Roboguru

diketahuivektoru=⎝⎛​7−41​⎠⎞​danv=⎝⎛​−210​⎠⎞​.proyeksivektororthogonaludanvadalah...

Pertanyaan

d i k e t a h u i space v e k t o r space u with rightwards arrow on top equals open parentheses table row 7 row cell negative 4 end cell row 1 end table close parentheses space d a n space v with rightwards arrow on top equals open parentheses table row cell negative 2 end cell row 1 row 0 end table close parentheses. 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 stack u space with rightwards arrow on top d a n space stack v space with rightwards arrow on top a d a l a h...

  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

Pembahasan Soal:

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 ⃗ p a d a space v ⃗ equals fraction numerator u with rightwards arrow on top. space v with rightwards arrow on top over denominator vertical line v ⃗ vertical line squared end fraction. v ⃗  equals fraction numerator 7. left parenthesis negative 2 right parenthesis plus left parenthesis negative 4 right parenthesis. left parenthesis negative 1 right parenthesis plus 1.0 over denominator left parenthesis negative 2 right parenthesis squared plus left parenthesis negative 1 right parenthesis squared plus 0 squared end fraction open parentheses table row cell negative 2 end cell row cell negative 1 end cell row 0 end table close parentheses  equals fraction numerator negative 10 over denominator 5 end fraction open parentheses table row cell negative 2 end cell row cell negative 1 end cell row 0 end table close parentheses  equals negative 2 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

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

N. Mustikowati

Mahasiswa/Alumni Universitas Negeri Jakarta

Terakhir diupdate 04 Oktober 2021

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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Pertanyaan yang serupa

Diketahui segitiga ABC dengan koordinat titik A(5,−3,−6), B(1,1,1), dan C(−2,4,−6). Tentukan: a. proyeksi vektor AB⇀ pada AC⇀; b. tinggi segitiga ; dan c. luas segitiga .

Pembahasan Soal:

Perhatikan perhitungan berikut.

Ingat, proyeksi vektor straight A with rightwards harpoon with barb upwards on top pada straight B with rightwards harpoon with barb upwards on top:

Proyeksi space vektor equals fraction numerator straight A with rightwards harpoon with barb upwards on top times straight B with rightwards harpoon with barb upwards on top over denominator open vertical bar straight B with rightwards harpoon with barb upwards on top close vertical bar squared end fraction times straight B with rightwards harpoon with barb upwards on top

Perhatikan gambar berikut.

a. proyeksi vektor AB with rightwards harpoon with barb upwards on top pada AC with rightwards harpoon with barb upwards on top

table attributes columnalign right center left columnspacing 0px end attributes row cell AB with rightwards harpoon with barb upwards on top end cell equals cell straight B with rightwards harpoon with barb upwards on top minus straight A with rightwards harpoon with barb upwards on top end cell row blank equals cell open parentheses table row 1 row 1 row 1 end table close parentheses minus open parentheses table row 5 row cell negative 3 end cell row cell negative 6 end cell end table close parentheses end cell row blank equals cell open parentheses table row cell 1 minus 5 end cell row cell 1 minus open parentheses negative 3 close parentheses end cell row cell 1 minus open parentheses negative 6 close parentheses end cell end table close parentheses end cell row blank equals cell open parentheses table row cell negative 4 end cell row cell 1 plus 3 end cell row cell 1 plus 6 end cell end table close parentheses end cell row blank equals cell open parentheses table row cell negative 4 end cell row 4 row 7 end table close parentheses end cell end table

table attributes columnalign right center left columnspacing 0px end attributes row cell AC with rightwards harpoon with barb upwards on top end cell equals cell straight C with rightwards harpoon with barb upwards on top minus straight A with rightwards harpoon with barb upwards on top end cell row blank equals cell open parentheses table row cell negative 2 end cell row 4 row cell negative 6 end cell end table close parentheses minus open parentheses table row 5 row cell negative 3 end cell row cell negative 6 end cell end table close parentheses end cell row blank equals cell open parentheses table row cell negative 2 minus 5 end cell row cell 4 minus open parentheses negative 3 close parentheses end cell row cell negative 6 minus open parentheses negative 6 close parentheses end cell end table close parentheses end cell row blank equals cell open parentheses table row cell negative 7 end cell row cell 4 plus 3 end cell row cell negative 6 plus 6 end cell end table close parentheses end cell row blank equals cell open parentheses table row cell negative 7 end cell row 7 row 0 end table close parentheses end cell end table

Misalkan AD with rightwards harpoon with barb upwards on top merupakan proyeksi vektor vektor AB with rightwards harpoon with barb upwards on top pada AC with rightwards harpoon with barb upwards on top, maka:

table attributes columnalign right center left columnspacing 0px end attributes row cell AD with rightwards harpoon with barb upwards on top end cell equals cell fraction numerator AB with rightwards harpoon with barb upwards on top times AC with rightwards harpoon with barb upwards on top over denominator open vertical bar AC with rightwards harpoon with barb upwards on top close vertical bar squared end fraction times AC with rightwards harpoon with barb upwards on top end cell row blank equals cell fraction numerator negative 4 times negative 7 plus 4 times 7 plus 7 times 0 over denominator open parentheses square root of negative 7 squared plus 7 squared plus 0 squared end root close parentheses squared end fraction times open parentheses negative 7 comma space 7 comma space 0 close parentheses end cell row blank equals cell fraction numerator 28 plus 28 plus 0 over denominator open parentheses square root of 49 plus 49 plus 0 end root close parentheses squared end fraction times open parentheses negative 7 comma space 7 comma space 0 close parentheses end cell row blank equals cell 56 over open parentheses square root of 98 close parentheses squared times open parentheses negative 7 comma space 7 comma space 0 close parentheses end cell row blank equals cell 56 over 98 times open parentheses negative 7 comma space 7 comma space 0 close parentheses end cell row blank equals cell open parentheses fraction numerator negative 7 times 56 over denominator 98 end fraction comma space fraction numerator 7 times 56 over denominator 98 end fraction comma space fraction numerator 0 times 56 over denominator 98 end fraction close parentheses end cell row blank equals cell open parentheses negative 4 comma space 4 comma space 0 close parentheses end cell end table

Jadi, proyeksi vektor vektor AB with rightwards harpoon with barb upwards on top pada AC with rightwards harpoon with barb upwards on top adalah AD with rightwards harpoon with barb upwards on top equals open parentheses negative 4 comma space 4 comma space 0 close parentheses.

b. tinggi segitiga ABC

Diketahui AB with rightwards harpoon with barb upwards on top equals open parentheses negative 4 comma space 4 comma space 7 close parentheses dan AD with rightwards harpoon with barb upwards on top equals open parentheses negative 4 comma space 4 comma space 0 close parentheses, maka untuk mencari tinggi segitiga tersebut dapat menggunakan teorema Pythagoras dengan terlebih dahulu mencari panjang AB with rightwards harpoon with barb upwards on top dan AD with rightwards harpoon with barb upwards on top, yaitu:

table attributes columnalign right center left columnspacing 0px end attributes row cell open vertical bar AB with rightwards harpoon with barb upwards on top close vertical bar end cell equals cell square root of negative 4 squared plus 4 squared plus 7 end root end cell row blank equals cell square root of 16 plus 16 plus 49 end root end cell row blank equals cell square root of 81 end cell row blank equals 9 end table

table attributes columnalign right center left columnspacing 0px end attributes row cell open vertical bar AD with rightwards harpoon with barb upwards on top close vertical bar end cell equals cell square root of negative 4 squared plus 4 squared plus 0 squared end root end cell row blank equals cell square root of 16 plus 16 plus 0 end root end cell row blank equals cell square root of 32 end cell row blank equals cell square root of 16 times 2 end root end cell row blank equals cell 4 square root of 2 end cell end table

Sehingga tinggi segitiga tersebut yaitu:

table attributes columnalign right center left columnspacing 0px end attributes row t equals cell square root of 9 squared minus open parentheses 4 square root of 2 close parentheses squared end root end cell row blank equals cell square root of 81 minus 16 times 2 end root end cell row blank equals cell square root of 81 minus 32 end root end cell row blank equals cell square root of 49 end cell row blank equals 7 end table

Didapatkan panjang tinggi segitiga tersebut adalah 7 satuan.

c. luas segitiga ABC

Terlebih dahulu dicari panjang AC with rightwards harpoon with barb upwards on top, yaitu:

table attributes columnalign right center left columnspacing 0px end attributes row cell open vertical bar AC with rightwards harpoon with barb upwards on top close vertical bar end cell equals cell square root of negative 7 squared plus 7 squared plus 0 squared end root end cell row blank equals cell square root of 49 plus 49 plus 0 end root end cell row blank equals cell square root of 98 end cell row blank equals cell square root of 2 times 49 end root end cell row blank equals cell 7 square root of 2 end cell end table

Maka:

table attributes columnalign right center left columnspacing 0px end attributes row Luas equals cell fraction numerator open vertical bar AC with rightwards harpoon with barb upwards on top close vertical bar times straight t over denominator 2 end fraction end cell row blank equals cell fraction numerator 7 square root of 2 times 7 over denominator 2 end fraction end cell row blank equals cell 49 over 2 square root of 2 end cell end table

Jadi, luas segitiga tersebut adalah 49 over 2 square root of 2 satuan persegi.

0

Roboguru

Perhatikan gambar berikut!    Proyeksi vektor ortogonal AC pada AB adalah ...

Pembahasan Soal:

Berdasarkan gambar AB with rightwards arrow on top equals open parentheses table row 6 row cell negative 2 end cell end table close parentheses dan AC with rightwards arrow on top equals open parentheses table row 4 row 2 end table close parentheses, maka 

table attributes columnalign right center left columnspacing 0px end attributes row cell AC with rightwards arrow on top times AB with rightwards arrow on top end cell equals cell open parentheses table row 6 row cell negative 2 end cell end table close parentheses times open parentheses table row 4 row 2 end table close parentheses end cell row blank equals cell 24 minus 4 end cell row blank equals 20 row blank blank blank row cell open vertical bar AB with rightwards arrow on top close vertical bar end cell equals cell square root of 6 squared plus open parentheses negative 2 close parentheses squared end root end cell row blank equals cell square root of 36 plus 4 end root end cell row blank equals cell square root of 40 end cell row blank equals cell 2 square root of 10 end cell end table 

 Sehingga proyeksi vektor ortogonal AC with rightwards arrow on top pada AB with rightwards arrow on top adalah 

table attributes columnalign right center left columnspacing 0px end attributes row cell straight c with rightwards arrow on top end cell equals cell open parentheses fraction numerator AC with rightwards arrow on top times AB with rightwards arrow on top over denominator open vertical bar AB with rightwards arrow on top close vertical bar squared end fraction close parentheses times AB with rightwards arrow on top end cell row blank equals cell open parentheses 20 over open parentheses 2 square root of 10 close parentheses squared close parentheses times open parentheses table row 6 row cell negative 2 end cell end table close parentheses end cell row blank equals cell open parentheses 20 over 40 close parentheses times open parentheses table row 6 row cell negative 2 end cell end table close parentheses end cell row blank equals cell 1 half times open parentheses table row 6 row cell negative 2 end cell end table close parentheses end cell row blank equals cell open parentheses table row 3 row cell negative 1 end cell end table close parentheses end cell end table 

Dengan demikian, jawaban yang tepat adalah C.

0

Roboguru

Diketahui titik-titik A(1,−3,2), B(1,3,10) dan C(4,−1,8). Jika D merupakan proyeksi titik C pada garis AB maka AD sama dengan ...

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.

0

Roboguru

Diketahui a=2i−3j​+6k dan b=i+5j​+3k, tentukanlah: a. proyeksi vektor orthogonal a pada b

Pembahasan Soal:

Vektor a with rightwards arrow on top equals open parentheses 2 comma negative 3 comma 6 close parentheses dan vektor b with rightwards arrow on top equals open parentheses 1 comma 5 comma 3 close parentheses.

Proyeksi vektor orthogonal a with rightwards arrow on top pada b with rightwards arrow on top sebagai berikut:

table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses fraction numerator a with rightwards harpoon with barb upwards on top times b with rightwards harpoon with barb upwards on top over denominator open vertical bar b with rightwards harpoon with barb upwards on top close vertical bar squared end fraction close parentheses times b with rightwards harpoon with barb upwards on top end cell equals cell open parentheses fraction numerator left parenthesis 2 comma negative 3 comma 6 right parenthesis times left parenthesis 1 comma 5 comma 3 right parenthesis over denominator open parentheses square root of 1 squared plus 5 squared plus 3 squared end root close parentheses squared end fraction close parentheses times left parenthesis 1 comma 5 comma 3 right parenthesis end cell row blank equals cell open parentheses fraction numerator open parentheses 2 times 1 close parentheses plus open parentheses negative 3 times 5 close parentheses plus open parentheses 6 times 3 close parentheses over denominator left parenthesis 1 plus 25 plus 9 right parenthesis end fraction close parentheses times left parenthesis 1 comma 5 comma 3 right parenthesis end cell row blank equals cell open parentheses fraction numerator 2 minus 15 plus 18 over denominator 35 end fraction close parentheses times left parenthesis 1 comma 5 comma 3 right parenthesis end cell row blank equals cell open parentheses 5 over 35 close parentheses times left parenthesis 1 comma 5 comma 3 right parenthesis end cell row blank equals cell 1 over 7 times left parenthesis 1 comma 5 comma 3 right parenthesis end cell row blank equals cell open parentheses 1 over 7 comma 5 over 7 comma 3 over 7 close parentheses end cell end table

Jadi, proyeksi vektor orthogonal a with rightwards arrow on top pada b with rightwards arrow on top adalah open parentheses 1 over 7 comma 5 over 7 comma 3 over 7 close parentheses.

0

Roboguru

Deketahui segitiga ABCD dengan koordinat A=(2,−1,−1), B(−1,4,−2) dan C(5,0,−3). Proyeksi vektor orthognal AB pada AC adalah ...

Pembahasan Soal:

Ingat konsep proyeksi vektor orthognal AB with rightwards arrow on top pada AC with rightwards arrow on top adalah 

Proyeksi space vektor space AC with rightwards arrow on top equals fraction numerator stack AB times with rightwards arrow on top AC with rightwards arrow on top over denominator open vertical bar AC with rightwards arrow on top close vertical bar squared end fraction stack times AC with rightwards arrow on top  

Diketahui straight A equals open parentheses 2 comma space minus 1 comma space minus 1 close parenthesesstraight B open parentheses negative 1 comma space 4 comma space minus 2 close parentheses dan straight C open parentheses 5 comma space 0 comma space minus 3 close parentheses maka:

table attributes columnalign right center left columnspacing 0px end attributes row cell AB with rightwards arrow on top end cell equals cell straight B with rightwards arrow on top minus straight A with rightwards arrow on top end cell row blank equals cell open parentheses negative 1 comma space 4 comma space minus 2 close parentheses minus open parentheses 2 comma space minus 1 comma space minus 1 close parentheses end cell row blank equals cell open parentheses negative 3 comma space 5 comma space minus 1 close parentheses end cell row cell AC with rightwards arrow on top end cell equals cell straight C with rightwards arrow on top minus straight A with rightwards arrow on top end cell row blank equals cell open parentheses 5 comma space 0 comma space minus 3 close parentheses minus open parentheses 2 comma space minus 1 comma space minus 1 close parentheses end cell row blank equals cell open parentheses 3 comma space 1 comma space minus 2 close parentheses end cell end table

maka 

table attributes columnalign right center left columnspacing 0px end attributes row cell stack AB times with rightwards arrow on top AC with rightwards arrow on top end cell equals cell open parentheses negative 3 comma space 5 comma space minus 1 close parentheses times open parentheses 3 comma space 1 comma space minus 2 close parentheses end cell row blank equals cell negative 9 plus 5 plus 2 end cell row blank equals cell negative 2 end cell row cell open vertical bar AC with rightwards arrow on top close vertical bar end cell equals cell square root of 3 squared plus 1 squared plus open parentheses negative 2 close parentheses squared end root end cell row blank equals cell square root of 9 plus 1 plus 4 end root end cell row blank equals cell square root of 14 end cell end table 

sehingga

table attributes columnalign right center left columnspacing 0px end attributes row cell Proyeksi space vektor space AC with rightwards arrow on top end cell equals cell fraction numerator AB 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 negative 2 over denominator open parentheses square root of 14 close parentheses squared end fraction open parentheses 3 comma space 1 comma space minus 2 close parentheses end cell row blank equals cell fraction numerator negative 2 over denominator 14 end fraction open parentheses 3 comma space 1 comma space minus 2 close parentheses end cell row blank equals cell negative 1 over 7 open parentheses 3 comma space 1 comma space minus 2 close parentheses end cell row blank equals cell open parentheses negative 3 over 7 comma space minus 1 over 7 comma space 2 over 7 close parentheses end cell end table  

Dengan demikian proyeksi vektor orthognal AB with rightwards arrow on top pada AC with rightwards arrow on top adalah table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell open parentheses negative 3 over 7 comma space minus 1 over 7 comma space 2 over 7 close parentheses end cell end table.

0

Roboguru

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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