Diketahui titik-titik , dan . Jika merupakan proyeksi titik pada garis maka sama dengan ...

Pertanyaan

Diketahui titik-titik $text A end text open parentheses 1 comma space minus 3 comma space 2 close parentheses$ , $text B end text open parentheses 1 comma space 3 comma space 10 close parentheses$ dan $text C end text open parentheses 4 comma space minus 1 comma space 8 close parentheses$ . Jika $text D end text$ merupakan proyeksi titik $text C end text$ pada garis $text AB end text$ maka $stack text AD end text with rightwards arrow on top$ sama dengan ...

$36 stack text i end text with rightwards arrow on top plus 48 stack text j end text with rightwards arrow on top$
$stack text i end text with rightwards arrow on top plus 36 stack text j end text with rightwards arrow on top minus 48 stack text k end text with rightwards arrow on top$
$stack text i end text with rightwards arrow on top plus 36 stack text j end text with rightwards arrow on top plus 48 stack text k end text with rightwards arrow on top$
$36 stack text j end text with rightwards arrow on top plus 48 stack text k end text with rightwards arrow on top$
$36 stack text j end text with rightwards arrow on top minus 48 stack text k end text with rightwards arrow on top$

Pembahasan Soal:

Panjang vektor $a with rightwards arrow on top equals x i with rightwards arrow on top plus y j with rightwards arrow on top plus z k with rightwards arrow on top$ , yaitu $open vertical bar a with rightwards arrow on top close vertical bar equals square root of x squared plus y squared plus z squared end root$

Jika vektor $a with rightwards arrow on top equals a subscript 1 i with rightwards arrow on top plus a subscript 2 j with rightwards arrow on top plus a subscript 3 k with rightwards arrow on top$ dan $b with rightwards arrow on top equals b subscript 1 i with rightwards arrow on top plus b subscript 2 j with rightwards arrow on top plus b subscript 3 k with rightwards arrow on top$ , maka

$a with rightwards arrow on top times b with rightwards arrow on top equals a subscript 1 times b subscript 1 plus a subscript 2 times b subscript 2 plus a subscript 3 times b subscript 3$

Proyeksi vektor orthogonal vektor $a with rightwards arrow on top$ pada $b with rightwards arrow on top$ dapat ditentukan menggunakan rumus berikut.

$c with rightwards arrow on top equals fraction numerator a with rightwards arrow on top times b with rightwards arrow on top over denominator open vertical bar b with rightwards arrow on top close vertical bar squared end fraction times b with rightwards arrow on top$

Komponen vektor $stack A C with rightwards arrow on top$ dan $stack A B with rightwards arrow on top$ dapat ditentukan sebagai berikut.

$table attributes columnalign right center left columnspacing 0px end attributes row cell stack A C with rightwards arrow on top end cell equals cell text C-A end text end cell row blank equals cell open parentheses table row 4 row cell negative 1 end cell row 8 end table close parentheses minus open parentheses table row 1 row cell negative 3 end cell row 2 end table close parentheses end cell row blank equals cell open parentheses table row 3 row 2 row 6 end table close parentheses end cell end table$

$table attributes columnalign right center left columnspacing 0px end attributes row cell stack A B with rightwards arrow on top end cell equals cell text B-A end text end cell row blank equals cell open parentheses table row 1 row 3 row 10 end table close parentheses minus open parentheses table row 1 row cell negative 3 end cell row 2 end table close parentheses end cell row blank equals cell open parentheses table row 0 row 6 row 8 end table close parentheses end cell end table$

Proyeksi vektor $stack A C with rightwards harpoon with barb upwards on top$ pada $stack A B with rightwards arrow on top$ adalah sebagai berikut.

$table attributes columnalign right center left columnspacing 0px end attributes row cell stack A D with rightwards arrow on top end cell equals cell fraction numerator stack A C with rightwards arrow on top times stack A B with rightwards arrow on top over denominator open vertical bar stack A B with rightwards arrow on top close vertical bar squared end fraction times stack A B with rightwards arrow on top end cell row blank equals cell fraction numerator open parentheses table row 3 row 2 row 6 end table close parentheses open parentheses table row 0 row 6 row 8 end table close parentheses over denominator open parentheses square root of 0 squared plus 6 squared plus 8 squared end root close parentheses squared end fraction times open parentheses table row 0 row 6 row 8 end table close parentheses end cell row blank equals cell fraction numerator 3 times 0 plus 2 times 6 plus 6 times 8 over denominator open parentheses square root of 100 close parentheses squared end fraction times open parentheses table row 0 row 6 row 8 end table close parentheses end cell row blank equals cell 60 over 100 times open parentheses table row 0 row 6 row 8 end table close parentheses end cell row blank equals cell 3 over 5 times open parentheses table row 0 row 6 row 8 end table close parentheses end cell row blank equals cell times open parentheses table row 0 row cell 18 over 5 end cell row cell 24 over 5 end cell end table close parentheses end cell end table$

Diperoleh vektor $stack A D with rightwards arrow on top equals 18 over 5 j with rightwards arrow on top plus 24 over 5 k with rightwards arrow on top$

Oleh karena itu, tidak terdapat pilihan jawaban yang tepat.

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

H. Eka

Mahasiswa/Alumni Universitas Pendidikan Indonesia

Terakhir diupdate 11 Juli 2021

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$a with rightwards arrow on top times b with rightwards arrow on top equals a subscript 1 times b subscript 1 plus a subscript 2 times b subscript 2 plus a subscript 3 times b subscript 3$

Proyeksi vektor orthogonal vektor $a with rightwards arrow on top$ pada $b with rightwards arrow on top$ dapat ditentukan menggunakan rumus berikut.

Berdasarkan konsep tersebut dapat ditentukan penyelesaian soal sebagai berikut.

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sehingga

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Pembahasan Soal:

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$open vertical bar c with rightwards arrow on top close vertical bar equals open vertical bar fraction numerator a with rightwards arrow on top times b with rightwards arrow on top over denominator open vertical bar b with rightwards arrow on top close vertical bar end fraction close vertical bar$

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$table attributes columnalign right center left columnspacing 0px end attributes row cell stack A B with rightwards arrow on top end cell equals cell text B-A end text end cell row blank equals cell open parentheses table row 10 row 1 row 3 end table close parentheses minus open parentheses table row 2 row 1 row cell negative 3 end cell end table close parentheses end cell row blank equals cell open parentheses table row 8 row 0 row 6 end table close parentheses end cell end table$

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Oleh karena itu, jawaban yang tepat adalah C.

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$c with rightwards arrow on top space a d a l a h space p r o y e k s i space v e k t o r space o r t h o g o n a l space b with rightwards arrow on top space p a d a space a with rightwards arrow on top comma c with ⃗ on top equals fraction numerator a with ⃗ on top straight space. b with ⃑ on top over denominator open vertical bar a with ⃗ on top close vertical bar squared end fraction. a with ⃗ on top equals fraction numerator open parentheses table row 1 row cell negative 2 end cell row 1 end table close parentheses straight space. open parentheses table row 3 row 1 row cell negative 2 end cell end table close parentheses over denominator open parentheses 1 close parentheses squared plus open parentheses negative 2 close parentheses squared plus open parentheses 1 close parentheses squared end fraction. open parentheses table row 1 row cell negative 2 end cell row 1 end table close parentheses equals fraction numerator open parentheses 1 close parentheses. open parentheses 3 close parentheses plus open parentheses negative 2 close parentheses. open parentheses 1 close parentheses plus 1 open parentheses negative 2 close parentheses over denominator 6 end fraction. open parentheses table row 1 row cell negative 2 end cell row 1 end table close parentheses equals fraction numerator negative 1 over denominator 6 end fraction open parentheses table row 1 row cell negative 2 end cell row 1 end table close parentheses a t a u space c with ⃗ on top equals fraction numerator negative 1 over denominator 6 end fraction open parentheses i minus 2 j plus k close parentheses$

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