Roboguru

Diketahui vektor  dan  panjang proyeksi vektor skalar  pada vektor  adalah ...

Pertanyaan

Diketahui vektor straight u with rightwards arrow on top equals open parentheses 2 comma space minus 1 comma space 3 close parentheses dan straight v with rightwards arrow on top equals open parentheses negative 3 comma space 2 comma space 6 close parentheses panjang proyeksi vektor skalar 3 straight u with rightwards arrow on top plus 2 straight v with rightwards arrow on top pada vektor straight v with rightwards arrow on top adalah ...

Pembahasan Soal:

Diketahui vektor straight u with rightwards arrow on top equals open parentheses 2 comma space minus 1 comma space 3 close parentheses dan straight v with rightwards arrow on top equals open parentheses negative 3 comma space 2 comma space 6 close parentheses, maka 

table attributes columnalign right center left columnspacing 0px end attributes row cell 3 u with rightwards arrow on top plus 2 v with rightwards arrow on top end cell equals cell 3 open parentheses 2 comma space minus 1 comma space 3 close parentheses plus 2 open parentheses negative 3 comma space 2 comma space 6 close parentheses end cell row blank equals cell open parentheses 6 comma space minus 3 comma space 9 close parentheses plus open parentheses negative 6 comma space 4 comma space 12 close parentheses end cell row blank equals cell open parentheses 0 comma space 1 comma space 21 close parentheses end cell end table 

Hasil kali titik 3 straight u with rightwards arrow on top plus 2 straight v with rightwards arrow on top dengan straight v with rightwards arrow on top adalah 

table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses 3 u with rightwards harpoon with barb upwards on top plus 2 v with rightwards arrow on top close parentheses times open parentheses v with rightwards arrow on top close parentheses end cell equals cell open parentheses 0 comma space 1 comma space 21 close parentheses times open parentheses negative 3 comma space 2 comma space 6 close parentheses end cell row blank equals cell open parentheses 0 plus 2 plus 126 close parentheses end cell row blank equals 128 end table 

Panjang vektor v with rightwards arrow on top adalah 

table attributes columnalign right center left columnspacing 0px end attributes row cell open vertical bar v with rightwards arrow on top close vertical bar end cell equals cell square root of open parentheses negative 3 close parentheses squared plus 2 squared plus 6 squared end root end cell row blank equals cell square root of 9 plus 4 plus 36 end root end cell row blank equals cell square root of 49 end cell row blank equals 7 end table  

Sehingga, panjang proyeksi vektor skalar 3 u with rightwards arrow on top plus 2 v with rightwards arrow on top pada vektor v with rightwards arrow on top adalah 

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 open parentheses 3 u with rightwards arrow on top plus 2 v with rightwards arrow on top close parentheses times v with rightwards arrow on top over denominator open vertical bar v with rightwards arrow on top close vertical bar end fraction end cell row blank equals cell 128 over 7 end cell end table 

Dengan demikian, panjang proyeksi vektor skalar 3 u with rightwards arrow on top plus 2 v with rightwards arrow on top pada vektor v with rightwards arrow on top adalah 128 over 7 satuan panjang.  

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

L. Rante

Mahasiswa/Alumni Universitas Negeri Makassar

Terakhir diupdate 04 Juni 2021

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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Pembahasan Soal:

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 u space ⃗ space p a d a space v space ⃑ comma space m i s a l k a n space c space ⃗  c ⃗ equals fraction numerator u ⃑. v ⃑ over denominator vertical line v ⃑ vertical line squared end fraction. v ⃑ equals fraction numerator open parentheses table row 0 row 2 row 2 end table close parentheses. open parentheses table row cell negative 2 end cell row 0 row 2 end table close parentheses over denominator left parenthesis negative 2 right parenthesis squared plus left parenthesis 0 right parenthesis squared plus left parenthesis 2 right parenthesis squared end fraction open parentheses table row cell negative 2 end cell row 0 row 2 end table close parentheses  equals fraction numerator left parenthesis 0 right parenthesis. left parenthesis negative 2 right parenthesis plus left parenthesis 2 right parenthesis. left parenthesis 0 right parenthesis plus 2 left parenthesis 2 right parenthesis over denominator 8 end fraction open parentheses table row cell negative 2 end cell row 0 row 2 end table close parentheses  equals 4 over 8 open parentheses table row cell negative 2 end cell row 0 row 2 end table close parentheses  c space ⃗ equals 1 half open parentheses table row cell negative 2 end cell row 0 row 2 end table close parentheses equals open parentheses table row cell negative 1 end cell row 0 row 1 end table close parentheses space a t a u space c space ⃗ equals negative i plus k

0

Roboguru

Jika vector dan , maka proyeksi scalar orthogonal vector pada vector adalah ...

Pembahasan Soal:

Proyeksi skalar ortogonal suatu vektor pada vektor lain adalah :

  • Proyeksi vektor orthogonal v with rightwards arrow on top pada u with rightwards arrow on top space equals fraction numerator u with rightwards arrow on top times v with rightwards arrow on top over denominator open vertical bar u with rightwards harpoon with barb upwards on top close vertical bar end fraction
  • Proyeksi vektor orthogonal u with rightwards arrow on top pada v with rightwards arrow on top space equals fraction numerator v with rightwards arrow on top times u with rightwards arrow on top over denominator open vertical bar v with rightwards arrow on top close vertical bar end fraction

Diketahui vector u equals left parenthesis 3 comma negative 4 right parenthesis dan v equals left parenthesis 5 comma 2 right parenthesis, cari dulu vector left parenthesis u minus v right parenthesis dan vector u plus v,

table attributes columnalign right center left columnspacing 0px end attributes row cell left parenthesis u minus v right parenthesis end cell equals cell open square brackets table row 3 row cell negative 4 end cell end table close square brackets minus open square brackets table row 5 row 2 end table close square brackets end cell row blank equals cell open square brackets table row cell 3 minus 5 end cell row cell negative 4 minus 2 end cell end table close square brackets end cell row blank equals cell open square brackets table row cell negative 2 end cell row cell negative 6 end cell end table close square brackets end cell end table

table attributes columnalign right center left columnspacing 0px end attributes row cell left parenthesis u plus v right parenthesis end cell equals cell open square brackets table row 3 row cell negative 4 end cell end table close square brackets plus open square brackets table row 5 row 2 end table close square brackets end cell row blank equals cell open square brackets table row cell 3 plus 5 end cell row cell negative 4 plus 2 end cell end table close square brackets end cell row blank equals cell open square brackets table row 8 row cell negative 2 end cell end table close square brackets end cell end table

Maka di dapatkan u plus v equals left parenthesis 8 comma negative 2 right parenthesis dan u minus v equals left parenthesis negative 2 comma negative 6 right parenthesis.

Proyeksi scalar orthogonal vector left parenthesis u minus v right parenthesis pada vector u plus v,

table attributes columnalign right center left columnspacing 0px end attributes row blank equals cell fraction numerator open parentheses u plus v close parentheses open parentheses u minus v close parentheses over denominator open vertical bar u minus v close vertical bar end fraction end cell row blank equals cell fraction numerator open square brackets table row 8 row cell negative 2 end cell end table close square brackets times open square brackets table row cell negative 2 end cell row cell negative 6 end cell end table close square brackets over denominator square root of left parenthesis negative 2 right parenthesis squared plus left parenthesis negative 6 right parenthesis squared end root end fraction end cell row blank equals cell fraction numerator open square brackets table row cell 8 times negative 2 end cell row cell negative 2 times negative 6 end cell end table close square brackets over denominator square root of 40 end fraction end cell row blank equals cell fraction numerator open square brackets table row cell negative 16 end cell row 12 end table close square brackets over denominator square root of 40 end fraction end cell row blank equals cell fraction numerator open square brackets table row cell negative 16 end cell row 12 end table close square brackets over denominator square root of 4 cross times 10 end root end fraction end cell row blank equals cell fraction numerator open square brackets table row cell negative 16 end cell row 12 end table close square brackets over denominator 2 square root of 10 end fraction end cell row blank equals cell open square brackets table row cell fraction numerator negative 16 over denominator 2 square root of 10 end fraction end cell row cell fraction numerator 12 over denominator 2 square root of 10 end fraction end cell end table close square brackets space rasionalkan end cell row blank equals cell open square brackets table row cell fraction numerator negative 16 over denominator 2 square root of 10 end fraction cross times fraction numerator square root of 10 over denominator square root of 10 end fraction end cell row cell fraction numerator 12 over denominator 2 square root of 10 end fraction cross times fraction numerator square root of 10 over denominator square root of 10 end fraction end cell end table close square brackets end cell row blank equals cell open square brackets table row cell fraction numerator negative 16 square root of 10 over denominator 2 times 10 end fraction end cell row cell fraction numerator 12 square root of 10 over denominator 2 times 10 end fraction end cell end table close square brackets end cell row blank equals cell open square brackets table row cell fraction numerator negative 16 square root of 10 over denominator 20 end fraction end cell row cell fraction numerator 12 square root of 10 over denominator 20 end fraction end cell end table close square brackets space sederhanakan end cell row blank equals cell open square brackets table row cell negative 4 over 5 square root of 10 end cell row cell 3 over 5 square root of 10 end cell end table close square brackets end cell end table


Jadi, proyeksi scalar orthogonal vector left parenthesis u minus v right parenthesis pada vector u plus v adalah open parentheses negative 4 over 5 square root of 10 comma space 3 over 5 square root of 10 close parentheses.

 

0

Roboguru

Jika  dan . Proyeksi skalar  pada  adalah ....

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.

0

Roboguru

Diketahui titik . Tentukan bentuk segitiga tersebut!

Pembahasan Soal:

Diberikan titik koordinat begin mathsize 14px style text A(5,1,3), B(2,-1,-1), C(4,2,-4) end text end style. Untuk mengetahui bentuk segitiga yang terbentuk dari ketiga titik tersebut, akan dicari besar salah satu sudutnya.

Apabila titik begin mathsize 14px style text A end text end style dan begin mathsize 14px style text B end text end style dihubungkan maka akan diperoleh vektor

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell top enclose BA end cell equals cell top enclose OA minus top enclose OB end cell row blank equals cell open parentheses table row 5 row 1 row 3 end table close parentheses minus open parentheses table row 2 row cell negative 1 end cell row cell negative 1 end cell end table close parentheses end cell row blank equals cell open parentheses table row 3 row 2 row 4 end table close parentheses end cell end table end style 

Didapatkan panjang vektor

 begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell top enclose BA end cell equals cell square root of 3 squared plus 2 squared plus 4 squared end root end cell row blank equals cell square root of 9 plus 4 plus 16 end root end cell row blank equals cell square root of 29 end cell end table end style   

Apabila titik begin mathsize 14px style text B end text end style dan begin mathsize 14px style text C end text end style dihubungkan maka akan diperoleh vektor

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell top enclose BC end cell equals cell top enclose OC minus top enclose OB end cell row blank equals cell open parentheses table row 4 row 2 row cell negative 4 end cell end table close parentheses minus open parentheses table row 2 row cell negative 1 end cell row cell negative 1 end cell end table close parentheses end cell row blank equals cell open parentheses table row 2 row 3 row cell negative 3 end cell end table close parentheses end cell end table end style 

Didapatkan panjang vektor

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell top enclose BC end cell equals cell square root of 2 squared plus 3 squared plus left parenthesis negative 3 right parenthesis squared end root end cell row blank equals cell square root of 4 plus 9 plus 9 end root end cell row blank equals cell square root of 22 end cell end table end style 

Vektor begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell top enclose BA end cell end table space dan space table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell top enclose BC end cell end table end style mengapit begin mathsize 14px style text ∠ABC end text end style sehingga untuk mencari besar sudutnya dapat digunakan aturan Cosinus seperti berikut,

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell cos space alpha end cell equals cell fraction numerator top enclose BA times top enclose B C end enclose over denominator open vertical bar top enclose BA close vertical bar times open vertical bar top enclose B C end enclose close vertical bar end fraction end cell row cell cos space alpha end cell equals cell fraction numerator open parentheses table row 3 row 2 row 4 end table close parentheses times open parentheses table row 2 row 3 row cell negative 3 end cell end table close parentheses over denominator square root of 29 times square root of 22 end fraction end cell row cell cos space alpha end cell equals cell fraction numerator 6 plus 6 minus 12 over denominator square root of 29 times square root of 22 end fraction end cell row cell cos space alpha end cell equals 0 row alpha equals cell 90 degree end cell end table end style 

Didapatkan besar salah satu sudutnya adalah begin mathsize 14px style 90 degree end style. Jadi, segitiga begin mathsize 14px style text ABC end text end style merupakan segitiga siku-siku.

2

Roboguru

1. Jika , buktikan bahwa ketiga vektor tersebut saling tegak lurus.

Pembahasan Soal:

Vektor a with rightwards arrow on top comma space b with rightwards arrow on top comma space dan space c with rightwards arrow on top dikatakan saling tegak lurus jika

 a with rightwards arrow on top times b with rightwards arrow on top equals 0 a with rightwards arrow on top times c with rightwards arrow on top equals 0 b with rightwards arrow on top times c with rightwards arrow on top equals 0 

Perhatikan penghitungan berikut!

 a with rightwards arrow on top times b with rightwards arrow on top equals open parentheses table row 2 row 3 row 1 end table close parentheses times open parentheses table row 5 row cell negative 2 end cell row cell negative 4 end cell end table close parentheses equals 2 open parentheses 5 close parentheses plus 3 open parentheses negative 2 close parentheses plus 1 open parentheses negative 4 close parentheses equals 0 a with rightwards arrow on top times c with rightwards arrow on top equals open parentheses table row 2 row 3 row 1 end table close parentheses times open parentheses table row 10 row cell negative 13 end cell row 19 end table close parentheses equals 2 open parentheses 10 close parentheses plus 3 open parentheses negative 13 close parentheses plus 1 open parentheses 19 close parentheses equals 0 b with rightwards arrow on top times c with rightwards arrow on top equals open parentheses table row 5 row cell negative 2 end cell row cell negative 4 end cell end table close parentheses times open parentheses table row 10 row cell negative 13 end cell row 19 end table close parentheses equals 5 open parentheses 10 close parentheses plus open parentheses negative 2 close parentheses open parentheses negative 13 close parentheses plus open parentheses negative 4 close parentheses open parentheses 19 close parentheses equals 0  

Jadi, terbukti bahwa vektor a with rightwards arrow on top comma space b with rightwards arrow on top comma space dan space c with rightwards arrow on top saling tegak lurus.space 

0

Roboguru

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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