Roboguru

Diketahui: a=i+j​,b=2i−3j​+kdanc=4j​+3k Tentukan a×b !

Pertanyaan

Diketahui:

straight a with rightwards arrow on top equals straight i with rightwards arrow on top plus straight j with rightwards arrow on top comma space straight b equals 2 straight i with rightwards arrow on top minus 3 straight j with rightwards arrow on top plus stack straight k space with rightwards arrow on top dan stack space straight c with rightwards arrow on top equals 4 straight j with rightwards arrow on top plus 3 straight k with rightwards arrow on top

Tentukan straight a with rightwards arrow on top cross times straight b with rightwards arrow on top !

Pembahasan Soal:

Diketahui :

straight a with rightwards arrow on top equals straight i with rightwards arrow on top plus straight j with rightwards arrow on top comma space straight b equals 2 straight i with rightwards arrow on top minus 3 straight j with rightwards arrow on top plus stack straight k space with rightwards arrow on top dan stack space straight c with rightwards arrow on top equals 4 straight j with rightwards arrow on top plus 3 straight k with rightwards arrow on top

Sehingga, dengan menggunakan konsep cross product pada vektor:

table attributes columnalign right center left columnspacing 0px end attributes row cell straight a with rightwards arrow on top cross times straight b with rightwards arrow on top end cell equals cell open vertical bar table row straight i straight j straight k row 1 1 0 row 2 cell negative 3 end cell 1 end table close vertical bar end cell row blank equals cell open vertical bar table row 1 0 row cell negative 3 end cell 1 end table close vertical bar straight i plus open vertical bar table row 0 1 row 1 2 end table close vertical bar straight j plus open vertical bar table row 1 1 row 2 cell negative 3 end cell end table close vertical bar straight k end cell row cell straight a with rightwards arrow on top cross times straight b with rightwards arrow on top end cell equals cell 4 straight i minus straight j minus 5 straight k end cell end table

Jadi, table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell straight a with rightwards arrow on top end cell end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cross times end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell straight b 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 4 end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank straight i end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank minus end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank straight j end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank minus end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank 5 end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank straight k end table

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

A. Putri

Mahasiswa/Alumni Institut Pertanian Bogor

Terakhir diupdate 06 Oktober 2021

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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Pertanyaan yang serupa

Diketahui: a=i+j​,b=2i−3j​+kdanc=4j​+3k Tentukan a×(b+c) !

Pembahasan Soal:

Diketahui :

straight a with rightwards arrow on top equals straight i with rightwards arrow on top plus straight j with rightwards arrow on top comma space straight b equals 2 straight i with rightwards arrow on top minus 3 straight j with rightwards arrow on top plus stack straight k space with rightwards arrow on top dan stack space straight c with rightwards arrow on top equals 4 straight j with rightwards arrow on top plus 3 straight k with rightwards arrow on top

Sehingga, dengan menggunakan konsep cross product pada vektor:

table attributes columnalign right center left columnspacing 0px end attributes row cell straight b with rightwards arrow on top plus straight c with rightwards arrow on top end cell equals cell 2 straight i with rightwards arrow on top plus straight j with rightwards arrow on top plus 4 straight k with rightwards arrow on top end cell row cell straight a with rightwards arrow on top cross times open parentheses straight b with rightwards arrow on top plus straight c with rightwards arrow on top close parentheses end cell equals cell open vertical bar table row straight i straight j straight k row 1 1 0 row 2 1 4 end table close vertical bar end cell row blank equals cell open vertical bar table row 1 0 row 1 4 end table close vertical bar straight i plus open vertical bar table row 0 1 row 4 2 end table close vertical bar straight j plus open vertical bar table row 1 1 row 2 1 end table close vertical bar straight k end cell row cell straight a with rightwards arrow on top cross times open parentheses straight b with rightwards arrow on top plus straight c with rightwards arrow on top close parentheses end cell equals cell 4 straight i minus 4 straight j minus straight k end cell end table


Jadi, table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell straight a with rightwards arrow on top end cell end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cross times end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell straight b 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 4 end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank straight i end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank minus end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank straight j end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank minus end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank 5 end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank straight k end table

0

Roboguru

Jika vektor a=⎝⎛​201​⎠⎞​,b=⎝⎛​13−1​⎠⎞​,c=⎝⎛​−5−11​⎠⎞​,maka2b+a−3cadalah...

Pembahasan Soal:

stack 2 straight b with rightwards arrow on top space equals space open parentheses table row 2 row 6 row cell negative 2 end cell end table close parentheses straight a with rightwards arrow on top space equals open parentheses table row 2 row 0 row 1 end table close parentheses 3 straight c with rightwards arrow on top space equals space open parentheses table row cell negative 15 end cell row cell negative 3 end cell row 3 end table close parentheses Maka comma space space space stack 2 straight b with rightwards arrow on top plus straight a with rightwards arrow on top minus 3 straight c with rightwards arrow on top space adalah space open parentheses table row 19 row 9 row 2 end table close parentheses 

0

Roboguru

Diketahui vektor p​=i+j−4k,danq​=−2i−j. Nilai sinus sudut antara vektor p​dan q​=....

Pembahasan Soal:

c o s space theta equals fraction numerator p ⃗. q with rightwards arrow on top over denominator vertical line p ⃗ vertical line. vertical line q ⃗ vertical line end fraction equals fraction numerator i plus j minus 4 k right parenthesis. left parenthesis negative 2 i minus j right parenthesis over denominator square root of 1 squared plus 1 squared plus left parenthesis negative 4 right parenthesis squared end root. square root of left parenthesis negative 2 right parenthesis squared plus left parenthesis negative 1 right parenthesis squared plus 0 squared end root end fraction  equals fraction numerator negative 2 minus 1 minus 0 over denominator square root of 18. square root of 5 end fraction equals fraction numerator negative 3 over denominator square root of 90 end fraction equals fraction numerator negative 3 over denominator 3 square root of 10 end fraction equals fraction numerator negative 1 over denominator square root of 10 end fraction equals negative 1 over 10 square root of 10  M a k a comma D e n g a n space i d e n t i t a s space s i n space theta equals  square root of left parenthesis 1 minus cos squared theta end root  equals square root of 1 minus open parentheses negative 1 over 10 square root of 10 right parenthesis squared close parentheses end root  equals 3 over 10 square root of 10

0

Roboguru

Diketahui vektor c=4i+2j dan vektor e=6i−4j. Tentukan vektor 3c−e.

Pembahasan Soal:

begin mathsize 14px style c with rightwards arrow on top equals 4 i plus 2 j equals open square brackets table row 4 row 2 end table close square brackets e with rightwards arrow on top equals 6 i minus 4 j equals open square brackets table row 6 row cell negative 4 end cell end table close square brackets end style 

Maka,

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell 3 c with rightwards arrow on top minus e with rightwards arrow on top end cell equals cell 3 open square brackets table row 4 row 2 end table close square brackets minus open square brackets table row 6 row cell negative 4 end cell end table close square brackets end cell row blank equals cell open square brackets table row 12 row 6 end table close square brackets minus open square brackets table row 6 row cell negative 4 end cell end table close square brackets end cell row blank equals cell open square brackets table row 6 row 10 end table close square brackets end cell end table end style 

begin mathsize 14px style 3 c with rightwards arrow on top minus e with rightwards arrow on top equals open square brackets table row 6 row 10 end table close square brackets equals 6 i plus 10 j end style 

Jadi, vektor begin mathsize 14px style bold 3 bold c with bold rightwards arrow on top bold minus bold e with bold rightwards arrow on top bold equals bold 6 bold italic i bold plus bold 10 bold italic j end style.

0

Roboguru

Vektor v yang digambarkan di bawah ini sama dengan ...

Pembahasan Soal:

Vektor v with rightwards arrow on top bergeser sejauh 7 satuan ke arah kiri, berarti nilai komponen x equals negative 7 

Vektor v with rightwards arrow on top bergeser sejauh 3 satuan ke arah atas, berati nilai komponen y equals 3 

sehingga diperoleh nilai vektor v with rightwards arrow on top, yaitu

table attributes columnalign right center left columnspacing 0px end attributes row cell v with rightwards arrow on top end cell equals cell open parentheses table row cell komponen space x end cell row cell komponen space y end cell end table close parentheses end cell row blank equals cell open parentheses table row cell negative 7 end cell row 3 end table close parentheses end cell end table 

Jadi, jawaban yang tepat adalah B.

0

Roboguru

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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