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Diketahui koordinat . Tentukan k jika vektor  tegak lurus dengan vektor .

Pertanyaan

Diketahui koordinat begin mathsize 14px style straight A open parentheses 7 comma 1 comma negative 2 close parentheses space dan space straight B open parentheses 10 comma negative 2 comma negative 5 close parentheses end style. Tentukan k jika vektor begin mathsize 14px style AB with rightwards arrow on top end style tegak lurus dengan vektor begin mathsize 14px style straight u with rightwards arrow on top equals open parentheses table row cell straight k minus 3 end cell row cell straight k plus 1 end cell row cell straight k minus 2 end cell end table close parentheses end style.

Pembahasan Soal:

1. Tentukan vektor begin mathsize 14px style AB with rightwards arrow on top end style 

begin mathsize 14px style AB with rightwards arrow on top equals open parentheses table row 10 row cell negative 2 end cell row cell negative 5 end cell end table close parentheses minus open parentheses table row 7 row 1 row cell negative 2 end cell end table close parentheses AB with rightwards arrow on top equals open parentheses table row 3 row cell negative 3 end cell row cell negative 3 end cell end table close parentheses end style

2. Tentukan nilai k

begin mathsize 14px style space space space space space space space space space space space space space space space space space space space space space space space AB with rightwards arrow on top times straight u with rightwards arrow on top equals 0 space space space space space space space space space space space space space space space space space space space space open parentheses table row 3 row cell negative 3 end cell row cell negative 3 end cell end table close parentheses times open parentheses table row cell straight k minus 3 end cell row cell straight k plus 1 end cell row cell straight k minus 2 end cell end table close parentheses equals 0 3 open parentheses straight k minus 3 close parentheses minus 3 open parentheses straight k plus 1 close parentheses minus 3 open parentheses straight k minus 2 close parentheses equals 0 space space space space space space space space 3 straight k minus 9 minus 3 straight k minus 3 minus 3 straight k plus 6 equals 0 space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space minus 3 straight k minus 6 equals 0 space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space 3 straight k equals negative 6 space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space straight k equals negative 2 end style

Jadi, nilai k adalah begin mathsize 14px style negative 2 end style.

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

R. Azizatul

Mahasiswa/Alumni Universitas Negeri Yogyakarta

Terakhir diupdate 29 Maret 2021

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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Pertanyaan yang serupa

Jika  dan sudut antara  adalah , tentukanlah !

Pembahasan Soal:

Diketahui:

table attributes columnalign right center left columnspacing 0px end attributes row cell open vertical bar u with rightwards arrow on top close vertical bar end cell equals 6 row cell open vertical bar v with rightwards arrow on top close vertical bar end cell equals 4 row theta equals cell 60 degree end cell end table 

maka dapat ditentukan:

table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses u with rightwards arrow on top plus v with rightwards arrow on top close parentheses times v with rightwards arrow on top end cell equals cell u with rightwards arrow on top times v with rightwards arrow on top plus v with rightwards arrow on top times v with rightwards arrow on top end cell row blank equals cell open vertical bar u with rightwards arrow on top close vertical bar open vertical bar v with rightwards arrow on top close vertical bar cos space theta plus open vertical bar v with rightwards arrow on top close vertical bar squared end cell row blank equals cell 6 times 4 times cos space 60 degree plus 4 squared end cell row blank equals cell 24 times 1 half plus 16 end cell row blank equals cell 12 plus 16 end cell row blank equals 28 end table 

Dengan demikian, diperoleh table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell open parentheses u with rightwards arrow on top plus v with rightwards arrow on top close parentheses end cell end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank times end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell v with rightwards arrow on top end cell end table table attributes columnalign right center left columnspacing 0px end attributes row blank equals blank end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank 28 end table.

0

Roboguru

Diberikan titik-titik dan . Jika ruas garis berarah  berturut-turut mewakili vektor-vektor [endif]--> dan [endif]-->. Maka nilai dari

Pembahasan Soal:

Diketahui

u with rightwards arrow on top equals stack A B with rightwards arrow on top equals open parentheses table row cell x subscript B minus x subscript A end cell row cell y subscript B minus y subscript A end cell row cell z subscript B minus z subscript A end cell end table close parentheses equals open parentheses table row cell negative 1 minus 2 end cell row cell 3 minus 6 end cell row cell 7 minus open parentheses negative 5 close parentheses end cell end table close parentheses equals open parentheses table row cell negative 3 end cell row cell negative 3 end cell row 12 end table close parentheses    v with rightwards arrow on top equals stack B C with rightwards arrow on top equals open parentheses table row cell x subscript C minus x subscript B end cell row cell y subscript C minus y subscript B end cell row cell z subscript C minus z subscript B end cell end table close parentheses equals open parentheses table row cell 4 minus open parentheses negative 1 close parentheses end cell row cell negative 1 minus 3 end cell row cell 8 minus 7 end cell end table close parentheses equals open parentheses table row 5 row cell negative 4 end cell row 1 end table close parentheses    u with rightwards arrow on top minus v with rightwards arrow on top equals open parentheses table row cell negative 3 end cell row cell negative 3 end cell row 12 end table close parentheses minus open parentheses table row 5 row cell negative 4 end cell row 1 end table close parentheses equals open parentheses table row cell negative 8 end cell row 1 row 11 end table close parentheses    open parentheses u with rightwards arrow on top bullet open parentheses u with rightwards arrow on top plus v with rightwards arrow on top close parentheses close parentheses minus open parentheses 2 u with rightwards arrow on top bullet v with rightwards arrow on top close parentheses equals u with rightwards arrow on top bullet u with rightwards arrow on top plus u with rightwards arrow on top bullet v with rightwards arrow on top minus 2 u with rightwards arrow on top bullet v with rightwards arrow on top equals u with rightwards arrow on top bullet u with rightwards arrow on top minus u with rightwards arrow on top bullet v with rightwards arrow on top equals u with rightwards arrow on top bullet open parentheses u with rightwards arrow on top minus v with rightwards arrow on top close parentheses    equals open parentheses table row cell negative 3 end cell row cell negative 3 end cell row 12 end table close parentheses bullet open parentheses table row cell negative 8 end cell row 1 row 11 end table close parentheses equals open parentheses negative 3 close parentheses bullet open parentheses negative 8 close parentheses plus open parentheses negative 3 close parentheses bullet 1 plus 12 bullet 11 equals 24 minus 3 plus 132 equals 153

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Roboguru

Pembahasan Soal:

K a r e n a space v e k t o r space a with rightwards arrow on top space d a n space v e k t o r space b with rightwards arrow on top space s a l i n g space t e g a k space l u r u s comma space m a k a  a with rightwards arrow on top space times b with rightwards arrow on top space equals 0  left parenthesis m i space plus 3 j space space right parenthesis times left parenthesis m i space minus 4 j space space right parenthesis equals 0  m times m plus 3 times left parenthesis negative 4 right parenthesis equals 0  m squared minus 12 equals 0  m squared equals 12  m equals 2 square root of 3 space a t a u space m equals negative 2 square root of 3

0

Roboguru

Andi dan Bayu menghitung perkalian skalar dua buah vektor  dan . Andi menghitung perkalian tersebut dengan urutan , sedangkan Bayu . Pernyataan yang benar berkaitan dengan perhitungan mereka adalah .....

Pembahasan Soal:

Misalkan:

A with rightwards arrow on top equals open parentheses table row cell x subscript A end cell row cell y subscript A end cell end table close parentheses B with rightwards arrow on top equals open parentheses table row cell x subscript B end cell row cell y subscript B end cell end table close parentheses 

perkalian skalar dua vektor tersebut dapat ditentukan sebagai berikut.

table attributes columnalign right center left columnspacing 0px end attributes row cell A with rightwards arrow on top times B with rightwards arrow on top end cell equals cell open parentheses table row cell x subscript A end cell row cell y subscript A end cell end table close parentheses times open parentheses table row cell x subscript B end cell row cell y subscript B end cell end table close parentheses end cell row blank equals cell x subscript A times x subscript B plus y subscript A times y subscript B end cell row blank equals cell x subscript B times x subscript A plus y subscript B times y subscript A end cell row blank equals cell open parentheses table row cell x subscript B end cell row cell y subscript B end cell end table close parentheses times open parentheses table row cell x subscript A end cell row cell y subscript A end cell end table close parentheses end cell row cell A with rightwards arrow on top times B with rightwards arrow on top end cell equals cell B with rightwards arrow on top times A with rightwards arrow on top end cell end table 

Dengan demikian, dapat disimpulkan bahwa hasil perhitungan operasi skalar dua vektor yang dilakukan Andi dan Bayu adalah sama.

0

Roboguru

Jika vektor saling tegak lurus, maka nilai dari adalah ....

Pembahasan Soal:

Ingat bahwa jika vektor begin mathsize 14px style a with rightwards arrow on top space d a n space b with rightwards arrow on top end stylesaling tegak lurus, maka begin mathsize 14px style a with rightwards arrow on top times b with rightwards arrow on top equals 0 end style.
Oleh karena itu, jika vektor begin mathsize 14px style u with rightwards arrow on top space d a n space v with rightwards arrow on top end style saling tegak lurus, maka begin mathsize 14px style u with rightwards arrow on top times v with rightwards arrow on top end style = 0.

Jadi, jawaban yang tepat adalah C.

 

0

Roboguru

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