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Tentukan modulus dari vektor berikut:  dengan titik  dan titik

Pertanyaan

Tentukan modulus dari vektor berikut:

AB with rightwards arrow on top dengan titik A left parenthesis minus sign 2 comma space 3 comma minus sign 1 right parenthesis dan titik B left parenthesis 2 comma space 1 comma minus sign 4 right parenthesis

Pembahasan Soal:

Diberikan vektor AB with rightwards arrow on top dengan titik A left parenthesis minus sign 2 comma space 3 comma minus sign 1 right parenthesis dan titik B left parenthesis 2 comma space 1 comma minus sign 4 right parenthesis.

table attributes columnalign right center left columnspacing 0px end attributes row cell AB with rightwards arrow on top end cell equals cell OB with rightwards arrow on top minus OA with rightwards arrow on top end cell row blank equals cell open parentheses table row 2 row 1 row cell negative 4 end cell end table close parentheses minus open parentheses table row cell negative 2 end cell row 3 row cell negative 1 end cell end table close parentheses end cell row blank equals cell open parentheses table row cell 2 plus 2 end cell row cell 1 minus 3 end cell row cell negative 4 plus 1 end cell end table close parentheses end cell row blank equals cell open parentheses table row 4 row cell negative 2 end cell row cell negative 3 end cell end table close parentheses end cell end table 

Modulus atau panjang dari vektor AB with rightwards arrow on top adalah

table attributes columnalign right center left columnspacing 0px end attributes row cell open vertical bar AB with rightwards arrow on top close vertical bar end cell equals cell square root of 4 squared plus left parenthesis negative 2 right parenthesis squared plus left parenthesis negative 3 right parenthesis squared end root end cell row blank equals cell square root of 16 plus 4 plus 9 end root end cell row blank equals cell square root of 29 end cell end table 

Jadi, modulus vektor tersebut adalah square root of 29 satuan.

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

I. Kumaralalita

Mahasiswa/Alumni Universitas Gadjah Mada

Terakhir diupdate 06 Juni 2021

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Pertanyaan yang serupa

Tentukan modulus dari vektor-vektor berikut. b.  dengan titik  dan titik B

Pembahasan Soal:

Vektor begin mathsize 14px style stack A B with rightwards arrow on top end style dapat ditentukan sebagai berikut.

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell stack A B with rightwards arrow on top end cell equals cell B minus A end cell row blank equals cell open parentheses table row 2 row 1 row cell negative 4 end cell end table close parentheses minus open parentheses table row cell negative 2 end cell row 3 row cell negative 1 end cell end table close parentheses end cell row blank equals cell open parentheses table row 4 row cell negative 2 end cell row cell negative 3 end cell end table close parentheses end cell end table end style

Modulus vektor tersebut dapat ditentukan sebagai berikut.

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell open vertical bar stack A B with rightwards arrow on top close vertical bar end cell equals cell square root of x squared plus y squared plus z squared end root end cell row blank equals cell square root of 4 squared plus open parentheses negative 2 close parentheses squared plus open parentheses negative 3 close parentheses squared end root end cell row blank equals cell square root of 16 plus 4 plus 9 end root end cell row blank equals cell square root of 29 end cell end table end style

Dengan demikian, modulus vektor begin mathsize 14px style stack A B with rightwards arrow on top end style adalah begin mathsize 14px style square root of 29 end style 

0

Roboguru

Tentukan modulus dari vektor-vektor berikut. b.  dengan titik  dan titik .

Pembahasan Soal:

Modulus vektor adalah besar atau nilai vektor. Jika diketahui titik begin mathsize 14px style straight A open parentheses x subscript 1 comma space y subscript 1 comma space z subscript 1 close parentheses end style dan titik begin mathsize 14px style straight B open parentheses x subscript 2 comma space y subscript 2 comma space z subscript 2 close parentheses end style, maka besar nilai vektor begin mathsize 14px style AB with rightwards arrow on top end style adalah 

begin mathsize 14px style AB with rightwards arrow on top equals square root of open parentheses x subscript 2 minus x subscript 1 close parentheses squared plus open parentheses y subscript 2 minus y subscript 1 close parentheses squared plus open parentheses z subscript 2 minus z subscript 1 close parentheses squared end root end style .

Akan ditentukan modulus dari begin mathsize 14px style AB with rightwards arrow on top end style dengan titik begin mathsize 14px style straight A open parentheses negative 2 comma space 3 comma space minus 1 close parentheses end style dan titik begin mathsize 14px style straight B open parentheses 2 comma space 1 comma space minus 4 close parentheses end style:

begin mathsize 14px style table attributes columnalign right center left columnspacing 2px end attributes row cell AB with rightwards arrow on top end cell equals cell square root of open parentheses 2 minus open parentheses negative 2 close parentheses close parentheses squared plus open parentheses 1 minus 3 close parentheses squared plus open parentheses negative 4 minus open parentheses negative 1 close parentheses close parentheses squared end root end cell row blank equals cell square root of 4 squared plus open parentheses negative 2 close parentheses squared plus open parentheses negative 3 close parentheses squared end root end cell row blank equals cell square root of 16 plus 4 plus 9 end root end cell row cell AB with rightwards arrow on top end cell equals cell square root of 29 end cell end table end style  

Jadi, diperoleh modulus dari vektor tersebut adalah begin mathsize 14px style square root of 29 end style.

0

Roboguru

Tentukan modulus dari vektor-vektor berikut. b.    dengan titik  dan titik .

Pembahasan Soal:

begin mathsize 14px style stack A B with rightwards arrow on top equals b with rightwards arrow on top minus a with rightwards arrow on top end style 

     begin mathsize 14px style equals open parentheses table row 2 row 1 row cell negative 4 end cell end table close parentheses minus open parentheses table row cell negative 2 end cell row 3 row cell negative 1 end cell end table close parentheses equals open parentheses table row 4 row cell negative 2 end cell row cell negative 3 end cell end table close parentheses end style 

Modulus artinya panjang vektor.

begin mathsize 14px style open vertical bar stack A B with rightwards arrow on top close vertical bar equals square root of 4 squared plus open parentheses negative 2 close parentheses squared plus open parentheses negative 3 close parentheses squared space end root equals square root of 29 end style 

Jadi modulus vektor AB adalah 29.

0

Roboguru

Diketahui  dan . Tentukan panjang vektor .

Pembahasan Soal:

Gunakan konsep panjang vektor pada dimensi tiga.

m with rightwards arrow on top equals open parentheses table row x row y row z end table close parentheses rightwards arrow open vertical bar m with rightwards arrow on top close vertical bar equals square root of x squared plus y squared plus z squared end root

Diketahui:
begin mathsize 14px style straight A open parentheses 2 comma space 1 comma space minus 4 close parentheses end style dan begin mathsize 14px style straight B open parentheses 3 comma space minus 5 comma space 1 close parentheses end style, sehingga vektor posisinya adalah a with rightwards arrow on top equals open parentheses table row 2 row 1 row cell negative 4 end cell end table close parentheses dan b with rightwards arrow on top equals open parentheses table row 3 row cell negative 5 end cell row 1 end table close parentheses

Akan ditentukan panjang vektor begin mathsize 14px style 2 a minus b end style.

table attributes columnalign right center left columnspacing 2px end attributes row cell 2 a minus b end cell equals cell 2 open parentheses table row 2 row 1 row cell negative 4 end cell end table close parentheses minus open parentheses table row 3 row cell negative 5 end cell row 1 end table close parentheses end cell row blank equals cell open parentheses table row 4 row 2 row cell negative 8 end cell end table close parentheses minus open parentheses table row 3 row cell negative 5 end cell row 1 end table close parentheses end cell row cell 2 a minus b end cell equals cell open parentheses table row 1 row 7 row cell negative 9 end cell end table close parentheses end cell row cell open vertical bar 2 a minus b close vertical bar end cell equals cell square root of 1 squared plus 7 squared plus open parentheses negative 9 close parentheses squared end root end cell row blank equals cell square root of 1 plus 49 plus 81 end root end cell row cell open vertical bar 2 a minus b close vertical bar end cell equals cell square root of 131 end cell end table

Jadi, diperoleh panjang vektor begin mathsize 14px style 2 a minus b end style adalah square root of 131.

0

Roboguru

Diketahui vektor  dan vektor . Besar sudut antara vektor  adalah..

Pembahasan Soal:

Diketahui:

a with rightwards arrow on top equals 4 i with rightwards arrow on top minus 2 j with rightwards arrow on top plus 2 k with rightwards arrow on top

b with rightwards arrow on top equals i with rightwards arrow on top plus j with rightwards arrow on top plus 2 k with rightwards arrow on top

Ditanya: Besar sudut antara vektor a with rightwards arrow on top space d a n space b with rightwards arrow on top?

Jawab: 

Besar sudut antara 2 vektor dapat dicari dengan menggunakan rumus:

cos space theta equals fraction numerator a with rightwards arrow on top times b with rightwards arrow on top over denominator open vertical bar a with rightwards arrow on top close vertical bar open vertical bar b with rightwards arrow on top close vertical bar end fraction

Sehingga,

table attributes columnalign right center left columnspacing 0px end attributes row cell cos space theta end cell equals cell fraction numerator 4 cross times 1 plus left parenthesis negative 2 right parenthesis cross times 1 plus 2 cross times 2 over denominator square root of 4 squared plus left parenthesis negative 2 right parenthesis squared plus 2 to the power of 2 end exponent end root square root of 1 squared plus 1 squared plus 2 squared end root end fraction end cell row cell cos space theta end cell equals cell fraction numerator 4 minus 2 plus 4 over denominator square root of 16 plus 4 plus 4 end root square root of 1 plus 1 plus 4 end root end fraction end cell row cell cos space theta end cell equals cell fraction numerator 6 over denominator square root of 24 square root of 6 end fraction end cell row cell cos space theta end cell equals cell fraction numerator 6 over denominator square root of 144 end fraction end cell row cell cos space theta end cell equals cell 6 over 12 end cell row cell cos space theta end cell equals cell 1 half end cell row theta equals cell 60 degree end cell end table

Oleh karena itu, jawaban yang benar adalah B.

 

0

Roboguru

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