Segitiga ABC dengan titik koordinat A(2, −1, −3), B(−1, 1, −11), dan C(4, −3, −2). p​ adalah proyeksi vektor ortogonal dari vektor AB pada AC. Vektor p​ dalam i, j​, dan k adalah ....

Pertanyaan

Segitiga ABC dengan titik koordinat straight A open parentheses 2 comma space minus 1 comma space minus 3 close parenthesesstraight B open parentheses negative 1 comma space 1 comma space minus 11 close parentheses, dan straight C open parentheses 4 comma space minus 3 comma space minus 2 close parenthesesp with rightwards arrow on top adalah proyeksi vektor ortogonal dari vektor AB with rightwards arrow on top pada AC with rightwards arrow on top. Vektor p with rightwards arrow on top dalam i with hat on top comma space j with hat on top comma space dan space k with hat on top adalah ....

  1. negative 12 i with hat on top plus 12 j with hat on top minus 6 k with hat on top

  2. negative 6 i with hat on top plus 4 j with hat on top minus 16 k with hat on top

  3. negative 4 i with hat on top plus 4 j with hat on top minus 2 k with hat on top

  4. negative 6 i with hat on top minus 4 j with hat on top plus 16 k with hat on top

  5. 12 i with hat on top minus 12 j with hat on top plus 6 k with hat on top

D. Rajib

Master Teacher

Mahasiswa/Alumni Universitas Muhammadiyah Malang

Jawaban terverifikasi

Jawaban

jawaban yang benar adalah C.

Pembahasan

Jawaban yang benar untuk pertanyaan tersebut adalah C.

Diketahui segitiga ABC dengan titik koordinat straight A open parentheses 2 comma space minus 1 comma space minus 3 close parenthesesstraight B open parentheses negative 1 comma space 1 comma space minus 11 close parentheses, dan straight C open parentheses 4 comma space minus 3 comma space minus 2 close parentheses, serta p with rightwards arrow on top adalah proyeksi vektor ortogonal dari vektor AB with rightwards arrow on top pada AC with rightwards arrow on top. Ingat bahwa, table attributes columnalign right center left columnspacing 0px end attributes row cell p with rightwards arrow on top end cell equals cell fraction numerator stack A B with rightwards arrow on top times AC with rightwards arrow on top over denominator open vertical bar AC with rightwards arrow on top close vertical bar squared end fraction times AC with rightwards arrow on top end cell end table

table attributes columnalign right center left columnspacing 0px end attributes row cell p with rightwards arrow on top end cell equals cell fraction numerator stack A B with rightwards arrow on top times AC with rightwards arrow on top over denominator open vertical bar AC with rightwards arrow on top close vertical bar squared end fraction times AC with rightwards arrow on top end cell row blank equals cell fraction numerator open parentheses B minus A close parentheses times open parentheses C minus A close parentheses over denominator open vertical bar open parentheses C minus A close parentheses close vertical bar squared end fraction times open parentheses C minus A close parentheses end cell row blank equals cell fraction numerator open square brackets open parentheses table row cell negative 1 end cell row 1 row cell negative 11 end cell end table close parentheses minus open parentheses table row 2 row cell negative 1 end cell row cell negative 3 end cell end table close parentheses close square brackets times open square brackets open parentheses table row 4 row cell negative 3 end cell row cell negative 2 end cell end table close parentheses minus open parentheses table row 2 row cell negative 1 end cell row cell negative 3 end cell end table close parentheses close square brackets over denominator open vertical bar open square brackets open parentheses table row 4 row cell negative 3 end cell row cell negative 2 end cell end table close parentheses minus open parentheses table row 2 row cell negative 1 end cell row cell negative 3 end cell end table close parentheses close square brackets close vertical bar squared end fraction open square brackets open parentheses table row 4 row cell negative 3 end cell row cell negative 2 end cell end table close parentheses minus open parentheses table row 2 row cell negative 1 end cell row cell negative 3 end cell end table close parentheses close square brackets end cell row blank equals cell fraction numerator open parentheses table row cell negative 3 end cell row 2 row cell negative 8 end cell end table close parentheses times open parentheses table row 2 row cell negative 2 end cell row 1 end table close parentheses over denominator open vertical bar open parentheses table row 2 row cell negative 2 end cell row 1 end table close parentheses close vertical bar squared end fraction open parentheses table row 2 row cell negative 2 end cell row 1 end table close parentheses end cell row blank equals cell fraction numerator negative 6 minus 4 minus 8 over denominator 2 squared plus open parentheses negative 2 close parentheses squared plus 1 squared end fraction open parentheses table row 2 row cell negative 2 end cell row 1 end table close parentheses end cell row blank equals cell fraction numerator negative 18 over denominator 9 end fraction open parentheses table row 2 row cell negative 2 end cell row 1 end table close parentheses end cell row blank equals cell open parentheses table row cell negative 4 end cell row 4 row cell negative 2 end cell end table close parentheses end cell end table

maka, vektor p with rightwards arrow on top dalam i with hat on top comma space j with hat on top comma space dan space k with hat on top adalah negative 4 i with hat on top plus 4 j with hat on top minus 2 k with hat on top

Oleh karena itu, jawaban yang benar adalah C.

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Diketahui koordinat A(−4, 2, 3), B(7, 8, −1), dan C(1, 0, 7). Jika AB wakil dari vektor u, AC wakil dari vektor v, maka proyeksi vektor ortogonal dari u pada v adalah ....

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