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Pertanyaan

d ik e t ah u i v e k t or u = ⎝ ⎛ ​ 7 − 4 1 ​ ⎠ ⎞ ​ d an v = ⎝ ⎛ ​ − 2 1 0 ​ ⎠ ⎞ ​ . p roye k s i v e k t or or t h o g o na l u d an v a d a l ah ...

  1. negative 2 over 5 open parentheses table row 4 row 2 row 0 end table close parentheses

  2. negative 1 fifth open parentheses table row 4 row 2 row 0 end table close parentheses

  3. 1 fifth open parentheses table row 4 row 2 row 0 end table close parentheses

  4. 2 over 5 open parentheses table row 4 row 2 row 0 end table close parentheses

  5. open parentheses table row 4 row 2 row 0 end table close parentheses

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N. Mustikowati

Master Teacher

Mahasiswa/Alumni Universitas Negeri Jakarta

Jawaban terverifikasi

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Pembahasan

P r o y e k s i space v e k t o r space o r t h o g o n a l space u ⃗ p a d a space v ⃗ equals fraction numerator u with rightwards arrow on top. space v with rightwards arrow on top over denominator vertical line v ⃗ vertical line squared end fraction. v ⃗  equals fraction numerator 7. left parenthesis negative 2 right parenthesis plus left parenthesis negative 4 right parenthesis. left parenthesis negative 1 right parenthesis plus 1.0 over denominator left parenthesis negative 2 right parenthesis squared plus left parenthesis negative 1 right parenthesis squared plus 0 squared end fraction open parentheses table row cell negative 2 end cell row cell negative 1 end cell row 0 end table close parentheses  equals fraction numerator negative 10 over denominator 5 end fraction open parentheses table row cell negative 2 end cell row cell negative 1 end cell row 0 end table close parentheses  equals negative 2 open parentheses table row cell negative 2 end cell row cell negative 1 end cell row 0 end table close parentheses  equals open parentheses table row 4 row 2 row 0 end table close parentheses

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wiwin nurhayati

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