Roboguru

Diketahui a=(311​), b=4i−10j, dan c=21​b. Jika vektor satuan dari hasil a+b−c adalah n​1​(m6​), maka nilai m dan n adalah ...

Pertanyaan

Diketahui begin mathsize 14px style straight a with bar on top equals open parentheses table row 3 row 11 end table close parentheses end stylebegin mathsize 14px style straight b with bar on top equals 4 straight i minus 10 straight j end style, dan begin mathsize 14px style straight c with bar on top equals 1 half straight b with bar on top end style. Jika vektor satuan dari hasil begin mathsize 14px style a with bar on top plus b with bar on top minus c with bar on top end style adalah begin mathsize 14px style fraction numerator 1 over denominator square root of n end fraction open parentheses table row m row 6 end table close parentheses end style, maka nilai begin mathsize 14px style m end style dan begin mathsize 14px style n end style adalah ...

  1. begin mathsize 14px style negative 5 end style dan begin mathsize 14px style 59 end style 

  2. begin mathsize 14px style 5 end style dan begin mathsize 14px style 59 end style 

  3. begin mathsize 14px style 5 end style dan begin mathsize 14px style 61 end style 

  4. begin mathsize 14px style 59 end style dan undefined 

  5. undefined dan begin mathsize 14px style 5 end style

Pembahasan Soal:

Diketahui  vektor begin mathsize 14px style straight a with bar on top equals open parentheses table row 3 row 11 end table close parentheses end stylebegin mathsize 14px style straight b with bar on top equals 4 straight i minus 10 straight j end style, dan begin mathsize 14px style straight c with bar on top equals 1 half straight b with bar on top end style maka:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell a with bar on top end cell equals cell open parentheses table row 3 row 11 end table close parentheses end cell row cell b with bar on top end cell equals cell open parentheses table row 4 row cell negative 10 end cell end table close parentheses end cell row cell c with bar on top end cell equals cell 1 half b with bar on top end cell row blank equals cell open parentheses table row 2 row cell negative 5 end cell end table close parentheses end cell row blank blank blank end table end style

Dengan menggunakan konsep penjumlahan vektor maka:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell straight a with bar on top plus straight b with bar on top minus straight c with bar on top end cell equals cell open parentheses table row 3 row 11 end table close parentheses plus open parentheses table row 4 row cell negative 10 end cell end table close parentheses minus open parentheses table row 2 row cell negative 5 end cell end table close parentheses end cell row blank equals cell open parentheses table row cell 3 plus 4 minus 2 end cell row cell 11 minus 10 plus 5 end cell end table close parentheses end cell row blank equals cell open parentheses table row 5 row 6 end table close parentheses end cell end table end style 

 Jika vektor satuan dari hasil begin mathsize 14px style a with bar on top plus b with bar on top minus c with bar on top end style adalah begin mathsize 14px style fraction numerator 1 over denominator square root of n end fraction open parentheses table row m row 6 end table close parentheses end style maka:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell Vektor space satuan space straight a with bar on top plus straight b with bar on top minus straight c with bar on top end cell equals cell fraction numerator straight a with bar on top plus straight b with bar on top minus straight c with bar on top over denominator open vertical bar straight a with bar on top plus straight b with bar on top minus straight c with bar on top close vertical bar end fraction end cell row cell fraction numerator 1 over denominator square root of n end fraction open parentheses table row m row 6 end table close parentheses end cell equals cell fraction numerator open parentheses table row 5 row 6 end table close parentheses over denominator square root of 5 squared plus 6 squared end root end fraction end cell row cell fraction numerator 1 over denominator square root of n end fraction open parentheses table row m row 6 end table close parentheses end cell equals cell fraction numerator open parentheses table row 5 row 6 end table close parentheses over denominator square root of 25 plus 36 end root end fraction end cell row cell fraction numerator 1 over denominator square root of n end fraction open parentheses table row m row 6 end table close parentheses end cell equals cell fraction numerator open parentheses table row 5 row 6 end table close parentheses over denominator square root of 25 plus 36 end root end fraction end cell row cell fraction numerator 1 over denominator square root of n end fraction open parentheses table row m row 6 end table close parentheses end cell equals cell fraction numerator open parentheses table row 5 row 6 end table close parentheses over denominator square root of 61 end fraction end cell row cell fraction numerator 1 over denominator square root of n end fraction open parentheses table row m row 6 end table close parentheses end cell equals cell fraction numerator 1 over denominator square root of 61 end fraction open parentheses table row 5 row 6 end table close parentheses end cell end table end style 

Dengan demikian diperoleh nilai begin mathsize 14px style m equals 5 end style dan begin mathsize 14px style n equals 61 end style.

Oleh karena itu, jawaban yang benar adalah C. 

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

S. Amamah

Mahasiswa/Alumni Universitas Negeri Malang

Terakhir diupdate 06 Oktober 2021

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Pertanyaan yang serupa

Diberikan u=(3−1​),v=(3−1​)danw=(−7−7​). a=u+v+w, Tentukan: b. Vektor satuan a

Pembahasan Soal:

undefined

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell open vertical bar a with rightwards arrow on top close vertical bar end cell equals cell square root of left parenthesis negative 3 right parenthesis squared plus left parenthesis negative 4 right parenthesis squared end root end cell row blank equals cell square root of 9 plus 16 end root end cell row blank equals cell square root of 25 end cell row blank equals 5 end table end style

 begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell Vektor space satuan space a with rightwards arrow on top end cell equals cell fraction numerator a with rightwards arrow on top over denominator stack open vertical bar a close vertical bar with rightwards arrow on top end fraction end cell row blank equals cell fraction numerator 1 over denominator stack vertical line a vertical line with rightwards arrow on top end fraction a with rightwards arrow on top end cell row blank equals cell 1 fifth open parentheses table row cell negative 3 end cell row cell negative 4 end cell end table close parentheses end cell row blank equals cell open parentheses table row cell bevelled fraction numerator negative 3 over denominator 4 end fraction end cell row cell bevelled fraction numerator negative 4 over denominator 5 end fraction end cell end table close parentheses end cell end table end style  

0

Roboguru

Diketahui vektor satuan u=ai+53​j​. Jika vektor v=i+bj​ tegak lurus vektor u maka ab 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.

0

Roboguru

Diketahui vektor a=i−2j dan b=2i−7j. Perhatikan pernyataan-pernyataan berikut! (i) Modulus vektor a adalah 3​. (ii) Modulus vektor b adalah 53​. (iii) Vektor satuan dari vektor  adalah 51​5​i−52​5​j...

Pembahasan Soal:

Jawaban yang tepat adalah nomor (iii) dan (iv):

(iii) Vektor satuan dari vektor a with rightwards arrow on top:

table attributes columnalign right center left columnspacing 0px end attributes row cell vektor space satuan space a with rightwards arrow on top end cell equals cell fraction numerator a with rightwards arrow on top over denominator open vertical bar a with rightwards arrow on top close vertical bar end fraction end cell row blank equals cell fraction numerator open parentheses table row 1 row cell thin space thin space thin space minus 2 end cell end table close parentheses over denominator square root of 1 plus open parentheses negative 2 close parentheses squared end root end fraction end cell row blank equals cell open parentheses table row cell fraction numerator 1 over denominator square root of 1 plus open parentheses negative 2 close parentheses squared end root end fraction times thin space 1 end cell row cell fraction numerator 1 over denominator square root of 1 plus open parentheses negative 2 close parentheses squared end root end fraction open parentheses negative 2 close parentheses end cell end table close parentheses end cell row blank equals cell open parentheses table row cell fraction numerator 1 over denominator square root of 5 end fraction end cell row cell negative fraction numerator 2 over denominator square root of 5 end fraction end cell end table close parentheses end cell row blank equals cell open parentheses table row cell 1 fifth square root of 5 end cell row cell negative 2 over 5 square root of 5 end cell end table close parentheses end cell row blank equals cell 1 fifth square root of 5 i minus 2 over 5 square root of 5 end cell end table 

(iv) Vektor satuan dari vektor b with rightwards arrow on top:

table attributes columnalign right center left columnspacing 0px end attributes row cell vektor space satuan space b with rightwards arrow on top end cell equals cell fraction numerator 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 blank equals cell fraction numerator open parentheses table row 2 row cell negative 7 end cell end table close parentheses over denominator square root of 2 plus open parentheses negative 7 close parentheses squared end root end fraction end cell row blank equals cell open parentheses table row cell fraction numerator 1 over denominator square root of 2 plus open parentheses negative 7 close parentheses squared end root end fraction times 2 end cell row cell fraction numerator 1 over denominator square root of 2 plus open parentheses negative 7 close parentheses squared end root end fraction open parentheses negative 7 close parentheses end cell end table close parentheses end cell row blank equals cell open parentheses table row cell fraction numerator 2 over denominator square root of 51 end fraction end cell row cell negative fraction numerator 7 over denominator square root of 51 end fraction end cell end table close parentheses end cell row blank equals cell open parentheses table row cell 2 over 51 square root of 51 end cell row cell negative 7 over 51 square root of 51 end cell end table close parentheses end cell row blank equals cell 2 over 51 square root of 51 i minus 7 over 51 square root of 51 space j end cell end table 

Jadi, pernyataan yang benar ditunjukkan oleh (iii) dan (iv).

Dengan demikian, jawaban yang tepat adalah E.

0

Roboguru

Diketahui vektor-vektor: a=(60​),b=(−3−4​),c=(52​),dand=(4−1​) Tentukan vektor satuan dari 31​a+85​(2a+7b) !

Pembahasan Soal:

Ingat bahwa vektor satuan adalah vektor yang panjangnya sama dengan satu. Berdasarkan soal diketahui straight a with rightwards arrow on top equals open parentheses table row 6 row 0 end table close parentheses comma space straight b with rightwards arrow on top equals open parentheses table row cell negative 3 end cell row cell negative 4 end cell end table close parentheses comma space straight c with rightwards arrow on top equals open parentheses table row 5 row 2 end table close parentheses comma space dan space straight d with rightwards arrow on top equals open parentheses table row 4 row cell negative 1 end cell end table close parentheses, sehingga panjang vektor 1 third straight a with rightwards arrow on top plus 5 over 8 left parenthesis 2 straight a with rightwards arrow on top plus 7 straight b with rightwards arrow on top right parenthesis diperoleh:

table attributes columnalign right center left columnspacing 0px end attributes row cell 1 third straight a with rightwards arrow on top plus 5 over 8 left parenthesis 2 straight a with rightwards arrow on top plus 7 straight b with rightwards arrow on top right parenthesis end cell equals cell 1 third open parentheses table row 6 row 0 end table close parentheses plus 5 over 8 open parentheses table row cell 12 plus left parenthesis negative 21 right parenthesis end cell row cell 0 minus 28 end cell end table close parentheses end cell row blank equals cell open parentheses table row 2 row 0 end table close parentheses plus 5 over 8 open parentheses table row cell negative 9 end cell row cell negative 28 end cell end table close parentheses end cell row blank equals cell open parentheses table row 2 row 0 end table close parentheses plus open parentheses table row cell fraction numerator negative 45 over denominator 8 end fraction end cell row cell fraction numerator negative 140 over denominator 8 end fraction end cell end table close parentheses end cell row blank equals cell open parentheses table row cell negative 3 comma 625 end cell row cell negative 17 comma 5 end cell end table close parentheses end cell row cell open vertical bar 1 third straight a plus 5 over 8 left parenthesis 2 straight a plus 7 straight b right parenthesis close vertical bar end cell equals cell square root of left parenthesis negative 3 comma 625 right parenthesis squared plus left parenthesis negative 17 comma 5 right parenthesis squared end root end cell row blank equals cell 17 comma 871 end cell end table


Sehingga, vektor satuan 1 third straight a with rightwards arrow on top plus 5 over 8 left parenthesis 2 straight a with rightwards arrow on top plus 7 straight b with rightwards arrow on top right parenthesis diperoleh:

31a+85(2a+7b)===31a+85(2a+7b)1(31a+85(2a+7b))17,8711(3,62517,5)(0,2020,979) 

Dengan demikian, vektor satuan dari 31a+85(2a+7b) adalah (0,2020,979).

0

Roboguru

Diketahui dua vektor yang sama, yaitu vektor u=(2n+110​) dan v=15i+(m2+6)j​. c. Tentukan vektor satuan dari vektor v?

Pembahasan Soal:

Diketahui dua vektor yang sama, yaitu vektor begin mathsize 14px style u with rightwards arrow on top equals open parentheses table row cell 2 n plus 1 end cell row 10 end table close parentheses end style dan begin mathsize 14px style v with rightwards arrow on top equals 15 i with rightwards arrow on top plus left parenthesis m squared plus 6 right parenthesis j with rightwards arrow on top end style, maka:

table attributes columnalign right center left columnspacing 0px end attributes row cell v with rightwards arrow on top end cell equals cell u with rightwards arrow on top end cell row cell open parentheses table row 15 row cell m squared plus 6 end cell end table close parentheses end cell equals cell open parentheses table row cell 2 n plus 1 end cell row 10 end table close parentheses end cell row cell open parentheses table row 15 row cell 2 squared plus 6 end cell end table close parentheses end cell equals cell open parentheses table row cell 2 n plus 1 end cell row 10 end table close parentheses end cell row cell v with rightwards arrow on top end cell equals cell open parentheses table row 15 row 10 end table close parentheses end cell end table 

Dengan rumus mencari vektor satuan pada dua dimensi diperoleh:

table attributes columnalign right center left columnspacing 0px end attributes row cell e subscript v with rightwards arrow on top end subscript end cell equals cell fraction numerator 1 over denominator open vertical bar v with rightwards arrow on top close vertical bar end fraction times v with rightwards arrow on top end cell row cell e subscript v with rightwards arrow on top end subscript end cell equals cell fraction numerator 1 over denominator square root of 15 squared plus 10 squared end root end fraction times open parentheses table row 15 row 10 end table close parentheses end cell row cell e subscript v with rightwards arrow on top end subscript end cell equals cell fraction numerator 1 over denominator square root of 325 end fraction times open parentheses table row 15 row 10 end table close parentheses end cell row cell e subscript v with rightwards arrow on top end subscript end cell equals cell fraction numerator 1 over denominator 5 square root of 13 end fraction cross times fraction numerator square root of 13 over denominator square root of 13 end fraction times open parentheses table row 15 row 10 end table close parentheses end cell row cell e subscript v with rightwards arrow on top end subscript end cell equals cell 1 over 65 square root of 13 times open parentheses table row 15 row 10 end table close parentheses end cell row cell e subscript v with rightwards arrow on top end subscript end cell equals cell open parentheses table row cell 15 over 65 square root of 13 end cell row cell 10 over 65 square root of 13 end cell end table close parentheses end cell row cell e subscript v with rightwards arrow on top end subscript end cell equals cell open parentheses table row cell 3 over 13 square root of 13 end cell row cell 2 over 13 square root of 13 end cell end table close parentheses end cell end table 

Dengan demikian, vektor satuan dari vektor begin mathsize 14px style v with rightwards arrow on top end style adalah v with rightwards arrow on top equals open parentheses table row cell 3 over 13 square root of 13 end cell row cell 2 over 13 square root of 13 end cell end table close parentheses.

0

Roboguru

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