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Jika titik P membagi ruas garis yang menghubungkan titik A ( x 1 ​ , y 1 ​ , z 1 ​ ) dan titik B ( x 2 ​ , y 2 ​ , z 2 ​ ) dengan perbandingan AP : PB = m : n , tunjukkan bahwa koordinat titik adalah ( x p ​ , y p ​ , z p ​ ) dengan: x p ​ = m + n m x 2 ​ + n x 1 ​ ​ , y p ​ = m + n m y 2 ​ + n y 1 ​ ​ , z p ​ = m + n m z 2 ​ + n z 1 ​ ​

Jika titik  membagi ruas garis yang menghubungkan titik  dan titik  dengan perbandingan , tunjukkan bahwa koordinat titik undefined adalah  dengan:

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Diketahui Sehingga Dengan demikian terbukti bahwa dimana .

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begin mathsize 14px style A open parentheses x subscript 1 comma y subscript 1 comma z subscript 1 close parentheses B open parentheses x subscript 2 comma y subscript 2 comma z subscript 2 close parentheses P open parentheses x subscript p comma y subscript p comma z subscript p close parentheses AP colon PB equals m colon n end style

Sehingga

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell A P colon P B end cell equals cell m colon n end cell row cell fraction numerator A P over denominator P B end fraction end cell equals cell m over n end cell row cell n times A P end cell equals cell m times P B end cell row cell n times open parentheses P minus A close parentheses end cell equals cell m times open parentheses B minus P close parentheses end cell row cell n times P minus n times A end cell equals cell m times B minus m times P end cell row cell n times P plus m times. P end cell equals cell n times A plus m times B end cell row cell P times open parentheses n plus m close parentheses end cell equals cell n times A plus m times B end cell row P equals cell fraction numerator n times A plus m times B over denominator n plus m end fraction end cell row P equals cell fraction numerator n times A plus m times B over denominator m plus n end fraction end cell row cell open parentheses x subscript p comma y subscript p comma z subscript p close parentheses end cell equals cell fraction numerator n times open parentheses x subscript 1 comma y subscript 1 comma z subscript 1 close parentheses plus m times open parentheses x subscript 2 comma y subscript 2 comma z subscript 2 close parentheses over denominator m plus n end fraction end cell row cell open parentheses x subscript p comma y subscript p comma z subscript p close parentheses end cell equals cell fraction numerator open parentheses n x subscript 1 comma n y subscript 1 comma n z subscript 1 close parentheses plus open parentheses m x subscript 2 comma m y subscript 2 comma m z subscript 2 close parentheses over denominator m plus n end fraction end cell row cell open parentheses x subscript p comma y subscript p comma z subscript p close parentheses end cell equals cell fraction numerator open parentheses n x subscript 1 plus m x subscript 2 comma n y subscript 1 plus m y subscript 2 comma n z subscript 1 plus m z subscript 2 close parentheses over denominator m plus n end fraction end cell row cell open parentheses x subscript p comma y subscript p comma z subscript p close parentheses end cell equals cell open parentheses fraction numerator n x subscript 1 plus m x subscript 2 over denominator m plus n end fraction comma fraction numerator n y subscript 1 plus m y subscript 2 over denominator m plus n end fraction comma fraction numerator n z subscript 1 plus m z subscript 2 over denominator m plus n end fraction close parentheses end cell row cell left parenthesis x subscript p comma y subscript p comma z subscript p right parenthesis end cell equals cell open parentheses fraction numerator m x subscript 2 plus n x subscript 1 over denominator m plus n end fraction comma fraction numerator m y subscript 2 plus n y subscript 1 over denominator m plus n end fraction comma fraction numerator m z subscript 2 plus n z subscript 1 over denominator m plus n end fraction close parentheses end cell end table end style

Dengan demikian terbukti bahwa begin mathsize 14px style P open parentheses x subscript p comma y subscript p comma z subscript p close parentheses end style dimana 

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Diketahui titik A ( 2 , 1 , 0 ) , B ( 2 , 4 , 3 ) , dan C ( 1 , 3 , 2 ) . Titik D terletak pada AB sehingga BD = 2 1 ​ DA . Tentukan panjang proyeksi vektor berikut. b. CA pada CD

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