Pertanyaan

Diketahui a = 2 i − 2 j − k dan b = 3 i + 2 j + k . Luasjajaran genjang yang dibentuk oleh a + b dan b adalah ....

Diketahui $a = 2 i - 2 j - k dan b = 3 i + 2 j + k$ . Luas jajaran genjang yang dibentuk oleh $a + b dan b$ adalah ....

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A. Rizky

Master Teacher

Mahasiswa/Alumni Universitas Negeri Malang

Jawaban terverifikasi

Pembahasan

$begin mathsize 14px style straight a with rightwards arrow on top plus straight b with rightwards arrow on top equals open parentheses table row 2 row cell negative 2 end cell row cell negative 1 end cell end table close parentheses plus open parentheses table row 3 row 2 row 1 end table close parentheses equals open parentheses table row 5 row 0 row 0 end table close parentheses Pertama short dash tama space cari space cosinus space sudut space antara space vektor space straight a with rightwards arrow on top plus straight b with rightwards arrow on top space dan space vektor space straight b with rightwards arrow on top cos invisible function application space straight alpha equals fraction numerator open parentheses straight a with rightwards arrow on top plus straight b with rightwards arrow on top close parentheses. straight b with rightwards arrow on top over denominator open vertical bar straight a with rightwards arrow on top plus straight b with rightwards arrow on top close vertical bar open vertical bar straight b with rightwards arrow on top close vertical bar end fraction cos invisible function application space straight alpha equals fraction numerator open parentheses table row 5 row 0 row 0 end table close parentheses. open parentheses table row 3 row 2 row 1 end table close parentheses over denominator square root of 5 squared plus 0 squared plus 0 squared end root square root of 3 squared plus 2 squared plus 1 squared end root end fraction cos invisible function application space straight alpha equals fraction numerator 15 over denominator 5 square root of 14 end fraction equals fraction numerator 3 over denominator square root of 14 end fraction Kemudian end style$

begin mathsize 14px style square root of 14 minus 3 squared end root space space space space space space space rightwards arrow for blank of equals square root of 5 end style

$begin mathsize 14px style Maka space sin space invisible function application straight alpha equals fraction numerator square root of 5 over denominator square root of 14 end fraction Sehingga space luas space jajargenjang space yang space dibatasi space oleh space vektor space straight a with rightwards arrow on top plus straight b with rightwards arrow on top space dan space vektor space straight b with rightwards arrow on top space adalah open vertical bar straight a with rightwards arrow on top plus straight b with rightwards arrow on top close vertical bar open vertical bar straight b with rightwards arrow on top close vertical bar sin space invisible function application straight alpha 5 square root of 14. fraction numerator square root of 5 over denominator square root of 14 end fraction equals 5 square root of 5 end style$