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Diketahui titik P(2,3) dan titik Q(−5,4). Tentukan: c. ∣∣​PQ​∣∣​

Pertanyaan

Diketahui titik begin mathsize 14px style straight P left parenthesis 2 comma 3 right parenthesis end style dan titik begin mathsize 14px style straight Q left parenthesis negative 5 comma 4 right parenthesis end styleTentukan:

c. begin mathsize 14px style open vertical bar stack P Q with rightwards arrow on top close vertical bar end style 

Pembahasan Soal:

begin mathsize 14px style open vertical bar stack P Q with rightwards arrow on top close vertical bar equals square root of open parentheses negative 7 close parentheses squared plus 1 squared end root equals 5 square root of 2 end style 

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

L. Desia

Mahasiswa/Alumni UIN Sunan Gunung Djati Bandung

Terakhir diupdate 05 Oktober 2021

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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Pertanyaan yang serupa

Diketahui A(2,6,−5), B(−1,3,7), dan C(4,−1,8). AB dan BC berturut-turut mewakili vektor u dan v. Tentukan: d. (2u−3v)2

Pembahasan Soal:

Jika a with rightwards arrow on top equals open parentheses table row cell a subscript 1 end cell row cell a subscript 2 end cell end table close parentheses dan b with rightwards arrow on top equals open parentheses table row cell b subscript 1 end cell row cell b subscript 2 end cell end table close parentheses, maka a with rightwards arrow on top times b with rightwards arrow on top adalah:

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 table row cell a subscript 1 end cell row cell a subscript 2 end cell end table close parentheses times open parentheses table row cell b subscript 1 end cell row cell b subscript 2 end cell end table close parentheses end cell row blank equals cell a subscript 1 b subscript 1 plus a subscript 2 b subscript 2 end cell end table 

Vektor Posisi adalah vektor yang berpangkal di pusat koordinat open parentheses 0 comma space 0 close parentheses dan berujung di suatu titik open parentheses x comma space y close parentheses.

Diketahui straight A open parentheses 2 comma space 6 comma space minus 5 close parentheses, straight B open parentheses negative 1 comma space 3 comma space 7 close parentheses, dan straight C open parentheses 4 comma space minus 1 comma space 8 close parentheses.

Nilai vektor posisi akan sama dengan koordinat titik ujungnya, maka tentukan AB with rightwards arrow on top dan BC with rightwards arrow on top.

table attributes columnalign right center left columnspacing 0px end attributes row cell AB with rightwards arrow on top end cell equals cell b with rightwards arrow on top minus a with rightwards arrow on top end cell row cell u with rightwards arrow on top end cell equals cell open parentheses table row cell negative 1 end cell row 3 row 7 end table close parentheses minus open parentheses table row 2 row 6 row cell negative 5 end cell end table close parentheses end cell row cell u with rightwards arrow on top end cell equals cell open parentheses table row cell negative 1 minus 2 end cell row cell 3 minus 6 end cell row cell 7 plus 5 end cell end table close parentheses end cell row cell u with rightwards arrow on top end cell equals cell open parentheses table row cell negative 3 end cell row cell negative 3 end cell row 12 end table close parentheses end cell row cell BC with rightwards arrow on top end cell equals cell c with rightwards arrow on top minus b with rightwards arrow on top end cell row cell v with rightwards arrow on top end cell equals cell open parentheses table row 4 row cell negative 1 end cell row 8 end table close parentheses minus open parentheses table row cell negative 1 end cell row 3 row 7 end table close parentheses end cell row cell v with rightwards arrow on top end cell equals cell open parentheses table row cell 4 plus 1 end cell row cell negative 1 minus 3 end cell row cell 8 minus 7 end cell end table close parentheses end cell row cell v with rightwards arrow on top end cell equals cell open parentheses table row 5 row cell negative 4 end cell row 1 end table close parentheses end cell end table   

d. open parentheses 2 u with rightwards arrow on top minus 3 v with rightwards arrow on top close parentheses squared

begin mathsize 12px style table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell open parentheses 2 u with rightwards arrow on top minus 3 v with rightwards arrow on top close parentheses squared end cell row blank equals cell 4 open parentheses u with rightwards arrow on top close parentheses squared minus 12 open parentheses u with rightwards arrow on top times v with rightwards arrow on top close parentheses plus 9 open parentheses v with rightwards arrow on top close parentheses squared end cell row blank equals cell 4 open parentheses table row cell negative 3 end cell row cell negative 3 end cell row 12 end table close parentheses squared minus 12 open parentheses open parentheses table row cell negative 3 end cell row cell negative 3 end cell row cell negative 3 end cell end table close parentheses times open parentheses table row 5 row cell negative 4 end cell row 1 end table close parentheses close parentheses plus 9 open parentheses table row 5 row cell negative 4 end cell row 1 end table close parentheses squared end cell row blank equals cell 4 open parentheses open parentheses negative 3 close parentheses squared plus open parentheses negative 3 close parentheses squared plus 12 squared close parentheses minus 12 open parentheses negative 15 plus 12 plus 12 close parentheses plus 9 open parentheses 5 squared plus open parentheses negative 4 close parentheses squared plus 1 squared close parentheses end cell row blank equals cell 4 times 162 minus 12 times 9 plus 9 times 42 end cell row blank equals cell 648 minus 108 plus 378 end cell row blank equals 918 end table end style 

Jadi, Error converting from MathML to accessible text..

0

Roboguru

Diketahui koordinat titik A(4,−3), B(−1,−5) dan C(−2,3). Jika a⇀, b⇀ dan c⇀ berturut-turut vektor posisi titik A, B, dan C, hasil 2a⇀−3b⇀+c⇀ adalah …

Pembahasan Soal:

Vektor posisi titik A, B, dan C yaitu:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell a with rightwards harpoon with barb upwards on top end cell equals cell 4 i with hat on top minus 3 j with hat on top end cell row cell b with rightwards harpoon with barb upwards on top end cell equals cell negative i with hat on top minus 5 j with hat on top end cell row cell c with rightwards harpoon with barb upwards on top end cell equals cell negative 2 i with hat on top plus 3 j with hat on top end cell end table end style 

Dengan konsep perkalian vektor dengan skalar diperoleh:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell 2 a with rightwards harpoon with barb upwards on top end cell equals cell 2 open parentheses 4 i with hat on top minus 3 j with hat on top close parentheses end cell row blank equals cell 8 i with hat on top minus 6 j with hat on top end cell row cell 3 b with rightwards harpoon with barb upwards on top end cell equals cell 3 open parentheses negative i with hat on top minus 5 j with hat on top close parentheses end cell row blank equals cell negative 3 i with hat on top minus 15 j with hat on top end cell end table end style

Operasi penjumlahan dan pengurangan vektor secara aljabar dari begin mathsize 14px style 2 a with rightwards harpoon with barb upwards on top minus 3 b with rightwards harpoon with barb upwards on top plus c with rightwards harpoon with barb upwards on top end style yaitu:

2a3b+c====(8i6j)(3i15j)+(2i+3j)8i6j+3i+15j2i+3j(8+32)i+(6+15+3)j9i+12j

 

Jadi, hasil begin mathsize 14px style 2 a with rightwards harpoon with barb upwards on top minus 3 b with rightwards harpoon with barb upwards on top plus c with rightwards harpoon with barb upwards on top end style adalah begin mathsize 14px style 9 i with hat on top plus 12 j with hat on top end style.

0

Roboguru

Perhatikan gambar di bawah ini Jika p​ menyatakan vektor posisi titik P terhadap titik O, maka ....

Pembahasan Soal:

Perhatikan bahwa begin mathsize 14px style p with rightwards arrow on top equals stack O P with rightwards arrow on top end style. Sehingga titik awalnya adalah O dan titik akhirnya adalah P.
Karena koordinat titik O adalah (0,0,0), maka begin mathsize 14px style x subscript 1 equals 0 comma space y subscript 1 equals 0 comma space d a n space z subscript 1 equals 0. end style

Kemudian karena koordinat titik P adalah (1,3,2), maka begin mathsize 14px style x subscript 2 equals 1 comma space y subscript 2 equals 3 comma space d a n space z subscript 2 equals 2. end style

Maka

begin mathsize 14px style p with rightwards arrow on top equals open parentheses table row cell x subscript 2 minus x subscript 1 end cell row cell y subscript 2 minus y subscript 1 end cell row cell z subscript 2 minus z subscript 1 end cell end table close parentheses equals open parentheses table row cell 1 minus 0 end cell row cell 3 minus 0 end cell row cell 2 minus 0 end cell end table close parentheses equals open parentheses table row 1 row 3 row 2 end table close parentheses end style  

0

Roboguru

Jika p​, q​, r dan t adalah vektor posisi dari P(−1,2,21​); Q(3,0,2); R(2,3,1) dan T(9,−14,2), maka nilai k yang memenuhi kp​+3q​−2r=t adalah....

Pembahasan Soal:

Diketahui

table attributes columnalign right center left columnspacing 0px end attributes row cell straight P open parentheses negative 1 comma 2 comma 1 half close parentheses end cell rightwards arrow cell top enclose p equals open parentheses table row cell negative 1 end cell row 2 row cell 1 half end cell end table close parentheses end cell row cell straight Q open parentheses 3 comma 0 comma 2 close parentheses end cell rightwards arrow cell top enclose q equals open parentheses table row 3 row 0 row 2 end table close parentheses end cell row cell straight R open parentheses 2 comma 3 comma 1 close parentheses end cell rightwards arrow cell top enclose r equals open parentheses table row 2 row 3 row 1 end table close parentheses end cell row cell straight T open parentheses 9 comma negative 14 comma 2 close parentheses end cell rightwards arrow cell top enclose t equals open parentheses table row 9 row cell negative 14 end cell row 2 end table close parentheses end cell row blank blank blank end table

Perhatikan perhitungan berikut ini

table attributes columnalign right center left columnspacing 0px end attributes row cell k top enclose p plus 3 top enclose q minus 2 top enclose r end cell equals cell top enclose t end cell row cell k open parentheses table row cell negative 1 end cell row 2 row cell 1 half end cell end table close parentheses plus 3 open parentheses table row 3 row 0 row 2 end table close parentheses minus 2 open parentheses table row 2 row 3 row 1 end table close parentheses end cell equals cell open parentheses table row 9 row cell negative 14 end cell row 2 end table close parentheses end cell row cell open parentheses table row cell negative k end cell row cell 2 k end cell row cell 1 half k end cell end table close parentheses plus open parentheses table row 9 row 0 row 6 end table close parentheses minus open parentheses table row 4 row 6 row 2 end table close parentheses end cell equals cell open parentheses table row 9 row cell negative 14 end cell row 2 end table close parentheses end cell row cell open parentheses table row cell negative k plus 9 minus 4 end cell row cell 2 k plus 0 minus 6 end cell row cell k plus 6 minus 2 end cell end table close parentheses end cell equals cell open parentheses table row 9 row cell negative 14 end cell row 2 end table close parentheses end cell row cell open parentheses table row cell negative k plus 5 end cell row cell 2 k minus 6 end cell row cell 1 half k plus 4 end cell end table close parentheses end cell equals cell open parentheses table row 9 row cell negative 14 end cell row 2 end table close parentheses end cell end table

Dengan menggunakan kesamaan vektor diperoleh

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

Dengan demikian nilai k yang memenuhi adalah negative 4.

Oleh karena itu, jawaban yang benar adalah E.

0

Roboguru

Diketahui titik A(5,-3), B(-2,-7), dan C(-6,1). Tentukan AB,BC, dan AC!

Pembahasan Soal:

vektor posisi dari titik A, B, dan C masing-masing sebagai berikut.

stack size 14px O size 14px A with size 14px rightwards arrow on top size 14px equals size 14px a with size 14px rightwards arrow on top size 14px equals begin mathsize 14px style left parenthesis 5 comma negative 3 right parenthesis end style stack size 14px O size 14px B with size 14px rightwards arrow on top size 14px equals size 14px b with size 14px rightwards arrow on top size 14px equals begin mathsize 14px style left parenthesis negative 2 comma negative 7 right parenthesis end style size 14px space stack size 14px O size 14px C with size 14px rightwards arrow on top size 14px equals size 14px c with size 14px rightwards arrow on top size 14px equals begin mathsize 14px style left parenthesis negative 6 comma 1 right parenthesis end style 

maka  begin mathsize 14px style AB with rightwards arrow on top comma space BC with rightwards arrow on top comma space end style dan begin mathsize 14px style AC with rightwards arrow on top end style

begin mathsize 14px style AB with rightwards arrow on top equals b with rightwards arrow on top minus a with rightwards arrow on top space space space space space space equals left parenthesis negative 2 comma negative 7 right parenthesis minus left parenthesis 5 comma negative 3 right parenthesis space space space space space space equals left parenthesis negative 2 minus 5 comma negative 7 plus 3 right parenthesis space space space space space space equals left parenthesis negative 7 comma negative 4 right parenthesis BC with rightwards arrow on top equals c with rightwards arrow on top minus b with rightwards arrow on top space space space space space space equals left parenthesis negative 6 comma 1 right parenthesis minus left parenthesis negative 2 comma negative 7 right parenthesis space space space space space space equals left parenthesis negative 6 plus 2 comma 1 plus 7 right parenthesis space space space space space space equals left parenthesis negative 4 comma 8 right parenthesis stack A C with rightwards arrow on top equals c with rightwards arrow on top minus a with rightwards arrow on top space space space space space equals open parentheses negative 6 comma 1 close parentheses minus open parentheses 5 comma negative 3 close parentheses space space space space space equals open parentheses negative 6 minus 5 comma 1 plus 3 close parentheses space space space space space equals open parentheses negative 11 comma 4 close parentheses end style 

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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