Pertanyaan

Diketahui vektor v = xi + j dan u = i − 3 j . Tentukan nilai yang mungkin jika vektor satuan dari vektor adalah ( 13 2 13 − 2 ) .

Diketahui vektor $v = xi + j$ dan $u = i - 3 j .$ Tentukan nilai begin mathsize 14px style x end style yang mungkin jika vektor satuan dari vektor begin mathsize 14px style straight u with rightwards harpoon with barb upwards on top plus straight v with rightwards harpoon with barb upwards on top end style adalah $(\frac{2}{13} \frac{- 2}{13}) .$

R. RGFLLIMA

Master Teacher

Jawaban terverifikasi

Jawaban

jika dan maka tidak adanilai yang mungkin jika vektor satuan dari vektor .

jika

begin mathsize 14px style straight v equals xi plus straight j end style

dan

begin mathsize 14px style straight u equals straight i minus 3 straight j end style

maka tidak ada nilai begin mathsize 14px style x end style

yang mungkin jika vektor satuan dari vektor $begin mathsize 14px style straight u with rightwards harpoon with barb upwards on top plus straight v with rightwards harpoon with barb upwards on top equals open parentheses table row cell fraction numerator 2 over denominator square root of 13 end fraction end cell row cell fraction numerator negative 2 over denominator square root of 13 end fraction end cell end table close parentheses end style$ .

Pembahasan

Operasi penjumlahan vektor. vektor atau vektor atau maka: vektor satuan sehingga diperoleh: dari kesamaan 2 vektor maka diperoleh nilai harus : 1) 2) syarat 1 dan 2 nilai tidak ada yang sama sehingga tidak ada nilai yang memenuhi. Jadi, jika dan maka tidak adanilai yang mungkin jika vektor satuan dari vektor .

Operasi penjumlahan vektor.

vektor begin mathsize 14px style straight v equals xi plus straight j end style atau

vektor begin mathsize 14px style straight u equals straight i minus 3 straight j end style atau

maka:

vektor satuan $begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell straight u with rightwards harpoon with barb upwards on top end cell end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank plus end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell straight v with rightwards harpoon with barb upwards on top end cell end table table attributes columnalign right center left columnspacing 0px end attributes row blank equals blank end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell fraction numerator straight u with rightwards harpoon with barb upwards on top plus straight v with rightwards harpoon with barb upwards on top over denominator open vertical bar straight u with rightwards harpoon with barb upwards on top plus straight v with rightwards harpoon with barb upwards on top close vertical bar end fraction end cell end table end style$ sehingga diperoleh:

$begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses table row cell fraction numerator 2 over denominator square root of 13 end fraction end cell row cell fraction numerator negative 2 over denominator square root of 13 end fraction end cell end table close parentheses end cell equals cell fraction numerator open parentheses table row cell straight x plus 1 end cell row cell negative 2 end cell end table close parentheses over denominator square root of left parenthesis straight x plus 1 right parenthesis squared plus left parenthesis negative 2 right parenthesis squared end root end fraction end cell row cell open parentheses table row cell fraction numerator 2 over denominator square root of 13 end fraction end cell row cell fraction numerator negative 2 over denominator square root of 13 end fraction end cell end table close parentheses end cell equals cell fraction numerator open parentheses table row cell straight x plus 1 end cell row cell negative 2 end cell end table close parentheses over denominator square root of straight x squared plus 2 straight x plus 1 plus 4 end root end fraction end cell row cell open parentheses table row cell fraction numerator 2 over denominator square root of 13 end fraction end cell row cell fraction numerator negative 2 over denominator square root of 13 end fraction end cell end table close parentheses end cell equals cell fraction numerator open parentheses table row cell straight x plus 1 end cell row cell negative 2 end cell end table close parentheses over denominator square root of straight x squared plus 2 straight x plus 5 end root end fraction end cell row cell open parentheses table row cell fraction numerator 2 over denominator square root of 13 end fraction end cell row cell fraction numerator negative 2 over denominator square root of 13 end fraction end cell end table close parentheses end cell equals cell open parentheses table row cell fraction numerator straight x plus 1 over denominator square root of straight x squared plus 2 straight x plus 5 end root end fraction end cell row cell fraction numerator negative 2 over denominator square root of straight x squared plus 2 straight x plus 5 end root end fraction end cell end table close parentheses end cell end table end style$

dari kesamaan 2 vektor maka diperoleh nilai begin mathsize 14px style x end style harus :

$begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell fraction numerator 2 over denominator square root of 13 end fraction end cell equals cell fraction numerator left parenthesis straight x plus 1 right parenthesis over denominator square root of straight x squared plus 2 straight x plus 5 end root end fraction end cell row cell 4 over 13 end cell equals cell fraction numerator left parenthesis straight x plus 1 right parenthesis squared over denominator straight x squared plus 2 straight x plus 5 end fraction end cell row cell 4 over 13 end cell equals cell fraction numerator straight x squared plus 2 straight x plus 1 over denominator straight x squared plus 2 straight x plus 5 end fraction end cell row cell 4 left parenthesis straight x squared plus 2 straight x plus 5 right parenthesis end cell equals cell 13 left parenthesis straight x squared plus 2 straight x plus 1 right parenthesis end cell row cell 4 straight x squared plus 8 straight x plus 20 end cell equals cell 13 squared plus 26 straight x plus 13 end cell row cell negative 9 straight x squared minus 18 straight x plus 7 end cell equals 0 row cell 9 straight x squared plus 18 straight x minus 7 end cell equals 0 row cell straight x subscript 1 comma 2 end subscript end cell equals cell fraction numerator negative straight b plus-or-minus square root of straight b squared minus 4 ac end root over denominator 2 straight a end fraction end cell row blank equals cell fraction numerator negative 18 plus-or-minus square root of 18 squared minus 4 times 9 times left parenthesis negative 7 right parenthesis end root over denominator 2 left parenthesis 9 right parenthesis end fraction end cell row blank equals cell fraction numerator negative 18 plus-or-minus square root of 324 plus 252 end root over denominator 18 end fraction end cell row blank equals cell fraction numerator negative 18 plus-or-minus square root of 576 over denominator 18 end fraction end cell row blank equals cell fraction numerator negative 18 plus-or-minus 24 over denominator 18 end fraction end cell row cell straight x subscript 1 end cell equals cell fraction numerator negative 18 plus 24 over denominator 18 end fraction end cell row blank equals cell 6 over 18 end cell row blank equals cell 1 third end cell row cell straight x subscript 2 end cell equals cell fraction numerator negative 18 minus 24 over denominator 18 end fraction end cell row blank equals cell fraction numerator negative 42 over denominator 18 end fraction end cell row blank equals cell fraction numerator negative 7 over denominator 3 end fraction end cell end table end style$

syarat 1 dan 2 nilai begin mathsize 14px style x end style tidak ada yang sama sehingga tidak ada nilai yang memenuhi.

Jadi, jika begin mathsize 14px style straight v equals xi plus straight j end style dan maka tidak ada nilai begin mathsize 14px style x end style yang mungkin jika vektor satuan dari vektor $begin mathsize 14px style straight u with rightwards harpoon with barb upwards on top plus straight v with rightwards harpoon with barb upwards on top equals open parentheses table row cell fraction numerator 2 over denominator square root of 13 end fraction end cell row cell fraction numerator negative 2 over denominator square root of 13 end fraction end cell end table close parentheses end style$ .