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Tentukan vektor posisi titik C dan nyatakan bentuknya dalam vektor basis, jika C terletak pada ruas garis AB dengan perbandingan sbb: a. AC:CB=2:1, dengan A(4,−3,7)danB(1,4,1).

Pertanyaan

Tentukan vektor posisi titik C dan nyatakan bentuknya dalam vektor basis, jika C terletak pada ruas garis AB dengan perbandingan sbb:

a. begin mathsize 14px style AC space colon space CB equals 2 space colon space 1 end style, dengan begin mathsize 14px style straight A left parenthesis 4 comma space minus 3 comma space 7 right parenthesis space dan space straight B left parenthesis 1 comma space 4 comma space 1 right parenthesis. end style

Pembahasan Soal:

Mencari dengan perbandingan vektor:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell c with rightwards arrow on top end cell equals cell fraction numerator 1 cross times a with rightwards arrow on top plus 2 cross times b with rightwards arrow on top over denominator 1 plus 2 end fraction end cell row blank equals cell 1 third open square brackets 1 open parentheses table row 4 row cell negative 3 end cell row 7 end table close parentheses plus 2 open parentheses table row 1 row 4 row 1 end table close parentheses close square brackets end cell row blank equals cell 1 third open parentheses table row cell 4 plus 2 end cell row cell negative 3 plus 8 end cell row cell 7 plus 2 end cell end table close parentheses end cell row blank equals cell open parentheses table row 2 row cell 5 over 3 end cell row 3 end table close parentheses end cell end table end style 

Jadi, begin mathsize 14px style 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 2 row cell 5 over 3 end cell row 3 end table close parentheses end cell end table end style.

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

S. Yoga

Mahasiswa/Alumni Universitas Pendidikan Indonesia

Terakhir diupdate 06 Oktober 2021

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Pertanyaan yang serupa

Diketahui tiga garis yang segaris yaitu titik A(4,−1,3) dan B(−2,2,−6) serta AC=32​AB , maka koordinat titik C adalah ....

Pembahasan Soal:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell top enclose A C end enclose end cell equals cell left parenthesis left parenthesis 0 minus 4 right parenthesis comma left parenthesis 1 minus left parenthesis negative 1 right parenthesis right parenthesis comma left parenthesis negative 3 minus 3 right parenthesis right parenthesis end cell row cell 2 over 3 top enclose A B end enclose end cell equals cell 2 over 3 left parenthesis left parenthesis negative 2 minus 4 right parenthesis comma left parenthesis 2 minus left parenthesis negative 1 right parenthesis right parenthesis comma left parenthesis negative 6 minus 3 right parenthesis right parenthesis end cell row blank equals cell left parenthesis negative 4 comma space 2 comma space minus 6 right parenthesis end cell end table end style 

Maka titik C yang tepat adalah begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell C left parenthesis 0 comma space 1 comma negative 3 right parenthesis end cell end table end style.

Jadi, jawaban yang tepat adalah B.

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Roboguru

Diketahui titik K(3,1,−4),L(2,−6,8)danM(2,8,4). Jika titik P membagi KL sehingga KP:PL=2:3, vektor yang diwakili oleh PM adalah …

Pembahasan Soal:

Diketahui:

begin mathsize 14px style straight K left parenthesis 3 comma space 1 comma space minus 4 right parenthesis comma space straight L left parenthesis 2 comma space minus 6 comma space 8 right parenthesis space dan space straight M left parenthesis 2 comma space 8 comma space 4 right parenthesis end style
begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell KP with rightwards arrow on top space colon space PL with rightwards arrow on top end cell equals cell 2 space colon space 3 end cell row cell straight m space colon space straight n end cell equals cell 2 space colon space 3 end cell end table end style 

Mencari koordinat P:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row straight P equals cell open parentheses fraction numerator straight m times Xl plus straight n times Xk over denominator straight m plus straight n end fraction comma blank fraction numerator straight m times Yl plus straight n times Yk over denominator straight m plus straight n end fraction comma blank fraction numerator straight m times Zl plus straight n times Zk over denominator straight m plus straight n end fraction close parentheses end cell row blank equals cell open parentheses fraction numerator 2 times 2 plus 3 times 3 over denominator 2 plus 3 end fraction comma blank fraction numerator 2 times left parenthesis negative 6 right parenthesis plus 3 times 1 over denominator 2 plus 3 end fraction comma fraction numerator 2 times 8 plus 3 times left parenthesis negative 4 right parenthesis over denominator 2 plus 3 end fraction close parentheses blank end cell row blank equals cell open parentheses fraction numerator 4 plus 9 over denominator 5 end fraction comma blank fraction numerator negative 12 plus 3 over denominator 5 end fraction comma blank fraction numerator 16 plus left parenthesis negative 12 right parenthesis over denominator 5 end fraction close parentheses end cell row blank equals cell open parentheses 13 over 5 comma blank minus 9 over 5 comma blank 4 over 5 close parentheses end cell end table end style 

Mencari vektor begin mathsize 14px style PM with rightwards arrow on top end style:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell PM with rightwards arrow on top end cell equals cell straight m with rightwards arrow on top minus straight p with rightwards arrow on top end cell row blank equals cell open parentheses 2 comma space 8 comma space 4 close parentheses minus open parentheses 13 over 5 comma blank minus 9 over 5 comma blank 4 over 5 close parentheses end cell row blank equals cell open parentheses 2 minus 13 over 5 comma blank 8 minus open parentheses negative 9 over 5 close parentheses comma blank 4 minus 4 over 5 close parentheses end cell row blank equals cell open parentheses negative 3 over 5 comma blank 49 over 5 comma blank 16 over 5 close parentheses end cell end table end style 

Jadi, vektor yang diwakili oleh begin mathsize 14px style PM with rightwards arrow on top end style adalah begin mathsize 14px style open parentheses table row cell negative 3 over 5 end cell row cell 49 over 5 end cell row cell 16 over 5 end cell end table close parentheses end style.

Dengan demikian, jawaban yang tepat adalah D.

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Roboguru

Vektor posisi titik A dan B berturut-turut adalah a→danb. Titik P dan Q terletak pada ruas garis AB sehingga AP:PB=3:4 dan AQ:QB=2:−5. Tentukan: a. vektor posisi titik P.

Pembahasan Soal:

Menentukan posisi titik P:

table attributes columnalign right center left columnspacing 0px end attributes row cell p with rightwards arrow on top end cell equals cell fraction numerator m b with rightwards arrow on top plus n a with rightwards arrow on top over denominator m plus n end fraction end cell row blank equals cell fraction numerator 3 b with rightwards arrow on top plus 4 a with rightwards arrow on top over denominator 3 plus 4 end fraction end cell row blank equals cell fraction numerator 3 b with rightwards arrow on top plus 4 a with rightwards arrow on top over denominator 7 end fraction end cell end table   

Jadi, vektor posisi titik P adalah table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell p 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 fraction numerator 3 b with rightwards arrow on top plus 4 a with rightwards arrow on top over denominator 7 end fraction end cell end table.

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Roboguru

Titik-titik sudut segitiga ABC adalah A(3,1,6),B(0,4,−3),danC(1,1,−4). Titik D membagi AB di dalam dengan perbandingan 1:2. Titik E adalah titik tengah AC dan titik F membagi BC  di luar dengan perban...

Pembahasan Soal:

Ingat kembali,

Rumus perbandingan pada vektor

Titik straight B membagi vektor top enclose AC dengan perbandingan m colon n

AB with bar on top space colon space BC with bar on top space equals space m space colon space n

OB with bar on top equals fraction numerator n OA with bar on top space plus space m OC with bar on top over denominator n plus m end fraction

Titik straight B membagi di luar vektor top enclose AC dengan perbandingan m colon n

AB with bar on top space colon space BC with bar on top space equals space minus m space colon space n space equals space m space colon space minus n

Titik segaris

AB with bar on top space equals space k BC with bar on top

Berdasarkan penjelasan tersebut, diperoleh sebagai berikut

a. Titik straight D membagi AB di dalam dengan perbandingan 1 colon 2

stack A D with bar on top space colon space stack D B with bar on top space equals space 1 space colon space 2 stack O D with bar on top space equals fraction numerator 2 open parentheses 3 comma 1 comma 6 close parentheses plus 1 open parentheses 0 comma 4 comma negative 3 close parentheses over denominator 2 plus 1 end fraction equals fraction numerator open parentheses 6 comma 2 comma 12 close parentheses plus open parentheses 0 comma 4 comma negative 3 close parentheses over denominator 3 end fraction equals fraction numerator open parentheses 6 comma 6 comma 9 close parentheses over denominator 3 end fraction equals open parentheses 2 comma 2 comma 3 close parentheses

Sehingga koordinat straight D left parenthesis 2 comma 2 comma 3 right parenthesis

Titik straight E adalah titik tengah top enclose AC

stack O E with bar on top equals fraction numerator open parentheses 3 comma 1 comma 6 close parentheses plus open parentheses 1 comma 1 comma negative 4 close parentheses over denominator 2 end fraction equals fraction numerator open parentheses 4 comma 2 comma 2 close parentheses over denominator 2 end fraction equals open parentheses 2 comma 1 comma 1 close parentheses

Sehingga koordinat straight E left parenthesis 2 comma 1 comma 1 right parenthesis

Titik straight F membagi top enclose BC di luar dengan perbandingan 2 colon 1 

stack B F with bar on top space colon space stack C F with bar on top space equals space 2 space colon space 1 stack B F with bar on top space colon space stack F C with bar on top space equals space 2 space colon space minus 1 stack O F with bar on top space equals fraction numerator negative 1 open parentheses 0 comma 4 comma negative 3 close parentheses plus left parenthesis 2 right parenthesis open parentheses 1 comma 1 comma negative 4 close parentheses over denominator negative 1 plus 2 end fraction equals fraction numerator open parentheses 0 comma negative 4 comma 3 close parentheses plus open parentheses 2 comma 2 comma negative 8 close parentheses over denominator 1 end fraction equals open parentheses 2 comma negative 2 comma negative 5 close parentheses

Sehingga koordinat F left parenthesis 2 comma negative 2 comma negative 5 right parenthesis

b. Tunjukkan straight D comma space straight E comma space dan space straight F segaris

Apabila straight D comma space straight E comma space dan space straight F segaris, maka stack D E with bar on top space equals k space stack E F with bar on top

stack D E with bar on top equals stack O E with bar on top minus stack O D with bar on top equals left parenthesis 2 comma 1 comma 1 right parenthesis minus open parentheses 2 comma 2 comma 3 close parentheses equals open parentheses 0 comma negative 1 comma negative 2 close parentheses stack E F with bar on top equals stack O F with bar on top minus stack O E with bar on top equals open parentheses 2 comma negative 2 comma negative 5 close parentheses minus left parenthesis 2 comma 1 comma 1 right parenthesis equals open parentheses 0 comma negative 3 comma negative 6 close parentheses

table attributes columnalign right center left columnspacing 0px end attributes row cell stack D E with bar on top space end cell equals cell k space stack E F with bar on top end cell row cell open parentheses 0 comma negative 1 comma negative 2 close parentheses end cell equals cell 1 third space open parentheses 0 comma negative 3 comma negative 6 close parentheses end cell end table

Jadi, terbukti bahwa straight D comma space straight E comma space dan space straight F segaris

c. Tentukan top enclose DE space end enclose colon space top enclose EF

stack D E with bar on top equals stack O E with bar on top minus stack O D with bar on top equals left parenthesis 2 comma 1 comma 1 right parenthesis minus open parentheses 2 comma 2 comma 3 close parentheses equals open parentheses 0 comma negative 1 comma negative 2 close parentheses stack E F with bar on top equals stack O F with bar on top minus stack O E with bar on top equals open parentheses 2 comma negative 2 comma negative 5 close parentheses minus left parenthesis 2 comma 1 comma 1 right parenthesis equals open parentheses 0 comma negative 3 comma negative 6 close parentheses

stack D E with bar on top space colon space stack E F with bar on top space equals space 1 space colon space 3space space 

0

Roboguru

Diketahui koordinat titik A(5,1)danB(−3,5). Jika AC:CB=1:3, koordinat titik C adalah ....

Pembahasan Soal:

Misalkan begin mathsize 14px style C left parenthesis x comma y right parenthesis end style. Maka:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell x minus 5 end cell equals cell 3 left parenthesis negative 3 minus x right parenthesis end cell row cell x minus 5 end cell equals cell negative 9 minus 3 x end cell row x equals cell negative 1 end cell end table end style

Jadi, jawaban yang tepat adalah E.

0

Roboguru

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