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Proyeksi vektor orthogonal dari a pada b adalah...

Pertanyaan

Proyeksi vektor orthogonal dari top enclose a pada top enclose b adalah...

  1. open vertical bar fraction numerator top enclose a times top enclose b over denominator open vertical bar top enclose a close vertical bar open vertical bar top enclose b close vertical bar end fraction close vertical bar      

  2. open vertical bar fraction numerator top enclose a times top enclose b over denominator open vertical bar top enclose a close vertical bar open vertical bar top enclose b close vertical bar end fraction close vertical bar top enclose b    space 

  3. open parentheses fraction numerator times top enclose a times top enclose b over denominator open vertical bar top enclose a close vertical bar open vertical bar top enclose b close vertical bar end fraction close parentheses top enclose b     

  4. open parentheses fraction numerator top enclose a times top enclose b over denominator top enclose b times top enclose b end fraction close parentheses top enclose b    

  5. open parentheses fraction numerator top enclose a times top enclose b over denominator top enclose b times top enclose b end fraction close parentheses times top enclose b  

Pembahasan Soal:

Perhatikan proyeksi vektor  top enclose a pada top enclose b berikut.

Dengan proyeksi tersebut akan diperoleh sebuah vektor baru misalnya vektor top enclose c. Vektor top enclose c ini disebut proyeksi vektor orthogonal vektor top enclose a pada vektor top enclose b. Perhatikan bahwa jika sudut antara top enclose a dan top enclose b lancip maka top enclose a searah dengan top enclose b, sedangkan jika tumpul maka top enclose c berlawanan arah dengan top enclose b.

Ini berarti bahwa vektor top enclose c dapat dinyatakan sebagai k top enclose b, dengan k adalah suatu bilangan yang sama dengan fraction numerator open vertical bar top enclose c close vertical bar over denominator open vertical bar top enclose b close vertical bar end fraction. Karena open vertical bar top enclose c close vertical bar equals fraction numerator top enclose a times top enclose b over denominator open vertical bar top enclose b close vertical bar end fraction 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 fraction numerator open vertical bar top enclose c close vertical bar over denominator open vertical bar top enclose b close vertical bar end fraction top enclose b end cell row blank equals cell fraction numerator fraction numerator top enclose a times top enclose b over denominator open vertical bar top enclose b close vertical bar end fraction over denominator open vertical bar top enclose b close vertical bar end fraction top enclose b end cell row blank equals cell left parenthesis fraction numerator top enclose a times top enclose b over denominator open vertical bar top enclose b close vertical bar squared end fraction right parenthesis top enclose b end cell row blank equals cell left parenthesis fraction numerator top enclose a times top enclose b over denominator top enclose b times top enclose b end fraction right parenthesis top enclose b end cell end table 

Jadi, proyeksi vektor orthogonal dari top enclose a pada top enclose b adalah open parentheses fraction numerator top enclose a times top enclose b over denominator top enclose b times top enclose b end fraction close parentheses top enclose b.

Oleh karena itu, jawaban yang benar adalah D.

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

E. Lestari

Mahasiswa/Alumni Universitas Sebelas Maret

Terakhir diupdate 06 Oktober 2021

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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Diketahuivektora=i+6j−kdanb=7i+4j+mk.Jikapanjangproyeksivektorakevektorbsamadengan31​panjangvektorb,makam=⋯

Pembahasan Soal:

M i s a l k a n space p r o y e k s i space v e k t o r space a with rightwards arrow on top space k e space v e k t o r space b with rightwards arrow on top space a d a l a h space v e k t o r space c with rightwards arrow on top space d a n space p a n j a n g n y a space s a m a space d e n g a n space 1 third p a n j a n g space v e k t o r space space b with rightwards arrow on top comma space m a k a  c with rightwards arrow on top equals 1 third space b with rightwards arrow on top  K a r e n a space  c with rightwards arrow on top equals open parentheses 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 squared end fraction close parentheses b with rightwards arrow on top  M a k a  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 squared end fraction space equals space 1 third  fraction numerator left parenthesis i plus 6 j minus k right parenthesis times left parenthesis 7 i plus 4 j plus m k right parenthesis over denominator vertical line 7 i plus 4 j plus m k vertical line squared end fraction equals 1 third  fraction numerator 1 times 7 plus 6 times 4 plus left parenthesis negative 1 right parenthesis times m over denominator 7 squared plus 4 squared plus m squared end fraction equals 1 third  fraction numerator 7 plus 24 minus m over denominator 49 plus 16 plus m squared end fraction space equals 1 third  fraction numerator 31 minus m over denominator 65 plus m squared end fraction equals 1 third  3 left parenthesis 31 minus m right parenthesis equals 1 left parenthesis 65 plus m squared right parenthesis  93 minus 3 m equals 65 plus m squared m squared plus 3 m plus 65 minus 93 equals 0 m squared plus 3 m minus 28 equals 0 left parenthesis m plus 7 right parenthesis left parenthesis m minus 4 right parenthesis equals 0 S e h i n g g a m equals negative 7 a t a u m equals 4.

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Diketahui vector u⇀=2i⇀+pj⇀​+2k⇀ dan v⇀=3i⇀−j⇀​+k⇀. Jika proyeksi scalar orthogonal vector v⇀ pada arah u⇀ sama dengan setengah panjang vector , maka nilai p adalah ....

Pembahasan Soal:

Diketahui:

  • u with rightwards harpoon with barb upwards on top equals 2 i with rightwards harpoon with barb upwards on top plus p j with rightwards harpoon with barb upwards on top plus 2 k with rightwards harpoon with barb upwards on top 
  • v with rightwards harpoon with barb upwards on top equals 3 i with rightwards harpoon with barb upwards on top minus j with rightwards harpoon with barb upwards on top plus space k with rightwards harpoon with barb upwards on top 

Sehingga, panjang proyeksi orthogonal v with rightwards harpoon with barb upwards on top pada arah u with rightwards harpoon with barb upwards on top:

table attributes columnalign right center left columnspacing 0px end attributes row cell fraction numerator v with rightwards harpoon with barb upwards on top times u with rightwards harpoon with barb upwards on top over denominator open vertical bar u with rightwards harpoon with barb upwards on top close vertical bar end fraction end cell equals cell 1 half open vertical bar u with rightwards arrow on top close vertical bar end cell row cell fraction numerator 3 times 2 plus open parentheses negative 1 close parentheses times p plus 1 times 2 over denominator open vertical bar u with rightwards arrow on top close vertical bar end fraction end cell equals cell 1 half open vertical bar u with rightwards arrow on top close vertical bar end cell row cell fraction numerator 6 minus p plus 2 over denominator open vertical bar u with rightwards arrow on top close vertical bar end fraction end cell equals cell 1 half open vertical bar u with rightwards arrow on top close vertical bar end cell row cell 8 minus p end cell equals cell 1 half open vertical bar u with rightwards arrow on top close vertical bar squared end cell row cell 8 minus p end cell equals cell 1 half open parentheses 2 squared plus p squared plus 2 squared close parentheses end cell row cell 8 minus p end cell equals cell 1 half open parentheses 4 plus p squared plus 4 close parentheses end cell row cell 2 open parentheses 8 minus p close parentheses end cell equals cell 8 plus p squared end cell row cell 16 minus 2 p end cell equals cell 8 plus p squared end cell row cell negative p squared minus 2 p plus 8 end cell equals 0 row cell p squared plus 2 p minus 8 end cell equals 0 row cell open parentheses p plus 4 close parentheses open parentheses p minus 2 close parentheses end cell equals 0 row blank blank cell p equals negative 4 space atau space p equals 2 end cell end table 

  Jadi, nilai p equals negative 4 atau p equals 2

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Roboguru

Diketahui vektor a=i−2j+k dan vektor b=3i+j−2k. Vektor c mewakili vektor hasil proyeksi orthogonal vektor b pada vektor a, maka vektor c=...

Pembahasan Soal:

c with rightwards arrow on top space a d a l a h space p r o y e k s i space v e k t o r space o r t h o g o n a l space b with rightwards arrow on top space p a d a space a with rightwards arrow on top comma  c with ⃗ on top equals fraction numerator a with ⃗ on top straight space. b with ⃑ on top over denominator open vertical bar a with ⃗ on top close vertical bar squared end fraction. a with ⃗ on top equals fraction numerator open parentheses table row 1 row cell negative 2 end cell row 1 end table close parentheses straight space. open parentheses table row 3 row 1 row cell negative 2 end cell end table close parentheses over denominator open parentheses 1 close parentheses squared plus open parentheses negative 2 close parentheses squared plus open parentheses 1 close parentheses squared end fraction. open parentheses table row 1 row cell negative 2 end cell row 1 end table close parentheses  equals fraction numerator open parentheses 1 close parentheses. open parentheses 3 close parentheses plus open parentheses negative 2 close parentheses. open parentheses 1 close parentheses plus 1 open parentheses negative 2 close parentheses over denominator 6 end fraction. open parentheses table row 1 row cell negative 2 end cell row 1 end table close parentheses  equals fraction numerator negative 1 over denominator 6 end fraction open parentheses table row 1 row cell negative 2 end cell row 1 end table close parentheses  a t a u space c with ⃗ on top equals fraction numerator negative 1 over denominator 6 end fraction open parentheses i minus 2 j plus k close parentheses

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Diketahui vector u=i+2j​−kdanv=i+j​+mk, panjang proyeksi upadavadalah32​3​. Bila m>0 maka nilai m+2=...

Pembahasan Soal:

Diketahui :

Diketahui vector top enclose straight u equals straight i with overbrace on top plus 2 straight j with overbrace on top minus straight k with overbrace on top space dan space straight v equals straight i with overbrace on top plus straight j with overbrace on top plus straight m straight k with overbrace on top,

panjang proyeksi straight u with rightwards arrow on top space pada space straight v with rightwards arrow on top space adalah space 2 over 3 square root of 3. Bila m>0 

Ditanya :

maka nilai m+2=...

Penyeesaian 

Misalkan c with rightwards arrow on top adalah straight u with rightwards arrow on top space pada space straight v with rightwards arrow on top space adalah space 2 over 3 square root of 3

table attributes columnalign right center left columnspacing 0px end attributes row cell straight c with rightwards harpoon with barb upwards on top end cell equals cell fraction numerator straight a with rightwards arrow on top straight b with rightwards arrow on top over denominator open vertical bar straight v with rightwards arrow on top close vertical bar squared end fraction end cell row cell 2 over 3 square root of 3 end cell equals cell fraction numerator open parentheses 1 comma 2 comma negative 1 close parentheses open parentheses 1 comma 1 comma straight m close parentheses over denominator square root of 1 squared plus 1 squared plus straight m squared end root end fraction end cell row cell 2 over 3 square root of 3 end cell equals cell fraction numerator 1 plus 2 minus straight m over denominator square root of 2 plus straight m squared end root end fraction end cell row cell 2 over 3 square root of 3 open parentheses square root of 2 plus straight m squared end root close parentheses end cell equals cell 3 minus straight m end cell row cell open parentheses 2 over 3 square root of 3 squared close parentheses open parentheses square root of 2 plus straight m squared end root close parentheses squared end cell equals cell open parentheses 3 minus straight m close parentheses squared end cell row cell 8 over 3 plus 4 over 3 straight m squared end cell equals cell 9 minus 6 straight m plus straight m squared end cell row cell 8 plus 4 straight m squared end cell equals cell 27 minus 18 straight m minus 3 straight m squared end cell row cell straight m squared plus 18 straight m minus 19 end cell equals 0 row cell open parentheses straight m plus 19 close parentheses open parentheses straight m minus 1 close parentheses end cell equals 0 row cell maka space straight m end cell equals cell negative 19 space dan space straight m equals 1 end cell end table

Nilai m = -19 tidak memenuhi, maka nilai m = 1, jadi nilia m space plus space 2 space equals space 1 plus 2 space equals space 3.

Oleh karena itu, jawaban yang benar adalah B.

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Diketahui vektor a=⎝⎛​−42−2​⎠⎞​ dan vektor b=⎝⎛​−330​⎠⎞​. Tentukan: c. Proyeksi vektor orthogonal a dan b

Pembahasan Soal:

Panjang vektor b with rightwards arrow on top adalah sebagai berikut.

table attributes columnalign right center left columnspacing 0px end attributes row cell open vertical bar b with rightwards arrow on top close vertical bar end cell equals cell square root of x squared plus y squared plus z squared end root end cell row blank equals cell square root of open parentheses negative 3 close parentheses squared plus 3 squared plus 0 squared end root end cell row blank equals cell square root of 9 plus 9 plus 0 end root end cell row blank equals cell square root of 18 end cell row blank equals cell 3 square root of 2 end cell end table

Perkalian vektor a with rightwards arrow on top dan b with rightwards arrow on top adalah 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 cell open parentheses table row cell negative 4 end cell row 2 row cell negative 2 end cell end table close parentheses open parentheses table row cell negative 3 end cell row 3 row 0 end table close parentheses end cell row blank equals cell open parentheses negative 4 close parentheses open parentheses negative 3 close parentheses plus 2 times 3 plus open parentheses negative 2 close parentheses open parentheses 0 close parentheses end cell row blank equals cell 12 plus 6 plus 0 end cell row blank equals 18 end table

Proyeksi vektor orthogonal a with rightwards arrow on top dan b with rightwards arrow on top dapat ditentukan sebagai berikut.

table attributes columnalign right center left columnspacing 0px end attributes row cell c with rightwards arrow on top end cell equals cell open parentheses 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 squared end fraction close parentheses times b with rightwards arrow on top end cell row blank equals cell open parentheses 18 over open parentheses 3 square root of 2 close parentheses squared close parentheses times open parentheses table row cell negative 3 end cell row 3 row 0 end table close parentheses end cell row blank equals cell open parentheses 18 over 18 close parentheses times open parentheses table row cell negative 3 end cell row 3 row 0 end table close parentheses end cell row blank equals cell open parentheses table row cell negative 3 end cell row 3 row 0 end table close parentheses end cell end table

Dengan demikian, proyeksi vektor orthogonal a with rightwards arrow on top pada b with rightwards arrow on top adalah 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 cell open parentheses table row cell negative 3 end cell row 3 row 0 end table close parentheses end cell end table 

0

Roboguru

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