Diketahui titik , , dan . Jika mewakili dan mewakili , maka proyeksi ortogonal vektor pada adalah ....

Pertanyaan

Diketahui titik $straight A open parentheses 2 comma space 7 comma space 8 close parentheses$ , $straight B open parentheses negative 1 comma space 1 comma space minus 1 close parentheses$ , dan $straight C open parentheses 0 comma space 3 comma space 2 close parentheses$ . Jika $u with rightwards arrow on top$ mewakili $AB with rightwards arrow on top$ dan $v with rightwards arrow on top$ mewakili $BC with rightwards arrow on top$ , maka proyeksi ortogonal vektor $u with rightwards arrow on top$ pada $v with rightwards arrow on top$ adalah ....

$negative 3 i with rightwards arrow on top minus 6 j with rightwards arrow on top minus 9 k with rightwards arrow on top$
$i with rightwards arrow on top plus 2 j with rightwards arrow on top plus 3 k with rightwards arrow on top$
$1 third i with rightwards arrow on top plus 2 over 3 j with rightwards arrow on top plus k with rightwards arrow on top$
$negative 9 i with rightwards arrow on top minus 18 j with rightwards arrow on top minus 27 k with rightwards arrow on top$
$3 i with rightwards arrow on top plus 6 j with rightwards arrow on top plus 9 k with rightwards arrow on top$

Pembahasan Soal:

Gunakan konsep proyeksi orthogonal vektor $u with rightwards arrow on top$ pada $v with rightwards arrow on top$ .

$Proyeksi space ortogonal equals open parentheses fraction numerator u with rightwards arrow on top times v with rightwards arrow on top over denominator open vertical bar v with rightwards arrow on top close vertical bar squared end fraction close parentheses v with rightwards arrow on top$

Diketahui:
Titik $straight A open parentheses 2 comma space 7 comma space 8 close parentheses$ , $straight B open parentheses negative 1 comma space 1 comma space minus 1 close parentheses$ , dan $straight C open parentheses 0 comma space 3 comma space 2 close parentheses$ . Jika $u with rightwards arrow on top$ mewakili $AB with rightwards arrow on top$ dan $v with rightwards arrow on top$ mewakili $BC with rightwards arrow on top$ .
Akan ditentukan proyeksi ortogonal vektor $u with rightwards arrow on top$ pada $v with rightwards arrow on top$ .

Tentukan vektor $u with rightwards arrow on top$ .

$table attributes columnalign right center left columnspacing 2px end attributes row cell u with rightwards arrow on top end cell equals cell AB with rightwards arrow on top end cell row blank equals cell straight B minus straight A end cell row blank equals cell open parentheses negative 1 minus 2 comma space 1 minus 7 comma space minus 1 minus 8 close parentheses end cell row cell u with rightwards arrow on top end cell equals cell open parentheses negative 3 comma space minus 6 comma space minus 9 close parentheses end cell end table$

Tentukan vektor $v with rightwards arrow on top$ .

$table attributes columnalign right center left columnspacing 2px end attributes row cell v with rightwards arrow on top end cell equals cell BC with rightwards arrow on top end cell row blank equals cell straight C minus straight B end cell row blank equals cell open parentheses 0 minus open parentheses negative 1 close parentheses comma space 3 minus 1 comma space 2 minus open parentheses negative 1 close parentheses close parentheses end cell row cell v with rightwards arrow on top end cell equals cell open parentheses 1 comma space 2 comma space 3 close parentheses end cell end table$

Sehingga proyeksi ortogonal vektor $u with rightwards arrow on top$ pada $v with rightwards arrow on top$ dapat diperoleh sebagai berikut.

$table attributes columnalign right center left columnspacing 2px end attributes row Proyeksi equals cell open parentheses fraction numerator u with rightwards arrow on top times v with rightwards arrow on top over denominator open vertical bar v with rightwards arrow on top close vertical bar squared end fraction close parentheses times v with rightwards arrow on top end cell row blank equals cell open parentheses fraction numerator open parentheses negative 3 close parentheses times 1 plus open parentheses negative 6 close parentheses times 2 plus open parentheses negative 9 close parentheses times 3 over denominator open parentheses square root of 1 squared plus 2 squared plus 3 squared end root close parentheses squared end fraction close parentheses open parentheses 1 comma space 2 comma space 3 close parentheses end cell row blank equals cell open parentheses fraction numerator negative 3 minus 12 minus 27 over denominator open parentheses square root of 1 plus 4 plus 9 end root close parentheses squared end fraction close parentheses open parentheses 1 comma space 2 comma space 3 close parentheses end cell row blank equals cell open parentheses 42 over open parentheses square root of 14 close parentheses squared close parentheses open parentheses 1 comma space 2 comma space 3 close parentheses end cell row blank equals cell open parentheses 42 over 14 close parentheses open parentheses 1 comma space 2 comma space 3 close parentheses end cell row blank equals cell 3 cross times open parentheses 1 comma space 2 comma space 3 close parentheses end cell row Proyeksi equals cell open parentheses 3 comma space 6 comma space 9 close parentheses end cell end table$

Sehingga diperoleh proyeksi ortogonalnya adalah $open parentheses 3 comma space 6 comma space 9 close parentheses$ atau $3 i with rightwards arrow on top plus 6 j with rightwards arrow on top plus 9 k with rightwards arrow on top$ .

Jadi, jawaban yang tepat adalah E.

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

Y. Fathoni

Mahasiswa/Alumni Universitas Negeri Yogyakarta.

Terakhir diupdate 01 Mei 2021

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Jika vektor , vektor , dan vektor proyeksi pada adalah , maka nilai adalah ....

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Dengan menerapkan rumus proyeksi ortogonal vektor, diperoleh perhitungan sebagai berikut.

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Jadi, nilai $x squared$ adalah 36.

Dengan demikian, jawaban yang tepat adalah E.

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Pembahasan Soal:

Misalkan vektor proyeksi ortogonal dari vektor $begin mathsize 14px style a with rightwards arrow on top end style$ terhadap vektor $begin mathsize 14px style b with rightwards arrow on top end style$ adalah $begin mathsize 14px style p with rightwards arrow on top end style$ . Oleh karena itu, didapatkan bahwa

$begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell p with rightwards arrow on top end cell equals cell open parentheses fraction numerator a with rightwards arrow on top ⋅ b with rightwards arrow on top over denominator open vertical bar b with rightwards arrow on top close vertical bar squared end fraction close parentheses b with rightwards arrow on top end cell row blank equals cell open parentheses fraction numerator open parentheses table row cell negative sign 4 end cell row cell negative sign 3 end cell row 5 end table close parentheses ⋅ open parentheses table row cell negative sign 3 end cell row 3 row cell negative sign 1 end cell end table close parentheses over denominator open parentheses square root of open parentheses negative sign 3 close parentheses squared plus 3 squared plus open parentheses negative sign 1 close parentheses squared end root close parentheses squared end fraction close parentheses open parentheses table row cell negative sign 3 end cell row 3 row cell negative sign 1 end cell end table close parentheses end cell row blank equals cell open parentheses fraction numerator open parentheses negative sign 4 close parentheses ⋅ open parentheses negative sign 3 close parentheses plus open parentheses negative sign 3 close parentheses ⋅ 3 plus 5 ⋅ open parentheses negative sign 1 close parentheses over denominator open parentheses square root of 9 plus 9 plus 1 end root close parentheses squared end fraction close parentheses open parentheses table row cell negative sign 3 end cell row 3 row cell negative sign 1 end cell end table close parentheses end cell row blank equals cell open parentheses fraction numerator 12 minus sign 9 minus sign 5 over denominator open parentheses square root of 19 close parentheses squared end fraction close parentheses open parentheses table row cell negative sign 3 end cell row 3 row cell negative sign 1 end cell end table close parentheses end cell row blank equals cell fraction numerator negative sign 2 over denominator 19 end fraction open parentheses table row cell negative sign 3 end cell row 3 row cell negative sign 1 end cell end table close parentheses end cell row blank equals cell open parentheses table row cell 6 over 19 end cell row cell negative sign 6 over 19 end cell row cell 2 over 19 end cell end table close parentheses end cell end table end style$

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Pembahasan Soal:

$begin mathsize 14px style straight p with rightwards arrow on top space proyeksi space straight u with rightwards arrow on top space pada space straight v with rightwards arrow on top space maka space straight p with rightwards arrow on top space dan space straight v with rightwards arrow on top space kolinear comma space sehingga colon straight p with rightwards arrow on top equals straight n. straight space straight v with rightwards arrow on top rightwards double arrow open parentheses table row 4 row cell negative 2 end cell row 4 end table close parentheses equals straight n straight space. straight space open parentheses table row straight a row cell negative straight b end cell row straight a end table close parentheses Jadi comma space minus 2 equals straight n open parentheses negative straight b close parentheses rightwards double arrow straight n equals 2 over straight b space dan space 4 equals na rightwards double arrow straight n equals 4 over straight a Sehingga colon 2 over straight b equals 4 over straight a rightwards double arrow straight a equals 2 straight b cos invisible function application space straight theta equals fraction numerator straight u with rightwards arrow on top straight space. straight space straight v with rightwards arrow on top over denominator open vertical bar straight u with rightwards arrow on top close vertical bar open vertical bar straight v with rightwards arrow on top close vertical bar end fraction 6 over 11 equals fraction numerator 9 straight a plus ab minus ab over denominator square root of 9 squared plus straight a squared plus straight b squared end root. straight space square root of straight a squared plus left parenthesis negative straight b right parenthesis squared plus straight a squared end root end fraction Substitusi space straight a equals 2 straight b 6 over 11 equals fraction numerator 18 straight b over denominator square root of 81 plus 5 straight b squared end root. straight space square root of 9 straight b squared end root end fraction 6 over 11 equals fraction numerator 6 over denominator square root of 81 plus 5 straight b squared end root end fraction square root of 81 plus 5 straight b squared end root equals 11 81 plus 5 straight b squared equals 121 5 straight b squared equals 121 minus 81 5 straight b squared equals 40 straight b squared equals 8 straight b equals plus-or-minus 2 square root of 2 Jadi comma space nilai space straight b equals 2 square root of 2 end style$

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