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

Pertanyaan

Diketahui titik straight C open parentheses 5 comma space minus 2 close parentheses dan titik straight D open parentheses 3 comma space 1 close parentheses. Kombinasi linear dari vektor CD with rightwards arrow on top adalah ...space 

  1. 2 i with rightwards arrow on top minus 3 j with rightwards arrow on top 

  2. negative 2 i with rightwards arrow on top minus 3 j with rightwards arrow on top 

  3. 2 i with rightwards arrow on top plus 3 j with rightwards arrow on top 

  4. 2 i with rightwards arrow on top plus 4 j with rightwards arrow on top 

  5. negative 2 i with rightwards arrow on top plus 3 j with rightwards arrow on top 

Pembahasan Soal:

Jika diketahui straight A open parentheses x subscript 1 comma space y subscript 1 close parentheses dan straight B open parentheses x subscript 2 comma space y subscript 2 close parentheses, maka BA with rightwards arrow on top dapat dirumuskan:

BA with rightwards arrow on top equals a with rightwards arrow on top minus b with rightwards arrow on top 

dimana:

table attributes columnalign right center left columnspacing 0px end attributes row cell a with rightwards arrow on top end cell equals cell titik space straight A end cell row cell b with rightwards arrow on top end cell equals cell space titik space straight B end cell end table 

Diketahui titik straight C open parentheses 5 comma space minus 2 close parentheses dan titik straight D open parentheses 3 comma space 1 close parentheses, maka kombinasi linear dari vektor CD with rightwards arrow on top adalah:

table attributes columnalign right center left columnspacing 0px end attributes row cell CD with rightwards arrow on top end cell equals cell d with rightwards arrow on top minus c with rightwards arrow on top end cell row blank equals cell open parentheses table row 3 row 1 end table close parentheses minus open parentheses table row 5 row cell negative 2 end cell end table close parentheses end cell row blank equals cell open parentheses table row cell 3 minus 5 end cell row cell 1 plus 2 end cell end table close parentheses end cell row blank equals cell open parentheses table row cell negative 2 end cell row 3 end table close parentheses end cell row blank equals cell negative 2 i with rightwards arrow on top plus 3 j with rightwards arrow on top end cell end table 

Sehingga, table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell CD with rightwards arrow on top end cell end table table attributes columnalign right center left columnspacing 0px end attributes row blank equals blank end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank minus end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank 2 end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell i with rightwards arrow on top end cell end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank plus end table 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 j with rightwards arrow on top end cell end table.

Jadi, jawaban yang tepat adalah E.

Pembahasan terverifikasi oleh Roboguru

Terakhir diupdate 10 Juni 2021

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