x plus 2 y plus z end cell equals 0 row cell negative open parentheses negative 3 z close parentheses plus 2 y plus z end cell equals 0 row cell 4 z plus 2 y end cell equals 0 row cell 2 y end cell equals cell negative 4 z end cell row y equals cell negative 2 z end cell end table

Dapat ditentukan perbandingan berikut.

table attributes columnalign right center left columnspacing 0px end attributes row cell x colon y colon z end cell equals cell negative 3 z colon negative 2 z colon z end cell row blank equals cell 3 colon 2 colon negative 1 end cell end table

Diperoleh text C end text open parentheses 3 comma space 2 comma space 1 close parentheses 

Oleh karena itu, tidak ada jawaban yang tepat.

Roboguru

Diketahui segitiga  dengan ,  dan . Besar sudut  adalah ...

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

Jika vektor a with rightwards arrow on top dan b with rightwards arrow on top membentuk sudut alpha, maka 

a with rightwards arrow on top times 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 space cos space alpha

Komponen dari vektor stack A B with rightwards arrow on top dan stack A C with rightwards arrow on top adalah 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 end text minus text A end text end cell row blank equals cell open parentheses table row 7 row 6 row 5 end table close parentheses minus open parentheses table row 3 row 1 row cell negative 2 end cell end table close parentheses end cell row blank equals cell open parentheses table row 4 row 5 row 7 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 1 row 6 row 2 end table close parentheses minus open parentheses table row 3 row 1 row cell negative 2 end cell end table close parentheses end cell row blank equals cell open parentheses table row cell negative 2 end cell row 5 row 4 end table close parentheses end cell end table

Berdasarkan konsep di atas, dapat ditentukan besar sudut text BAC end text sebagai berikut.

Misal: alpha equals angle text BAC end text

table attributes columnalign right center left columnspacing 0px end attributes row cell stack A B with rightwards arrow on top times stack A C with rightwards arrow on top end cell equals cell open vertical bar stack A B with rightwards arrow on top close vertical bar open vertical bar stack A C with rightwards arrow on top close vertical bar space cos space alpha end cell row cell cos space alpha end cell equals cell fraction numerator stack A B with rightwards arrow on top times stack A C with rightwards arrow on top over denominator open vertical bar stack A B with rightwards arrow on top close vertical bar times open vertical bar stack A C with rightwards arrow on top close vertical bar end fraction end cell row blank equals cell fraction numerator open parentheses table row 4 row 5 row 7 end table close parentheses open parentheses table row cell negative 2 end cell row 5 row 4 end table close parentheses over denominator square root of 4 squared plus 5 squared plus 7 squared end root times square root of open parentheses negative 2 close parentheses squared plus 5 squared plus 4 squared end root end fraction end cell row blank equals cell fraction numerator 4 times open parentheses negative 2 close parentheses plus 5 times 5 plus 7 times 4 over denominator square root of 16 plus 25 plus 49 end root times square root of 4 plus 25 plus 16 end root end fraction end cell row blank equals cell fraction numerator negative 8 plus 25 plus 28 over denominator square root of 90 times square root of 45 end fraction end cell row blank equals cell fraction numerator 45 over denominator 3 square root of 10 times 3 square root of 5 end fraction end cell row blank equals cell fraction numerator 45 over denominator 45 square root of 2 end fraction end cell row blank equals cell fraction numerator 1 over denominator square root of 2 end fraction cross times fraction numerator square root of 2 over denominator square root of 2 end fraction end cell row blank equals cell 1 half square root of 2 end cell end table

Diperoleh nilai cos space alpha equals 1 half square root of 2 sehingga alpha equals 45 degree 

Oleh karena itu, jawaban yang tepat adalah B.

Roboguru

Diketahui vektor satuan . Jika vektor  tegak lurus vektor  maka  sama dengan ...

Pembahasan Soal:

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 2 end cell row cell b subscript 1 end cell end table close parentheses, 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

Jika vektor a with rightwards arrow on top tegak lurus dengan vektor b with rightwards arrow on top, maka 

table attributes columnalign right center left columnspacing 0px end attributes row cell a with rightwards arrow on top times b with rightwards arrow on top end cell equals cell open vertical bar a with rightwards arrow on top close vertical bar open vertical bar b with rightwards arrow on top close vertical bar space cos space 90 degree end cell row cell a with rightwards arrow on top times b with rightwards arrow on top end cell equals 0 end table

Panjang vektor a with rightwards arrow on top, yaitu open vertical bar a with rightwards arrow on top close vertical bar equals square root of a subscript 1 squared plus a subscript 2 squared end root

Vektor satuan adalah vektor dengan panjang satu satuan.

Apabila diketahui vektor satuan u with rightwards arrow on top equals a i with rightwards arrow on top plus 3 over 5 j with rightwards arrow on top, maka nilai a dapat ditentukan sebagai berikut.

table attributes columnalign right center left columnspacing 0px end attributes row cell open vertical bar u with rightwards arrow on top close vertical bar end cell equals cell square root of a squared plus open parentheses 3 over 5 close parentheses squared end root end cell row 1 equals cell square root of a squared plus 9 over 25 end root end cell row 1 equals cell a squared plus 9 over 25 end cell row 25 equals cell 25 a squared plus 9 end cell row 0 equals cell 25 a squared minus 16 end cell row 0 equals cell open parentheses 5 a plus 4 close parentheses open parentheses 5 a minus 4 close parentheses end cell end table

a equals negative 4 over 5 space text atau end text space a equals 4 over 5

Berdasarkan konsep di atas, apabila vektor v with rightwards arrow on top tegak lurus u with rightwards arrow on top, maka dapat ditentukan hubungan berikut.

Untuk a equals negative 4 over 5

table attributes columnalign right center left columnspacing 0px end attributes row cell v with rightwards arrow on top times u with rightwards arrow on top end cell equals 0 row cell open parentheses table row 1 row b end table close parentheses open parentheses table row cell negative 4 over 5 end cell row cell 3 over 5 end cell end table close parentheses end cell equals 0 row cell negative 4 over 5 plus 3 over 5 b end cell equals 0 row cell 3 over 5 b end cell equals cell 4 over 5 end cell row cell 3 b end cell equals 4 row b equals cell 4 over 3 end cell end table

Diperoleh nilai a times b equals negative 4 over 5 times 4 over 3 equals negative 16 over 15

Untuk a equals 4 over 5

table attributes columnalign right center left columnspacing 0px end attributes row cell v with rightwards arrow on top times u with rightwards arrow on top end cell equals 0 row cell open parentheses table row 1 row b end table close parentheses open parentheses table row cell 4 over 5 end cell row cell 3 over 5 end cell end table close parentheses end cell equals 0 row cell 4 over 5 plus 3 over 5 b end cell equals 0 row cell 3 over 5 b end cell equals cell negative 4 over 5 end cell row b equals cell negative 4 over 3 end cell end table

Diperoleh nilai a times b equals 4 over 5 times open parentheses negative 4 over 3 close parentheses equals negative 16 over 15

Oleh karena itu, tidak terdapat jawaban yang tepat.

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

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