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Tentukan koordinat titik P yang terletak pada garis AB jika: a.

Pertanyaan

Tentukan koordinat titik P yang terletak pada garis AB jika:

a. straight A left parenthesis 3 comma space 1 right parenthesis comma space straight B left parenthesis 5 comma space – 1 right parenthesis comma space dan space AP space colon space PB equals 3 space colon space 1 

Pembahasan Soal:

Ingat kembali:

straight x subscript straight p equals fraction numerator straight m times straight x subscript straight B plus straight n times straight x subscript straight A over denominator straight m plus straight n end fraction comma space straight y subscript straight p equals fraction numerator straight m times straight y subscript straight B plus straight n times straight y subscript straight A over denominator straight m plus straight n end fraction 

Diketahui:

straight A open parentheses 3 comma space 1 close parentheses rightwards arrow straight x subscript straight A equals 3 comma space straight y subscript straight A equals 1 straight B open parentheses 5 comma space minus 1 close parentheses rightwards arrow straight x subscript straight B equals 5 comma space straight y subscript straight B equals negative 1 AP space colon space PB equals 3 space colon space 1 rightwards arrow straight m space colon space straight n equals 3 space colon space 1 

Maka:

straight x subscript straight p equals fraction numerator straight m times straight x subscript straight B plus straight n times straight x subscript straight A over denominator straight m plus straight n end fraction equals fraction numerator 3 times 5 plus 1 times 3 over denominator 3 plus 1 end fraction equals fraction numerator 15 plus 3 over denominator 4 end fraction equals 18 over 4 equals 9 over 2 straight y subscript straight p equals fraction numerator straight m times straight y subscript straight B plus straight n times straight y subscript straight A over denominator straight m plus straight n end fraction equals fraction numerator 3 times negative 1 plus 1 times 1 over denominator 3 plus 1 end fraction equals fraction numerator negative 3 plus 1 over denominator 4 end fraction equals fraction numerator negative 2 over denominator 4 end fraction equals fraction numerator negative 1 over denominator 2 end fraction 

Jadi, koordinat titik straight P yang terletak pada garis AB adalah open parentheses 9 over 2 comma space minus 1 half close parentheses.

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

S. Yoga

Mahasiswa/Alumni Universitas Pendidikan Indonesia

Terakhir diupdate 05 Juni 2021

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Pertanyaan yang serupa

Diketahui ,  dan . Titik  pada  sehingga , maka

Pembahasan Soal:

Diketahui A left parenthesis 0 comma space 1 comma space 5 right parenthesisB left parenthesis 0 comma space minus 4 comma space 5 right parenthesis dan C left parenthesis 3 comma space minus 1 comma space minus 4 right parenthesis. Titik P pada A B sehingga A P space colon space A B equals 2 space colon space 5, maka:

table attributes columnalign right center left columnspacing 0px end attributes row cell A P space colon space A B end cell equals cell 2 space colon space 5 end cell row cell fraction numerator P minus A over denominator B minus A end fraction end cell equals cell 2 over 5 end cell row cell 5 open parentheses P minus A close parentheses end cell equals cell 2 left parenthesis B minus A right parenthesis end cell row cell 5 P minus 5 A end cell equals cell 2 B minus 2 A end cell row cell 5 P end cell equals cell 2 B minus 2 A plus 5 A end cell row cell 5 P end cell equals cell 2 B plus 3 A end cell row P equals cell fraction numerator 2 B plus 3 A over denominator 5 end fraction end cell row P equals cell fraction numerator 2 left parenthesis 0 comma space minus 4 comma space 5 right parenthesis plus 3 left parenthesis 0 comma space 1 comma space 5 right parenthesis over denominator 5 end fraction end cell row P equals cell fraction numerator left parenthesis 0 comma space minus 8 comma space 10 right parenthesis plus left parenthesis 0 comma space 3 comma space 15 right parenthesis over denominator 5 end fraction end cell row P equals cell fraction numerator left parenthesis 0 comma space minus 5 comma space 25 right parenthesis over denominator 5 end fraction end cell row P equals cell left parenthesis 0 comma space minus 1 comma space 5 right parenthesis end cell end table 

Sehingga:

table attributes columnalign right center left columnspacing 0px end attributes row cell stack P C with rightwards arrow on top end cell equals cell C minus P end cell row blank equals cell left parenthesis 3 comma space minus 1 comma space minus 4 right parenthesis minus left parenthesis 0 comma space minus 1 comma space 5 right parenthesis end cell row blank equals cell left parenthesis 3 minus 0 comma space minus 1 minus left parenthesis negative 1 right parenthesis comma space minus 4 minus 5 right parenthesis end cell row blank equals cell left parenthesis 3 comma space 0 comma space minus 9 right parenthesis end cell row blank blank blank row cell open vertical bar stack P C with rightwards arrow on top close vertical bar end cell equals cell square root of 3 squared plus 0 squared plus left parenthesis negative 9 right parenthesis squared end root end cell row blank blank blank row blank equals cell square root of 9 plus 0 plus 81 end root end cell row blank equals cell square root of 90 end cell row blank equals cell 3 square root of 10 space satuan space panjang end cell end table 

Jadi, open vertical bar stack P C with rightwards arrow on top close vertical bar adalah table attributes columnalign right center left columnspacing 0px end attributes row blank blank 3 end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell square root of 10 end cell end table satuan panjang 

0

Roboguru

Diketahui titik . Jika titik P membagi KL sehingga , vektor yang diwakili oleh  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.

0

Roboguru

Jika  dan , tentukan koordinat titik R dan S yang membagi garis AB di dalam dengan perbandingan 2:1.

Pembahasan Soal:

Misalkan titik P membagi ruas garis AB dengan besar perbandingan m dan n. Kondisi ini terjadi saat titik P berada di antara titik A dan B. Cara menentukan koordinat P pada perbandingan vektor pada ruas garis dengan titik P berada di dalam dapat dilihat pada persamaan sebagai berikut.

p=m+nmB+nA 

Sehingga diperoleh:

r====2B+1A32(155)+1(121)3(3129)143 

Dengan demikian, koordinat titik R adalah (1,4,3).

0

Roboguru

Diketahui titik  dan titik . Titik C membagi ruas garis  dengan perbandingan . Ruas garis berarah  mewakili vektor  dan ruas garis berarah  mewakili vektor . a. Tentukan koordinat titik C, kemudian t...

Pembahasan Soal:

Diketahui titik straight A left parenthesis 5 comma space 2 comma space 1 right parenthesis dan titik straight B left parenthesis 9 comma space 10 comma space 13 right parenthesis. Titik C membagi ruas garis AB with rightwards arrow on top dengan perbandingan AC space colon space CB equals 1 space colon space 3, maka:

table attributes columnalign right center left columnspacing 0px end attributes row straight C equals cell fraction numerator 3 straight A plus 1 straight B over denominator 3 plus 1 end fraction end cell row blank equals cell fraction numerator 3 left parenthesis 5 comma space 2 comma space 1 right parenthesis plus 1 left parenthesis 9 comma space 10 comma space 13 right parenthesis over denominator 4 end fraction end cell row blank equals cell fraction numerator left parenthesis 15 comma space 6 comma space 3 right parenthesis plus left parenthesis 9 comma space 10 comma space 13 right parenthesis over denominator 4 end fraction end cell row blank equals cell fraction numerator left parenthesis 15 plus 9 comma space 6 plus 10 comma space 3 plus 13 right parenthesis over denominator 4 end fraction end cell row blank equals cell fraction numerator left parenthesis 24 comma space 16 comma space 16 right parenthesis over denominator 4 end fraction end cell row blank equals cell left parenthesis 6 comma space 4 comma space 4 right parenthesis end cell end table

Maka koordinat C adalah left parenthesis 6 comma space 4 comma space 4 right parenthesis, sehingga:

table attributes columnalign right center left columnspacing 0px end attributes row cell u with rightwards arrow on top end cell equals cell AC with rightwards arrow on top end cell row blank equals cell straight C minus straight A end cell row blank equals cell left parenthesis 6 comma space 4 comma space 4 right parenthesis minus left parenthesis 5 comma space 2 comma space 1 right parenthesis end cell row blank equals cell left parenthesis 1 comma space 2 comma space 3 right parenthesis end cell row blank blank blank row cell v with rightwards arrow on top end cell equals cell CB with rightwards arrow on top end cell row blank equals cell straight B minus straight C end cell row blank equals cell left parenthesis 9 comma space 10 comma space 13 right parenthesis minus left parenthesis 6 comma space 4 comma space 4 right parenthesis end cell row blank equals cell left parenthesis 9 minus 6 comma space 10 minus 4 comma space 13 minus 4 right parenthesis end cell row blank equals cell left parenthesis 3 comma space 6 comma space 9 right parenthesis end cell end table

Jadi, koordinat C adalah left parenthesis 6 comma space 4 comma space 4 right parenthesis, vektor u with rightwards arrow on top adalah left parenthesis 1 comma space 2 comma space 3 right parenthesis dan vektor v with rightwards arrow on top adalah left parenthesis 3 comma space 6 comma space 9 right parenthesis 

0

Roboguru

Diketahui titik , , dan . Titik  membagi ruas garis  dengan perbandingan . Tentukan: Koordinat titik C Proyeksi vektor orthogonal vektor  terhadap

Pembahasan Soal:

Diketahui vektor-vektor berikut:


a with bar on top equals open parentheses table row 1 row cell negative 1 end cell row 2 end table close parentheses b with bar on top equals open parentheses table row cell negative 5 end cell row cell negative 7 end cell row 20 end table close parentheses d with bar on top equals open parentheses table row 0 row cell negative 3 end cell row 0 end table close parentheses


Soal a.

Gunakan prinsip pembagian vektor untuk mencari vektor posisi c with bar on top.


stack A C with rightwards arrow on top space colon space stack C B with rightwards arrow on top space equals space 1 space colon space 2 c with bar on top equals fraction numerator b with bar on top plus 2 top enclose a over denominator 3 end fraction equals fraction numerator open parentheses table row cell negative 5 end cell row cell negative 7 end cell row 20 end table close parentheses plus 2 open parentheses table row 1 row cell negative 1 end cell row 2 end table close parentheses over denominator 3 end fraction equals fraction numerator open parentheses table row cell negative 5 end cell row cell negative 7 end cell row 20 end table close parentheses plus open parentheses table row 2 row cell negative 2 end cell row 4 end table close parentheses over denominator 3 end fraction equals fraction numerator open parentheses table row cell negative 3 end cell row cell negative 9 end cell row 24 end table close parentheses over denominator 3 end fraction equals open parentheses table row cell negative 1 end cell row cell negative 3 end cell row 8 end table close parentheses


Jadi koordinat titik C adalah left parenthesis negative 1 comma space minus 3 comma space 8 right parenthesis.

 

Soal b.

Langkah pertama adalah kita mencari vektor stack A C with rightwards arrow on top dan vektor stack A D with rightwards arrow on top  dari vektor posisi yang diketahui.

table attributes columnalign right center left columnspacing 0px end attributes row cell stack A C with rightwards arrow on top end cell equals cell c with bar on top minus a with bar on top end cell row blank equals cell open parentheses table row cell negative 1 end cell row cell negative 3 end cell row 8 end table close parentheses minus open parentheses table row 1 row cell negative 1 end cell row 2 end table close parentheses end cell row blank equals cell open parentheses table row 0 row cell negative 2 end cell row 6 end table close parentheses end cell end table

table attributes columnalign right center left columnspacing 0px end attributes row cell stack A D with rightwards arrow on top end cell equals cell d with bar on top minus a with bar on top end cell row blank equals cell open parentheses table row 0 row cell negative 3 end cell row 0 end table close parentheses minus open parentheses table row 1 row cell negative 1 end cell row 2 end table close parentheses end cell row blank equals cell open parentheses table row cell negative 1 end cell row cell negative 2 end cell row cell negative 2 end cell end table close parentheses end cell end table


Untuk mencari proyeksi vektor orthogonal vektor stack A C with rightwards arrow on top terhadap stack A D with rightwards arrow on top, kita harus mencari besar vektor stack A D with rightwards arrow on top.


table attributes columnalign right center left columnspacing 0px end attributes row cell open vertical bar stack A D with 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 2 close parentheses squared plus open parentheses negative 2 close parentheses squared end root end cell row blank equals cell square root of 1 plus 4 plus 4 end root end cell row blank equals cell square root of 9 end cell row blank equals 3 end table


Maka proyeksi vektor orthogonal vektor stack A C with rightwards arrow on top terhadap stack A D with rightwards arrow on top dinyatakan dengan rumus:


table attributes columnalign right center left columnspacing 0px end attributes row cell stack A C with rightwards arrow on top subscript stack A D with rightwards arrow on top end subscript end cell equals cell fraction numerator stack A C with rightwards arrow on top times stack A D with rightwards arrow on top over denominator open vertical bar stack A D with rightwards arrow on top close vertical bar squared end fraction times open vertical bar stack A D with rightwards arrow on top close vertical bar end cell row blank equals cell fraction numerator open parentheses table row 0 row cell negative 2 end cell row 6 end table close parentheses times open parentheses table row cell negative 1 end cell row cell negative 2 end cell row cell negative 2 end cell end table close parentheses over denominator 3 squared end fraction times open parentheses table row cell negative 1 end cell row cell negative 2 end cell row cell negative 2 end cell end table close parentheses end cell row blank equals cell fraction numerator 0 times negative 1 plus left parenthesis negative 2 right parenthesis times left parenthesis negative 2 right parenthesis plus 6 times left parenthesis negative 2 right parenthesis over denominator 9 end fraction times open parentheses table row cell negative 1 end cell row cell negative 2 end cell row cell negative 2 end cell end table close parentheses end cell row blank equals cell negative 8 over 9 times open parentheses table row cell negative 1 end cell row cell negative 2 end cell row cell negative 2 end cell end table close parentheses end cell row blank equals cell open parentheses table row cell 8 over 9 end cell row cell 16 over 9 end cell row cell 16 over 9 end cell end table close parentheses end cell end table


Jadi proyeksi vektor orthogonal vektor stack A C with rightwards arrow on top terhadap stack A D with rightwards arrow on top adalah table attributes columnalign right center left columnspacing 0px end attributes row cell stack A C with rightwards arrow on top subscript stack A D with rightwards arrow on top end subscript end cell equals cell open parentheses table row cell 8 over 9 end cell row cell 16 over 9 end cell row cell 16 over 9 end cell end table close parentheses end cell end table.

0

Roboguru

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