Roboguru

Diketahui koordinat titik P(2, -3, -8), Q(-1, 4, 3), dan R(5, -1, 6). Jika segiempat PQRS berbentuk jajargenjang, koordinat titik S adalah ....

Pertanyaan

Diketahui koordinat titik P(2, -3, -8), Q(-1, 4, 3), dan R(5, -1, 6). Jika segiempat PQRS berbentuk jajargenjang, koordinat titik S adalah ....
 

Pembahasan Soal:

P Q equals S R q minus p equals r minus s open parentheses table row cell negative 1 end cell row 4 row 3 end table close parentheses minus open parentheses table row 2 row cell negative 3 end cell row cell negative 8 end cell end table close parentheses equals open parentheses table row 5 row cell negative 1 end cell row 6 end table close parentheses minus open parentheses table row x row y row z end table close parentheses open parentheses table row cell negative 3 end cell row 7 row 11 end table close parentheses equals open parentheses table row 5 row cell negative 1 end cell row 6 end table close parentheses minus open parentheses table row x row y row z end table close parentheses open parentheses table row x row y row z end table close parentheses equals open parentheses table row 5 row cell negative 1 end cell row 6 end table close parentheses minus open parentheses table row cell negative 3 end cell row 7 row 11 end table close parentheses open parentheses table row x row y row z end table close parentheses equals open parentheses table row 8 row cell negative 8 end cell row cell negative 5 end cell end table close parentheses

Jadi, koordinat titik S (8, -8, -5).

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

R. Bella

Mahasiswa/Alumni UIN Syarif Hidayatullah Jakarta

Terakhir diupdate 13 September 2021

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Pertanyaan yang serupa

Diketahui vektor-vektor  dan . Kombinasi linear dari  adalah ....

Pembahasan Soal:

Jika diberikan vektor-vektor a with rightwards arrow on top equals open parentheses table row cell x subscript 1 end cell row cell y subscript 1 end cell end table close parentheses space dan space b with rightwards arrow on top equals open parentheses table row cell x subscript 2 end cell row cell y subscript 2 end cell end table close parentheses dan skalar m space dan space n maka kombinasi linear dari m a with rightwards arrow on top plus n b with rightwards arrow on top dapat ditentukan dengan rumus berikut.

table attributes columnalign right center left columnspacing 0px end attributes row cell m a with rightwards arrow on top plus n b with rightwards arrow on top end cell equals cell m open parentheses table row cell x subscript 1 end cell row cell y subscript 1 end cell end table close parentheses plus n open parentheses table row cell x subscript 2 end cell row cell y subscript 2 end cell end table close parentheses end cell row blank equals cell open parentheses m x subscript 1 plus n x subscript 2 close parentheses i with rightwards arrow on top plus open parentheses m y subscript 1 plus n y subscript 2 close parentheses j with rightwards arrow on top end cell end table 

Sehingga untuk vektor a with rightwards arrow on top equals open parentheses table row 4 row cell negative 2 end cell end table close parentheses space dan space b with rightwards arrow on top equals open parentheses table row cell negative 2 end cell row 3 end table close parentheses dapat ditentukan kombinasi linear 3 a with rightwards arrow on top plus 2 b with rightwards arrow on top sebagai berikut.

table attributes columnalign right center left columnspacing 0px end attributes row cell 3 a with rightwards arrow on top plus 2 b with rightwards arrow on top end cell equals cell 3 open parentheses table row 4 row cell negative 2 end cell end table close parentheses plus 2 open parentheses table row cell negative 2 end cell row 3 end table close parentheses end cell row blank equals cell open parentheses table row 12 row cell negative 6 end cell end table close parentheses plus open parentheses table row cell negative 4 end cell row 6 end table close parentheses end cell row blank equals cell open parentheses table row cell 12 minus 4 end cell row cell negative 6 plus 6 end cell end table close parentheses end cell row blank equals cell open parentheses table row 8 row 0 end table close parentheses end cell row blank equals cell 8 i with rightwards arrow on top plus 0 k with rightwards arrow on top end cell row blank equals cell 8 i with rightwards arrow on top end cell end table 

Jadi, jawaban yang benar adalah B.

0

Roboguru

Diketahui vektor posisi titik A adalah , dan vektor posisi titik B adalah . Jika 2AB+3AC=2BC, maka panjang vektor posisi titik C adalah ....

Pembahasan Soal:

Diketahui bahwa, 2AB+3AC=2BC , serta a with rightwards arrow on top equals open parentheses table row 4 row cell negative 4 end cell row 3 end table close parenthesesb with rightwards arrow on top equals open parentheses table row 4 row cell negative 5 end cell row 3 end table close parentheses dan c=c1c2c3, maka:

2AB+3AC2445[4]33+3c14c2[4]c332010+3c14c2+4c333c1123c2+103c39====2BC2c14c2[5]c332c14c2+5c332c182c2+102c36 

Berdasarkan kesamaan vektor, diperoleh:

3c112=2c183c12c1=8+12c1=43c2+10=2c2+103c22c2=0c2=03c39=2c363c32c3=6+9c3=3 

Sehingga panjang vektor C yaitu:

c====42+02+3216+9255

Dengan demikian, panjang vektor posisi titik C adalah 5.

Oleh karena itu, jawaban yang benar adalah D.

0

Roboguru

Diketahui titik-titik ,  dan . wakil dari , wakil dari . Kosinus sudut antara vektor dan vektor  adalah ...

Pembahasan Soal:

Misalkan terdapat dua buah titik straight A open parentheses x subscript 1 comma space y subscript 1 comma space z subscript 1 close parentheses dan straight B open parentheses x subscript 2 comma space y subscript 2 comma space z subscript 2 close parentheses. Vektor AB dapat didefinisikan sebagai:

AB with bar on top equals straight B minus straight A equals open parentheses table row cell straight x subscript 2 minus straight x subscript 1 end cell row cell straight y subscript 2 minus straight y subscript 1 end cell row cell straight z subscript 2 minus straight z subscript 1 end cell end table close parentheses

Sedangkan panjang vektor AB with rightwards arrow on top adalah:

open vertical bar AB with rightwards arrow on top close vertical bar equals square root of open parentheses x subscript 2 minus x subscript 1 close parentheses squared plus open parentheses y subscript 2 minus y subscript 1 close parentheses squared plus open parentheses z subscript 2 minus z subscript 1 close parentheses squared end root.

Lalu ingat juga rumus dot product antara dua vektor yaitu:

table attributes columnalign right center left columnspacing 0px end attributes row cell cos space straight theta end cell equals cell fraction numerator straight a with rightwards arrow on top times straight b with rightwards arrow on top over denominator open vertical bar straight a with rightwards arrow on top close vertical bar open vertical bar straight b with rightwards arrow on top close vertical bar end fraction end cell end table.

Pada soal diketahui straight A left parenthesis 2 comma space 4 comma space 1 right parenthesisstraight B left parenthesis 4 comma space 6 comma space 1 right parenthesis dan straight C left parenthesis 3 comma space 5 comma space 5 right parenthesis. AB with rightwards arrow on top wakil dari straight u with rightwards arrow on top, maka:

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

Panjang vektor straight u with rightwards arrow on top adalah:

table attributes columnalign right center left columnspacing 0px end attributes row cell open vertical bar straight u with rightwards arrow on top close vertical bar end cell equals cell square root of 2 squared plus 2 squared plus 0 squared end root end cell row blank equals cell square root of 4 plus 4 plus 0 end root end cell row blank equals cell square root of 8 end cell row blank equals cell 2 square root of 2 end cell end table

Sedangkan jika AC with rightwards arrow on top wakil dari straight v with rightwards arrow on top, maka:

table attributes columnalign right center left columnspacing 0px end attributes row cell AC with rightwards arrow on top end cell equals cell straight C minus straight A end cell row cell straight v with rightwards arrow on top end cell equals cell open parentheses table row 3 row 5 row 5 end table close parentheses minus open parentheses table row 2 row 4 row 1 end table close parentheses end cell row cell straight v with rightwards arrow on top end cell equals cell open parentheses table row 1 row 1 row 4 end table close parentheses end cell end table

Panjang vektor straight v with rightwards arrow on top adalah:

table attributes columnalign right center left columnspacing 0px end attributes row cell open vertical bar straight v with rightwards arrow on top close vertical bar end cell equals cell square root of 1 squared plus 1 squared plus 4 squared end root end cell row blank equals cell square root of 1 plus 1 plus 16 end root end cell row blank equals cell square root of 18 end cell row blank equals cell 3 square root of 2 end cell end table

Selanjutnya, kosinus sudut antara vektor straight u with rightwards arrow on top dan vektor straight v with rightwards arrow on top dapat dicari sebagai berikut.

table attributes columnalign right center left columnspacing 0px end attributes row cell cos space straight theta end cell equals cell fraction numerator straight u with rightwards arrow on top times straight v with rightwards arrow on top over denominator open vertical bar straight u with rightwards arrow on top close vertical bar open vertical bar straight v with rightwards arrow on top close vertical bar end fraction end cell row blank equals cell fraction numerator open parentheses 2 comma space 2 comma space 0 close parentheses times open parentheses 1 comma space 1 comma space 4 close parentheses over denominator 2 square root of 2 open parentheses 3 square root of 2 close parentheses end fraction end cell row blank equals cell fraction numerator 2 open parentheses 1 close parentheses plus 2 open parentheses 1 close parentheses plus 0 open parentheses 4 close parentheses over denominator 6 open parentheses 2 close parentheses end fraction end cell row blank equals cell 4 over 12 end cell row blank equals cell 1 third end cell end table

Dengan demikian, jawaban yang benar adalah B.

0

Roboguru

Diketahui , , dan . Jika  maka tentukan panjang !

Pembahasan Soal:

Berdasarkan konsep penjumlahan vektor tiga dimensi maka diperoleh vektor begin mathsize 14px style straight v with rightwards arrow on top end style:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell straight v with rightwards arrow on top end cell equals cell straight d with rightwards arrow on top minus 2 straight e with rightwards arrow on top plus 3 straight f with rightwards arrow on top end cell row blank equals cell open parentheses table row cell negative 7 end cell row 3 row 5 end table close parentheses minus 2 open parentheses table row 3 row cell negative 5 end cell row 4 end table close parentheses plus 3 open parentheses table row 1 row cell negative 7 end cell row 4 end table close parentheses end cell row blank equals cell open parentheses table row cell negative 7 end cell row 3 row 5 end table close parentheses minus open parentheses table row 6 row cell negative 10 end cell row 8 end table close parentheses plus open parentheses table row 3 row cell negative 21 end cell row 12 end table close parentheses end cell row blank equals cell open parentheses table row cell negative 7 minus 6 plus 3 end cell row cell 3 minus open parentheses negative 10 close parentheses plus open parentheses negative 21 close parentheses end cell row cell 5 minus 8 plus 12 end cell end table close parentheses end cell row blank equals cell open parentheses table row cell negative 10 end cell row cell negative 8 end cell row 9 end table close parentheses end cell end table end style 

Sehingga panjang vektor begin mathsize 14px style straight v with rightwards arrow on top end style adalah

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell open vertical bar straight v with rightwards arrow on top close vertical bar end cell equals cell square root of open parentheses negative 10 close parentheses squared plus open parentheses negative 8 close parentheses squared plus 9 squared end root end cell row blank equals cell square root of 100 plus 64 plus 81 end root end cell row blank equals cell square root of 245 end cell row blank equals cell square root of 49 cross times 5 end root end cell row blank equals cell 7 square root of 5 end cell end table end style 

Dengan demikian, panjang vektor begin mathsize 14px style straight v with rightwards arrow on top end style adalah begin mathsize 14px style 7 square root of 5 end style.

0

Roboguru

Diketahui vektor  dan titik . Jika panjang  sama dengan panjang  dan  berlawanan arah dengan , maka tentukan koordinat Q.

Pembahasan Soal:

Ingat vektor yang menghubungkan dua titik diperoleh dari koordinat titik akhir dikurangi titik awal.

panjang begin mathsize 14px style stack P Q with rightwards harpoon with barb upwards on top end style sama dengan panjang begin mathsize 14px style a with rightwards harpoon with barb upwards on top end style dan begin mathsize 14px style stack P Q with rightwards harpoon with barb upwards on top end style berlawanan arah dengan begin mathsize 14px style a with rightwards harpoon with barb upwards on top end style, artinya

begin mathsize 14px style stack P Q with rightwards harpoon with barb upwards on top equals negative a with bar on top end style 

begin mathsize 14px style q with bar on top minus p with bar on top equals negative a with bar on top end style 

begin mathsize 14px style q with bar on top equals p with bar on top minus a with bar on top end style 

begin mathsize 14px style q with bar on top equals open parentheses table row 2 row cell negative 1 end cell row 3 end table close parentheses minus open parentheses table row 4 row cell negative 5 end cell row 3 end table close parentheses equals open parentheses table row cell negative 2 end cell row 4 row 0 end table close parentheses end style 

Jadi koordinat titik Q adalah begin mathsize 14px style open parentheses negative 2 comma 4 comma 0 close parentheses end style.

0

Roboguru

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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