Jika vector u=(3,−4) dan v=(5,2), maka proyeksi scalar orthogonal vector (u−v) pada vector u+v adalah ...

Pertanyaan

Jika vector $u=(3,-4)$ dan $v=(5,2)$ , maka proyeksi scalar orthogonal vector $(u-v)$ pada vector $u+v$ adalah ...

Pembahasan Soal:

Proyeksi skalar ortogonal suatu vektor pada vektor lain adalah :

Proyeksi vektor orthogonal $v with rightwards arrow on top$ pada $u with rightwards arrow on top space equals fraction numerator u with rightwards arrow on top times v with rightwards arrow on top over denominator open vertical bar u with rightwards harpoon with barb upwards on top close vertical bar end fraction$
Proyeksi vektor orthogonal $u with rightwards arrow on top$ pada $v with rightwards arrow on top space equals fraction numerator v with rightwards arrow on top times u with rightwards arrow on top over denominator open vertical bar v with rightwards arrow on top close vertical bar end fraction$

Diketahui vector $u equals left parenthesis 3 comma negative 4 right parenthesis$ dan $v equals left parenthesis 5 comma 2 right parenthesis$ , cari dulu vector $left parenthesis u minus v right parenthesis$ dan vector $u plus v$ ,

$table attributes columnalign right center left columnspacing 0px end attributes row cell left parenthesis u minus v right parenthesis end cell equals cell open square brackets table row 3 row cell negative 4 end cell end table close square brackets minus open square brackets table row 5 row 2 end table close square brackets end cell row blank equals cell open square brackets table row cell 3 minus 5 end cell row cell negative 4 minus 2 end cell end table close square brackets end cell row blank equals cell open square brackets table row cell negative 2 end cell row cell negative 6 end cell end table close square brackets end cell end table$

$table attributes columnalign right center left columnspacing 0px end attributes row cell left parenthesis u plus v right parenthesis end cell equals cell open square brackets table row 3 row cell negative 4 end cell end table close square brackets plus open square brackets table row 5 row 2 end table close square brackets end cell row blank equals cell open square brackets table row cell 3 plus 5 end cell row cell negative 4 plus 2 end cell end table close square brackets end cell row blank equals cell open square brackets table row 8 row cell negative 2 end cell end table close square brackets end cell end table$

Maka di dapatkan $u plus v equals left parenthesis 8 comma negative 2 right parenthesis$ dan $u minus v equals left parenthesis negative 2 comma negative 6 right parenthesis$ .

Proyeksi scalar orthogonal vector $left parenthesis u minus v right parenthesis$ pada vector $u plus v$ ,

$table attributes columnalign right center left columnspacing 0px end attributes row blank equals cell fraction numerator open parentheses u plus v close parentheses open parentheses u minus v close parentheses over denominator open vertical bar u minus v close vertical bar end fraction end cell row blank equals cell fraction numerator open square brackets table row 8 row cell negative 2 end cell end table close square brackets times open square brackets table row cell negative 2 end cell row cell negative 6 end cell end table close square brackets over denominator square root of left parenthesis negative 2 right parenthesis squared plus left parenthesis negative 6 right parenthesis squared end root end fraction end cell row blank equals cell fraction numerator open square brackets table row cell 8 times negative 2 end cell row cell negative 2 times negative 6 end cell end table close square brackets over denominator square root of 40 end fraction end cell row blank equals cell fraction numerator open square brackets table row cell negative 16 end cell row 12 end table close square brackets over denominator square root of 40 end fraction end cell row blank equals cell fraction numerator open square brackets table row cell negative 16 end cell row 12 end table close square brackets over denominator square root of 4 cross times 10 end root end fraction end cell row blank equals cell fraction numerator open square brackets table row cell negative 16 end cell row 12 end table close square brackets over denominator 2 square root of 10 end fraction end cell row blank equals cell open square brackets table row cell fraction numerator negative 16 over denominator 2 square root of 10 end fraction end cell row cell fraction numerator 12 over denominator 2 square root of 10 end fraction end cell end table close square brackets space rasionalkan end cell row blank equals cell open square brackets table row cell fraction numerator negative 16 over denominator 2 square root of 10 end fraction cross times fraction numerator square root of 10 over denominator square root of 10 end fraction end cell row cell fraction numerator 12 over denominator 2 square root of 10 end fraction cross times fraction numerator square root of 10 over denominator square root of 10 end fraction end cell end table close square brackets end cell row blank equals cell open square brackets table row cell fraction numerator negative 16 square root of 10 over denominator 2 times 10 end fraction end cell row cell fraction numerator 12 square root of 10 over denominator 2 times 10 end fraction end cell end table close square brackets end cell row blank equals cell open square brackets table row cell fraction numerator negative 16 square root of 10 over denominator 20 end fraction end cell row cell fraction numerator 12 square root of 10 over denominator 20 end fraction end cell end table close square brackets space sederhanakan end cell row blank equals cell open square brackets table row cell negative 4 over 5 square root of 10 end cell row cell 3 over 5 square root of 10 end cell end table close square brackets end cell end table$

Jadi, proyeksi scalar orthogonal vector $left parenthesis u minus v right parenthesis$ pada vector $u plus v$ adalah $open parentheses negative 4 over 5 square root of 10 comma space 3 over 5 square root of 10 close parentheses$ .

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

G. Albiah

Mahasiswa/Alumni Universitas Galuh Ciamis

Terakhir diupdate 06 Oktober 2021

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$table attributes columnalign right center left columnspacing 0px end attributes row cell open vertical bar c close vertical bar end cell equals cell fraction numerator a times b over denominator open vertical bar b close vertical bar end fraction end cell row 2 equals cell fraction numerator open parentheses table row x row y end table close parentheses open parentheses table row 6 row 8 end table close parentheses over denominator square root of 6 squared plus 8 squared end root end fraction end cell row 2 equals cell fraction numerator 6 x plus 8 y over denominator 10 end fraction end cell row 20 equals cell 6 x plus 8 y end cell row 10 equals cell 3 x plus 4 y... left parenthesis 1 right parenthesis end cell end table$

Selanjutnya diketahui vektor $bold italic a$ dengan panjang $square root of 5 space$ , maka

$open vertical bar a close vertical bar equals square root of 5 square root of x squared plus y squared end root equals square root of 5 x squared plus y squared equals 5... left parenthesis 2 right parenthesis$

Substitusi persamaan (1) ke persamaan (2),

$3 x plus 4 y equals 10 rightwards arrow x equals fraction numerator 10 minus 4 y over denominator 3 end fraction x squared plus y squared equals 5 open parentheses fraction numerator 10 minus 4 y over denominator 3 end fraction close parentheses squared plus y squared equals 5 fraction numerator 100 minus 80 y plus 16 y squared over denominator 9 end fraction plus y squared equals 5 space left parenthesis k a l i space 9 right parenthesis 16 y squared minus 80 y plus 100 plus 9 y squared equals 45 25 y squared minus 80 y plus 55 equals 0 5 y squared minus 16 y plus 11 equals 0 left parenthesis 5 y minus 11 right parenthesis left parenthesis y minus 1 right parenthesis equals 0 y subscript 1 equals 11 over 5 space atau space y subscript 2 equals 1 x subscript 1 equals fraction numerator 10 minus 4 open parentheses begin display style 11 over 5 end style close parentheses over denominator 3 end fraction equals 2 over 5 x subscript 2 equals fraction numerator 10 minus 4 left parenthesis 1 right parenthesis over denominator 3 end fraction equals 2$

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Proyeksi skalar orthogonal a=(m,−18,6) pada b=(−3,0,4) adalah 6. Nilai m yang memenuhi adalah...

Pembahasan Soal:

Beberapa konsep yang dipakai :

$table attributes columnalign right center left columnspacing 0px end attributes row u equals cell open parentheses u subscript 1 comma u subscript 2 comma u subscript 3 comma..... comma u subscript n close parentheses end cell row v equals cell open parentheses v subscript 1 comma v subscript 2 comma v subscript 3 comma..... comma v subscript n close parentheses end cell row cell u times v end cell equals cell u subscript 1 v subscript 1 plus u subscript 2 v subscript 2 plus u subscript 3 v subscript 3 plus... plus u subscript n v subscript n end cell row cell open vertical bar u close vertical bar end cell equals cell square root of open parentheses u subscript 1 close parentheses squared plus open parentheses u subscript 2 close parentheses squared plus open parentheses u subscript 3 close parentheses squared plus... plus open parentheses u subscript n close parentheses squared end root end cell end table$

Sehingga:

$table attributes columnalign right center left columnspacing 0px end attributes row cell fraction numerator a times b over denominator open vertical bar b close vertical bar end fraction end cell equals 6 row cell fraction numerator open parentheses m cross times open parentheses negative 3 close parentheses close parentheses plus open parentheses negative 18 cross times 0 close parentheses plus left parenthesis 6 cross times 4 right parenthesis over denominator square root of open parentheses negative 3 close parentheses squared plus 0 squared plus 4 squared end root end fraction end cell equals 6 row cell fraction numerator negative 3 m plus 0 plus 24 over denominator square root of 9 plus 16 end root end fraction end cell equals 6 row cell fraction numerator negative 3 m plus 24 over denominator square root of 25 end fraction end cell equals 6 row cell fraction numerator negative 3 m plus 24 over denominator 5 end fraction end cell equals 6 row cell negative 3 m plus 24 end cell equals cell 6 cross times 5 end cell row cell negative 3 m plus 24 end cell equals 30 row cell negative 3 m end cell equals cell 30 minus 24 end cell row cell negative 3 m end cell equals 6 row m equals cell 6 divided by negative 3 end cell row m equals cell negative 2 end cell row blank blank blank end table$

Dengan demikian, nilai m yang memenuhi adalah $- 2$ .

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Pembahasan Soal:

Panjang vektor $a with rightwards arrow on top equals open parentheses table row x row y end table close parentheses$ , yaitu $open vertical bar a with rightwards arrow on top close vertical bar equals square root of x squared plus y squared end root$

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

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

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

Berdasarkan konsep di atas dapat ditentukan hubungan berikut.

$table attributes columnalign right center left columnspacing 0px end attributes row cell open vertical bar c with rightwards arrow on top close vertical bar end cell equals cell fraction numerator x with rightwards arrow on top times y with rightwards arrow on top over denominator open vertical bar y close vertical bar end fraction end cell row cell 4 over 5 square root of 10 end cell equals cell fraction numerator x with rightwards arrow on top times y with rightwards arrow on top over denominator square root of 3 squared plus 1 squared end root end fraction end cell row cell 4 over 5 square root of 10 end cell equals cell fraction numerator x with rightwards arrow on top times y with rightwards arrow on top over denominator square root of 10 end fraction end cell row cell x with rightwards arrow on top times y with rightwards arrow on top end cell equals cell fraction numerator 4 square root of 10 times square root of 10 over denominator 5 end fraction end cell row cell x with rightwards arrow on top times y with rightwards arrow on top end cell equals 8 end table$

Misal vektor $x with rightwards arrow on top equals open square brackets a comma space b close square brackets$ dapat ditentukan persamaan-persamaan berikut.

Persamaan 1:

$table attributes columnalign right center left columnspacing 0px end attributes row cell x with rightwards arrow on top times y with rightwards arrow on top end cell equals 8 row cell open parentheses table row a row b end table close parentheses open parentheses table row 3 row 1 end table close parentheses end cell equals 8 row cell 3 a plus b end cell equals 8 row b equals cell 8 minus 3 a end cell end table$

Persamaan 2:

$table attributes columnalign right center left columnspacing 0px end attributes row cell square root of a squared plus b squared end root end cell equals cell open vertical bar x with rightwards arrow on top close vertical bar end cell row cell square root of a squared plus b squared end root end cell equals cell 2 square root of 2 end cell row cell a squared plus b squared end cell equals 8 end table$

Substitusi persamaan 1 ke persamaan 2:

$table attributes columnalign right center left columnspacing 0px end attributes row cell a squared plus b squared end cell equals 8 row cell a squared plus open parentheses 8 minus 3 a close parentheses squared end cell equals 8 row cell a squared plus 64 minus 48 a plus 9 a squared end cell equals 8 row cell 10 a squared minus 48 a plus 56 end cell equals 0 row cell 5 a squared minus 24 a plus 28 end cell equals 0 row cell open parentheses a minus 2 close parentheses open parentheses 5 a minus 14 close parentheses end cell equals 0 end table$

$a equals 2$ atau $a equals 14 over 5$

Untuk $a equals 2$ diperoleh

$table attributes columnalign right center left columnspacing 0px end attributes row b equals cell 8 minus 3 a end cell row blank equals cell 8 minus 3 times 2 end cell row blank equals 2 end table$

Untuk $a equals 14 over 5$ diperoleh

$table attributes columnalign right center left columnspacing 0px end attributes row b equals cell 8 minus 3 a end cell row blank equals cell 8 minus 3 times 14 over 5 end cell row blank equals cell 40 over 5 minus 42 over 5 end cell row blank equals cell negative 2 over 5 end cell end table$

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Oleh karena itu, jawaban yang tepat adalah A.

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Pembahasan Soal:

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

Panjang proyeksi vektor $stack text a end text with rightwards arrow on top$ pada vektor $stack text b end text with rightwards arrow on top$ dapat ditentukan menggunakan rumus berikut.

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

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

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

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

Berdasarkan konsep di atas, panjang proyeksi vektor $stack A C with rightwards arrow on top$ pada vektor $stack A B with rightwards arrow on top$ adalah sebagai berikut.

$table attributes columnalign right center left columnspacing 0px end attributes row cell open vertical bar text AD end text close vertical bar end cell equals cell open vertical bar fraction numerator stack A C with rightwards arrow on top times stack A B with rightwards arrow on top over denominator open vertical bar stack A B with rightwards arrow on top close vertical bar end fraction close vertical bar end cell row blank equals cell open vertical bar fraction numerator open parentheses table row 6 row 3 row 2 end table close parentheses open parentheses table row 8 row 0 row 6 end table close parentheses over denominator square root of 8 squared plus 0 squared plus 6 squared end root end fraction close vertical bar end cell row blank equals cell open vertical bar fraction numerator 6 times 8 plus 3 times 0 plus 2 times 6 over denominator square root of 100 end fraction close vertical bar end cell row blank equals cell open vertical bar 60 over 10 close vertical bar end cell row blank equals 6 end table$

Diperoleh panjang $text AD=6 end text$

Oleh karena itu, jawaban yang tepat adalah C.

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Jika vektor a tegak lurus b maka proyeksi skalar orthogonal pada sama dengan...

Pembahasan Soal:

Ingat kembali proyeksi skalar orthogonal berikut.

Jika vektor $top enclose a$ diproyeksikan secara orthogonal pada $top enclose b$ , maka proyeksi skalar orthogonalnya adalah

$open vertical bar top enclose c close vertical bar equals open vertical bar open vertical bar top enclose a close vertical bar space cos space alpha close vertical bar$

Dari rumus di atas, diperoleh perhitungan sebagai berikut.

Vektor $top enclose a$ tegak lurus $top enclose b$ artinya $angle left parenthesis top enclose a comma space top enclose b right parenthesis equals 90 degree$ .

Sehingga proyeksi skalar orthogonal vektor $top enclose a$ pada $top enclose b$ adalah

$table attributes columnalign right center left columnspacing 0px end attributes row cell open vertical bar top enclose c close vertical bar end cell equals cell open vertical bar open vertical bar top enclose a close vertical bar space cos space alpha close vertical bar end cell row blank equals cell vertical line open vertical bar top enclose a close vertical bar space cos space 90 degree vertical line end cell row blank equals cell vertical line open vertical bar top enclose a close vertical bar space 0 vertical line end cell row blank equals cell vertical line 0 vertical line end cell row blank equals 0 row blank blank blank row blank blank blank end table$

Jadi, proyeksi skalar orthogonal vektor $top enclose a$ pada $top enclose b$ adalah 0.

Oleh karena itu, jawaban yang benar adalah B.

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