Roboguru

Jika  dan , maka:  (1)   (2)   (3)   (4)  sejajar   Pernyataan yang benar adalah ...

Pertanyaan

Jika a with rightwards arrow on top equals 3 i with rightwards arrow on top minus 2 j with rightwards arrow on top plus k dan b with rightwards arrow on top equals negative 3 i with rightwards arrow on top plus j with rightwards arrow on top minus 4 k with rightwards arrow on top, maka: 

(1) open vertical bar a with rightwards arrow on top plus b with rightwards arrow on top close vertical bar equals square root of 10 
(2) open vertical bar a with rightwards arrow on top close vertical bar space colon space open vertical bar b with rightwards arrow on top close vertical bar equals 1 space colon space 2 
(3) a with rightwards arrow on top times b with rightwards arrow on top equals negative 15 
(4) a with rightwards arrow on top sejajar b with rightwards arrow on top 

Pernyataan yang benar adalah ...

Pembahasan Soal:

Diketahui a with rightwards arrow on top equals 3 i with rightwards arrow on top minus 2 j with rightwards arrow on top plus k dan b with rightwards arrow on top equals negative 3 i with rightwards arrow on top plus j with rightwards arrow on top minus 4 k with rightwards arrow on top

(1) open vertical bar a with rightwards arrow on top plus b with rightwards arrow on top close vertical bar equals square root of 10  

table attributes columnalign right center left columnspacing 0px end attributes row cell a with rightwards arrow on top plus b with rightwards arrow on top end cell equals cell open parentheses 3 i with rightwards arrow on top minus 2 j with rightwards arrow on top plus k with rightwards arrow on top close parentheses plus open parentheses negative 3 i with rightwards arrow on top plus j with rightwards arrow on top minus 4 k with rightwards arrow on top close parentheses end cell row blank equals cell open parentheses negative j with rightwards arrow on top minus 3 k with rightwards arrow on top close parentheses end cell row blank blank cell maka space end cell row cell open vertical bar a with italic rightwards arrow on top plus b with italic rightwards arrow on top close vertical bar end cell equals cell square root of open parentheses negative 1 close parentheses squared plus open parentheses negative 3 close parentheses squared end root end cell row blank equals cell square root of 1 plus 9 end root end cell row blank equals cell square root of 10 end cell row blank blank blank end table 

(2) open vertical bar a with rightwards arrow on top close vertical bar space colon space open vertical bar b with rightwards arrow on top close vertical bar equals 1 space colon space 2

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 open parentheses 3 close parentheses squared plus open parentheses negative 2 close parentheses squared plus open parentheses 1 close parentheses squared end root end cell row blank equals cell square root of 9 plus 4 plus 1 end root end cell row blank equals cell square root of 14 end cell row blank blank blank row cell open vertical bar b with rightwards arrow on top close vertical bar end cell equals cell square root of open parentheses negative 3 close parentheses squared plus open parentheses 1 close parentheses squared plus open parentheses negative 4 close parentheses squared end root end cell row blank equals cell square root of 9 plus 1 plus 16 end root end cell row blank equals cell square root of 26 end cell row blank blank cell maka space end cell row cell open vertical bar a with rightwards arrow on top close vertical bar space colon space open vertical bar b with rightwards arrow on top close vertical bar end cell equals cell square root of 14 space colon space square root of 26 end cell end table 


(3) a with rightwards arrow on top times b with rightwards arrow on top equals negative 15 

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 3 i with rightwards arrow on top minus 2 j with rightwards arrow on top plus k with rightwards arrow on top close parentheses times open parentheses negative 3 i with rightwards arrow on top plus j with rightwards arrow on top minus 4 k with rightwards arrow on top close parentheses end cell row blank equals cell open parentheses negative 9 close parentheses plus open parentheses negative 2 close parentheses plus open parentheses negative 4 close parentheses end cell row blank equals cell negative 15 end cell end table 


(4) a with rightwards arrow on top sejajar b with rightwards arrow on top

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 negative 15 end cell row blank blank blank row cell open vertical bar a with rightwards arrow on top close vertical bar times open vertical bar b with rightwards arrow on top close vertical bar end cell equals cell square root of 14 times square root of 26 end cell row blank equals cell square root of 224 end cell row blank blank blank row cell Karena space open vertical bar a with rightwards arrow on top close vertical bar times open vertical bar b with rightwards arrow on top close vertical bar end cell not equal to cell a with rightwards arrow on top times b with rightwards arrow on top comma space maka space a with rightwards arrow on top space d a n space b with rightwards arrow on top space tidak space sejajar end cell end table 


Dengan demikian, pernyataan yang benar adalah open parentheses 1 close parentheses dan open parentheses 3 close parentheses

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

L. Rante

Mahasiswa/Alumni Universitas Negeri Makassar

Terakhir diupdate 07 Juni 2021

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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Pertanyaan yang serupa

Jika  dan  maka besar sudut yang dibentuk oleh vektor  dan  sama dengan...

Pembahasan Soal:

Ingat kembali  penjumlahan vektor, panjang vektor dan besar sudut pada vektor berikut.

  • Jika top enclose a equals a subscript 1 i plus a subscript 2 j plus a subscript 3 k dan top enclose b equals b subscript 1 i plus b subscript 2 j plus b subscript 3 k maka

top enclose a plus top enclose b equals left parenthesis a subscript 1 plus b subscript 1 right parenthesis i plus left parenthesis a subscript 2 plus b subscript 2 right parenthesis j plus left parenthesis a subscript 3 plus b subscript 3 right parenthesis k

top enclose a minus top enclose b equals left parenthesis a subscript 1 minus b subscript 1 right parenthesis i plus left parenthesis a subscript 2 minus b subscript 2 right parenthesis j plus left parenthesis a subscript 3 minus b subscript 3 right parenthesis k 

  • Jika theta adalah sudut antara vektor top enclose a space dan space top enclose b, maka cos space theta equals fraction numerator a bullet b over denominator open vertical bar top enclose a close vertical bar times open vertical bar top enclose b close vertical bar end fraction.

 

  • open vertical bar top enclose a close vertical bar equals square root of a subscript 1 squared plus a subscript 2 squared plus a subscript 3 squared end root 

 

  • a bullet b equals a subscript 1 b subscript 1 plus a subscript 2 b subscript 2 plus a subscript 3 b subscript 3

Dari rumus di atas, diperoleh perhitungan sebagai berikut.

table attributes columnalign right center left columnspacing 0px end attributes row cell left parenthesis top enclose a plus top enclose b right parenthesis end cell equals cell left parenthesis negative 1 plus 1 right parenthesis i plus left parenthesis 1 plus left parenthesis negative 1 right parenthesis right parenthesis j plus left parenthesis 2 plus 2 right parenthesis k end cell row blank equals cell 0 i plus 0 j plus 4 k end cell row cell left parenthesis top enclose a minus top enclose b right parenthesis end cell equals cell left parenthesis negative 1 minus 1 right parenthesis i plus left parenthesis 1 minus left parenthesis negative 1 right parenthesis right parenthesis j plus left parenthesis 2 minus 2 right parenthesis k end cell row blank equals cell negative 2 i plus 2 j plus 0 k end cell row blank blank blank end table 

Sehingga besar sudut  yang dibentuk oleh vektor left parenthesis top enclose a plus top enclose b right parenthesis dan left parenthesis top enclose a minus top enclose b right parenthesis adalah

table attributes columnalign right center left columnspacing 0px end attributes row cell cos space theta end cell equals cell fraction numerator left parenthesis top enclose a plus top enclose b right parenthesis bullet left parenthesis top enclose a minus top enclose b right parenthesis over denominator open vertical bar top enclose a plus top enclose b close vertical bar times open vertical bar top enclose a minus top enclose b close vertical bar end fraction end cell row cell cos space theta end cell equals cell fraction numerator left parenthesis 0 i plus 0 j plus 4 k right parenthesis bullet left parenthesis negative 2 i plus 2 j plus 0 k right parenthesis over denominator square root of 0 squared plus 0 squared plus 4 squared end root cross times square root of left parenthesis negative 2 right parenthesis squared plus 2 squared plus 0 squared end root end fraction end cell row cell cos space theta end cell equals cell fraction numerator left parenthesis 0 cross times 0 right parenthesis plus left parenthesis 0 cross times 2 right parenthesis plus left parenthesis 4 cross times 0 right parenthesis over denominator square root of 0 plus 0 plus 16 end root cross times square root of 4 plus 4 plus end root end fraction end cell row cell cos space theta end cell equals cell fraction numerator 0 over denominator square root of 16 cross times square root of 8 end fraction end cell row cell cos space theta end cell equals 0 row cell cos space theta end cell equals cell cos space 90 degree end cell row theta equals cell 90 degree end cell end table 

Jadi, besar sudut yang dibentuk oleh vektor left parenthesis top enclose a plus top enclose b right parenthesis dan left parenthesis top enclose a minus top enclose b right parenthesis  adalah 90 degree.

Oleh karena itu, jawaban yang benar adalah D.

 

0

Roboguru

Diketahui vektor ,  dan . Panjang proyeksi   pada  adalah...

Pembahasan Soal:

Ingat kembali penjumlahan vektor dan panjang proyeksi berikut.

  • top enclose a plus top enclose b equals left parenthesis a subscript 1 plus b subscript 1 right parenthesis top enclose i plus left parenthesis a subscript 2 plus b subscript 2 right parenthesis top enclose j plus left parenthesis a subscript 3 plus b subscript 3 right parenthesis top enclose k 
  • 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   

  • Bentuk akar fraction numerator a over denominator square root of b end fraction dapat dirasionalkan dengan cara fraction numerator a over denominator square root of b end fraction cross times fraction numerator square root of b over denominator square root of b end fraction 

Dari rumus di atas, diperoleh perhitungan sebagai berikut.

Menentukan top enclose b plus top enclose c 

table attributes columnalign right center left columnspacing 0px end attributes row cell top enclose b plus top enclose c end cell equals cell left parenthesis 1 plus 3 right parenthesis top enclose i plus left parenthesis 1 plus 0 right parenthesis top enclose j plus left parenthesis 2 plus left parenthesis negative 1 right parenthesis right parenthesis top enclose k end cell row blank equals cell 4 top enclose i plus top enclose j plus top enclose k end cell end table  

Panjang proyeksi  left parenthesis top enclose b plus top enclose c right parenthesis pada top enclose a adalah

table attributes columnalign right center left columnspacing 0px end attributes row cell top enclose d end cell equals cell open vertical bar fraction numerator left parenthesis top enclose b plus top enclose c right parenthesis bullet top enclose a over denominator open vertical bar top enclose a close vertical bar end fraction close vertical bar end cell row blank equals cell open vertical bar fraction numerator left parenthesis 4 top enclose i plus top enclose j plus top enclose k right parenthesis bullet left parenthesis top enclose i minus top enclose j plus top enclose k right parenthesis over denominator square root of 1 squared plus left parenthesis negative 1 right parenthesis squared plus 1 squared end root end fraction close vertical bar end cell row blank equals cell open vertical bar fraction numerator 4 left parenthesis 1 right parenthesis plus 1 left parenthesis negative 1 right parenthesis plus 1 left parenthesis 1 right parenthesis over denominator square root of 1 plus 1 plus 1 end root end fraction close vertical bar end cell row blank equals cell open vertical bar fraction numerator 4 minus 1 plus 1 over denominator square root of 3 end fraction close vertical bar end cell row blank equals cell open vertical bar fraction numerator 4 over denominator square root of 3 end fraction close vertical bar end cell row blank equals cell fraction numerator 4 over denominator square root of 3 end fraction cross times fraction numerator square root of 3 over denominator square root of 3 end fraction end cell row blank equals cell 4 over 3 square root of 3 end cell row blank equals cell 1 1 third square root of 3 end cell end table 

Jadi, panjang proyeksi  left parenthesis top enclose b plus top enclose c right parenthesis pada top enclose a adalah 1 1 third square root of 3.

Oleh karena itu, jawaban yang benar adalah D.

0

Roboguru

Diketahui   dan  . Jika  tegak lurus  dan   tegak lurus , maka  = ...

Pembahasan Soal:

Ingat kembali operasi pada vektor berikut.

Jika diketahui top enclose a equals a subscript 1 i plus a subscript 2 j plus a subscript 3 k space dan space top enclose b equals b subscript 1 i plus b subscript 2 j plus b subscript 3 k space, maka

  • top enclose a plus top enclose b equals left parenthesis a subscript 1 plus b subscript 1 right parenthesis i plus left parenthesis a subscript 2 plus b subscript 2 right parenthesis j plus left parenthesis a subscript 3 plus b subscript 3 right parenthesis k 
  • open vertical bar top enclose a close vertical bar equals square root of a subscript 1 squared plus a subscript 2 squared plus a subscript 3 squared end root
  • top enclose a bullet space top enclose b equals a subscript 1 b subscript 1 plus a subscript 2 b subscript 2 plus a subscript 3 b subscript 3
  • Vektor top enclose a space dan space top enclose b dikatakan saling tegak lurus jika top enclose a bullet space top enclose b equals 0 

Berdasarkan rumus di atas, diperoleh perhitungan sebagai berikut.

Karena vektor  top enclose a tegak lurus top enclose b , maka top enclose a bullet space top enclose b equals 0.

Sehingga

table attributes columnalign right center left columnspacing 0px end attributes row cell top enclose a bullet space top enclose b end cell equals 0 row cell 2 left parenthesis x right parenthesis plus left parenthesis negative 4 right parenthesis left parenthesis z right parenthesis plus 3 left parenthesis 4 right parenthesis end cell equals 0 row cell 2 x minus 4 z plus 12 end cell equals 0 row cell 2 x minus 4 z end cell equals cell negative 12 end cell end table  

Karena  top enclose c tegak lurus top enclose d, maka top enclose c bullet space top enclose d equals 0

Sehingga

 table attributes columnalign right center left columnspacing 0px end attributes row cell top enclose c bullet space top enclose d end cell equals 0 row cell 5 left parenthesis 2 right parenthesis plus left parenthesis negative 3 right parenthesis left parenthesis z right parenthesis plus 2 left parenthesis x right parenthesis end cell equals 0 row cell 10 minus 3 z plus 2 x end cell equals 0 row cell 2 x minus 3 z end cell equals cell negative 10 end cell end table 

Dari kedua persamaan di atas, akan ditentukan nilai z

 stack attributes charalign center stackalign right end attributes row 2 x minus 4 z equals negative 12 end row row 2 x minus 3 z equals negative 10 end row horizontal line row minus z equals negative 2 end row row z equals none 2 end row end stack minus  

Substitusi z equals 2 ke persamaan 2 x minus 4 z equals negative 12 maka diperoleh

table attributes columnalign right center left columnspacing 0px end attributes row cell 2 x minus 4 z end cell equals 12 row cell 2 x minus 4 left parenthesis 2 right parenthesis end cell equals cell negative 12 end cell row cell 2 x minus 8 end cell equals cell negative 12 end cell row cell 2 x end cell equals cell negative 12 plus 8 end cell row cell 2 x end cell equals cell negative 4 end cell row x equals cell fraction numerator negative 4 over denominator 2 end fraction end cell row x equals cell negative 2 end cell end table  

Karena x equals negative 2 dan z equals 2, maka top enclose b equals negative 2 top enclose i plus 2 top enclose j plus 4 k.

Vektor top enclose a plus top enclose b adalah

table attributes columnalign right center left columnspacing 0px end attributes row cell top enclose a plus top enclose b end cell equals cell left parenthesis 2 plus left parenthesis negative 2 right parenthesis right parenthesis top enclose i plus left parenthesis negative 4 plus 2 right parenthesis top enclose j plus left parenthesis 3 plus 4 right parenthesis k end cell row blank equals cell 0 i minus 2 j plus 7 k end cell end table 

Panjang vektor top enclose a plus top enclose b adalah

table attributes columnalign right center left columnspacing 0px end attributes row cell vertical line top enclose a plus top enclose b vertical line end cell equals cell square root of 0 squared plus left parenthesis negative 2 right parenthesis squared plus 7 to the power of 2 end exponent end root end cell row blank equals cell square root of 0 plus 4 plus 49 end root end cell row blank equals cell square root of 53 end cell end table 

Jadi, vertical line top enclose a plus top enclose b vertical line equals square root of 53.

Oleh karena itu, jawaban yang tepat adalah E.

0

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.

0

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.

0

Roboguru

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

Tapi Roboguru masih mau belajar. Menurut kamu pembahasan kali ini sudah membantu, belum?

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