Roboguru

Diketahui koordinat titik A(1,−4), B(p,2) dan C(5,q). Jika c merupakan vektor posisi titik C dan AB=c, nilai p+q adalah ....

Pertanyaan

Diketahui koordinat titik A left parenthesis 1 comma negative 4 right parenthesisB left parenthesis p comma space 2 right parenthesis dan C open parentheses 5 comma q close parentheses. Jika c with rightwards arrow on top merupakan vektor posisi titik C dan stack A B with rightwards arrow on top equals c with rightwards arrow on top, nilai p plus q adalah ....

  1. 12space 

  2. 10space 

  3. 6space 

  4. 2space 

  5. 0space 

Pembahasan Soal:

Diketahui A left parenthesis 1 comma negative 4 right parenthesisB left parenthesis p comma space 2 right parenthesis dan C open parentheses 5 comma q close parentheses, maka c with rightwards arrow on top equals open parentheses table row 5 row q end table close parentheses. Ingat bahwa table attributes columnalign right center left columnspacing 0px end attributes row cell stack A B with rightwards arrow on top end cell equals cell open parentheses table row cell b subscript 1 minus a subscript 1 end cell row cell b subscript 2 minus a subscript 2 end cell end table close parentheses end cell end table, maka

table attributes columnalign right center left columnspacing 0px end attributes row cell stack A B with rightwards arrow on top end cell equals cell open parentheses table row cell p minus 1 end cell row cell 2 plus 4 end cell end table close parentheses end cell row blank equals cell open parentheses table row cell p minus 1 end cell row 6 end table close parentheses end cell end table

Jika stack A B with rightwards arrow on top equals c with rightwards arrow on top, maka

table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses table row cell p minus 1 end cell row 6 end table close parentheses end cell equals cell open parentheses table row 5 row q end table close parentheses end cell row blank rightwards arrow cell p equals 6 space dan space q equals 6 end cell row cell p plus q end cell equals 12 end table

Jadi, jawaban yang tepat adalah A.

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

F. Kurnia

Mahasiswa/Alumni Universitas Jember

Terakhir diupdate 06 Oktober 2021

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Pertanyaan yang serupa

Diketahui koordinat titik M(−3,−4)danN(−2,6). Tentukan koordinat titik K jika NM=k.

Pembahasan Soal:

Diketahui M(3,4)danN(2,6)., serta stack N M with rightwards arrow on top equals k with rightwards arrow on top, maka

table attributes columnalign right center left columnspacing 0px end attributes row cell k with rightwards arrow on top end cell equals cell stack N M with rightwards arrow on top end cell row blank equals cell open parentheses table row cell negative 3 plus 2 end cell row cell negative 4 minus 6 end cell end table close parentheses end cell row blank equals cell open parentheses table row cell negative 1 end cell row cell negative 10 end cell end table close parentheses end cell end table

Oleh karena itu, koordinat titik K adalah K(1,10).

0

Roboguru

Diketahui dua vektor yang sama, yaitu vektor u=(2n+110​) dan v=15i+(m2+6)j​. a. Jika mdann adalah bilangan cacah, maka hitunglah hasil dari m−n! b. Gambarlah vektor u pada koordinat Cartesius dengan...

Pembahasan Soal:

a) Dua buah vektor disebut sama jika dan hanya jika besar dan arah kedua vektor tersebut sama. Vektor begin mathsize 14px style top enclose straight u space dan space top enclose straight v end style dikatakan sama, maka

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell top enclose u end cell equals cell top enclose v end cell row cell open parentheses table row cell 2 n plus 1 end cell row 10 end table close parentheses end cell equals cell open parentheses table row 15 row cell m squared plus 6 end cell end table close parentheses end cell end table end style

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell 2 n plus 1 end cell equals 15 row cell 2 n end cell equals 14 row n equals 7 end table end style

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell m squared plus 6 end cell equals 10 row cell m squared end cell equals 4 row m equals cell square root of 4 end cell row m equals cell plus-or-minus 2 end cell end table end style 

Karena begin mathsize 14px style m space text dan end text space n end style adalah bilangan cacah maka nilai

begin mathsize 14px style m equals 2 space dan space n equals 7 end style.

Dengan demikian hasil dari begin mathsize 14px style m minus n equals 2 minus 7 equals negative 5 end style

b) Substitusikan salah satu nilai untuk melengkapi vektor, sehingga

u=(27+110)=(1510).

Dengan demikian, gambar vektor begin mathsize 14px style top enclose u end style pada koordinat Cartesius dengan koordinat titik pangkalnya begin mathsize 14px style left parenthesis negative 5 comma 0 right parenthesis end style adalah seperti berikut,

c) Misalkan w adalah vektor satuan dari vektor begin mathsize 14px style top enclose v end style, maka

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell top enclose w end cell equals cell fraction numerator top enclose v over denominator open vertical bar top enclose v close vertical bar end fraction end cell row blank equals cell fraction numerator open parentheses table row 15 row 10 end table close parentheses over denominator square root of 15 squared plus 10 squared end root end fraction end cell row blank equals cell fraction numerator 1 over denominator square root of 225 plus 100 end root end fraction open parentheses table row 15 row 10 end table close parentheses end cell row blank equals cell fraction numerator 1 over denominator 5 square root of 13 end fraction open parentheses table row 15 row 10 end table close parentheses end cell row blank equals cell open parentheses table row cell fraction numerator 3 over denominator square root of 13 end fraction end cell row cell fraction numerator 2 over denominator square root of 13 end fraction end cell end table close parentheses end cell end table end style

Dengan demikian, vektor satuan dari vektor begin mathsize 14px style top enclose v end style yaitu (133132).

0

Roboguru

Perhatikan pernyataan-pernyataan berikut! i)     AC+DB=AB+DC ii)    AC+BC=BD+AD  Pernyataan yang bernilai benar ditunjukkan oleh nomor ....

Pembahasan Soal:

Perhatikan bahwa begin mathsize 14px style straight p with rightwards arrow on top equals straight q with rightwards arrow on top end style jika dan hanya jika begin mathsize 14px style straight p with rightwards arrow on top minus straight q with rightwards arrow on top equals 0 with rightwards arrow on top end style.

 

Kemudian perhatikan pernyataan pertama.

i)    begin mathsize 14px style AC with rightwards arrow on top plus DB with rightwards arrow on top equals AB with rightwards arrow on top plus DC with rightwards arrow on top end style

Misalkan begin mathsize 14px style straight p with rightwards arrow on top equals AC with rightwards arrow on top plus DB with rightwards arrow on top end style dan begin mathsize 14px style straight q with rightwards arrow on top equals AB with rightwards arrow on top plus DC with rightwards arrow on top end style.

Perhatikan bahwa

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell straight p with rightwards arrow on top minus straight q with rightwards arrow on top end cell equals cell open parentheses AC with rightwards arrow on top plus DB with rightwards arrow on top close parentheses minus open parentheses AB with rightwards arrow on top plus DC with rightwards arrow on top close parentheses end cell row blank equals cell AC with rightwards arrow on top plus DB with rightwards arrow on top minus AB with rightwards arrow on top minus DC with rightwards arrow on top end cell row blank equals cell AC with rightwards arrow on top minus DC with rightwards arrow on top plus DB with rightwards arrow on top minus AB with rightwards arrow on top end cell row blank equals cell AC with rightwards arrow on top minus open parentheses negative CD with rightwards arrow on top close parentheses plus DB with rightwards arrow on top minus open parentheses negative BA with rightwards arrow on top close parentheses end cell row blank equals cell AC with rightwards arrow on top plus CD with rightwards arrow on top plus DB with rightwards arrow on top plus BA with rightwards arrow on top end cell row blank equals cell AA with rightwards arrow on top end cell row blank equals cell 0 with rightwards arrow on top end cell end table end style

Karena begin mathsize 14px style straight p with rightwards arrow on top minus straight q with rightwards arrow on top equals 0 with rightwards arrow on top end style, maka begin mathsize 14px style straight p with rightwards arrow on top equals straight q with rightwards arrow on top end style atau begin mathsize 14px style AC with rightwards arrow on top plus DB with rightwards arrow on top equals AB with rightwards arrow on top plus DC with rightwards arrow on top end style.

Maka pernyataan i) bernilai benar.

 

Selanjutnya perhatikan pernyataan kedua.

ii)    begin mathsize 14px style AC with rightwards arrow on top plus BC with rightwards arrow on top equals BD with rightwards arrow on top plus AD with rightwards arrow on top end style

Misalkan begin mathsize 14px style straight p with rightwards arrow on top equals AC with rightwards arrow on top plus BC with rightwards arrow on top end style dan begin mathsize 14px style straight q with rightwards arrow on top equals BD with rightwards arrow on top plus AD with rightwards arrow on top end style.

Perhatikan bahwa

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell straight p with rightwards arrow on top minus straight q with rightwards arrow on top end cell equals cell open parentheses AC with rightwards arrow on top plus BC with rightwards arrow on top close parentheses minus open parentheses BD with rightwards arrow on top plus AD with rightwards arrow on top close parentheses end cell row blank equals cell AC with rightwards arrow on top plus BC with rightwards arrow on top minus BD with rightwards arrow on top minus AD with rightwards arrow on top end cell row blank equals cell AC with rightwards arrow on top minus AD with rightwards arrow on top plus BC with rightwards arrow on top minus BD with rightwards arrow on top end cell row blank equals cell AC with rightwards arrow on top minus open parentheses negative DA with rightwards arrow on top close parentheses plus BC with rightwards arrow on top minus open parentheses negative DB with rightwards arrow on top close parentheses end cell row blank equals cell AC with rightwards arrow on top plus DA with rightwards arrow on top plus BC with rightwards arrow on top plus DB with rightwards arrow on top end cell row blank equals cell DA with rightwards arrow on top plus AC with rightwards arrow on top plus DB with rightwards arrow on top plus BC with rightwards arrow on top end cell row blank equals cell open parentheses DA with rightwards arrow on top plus AC with rightwards arrow on top close parentheses plus open parentheses DB with rightwards arrow on top plus BC with rightwards arrow on top close parentheses end cell row blank equals cell DC with rightwards arrow on top plus DC with rightwards arrow on top end cell row blank equals cell 2 DC with rightwards arrow on top end cell end table end style 

Karena begin mathsize 14px style straight p with rightwards arrow on top minus straight q with rightwards arrow on top not equal to 0 end style, maka begin mathsize 14px style straight p with rightwards arrow on top not equal to straight q with rightwards arrow on top end style atau begin mathsize 14px style AC with rightwards arrow on top plus BC with rightwards arrow on top not equal to BD with rightwards arrow on top plus AD with rightwards arrow on top end style.

Maka pernyataan ii) bernilai salah.

 

Sehingga pernyataan yang bernilai benar ditunjukkan oleh nomor i) saja.

Jadi, jawaban yang tepat adalah A.

0

Roboguru

Tentukan komponen-komponen vektor AB dan BA, jika diketahui koordinat titik berikut! a. A(1,−2)danB(3,2)

Pembahasan Soal:

Diketahui koordinat titik begin mathsize 14px style straight A left parenthesis 1 comma negative 2 right parenthesis space dan space straight B left parenthesis 3 comma 2 right parenthesis end style. Vektor posisi dari titik A dan B tersebut sebagai berikut.

OA=(12)

OB=(32)

Kemudian, diperoleh vektor AB dan BA sebagai berikut.

AB====OBOA(32)(12)(312(2))(24)

BA====OAOB(12)(32)(1322)(24)

Dengan demikian, diperoleh komponen-komponen vektor AB dan BA berturut-turut adalah (24) dan (24).

0

Roboguru

Koordinat K(2,3) dan Vektor KL=(−54​). Koordinat titik L adalah ....

Pembahasan Soal:

Ingat rumus untuk menentukan vektor adalah sebagai berikut.

begin mathsize 14px style stack text KL end text with rightwards arrow on top equals stack text OL end text with rightwards arrow on top text-end text stack text OK end text with rightwards arrow on top end style  

Diketahui koordinat titik begin mathsize 14px style K equals open parentheses 2 comma space 3 close parentheses end style, maka  begin mathsize 14px style stack text OK end text with rightwards arrow on top equals open parentheses table row 2 row 3 end table close parentheses end style,

Dengan menggunakan rumus untuk menentukan vektor di atas, maka diperoleh perhitungan sebagai berikut.

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell stack text KL end text with rightwards arrow on top end cell equals cell stack text OL end text with rightwards arrow on top minus stack text OK end text with rightwards arrow on top end cell row cell open parentheses table row cell negative 5 end cell row 4 end table close parentheses end cell equals cell stack text OL end text with rightwards arrow on top minus open parentheses table row 2 row 3 end table close parentheses end cell row cell stack text OL end text with rightwards arrow on top end cell equals cell open parentheses table row cell negative 5 end cell row 4 end table close parentheses plus open parentheses table row 2 row 3 end table close parentheses end cell row cell stack text OL end text with rightwards arrow on top end cell equals cell open parentheses table row cell negative 3 end cell row 7 end table close parentheses end cell end table end style 

Dari perhitungan di atas diperoleh begin mathsize 14px style stack text OL end text with rightwards arrow on top equals open parentheses table row cell negative 3 end cell row 7 end table close parentheses end style, dengan demikian koordinat titik begin mathsize 14px style text L end text end style adalah begin mathsize 14px style open parentheses negative 3 comma space 7 close parentheses end style.

Oleh karena itu, jawaban yang benar adalah C.

0

Roboguru

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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