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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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F. Kurnia

Master Teacher

Mahasiswa/Alumni Universitas Jember

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Pembahasan

P r o y e k s i space v e k t o r space u space ⃗ space p a d a space v space ⃗ space a d a l a h semicolon  fraction numerator u ⃗. v ⃗ over denominator vertical line v ⃗ vertical line squared end fraction. v ⃗  M a k a semicolon  fraction numerator u ⃗. v ⃗ over denominator vertical line v ⃗ vertical line squared end fraction. v ⃗ equals fraction numerator open parentheses table row 7 row cell negative 4 end cell row 1 end table close parentheses open parentheses table row cell negative 2 end cell row cell negative 1 end cell row 0 end table close parentheses over denominator open parentheses square root of open parentheses negative 2 close parentheses squared plus open parentheses negative 1 close parentheses squared plus 0 squared end root close parentheses end fraction. open parentheses table row cell negative 2 end cell row cell negative 1 end cell row 0 end table close parentheses  equals open parentheses fraction numerator negative 14 plus 4 plus 0 over denominator open parentheses square root of 5 close parentheses squared end fraction close parentheses. 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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