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Diketahui vektor a = 5 i + 6 j ​ + k dan b = i − 2 j ​ − 2 k . Proyeksi ortogonal vektor a pada b adalah .... (UN 2012)

Diketahui vektor  dan . Proyeksi ortogonal vektor  pada  adalah ....(UN 2012)

  1. i with rightwards arrow on top plus 2 j with rightwards arrow on top plus 2 k with rightwards arrow on top

  2. i with rightwards arrow on top plus 2 j with rightwards arrow on top minus 2 k with rightwards arrow on top

  3. i with rightwards arrow on top minus 2 j with rightwards arrow on top plus 2 k with rightwards arrow on top

  4. i with rightwards arrow on top minus 2 j with rightwards arrow on top minus 2 k with rightwards arrow on top

  5. negative i with rightwards arrow on top plus 2 j with rightwards arrow on top plus 2 k with rightwards arrow on top

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A. Acfreelance

Master Teacher

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Pembahasan

Diketahui dan Misal, adalah Proyeksi ortogonal vektor pada b , maka :

Diketahui a with rightwards arrow on top equals 5 i with rightwards arrow on top plus 6 j with rightwards arrow on top plus k with rightwards arrow on top dan b with rightwards arrow on top equals i with rightwards arrow on top minus 2 j with rightwards arrow on top minus 2 k with rightwards arrow on top

Misal, c with rightwards arrow on topadalah Proyeksi ortogonal vektor a with rightwards arrow on top pada b with rightwards arrow on top , maka :

c with rightwards arrow on top equals 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 harpoon with barb upwards on top equals open parentheses fraction numerator 5.1 plus 6. left parenthesis negative 2 right parenthesis plus 1. left parenthesis negative 2 right parenthesis over denominator open parentheses square root of space 1 squared plus left parenthesis negative 2 right parenthesis squared plus left parenthesis negative 2 right parenthesis squared space end root close parentheses squared end fraction close parentheses space left parenthesis i with rightwards arrow on top minus 2 j with rightwards arrow on top minus 2 k with rightwards arrow on top right parenthesis

c with rightwards arrow on top equals fraction numerator 5 minus 12 minus 2 over denominator open parentheses square root of 9 close parentheses squared end fraction space left parenthesis i with rightwards arrow on top minus 2 j with rightwards arrow on top minus 2 k with rightwards arrow on top right parenthesis

c with rightwards arrow on top equals fraction numerator negative 9 over denominator 9 end fraction space left parenthesis i with rightwards arrow on top minus 2 j with rightwards arrow on top minus 2 k with rightwards arrow on top right parenthesis

c with rightwards arrow on top equals negative 1 space left parenthesis i with rightwards arrow on top minus 2 j with rightwards arrow on top minus 2 k with rightwards arrow on top right parenthesis

c with rightwards arrow on top equals negative i with rightwards arrow on top plus 2 j with rightwards arrow on top plus 2 k with rightwards arrow on top

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Wayan Dira

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Diketahui vektor a = 4 i − 2 j ​ + 2 k dan vektor b = 2 i − 6 j ​ + 4 k . Proyeksi vektor ortogonal vektor a pada b adalah .... (UN 2011)

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