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Jika vektor tak nol a dan b memiliki panjang yang sama dan memenuhi ∣ ∣ ​ 2 a + b ∣ ∣ ​ = ∣ ∣ ​ a − b ∣ ∣ ​ , maka sudut antara vektor dan adalah ... derajat.

Jika vektor tak nol  dan  memiliki panjang yang sama dan memenuhi , maka sudut antara vektor a with rightwards arrow on top dan b with rightwards arrow on top adalah ... derajat.

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D. Kusumawardhani

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diperolehsudut antara vektor dan adalah .

diperoleh sudut antara vektor a with rightwards arrow on top dan b with rightwards arrow on top adalah 120 degree.

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Ingat kembali menentukan sudut antara dua vektor. Sehingga, diperoleh perhitungan berikut. Misalkan dan . Vektor dan memiliki panjang yang sama, maka diperoleh Vektor dan memenuhi , maka diperoleh ∣ ∣ ​ 2 a + b ∣ ∣ ​ ( 2 x + m ) 2 + ( 2 y + n ) 2 ​ ( 2 x + m ) 2 + ( 2 y + n ) 2 4 x 2 + 4 m x + m 2 + 4 y 2 + 4 n y + n 2 ( 4 m x + 2 m x ) + ( 4 n y + 2 n y ) 6 m x + 6 n y 2 m x + 2 n y 2 ( m x + n y ) m x + n y ​ = = = = = = = = = ​ ∣ ∣ ​ a − b ∣ ∣ ​ ( x − m ) 2 + ( y − n ) 2 ​ ( x − m ) 2 + ( y − n ) 2 x 2 − 2 m x + m 2 + y 2 − 2 n y + n 2 ( x 2 − 4 x 2 ) + ( y 2 − 4 y 2 ) − 3 x 2 − 3 y 2 − x 2 − y 2 − ( x 2 + y 2 ) − 2 1 ​ ( x 2 + y 2 ) ​ Akan ditentukan sudut antara vektor dan , perhatikan perhitungan berikut. Jadi, diperolehsudut antara vektor dan adalah .

Ingat kembali menentukan sudut antara dua vektor.

cos space theta equals fraction numerator a with rightwards arrow on top times b with rightwards arrow on top over denominator open vertical bar a with rightwards arrow on top close vertical bar times open vertical bar b with rightwards arrow on top close vertical bar end fraction

Sehingga, diperoleh perhitungan berikut.

Misalkan a with rightwards arrow on top equals open parentheses table row x row y end table close parentheses dan b with rightwards arrow on top equals open parentheses table row m row n end table close parentheses.

a with rightwards arrow on top times b with rightwards arrow on top equals open parentheses table row x row y end table close parentheses times open parentheses table row m row n end table close parentheses equals m x plus n y 

Vektor a with rightwards arrow on top dan b with rightwards arrow on top memiliki panjang yang sama, maka diperoleh

table attributes columnalign right center left columnspacing 2px end attributes row cell open vertical bar a with rightwards arrow on top close vertical bar end cell equals cell open vertical bar b with rightwards arrow on top close vertical bar end cell row cell square root of x squared plus y squared end root end cell equals cell square root of m squared plus n squared end root end cell row cell x squared plus y squared end cell equals cell m squared plus n squared end cell end table

Vektor a with rightwards arrow on top dan b with rightwards arrow on top memenuhi open vertical bar 2 a with rightwards arrow on top plus b with rightwards arrow on top close vertical bar equals open vertical bar a with rightwards arrow on top minus b with rightwards arrow on top close vertical bar, maka diperoleh

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

 

Akan ditentukan sudut antara vektor a with rightwards arrow on top dan b with rightwards arrow on top, perhatikan perhitungan berikut.

table attributes columnalign right center left columnspacing 2px end attributes row cell cos space theta end cell equals cell fraction numerator a with rightwards arrow on top times b with rightwards arrow on top over denominator open vertical bar a with rightwards arrow on top close vertical bar times open vertical bar b with rightwards arrow on top close vertical bar end fraction end cell row blank equals cell fraction numerator m x plus n y over denominator square root of x squared plus y squared end root times square root of m squared plus n squared end root end fraction end cell row blank equals cell fraction numerator m x plus n y over denominator square root of x squared plus y squared end root times square root of x squared plus y squared end root end fraction end cell row blank equals cell fraction numerator negative begin display style 1 half end style up diagonal strike open parentheses x squared plus y squared close parentheses end strike over denominator up diagonal strike open parentheses x squared plus y squared close parentheses end strike end fraction end cell row cell cos space theta end cell equals cell negative 1 half end cell row theta equals cell arc space cos space open square brackets negative 1 half close square brackets end cell row theta equals cell 120 degree end cell end table 

Jadi, diperoleh sudut antara vektor a with rightwards arrow on top dan b with rightwards arrow on top adalah 120 degree.

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