Roboguru

Diketahui  dan . Jika , tentukanlah nilai .

Pertanyaan

Diketahui m with rightwards arrow on top equals open parentheses table row 4 row 3 row cell negative 2 end cell end table close parentheses dan n with rightwards arrow on top equals open parentheses table row 2 row cell negative 1 end cell row p end table close parentheses. Jika open vertical bar 2 m with rightwards arrow on top plus n with rightwards arrow on top close vertical bar equals 5 square root of 6, tentukanlah nilai p.

Pembahasan Soal:

Diketahui m with rightwards arrow on top equals open parentheses table row 4 row 3 row cell negative 2 end cell end table close parentheses dan n with rightwards arrow on top equals open parentheses table row 2 row cell negative 1 end cell row p end table close parentheses. Jika open vertical bar 2 m with rightwards arrow on top plus n with rightwards arrow on top close vertical bar equals 5 square root of 6, maka:

table attributes columnalign right center left columnspacing 0px end attributes row cell 2 m with rightwards arrow on top plus n with rightwards arrow on top end cell equals cell 2 open parentheses table row 4 row 3 row cell negative 2 end cell end table close parentheses plus open parentheses table row 2 row cell negative 1 end cell row p end table close parentheses end cell row blank equals cell open parentheses table row 8 row 6 row cell negative 4 end cell end table close parentheses plus open parentheses table row 2 row cell negative 1 end cell row p end table close parentheses end cell row blank equals cell open parentheses table row cell 8 plus 2 end cell row cell 6 plus left parenthesis negative 1 right parenthesis end cell row cell negative 4 plus p end cell end table close parentheses end cell row blank equals cell open parentheses table row 10 row 5 row cell negative 4 plus p end cell end table close parentheses end cell end table 

Sehingga:

table attributes columnalign right center left columnspacing 0px end attributes row cell open vertical bar 2 m with rightwards arrow on top plus n with rightwards arrow on top close vertical bar end cell equals cell square root of 10 squared plus 5 squared plus left parenthesis negative 4 plus p right parenthesis squared end root end cell row cell 5 square root of 6 end cell equals cell square root of 125 plus left parenthesis negative 4 plus p right parenthesis squared end root end cell row cell open parentheses 5 square root of 6 close parentheses squared end cell equals cell open parentheses square root of 125 plus left parenthesis negative 4 plus p right parenthesis squared end root close parentheses squared end cell row 150 equals cell 125 plus left parenthesis negative 4 plus p right parenthesis squared end cell row 25 equals cell left parenthesis negative 4 plus p right parenthesis squared end cell row cell plus-or-minus 5 end cell equals cell negative 4 plus p end cell end table 

table attributes columnalign right center left columnspacing 0px end attributes row cell negative 4 plus p end cell equals 5 row p equals cell 5 plus 4 end cell row p equals 9 end table 

table attributes columnalign right center left columnspacing 0px end attributes row cell negative 4 plus p end cell equals cell negative 5 end cell row p equals cell negative 5 plus 4 end cell row p equals cell negative 1 end cell end table 

Jadi, nilai p equals negative 1 atau p equals 9 

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

I. Sutiawan

Mahasiswa/Alumni Universitas Pasundan

Terakhir diupdate 02 Mei 2021

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

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   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 . Tentukan : b. panjang vektor  dan

Pembahasan Soal:

Perhatikan 

  • panjang vektor begin mathsize 14px style u with rightwards arrow on top end style 

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 3 squared plus 2 squared plus 2 squared end root end cell row blank equals cell square root of 9 plus 4 plus 4 end root end cell row blank equals cell square root of 17 end cell end table end style

  • panjang vektor begin mathsize 14px style v with rightwards arrow on top end style

begin mathsize 14px style 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 1 squared plus 1 squared plus open parentheses negative 4 close parentheses squared end root end cell row blank equals cell square root of 1 plus 1 plus 16 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 end style

  • panjang vektor

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell u with rightwards arrow on top plus 2 v with rightwards arrow on top end cell equals cell open parentheses table row 3 row 2 row 2 end table close parentheses plus 2 open parentheses table row 1 row 1 row cell negative 4 end cell end table close parentheses end cell row blank equals cell open parentheses table row 5 row 4 row cell negative 6 end cell end table close parentheses end cell row blank equals cell 5 i with rightwards arrow on top plus 4 j with rightwards arrow on top minus 6 k with rightwards arrow on top end cell end table end style

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 plus 2 v with rightwards arrow on top close vertical bar end cell equals cell square root of 5 squared plus 4 squared plus open parentheses negative 6 close parentheses squared end root end cell row blank equals cell square root of 25 plus 16 plus 36 end root end cell row blank equals cell square root of 77 end cell end table end style 

0

Roboguru

Jika vektor  dan , maka panjang vektor  yang memenuhi persamaan  adalah ....

Pembahasan Soal:

Gunakan konsep penjumlahan, pengurangan vektor dan panjang vektor.

begin mathsize 12px style m with rightwards arrow on top equals m subscript 1 i with rightwards arrow on top plus m subscript 2 j with rightwards arrow on top plus m subscript 2 k with rightwards arrow on top rightwards arrow open vertical bar m with rightwards arrow on top close vertical bar equals square root of open parentheses m subscript 1 close parentheses squared plus open parentheses m subscript 2 close parentheses squared plus open parentheses m subscript 3 close parentheses squared end root end style

Penjumlahan dan pengurangan vektor secara aljabar dengan menjumlahkan atau mengurangi unsur-unsur yang seletak.

Diketahui vektor a with rightwards arrow on top equals 3 i minus 4 j plus k dan b with rightwards arrow on top equals 5 i plus j plus 4 k, dan 2 a with rightwards arrow on top plus c with rightwards arrow on top equals 3 c with rightwards arrow on top plus 3 a with rightwards arrow on top minus b with rightwards arrow on top.

Akan ditentukan panjang vektor c with rightwards arrow on top.

*Terlebih dahulu tentukan vektor c with rightwards arrow on top.

table attributes columnalign right center left columnspacing 2px end attributes row cell 2 a with rightwards arrow on top plus c with rightwards arrow on top end cell equals cell 3 c with rightwards arrow on top plus 3 a with rightwards arrow on top minus b with rightwards arrow on top end cell row cell 2 a with rightwards arrow on top minus 3 a with rightwards arrow on top plus b with rightwards arrow on top end cell equals cell 3 c with rightwards arrow on top minus c with rightwards arrow on top end cell row cell negative a with rightwards arrow on top plus b with rightwards arrow on top end cell equals cell 2 c with rightwards arrow on top end cell row cell 2 c with rightwards arrow on top end cell equals cell negative a with rightwards arrow on top plus b with rightwards arrow on top end cell row cell c with rightwards arrow on top end cell equals cell 1 half open parentheses negative a with rightwards arrow on top plus b with rightwards arrow on top close parentheses end cell row cell c with rightwards arrow on top end cell equals cell 1 half open parentheses negative open parentheses 3 i minus 4 j plus k close parentheses plus open parentheses 5 i plus j plus 4 k close parentheses close parentheses end cell row cell c with rightwards arrow on top end cell equals cell 1 half open parentheses negative 3 i plus 4 j minus k plus 5 i plus j plus 4 k close parentheses end cell row cell c with rightwards arrow on top end cell equals cell 1 half open parentheses open parentheses negative 3 plus 5 close parentheses i plus open parentheses 4 plus 1 close parentheses j plus open parentheses negative 1 plus 4 close parentheses k close parentheses end cell row cell c with rightwards arrow on top end cell equals cell 1 half open parentheses 2 i plus 5 j plus 3 k close parentheses end cell row cell c with rightwards arrow on top end cell equals cell i plus 5 over 2 j plus 3 over 2 k end cell end table

*Menentukan panjang vektor c with rightwards arrow on top.

table attributes columnalign right center left columnspacing 2px end attributes row cell open vertical bar c with rightwards arrow on top close vertical bar end cell equals cell square root of open parentheses 1 close parentheses squared plus open parentheses 5 over 2 close parentheses squared plus open parentheses 3 over 2 close parentheses squared end root end cell row blank equals cell square root of 1 plus 25 over 4 plus 9 over 4 end root end cell row blank equals cell square root of 1 plus 34 over 4 end root end cell row blank equals cell square root of fraction numerator 4 plus 34 over denominator 4 end fraction end root end cell row blank equals cell square root of 38 over 4 end root end cell row blank equals cell 1 half square root of 38 end cell end table

Diperoleh panjang vektor c with rightwards arrow on top adalah 1 half square root of 38.

Oleh karena itu, tidak ada jawaban yang tepat, karena pilihan jawaban yang tepat adalah 1 half square root of 38.

0

Roboguru

Diberikan vektor-vektor  dengan . Jika  dan  sejajar, maka

Pembahasan Soal:

Karena kedua vektor sejajar, maka salah satu vektor adalah kelipatan dari vektor yang lainnya yang dapat kita tulis a with rightwards arrow on top equals k b with rightwards arrow on top. Nilai k dapat ditentukan sebagai berikut:

table attributes columnalign right center left columnspacing 0px end attributes row cell a with rightwards arrow on top end cell equals cell k b with rightwards arrow on top end cell row cell open parentheses table row x row cell negative 3 x end cell row cell 6 y end cell end table close parentheses end cell equals cell k open parentheses table row cell 1 minus y end cell row 3 row cell negative 1 minus x end cell end table close parentheses end cell end table  

Dari kesamaan dua vektor diperoleh:

table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell table row blank cell negative 3 x equals 3 k end cell left right double arrow cell negative x equals k end cell blank blank row blank cell x equals k open parentheses 1 minus y close parentheses end cell left right double arrow cell x equals negative x open parentheses 1 minus y close parentheses end cell left right double arrow cell negative 1 equals 1 minus y end cell row blank cell 6 y equals k open parentheses negative 1 minus x close parentheses end cell left right double arrow cell 6 times 2 equals negative x open parentheses negative 1 minus x close parentheses end cell left right double arrow cell 12 equals x plus x squared end cell end table table row blank row left right double arrow row left right double arrow end table table row blank row cell y equals 2 end cell row cell x squared plus x minus 12 equals 0 end cell end table end cell end table  

Diperoleh persamaan kuadrat, akar-akar dari persamaan kuadrat tersebut sebagai berikut:

table row cell x squared plus x minus 12 equals 0 end cell row cell open parentheses x plus 4 close parentheses open parentheses x minus 3 close parentheses equals 0 end cell end table table row cell x plus 4 equals 0 end cell logical or cell x minus 3 equals 0 end cell row cell x equals negative 4 end cell blank cell x equals 3 end cell end table 

Karena pada soal diketahui x greater than 0, maka nilai x equals 3.

Sehingga, vektor 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 a with rightwards arrow on top end cell equals cell x i with hat on top minus 3 x j with hat on top plus 6 y k with hat on top end cell row blank left right double arrow cell a with rightwards arrow on top equals 3 i with hat on top minus 3 times 3 j with hat on top plus 6 times 2 k with hat on top end cell row blank left right double arrow cell a with rightwards arrow on top equals 3 i with hat on top minus 9 j with hat on top plus 12 k with hat on top end cell row blank blank blank row cell b with rightwards arrow on top end cell equals cell open parentheses 1 minus y close parentheses i with hat on top plus 3 j with hat on top minus open parentheses 1 plus x close parentheses k with hat on top end cell row blank left right double arrow cell b with rightwards arrow on top equals open parentheses 1 minus 2 close parentheses i with hat on top plus 3 j with hat on top minus open parentheses 1 plus 3 close parentheses k with hat on top end cell row blank left right double arrow cell b with rightwards arrow on top equals negative i with hat on top plus 3 j with hat on top minus 4 k with hat on top end cell end table 

Dengan demikian:

a with rightwards arrow on top plus 3 b with rightwards arrow on top equals 3 i with hat on top minus 9 j with hat on top plus 12 k with hat on top plus 3 open parentheses negative i with hat on top plus 3 j with hat on top minus 4 k with hat on top close parentheses equals 3 i with hat on top minus 9 j with hat on top plus 12 k with hat on top minus 3 i with hat on top plus 9 j with hat on top minus 12 k with hat on top equals 0 

Oleh karena itu, jawaban yang benar adalah A.

0

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

RUANGGURU HQ

Jl. Dr. Saharjo No.161, Manggarai Selatan, Tebet, Kota Jakarta Selatan, Daerah Khusus Ibukota Jakarta 12860

Coba GRATIS Aplikasi Ruangguru

Produk Ruangguru

Produk Lainnya

Hubungi Kami

Ikuti Kami

©2021 Ruangguru. All Rights Reserved