Roboguru

Nilai u sehingga kedua garis dengan persamaan vektor  dan  saling berpotongan adalah ....

Pertanyaan

Nilai u sehingga kedua garis dengan persamaan vektor begin mathsize 14px style stack r subscript 1 with rightwards arrow on top equals open parentheses i with rightwards arrow on top plus k with rightwards arrow on top close parentheses plus s open parentheses i with rightwards arrow on top plus j with rightwards arrow on top plus 2 k with rightwards arrow on top close parentheses end style dan begin mathsize 14px style stack r subscript 2 with rightwards arrow on top equals open parentheses 2 i with rightwards arrow on top minus j with rightwards arrow on top plus 5 k with rightwards arrow on top close parentheses plus t open parentheses 4 j with rightwards arrow on top minus u k with rightwards arrow on top close parentheses end style saling berpotongan adalah ....

  1. -3

  2. -1

  3. 2

  4. 4

  5. 5

Pembahasan:

Misalkan titik P adalah titik potong kedua garis dengan vektor posisinya yaitu undefined.

Maka begin mathsize 14px style stack r subscript 1 with rightwards arrow on top equals stack r subscript 2 with rightwards arrow on top equals p with rightwards arrow on top end style.

Perhatikan bahwa

begin mathsize 14px style stack r subscript 1 with rightwards arrow on top equals open parentheses i with rightwards arrow on top plus k with rightwards arrow on top close parentheses plus s open parentheses i with rightwards arrow on top plus j with rightwards arrow on top plus 2 k with rightwards arrow on top close parentheses stack r subscript 1 with rightwards arrow on top equals i with rightwards arrow on top plus k with rightwards arrow on top plus s i with rightwards arrow on top plus s j with rightwards arrow on top plus 2 s k with rightwards arrow on top stack r subscript 1 with rightwards arrow on top equals open parentheses 1 plus s close parentheses i with rightwards arrow on top plus s j with rightwards arrow on top plus open parentheses 1 plus 2 s close parentheses k with rightwards arrow on top stack r subscript 1 with rightwards arrow on top equals open parentheses table row cell 1 plus s end cell row s row cell 1 plus 2 s end cell end table close parentheses end style

dan

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell stack r subscript 2 with rightwards arrow on top end cell equals cell open parentheses 2 i with rightwards arrow on top minus j with rightwards arrow on top plus 5 k with rightwards arrow on top close parentheses plus t open parentheses 4 j with rightwards arrow on top minus u k with rightwards arrow on top close parentheses end cell row cell stack r subscript 2 with rightwards arrow on top end cell equals cell 2 i with rightwards arrow on top minus j with rightwards arrow on top plus 5 k with rightwards arrow on top plus 4 t j with rightwards arrow on top minus u t k with rightwards arrow on top end cell row cell stack r subscript 2 with rightwards arrow on top end cell equals cell 2 i with rightwards arrow on top plus open parentheses negative 1 plus 4 t close parentheses j with rightwards arrow on top plus open parentheses 5 minus u t close parentheses k with rightwards arrow on top end cell row cell stack r subscript 2 with rightwards arrow on top end cell equals cell open parentheses table row 2 row cell negative 1 plus 4 t end cell row cell 5 minus u t end cell end table close parentheses end cell end table end style

Sehingga

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell stack r subscript 1 with rightwards arrow on top end cell equals cell stack r subscript 2 with rightwards arrow on top end cell row cell open parentheses table row cell 1 plus s end cell row s row cell 1 plus 2 s end cell end table close parentheses end cell equals cell open parentheses table row 2 row cell negative 1 plus 4 t end cell row cell 5 minus u t end cell end table close parentheses end cell end table end style

Maka didapatkan bahwa

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

Kemudian

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row s equals cell negative 1 plus 4 t end cell row 1 equals cell negative 1 plus 4 t end cell row 2 equals cell 4 t end cell row t equals cell 1 half end cell end table end style

Selanjutnya

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell 1 plus 2 s end cell equals cell 5 minus u t end cell row cell 1 plus 2 open parentheses 1 close parentheses end cell equals cell 5 minus u open parentheses 1 half close parentheses end cell row cell 1 plus 2 end cell equals cell 5 minus 1 half u end cell row 3 equals cell 5 minus 1 half u end cell row cell 1 half u end cell equals 2 row u equals 4 end table end style  

Jawaban terverifikasi

Dijawab oleh:

Terakhir diupdate 08 Januari 2021

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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Roboguru

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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