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Pertanyaan

D ik e t ah u i v e k t or u = ⎝ ⎛ ​ − 4 4 3 ​ ⎠ ⎞ ​ d an v = ⎝ ⎛ ​ − 3 − 6 0 ​ ⎠ ⎞ ​ . P roye k s i v e k t or u p a d a v a d a l ah ....

  1. 4 over 5 i minus 8 over 5 j

  2. negative 4 over 5 i minus 8 over 5 j

  3. 4 over 5 i plus 8 over 5 j

  4. 4 over 5 i minus 8 over 5 j plus 4 over 5 k

  5. negative 4 over 5 i minus 8 over 5 j plus 4 over 5 k

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F. Pratama

Master Teacher

Mahasiswa/Alumni Universitas Putra Indonesia YPTK Padang

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 with rightwards arrow on top space p a d a space v with rightwards arrow on top space a d a l a h space space fraction numerator u with rightwards arrow on top space space v with rightwards arrow on top over denominator open vertical bar space v with rightwards arrow on top close vertical bar squared end fraction space v with rightwards arrow on top  J a d i comma space p r o y e k s i space o r t h o g o n a l space v e k t o r space u with rightwards arrow on top space equals open parentheses table row cell negative 4 end cell row 4 row 3 end table close parentheses space p a d a space v e k t o r space space v with rightwards arrow on top equals open parentheses table row cell negative 3 end cell row cell negative 6 end cell row 0 end table close parentheses  equals fraction numerator open parentheses table row cell negative 4 end cell row 4 row 3 end table close parentheses space open parentheses table row cell negative 3 end cell row cell negative 6 end cell row 0 end table close parentheses over denominator open parentheses square root of left parenthesis negative 3 right parenthesis squared plus left parenthesis negative 6 right parenthesis squared plus 0 squared end root close parentheses squared end fraction open parentheses table row cell negative 3 end cell row cell negative 6 end cell row 0 end table close parentheses  equals fraction numerator 12 minus 24 plus 0 over denominator open parentheses square root of 9 plus 36 end root close parentheses squared end fraction open parentheses table row cell negative 3 end cell row cell negative 6 end cell row 0 end table close parentheses  equals negative 12 over 45 open parentheses table row cell negative 3 end cell row cell negative 6 end cell row 0 end table close parentheses  equals negative 4 over 15 open parentheses table row cell negative 3 end cell row cell negative 6 end cell row 0 end table close parentheses  equals open parentheses table row cell 4 over 5 end cell row cell 8 over 5 end cell row 0 end table close parentheses  equals 4 over 5 i with rightwards arrow on top plus 8 over 5 j with rightwards arrow on top

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Tentukan proyeksi ortogonal vektor r = 4 i − 2 j pada vektor s = 2 i − j .

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