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Pertanyaan

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Pernyataan berikut yang benar adalah ....space 

  1. KL with rightwards harpoon with barb upwards on top equals open parentheses table row 2 row 5 end table close parentheses semicolon space LM with rightwards harpoon with barb upwards on top equals open parentheses table row 4 row 1 end table close parentheses MK with rightwards harpoon with barb upwards on top equals open parentheses table row 3 row cell negative 3 end cell end table close parentheses 

  2. KL with rightwards harpoon with barb upwards on top equals open parentheses table row 7 row 2 end table close parentheses semicolon space LM with rightwards harpoon with barb upwards on top equals open parentheses table row cell negative 4 end cell row 1 end table close parentheses MK with rightwards harpoon with barb upwards on top equals open parentheses table row cell negative 3 end cell row cell negative 3 end cell end table close parentheses 

  3. KL with rightwards harpoon with barb upwards on top equals open parentheses table row 7 row cell negative 2 end cell end table close parentheses semicolon space LM with rightwards harpoon with barb upwards on top equals open parentheses table row 4 row cell negative 1 end cell end table close parentheses MK with rightwards harpoon with barb upwards on top equals open parentheses table row cell negative 3 end cell row 3 end table close parentheses 

  4. KL with rightwards harpoon with barb upwards on top equals open parentheses table row 7 row cell negative 2 end cell end table close parentheses semicolon space LM with rightwards harpoon with barb upwards on top equals open parentheses table row cell negative 4 end cell row 1 end table close parentheses MK with rightwards harpoon with barb upwards on top equals open parentheses table row cell negative 3 end cell row 3 end table close parentheses

  5. KL with rightwards harpoon with barb upwards on top equals open parentheses table row 7 row cell negative 2 end cell end table close parentheses semicolon space LM with rightwards harpoon with barb upwards on top equals open parentheses table row cell negative 4 end cell row cell negative 1 end cell end table close parentheses MK with rightwards harpoon with barb upwards on top equals open parentheses table row 3 row 3 end table close parentheses 

Pembahasan Soal:

Dari gambar di atas, diketahui koordinat dari titik K, L, dan M yaitu

straight K equals open parentheses table row 0 row 4 end table close parentheses semicolon space straight L equals open parentheses table row 7 row 2 end table close parentheses semicolon space straight M equals open parentheses table row 3 row 1 end table close parentheses 

Sehingga dapat diperoleh

table attributes columnalign right center left columnspacing 0px end attributes row cell KL with rightwards harpoon with barb upwards on top end cell equals cell open parentheses table row 7 row 2 end table close parentheses minus open parentheses table row 0 row 4 end table close parentheses equals open parentheses table row cell 7 minus 0 end cell row cell 2 minus 4 end cell end table close parentheses equals open parentheses table row 7 row cell negative 2 end cell end table close parentheses end cell row cell LM with rightwards harpoon with barb upwards on top end cell equals cell open parentheses table row 3 row 1 end table close parentheses minus open parentheses table row 7 row 2 end table close parentheses equals open parentheses table row cell 3 minus 7 end cell row cell 1 minus 2 end cell end table close parentheses equals open parentheses table row cell negative 4 end cell row cell negative 1 end cell end table close parentheses end cell row cell MK with rightwards harpoon with barb upwards on top end cell equals cell open parentheses table row 0 row 4 end table close parentheses minus open parentheses table row 3 row 1 end table close parentheses equals open parentheses table row cell 0 minus 3 end cell row cell 4 minus 1 end cell end table close parentheses equals open parentheses table row cell negative 3 end cell row 3 end table close parentheses end cell end table

Oleh karena itu, tidak ada pilihan jawaban yang tepat.space  

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

K. Prameswari

Mahasiswa/Alumni Universitas Gadjah Mada

Terakhir diupdate 07 Juni 2021

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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Pertanyaan yang serupa

Jika vektor satuan dari  adalah , maka st = ....

Pembahasan Soal:

Perhatikan bahwa begin mathsize 14px style e subscript a with rightwards arrow on top end subscript equals t i with rightwards arrow on top minus 4 over 5 j with rightwards arrow on top end style adalah vektor satuan. Sehingga

table attributes columnalign right center left columnspacing 0px end attributes row cell size 14px e subscript size 14px a with size 14px rightwards arrow on top end subscript end cell size 14px equals cell fraction numerator size 14px a with size 14px rightwards arrow on top over denominator begin mathsize 14px style vertical line a with rightwards arrow on top vertical line end style end fraction end cell row cell size 14px t size 14px i with size 14px rightwards arrow on top size 14px minus size 14px 4 over size 14px 5 size 14px j with size 14px rightwards arrow on top end cell size 14px equals cell fraction numerator size 14px 6 size 14px i with size 14px rightwards arrow on top size 14px plus size 14px s size 14px j with size 14px rightwards arrow on top over denominator square root of size 14px 6 to the power of size 14px 2 size 14px plus size 14px s to the power of size 14px 2 end root end fraction end cell row cell size 14px t size 14px i with size 14px rightwards arrow on top size 14px minus size 14px 4 over size 14px 5 size 14px j with size 14px rightwards arrow on top end cell size 14px equals cell fraction numerator size 14px 6 size 14px i with size 14px rightwards arrow on top size 14px plus size 14px s size 14px j with size 14px rightwards arrow on top over denominator square root of size 14px 36 size 14px plus size 14px s to the power of size 14px 2 end root end fraction end cell row cell size 14px t size 14px i with size 14px rightwards arrow on top size 14px minus size 14px 4 over size 14px 5 size 14px j with size 14px rightwards arrow on top end cell size 14px equals cell fraction numerator size 14px 6 over denominator square root of size 14px 36 size 14px plus size 14px s to the power of size 14px 2 end root end fraction size 14px i with size 14px rightwards arrow on top size 14px plus fraction numerator size 14px s over denominator square root of size 14px 36 size 14px plus size 14px s to the power of size 14px 2 end root end fraction size 14px j with size 14px rightwards arrow on top end cell end table

Sehingga

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row t equals cell fraction numerator 6 over denominator square root of 36 plus s squared end root end fraction end cell row cell negative 4 over 5 end cell equals cell fraction numerator s over denominator square root of 36 plus s squared end root end fraction end cell end table end style

Karena begin mathsize 14px style e subscript a with rightwards arrow on top end subscript end style adalah vektor satuan, maka

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell open vertical bar e subscript a with rightwards arrow on top end subscript close vertical bar end cell equals 1 row cell square root of t squared plus open parentheses negative 4 over 5 close parentheses squared end root end cell equals 1 row cell t squared plus open parentheses negative 4 over 5 close parentheses squared end cell equals 1 row cell t squared plus 16 over 25 end cell equals 1 row cell t squared end cell equals cell 9 over 25 end cell row t equals cell plus-or-minus 3 over 5 end cell end table end style

Jika diperhatikan, begin mathsize 14px style fraction numerator 6 over denominator square root of 36 plus s squared end root end fraction end style  tidak mungkin bernilai negatif. Sehingga t tidak mungkin bernilai negatif, maka begin mathsize 14px style t equals 3 over 5 end style.

Sehingga

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row t equals cell fraction numerator 6 over denominator square root of 36 plus s squared end root end fraction end cell row cell 3 over 5 end cell equals cell fraction numerator 6 over denominator square root of 36 plus s squared end root end fraction end cell row cell square root of 36 plus s squared end root end cell equals 10 row cell 36 plus s squared end cell equals 100 row cell s squared end cell equals 64 row s equals cell plus-or-minus 8 end cell end table end style

Kemudian perhatikan bahwa

begin mathsize 14px style negative 4 over 5 equals fraction numerator s over denominator square root of 36 plus s squared end root end fraction end style

Ruas kiri bernilai negatif, sehingga ruas kanan juga harus bernilai negatif. Hal ini terpenuhi jika s negatif, maka s = -8.

Maka

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell s t end cell equals cell negative 8 times 3 over 5 end cell row blank equals cell negative 24 over 5 end cell end table end style  

0

Roboguru

Diketahui vektor , maka vektor satuan dari vektor  adalah...

Pembahasan Soal:

Diketahui

a with rightwards arrow on top equals open parentheses negative 4 comma negative 3 close parentheses

Sehingga

table attributes columnalign right center left columnspacing 0px end attributes row cell open vertical bar a with rightwards arrow on top close vertical bar end cell equals cell square root of open parentheses negative 4 close parentheses squared plus open parentheses negative 3 close parentheses squared end root end cell row blank equals cell square root of 16 plus 9 end root end cell row blank equals cell square root of 25 end cell row blank equals 5 end table

Dengan demikian diperoleh vektor satuan dari vektor a adalah

table attributes columnalign right center left columnspacing 0px end attributes row cell fraction numerator a with rightwards arrow on top over denominator open vertical bar a with rightwards arrow on top close vertical bar end fraction end cell equals cell fraction numerator open parentheses negative 4 comma negative 3 close parentheses over denominator 5 end fraction end cell row blank equals cell open parentheses negative 4 over 5 comma negative 3 over 5 close parentheses end cell end table

Oleh karena itu, jawaban yang benar adalah B.

0

Roboguru

Diketahui vektor , , dan b. Tentukan vektor sauan dari vektor , vektor  dan vektor

Pembahasan Soal:

Untuk menyelesaikan soal di atas dapat kita selesaikan dengan menggunakan rumus:

begin mathsize 14px style a equals fraction numerator a with rightwards arrow on top over denominator open vertical bar a with rightwards arrow on top close vertical bar end fraction end style 

Pertama kita menentukan panjang dari masing-masing vektor di atas:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell vertical line a with rightwards arrow on top vertical line end cell end table equals square root of open parentheses a subscript 1 close parentheses squared plus open parentheses a subscript 2 close parentheses squared plus open parentheses a subscript 3 close parentheses squared plus... plus open parentheses a subscript n close parentheses squared end root comma space maka end style

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell vertical line straight a with rightwards arrow on top vertical line end cell equals cell square root of open parentheses negative 3 close parentheses squared plus open parentheses 4 close parentheses squared end root end cell row cell vertical line straight a with rightwards arrow on top vertical line end cell equals cell square root of 9 plus 16 end root end cell row cell vertical line straight a with rightwards arrow on top vertical line end cell equals cell square root of 25 end cell row cell vertical line straight a with rightwards arrow on top vertical line end cell equals 5 end table end style

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell vertical line b with rightwards arrow on top vertical line end cell equals cell square root of open parentheses 2 close parentheses squared plus open parentheses 3 close parentheses squared end root end cell row cell vertical line b with rightwards arrow on top vertical line end cell equals cell square root of 4 plus 9 end root end cell row cell vertical line b with rightwards arrow on top vertical line end cell equals cell square root of 13 end cell end table end style

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell vertical line c with rightwards arrow on top vertical line end cell equals cell square root of left parenthesis negative 1 right parenthesis squared plus left parenthesis 2 right parenthesis squared end root end cell row cell vertical line c with rightwards arrow on top vertical line end cell equals cell square root of 5 end cell end table end style

Maka vektor satuannya adalah:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row a equals cell fraction numerator a with rightwards arrow on top over denominator open vertical bar a with rightwards arrow on top close vertical bar end fraction end cell row a equals cell fraction numerator open parentheses table row cell negative 3 end cell row 4 end table close parentheses over denominator 5 end fraction end cell row a equals cell 1 fifth open parentheses table row cell negative 3 end cell row 4 end table close parentheses end cell row a equals cell open parentheses table row cell bevelled fraction numerator negative 3 over denominator 5 end fraction end cell row cell bevelled 4 over 5 end cell end table close parentheses end cell end table end style

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row b equals cell fraction numerator b with rightwards arrow on top over denominator open vertical bar b with rightwards arrow on top close vertical bar end fraction end cell row b equals cell fraction numerator open parentheses table row 2 row 3 end table close parentheses over denominator square root of 13 end fraction end cell row b equals cell fraction numerator 1 over denominator square root of 13 end fraction open parentheses table row 2 row 3 end table close parentheses end cell row b equals cell open parentheses table row cell bevelled fraction numerator 2 over denominator square root of 13 end fraction end cell row cell bevelled fraction numerator 3 over denominator square root of 13 end fraction end cell end table close parentheses end cell end table end style

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row c equals cell fraction numerator c with rightwards arrow on top over denominator vertical line c with rightwards arrow on top vertical line end fraction end cell row b equals cell fraction numerator blank over denominator square root of 5 end fraction end cell row b equals cell fraction numerator 1 over denominator square root of 5 end fraction open parentheses table row cell negative 1 end cell row 2 end table close parentheses end cell row b equals cell open parentheses table row cell bevelled fraction numerator negative 1 over denominator square root of 5 end fraction end cell row cell bevelled fraction numerator 2 over denominator square root of 5 end fraction end cell end table close parentheses end cell end table end style

 

0

Roboguru

Diketahui koordinat titik , , , dan . Jika hasil dari , maka nilai n adalah ...

Pembahasan Soal:

Ingat bagaimana cara menentukan suatu vektor dengan titik yang diketahui dan ingat operasi perkalian skalar dengan vektor serta operasi penjumlahan dan pengurangan vektor.

Menentukan vektor-vektor

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 2 , 5 close parentheses minus left parenthesis 4 , 2 right parenthesis end cell row blank equals cell left parenthesis negative 2 , 3 right parenthesis end cell row blank blank blank row cell stack D B with rightwards arrow on top end cell equals cell B minus D end cell row blank equals cell open parentheses 2 , 5 close parentheses minus left parenthesis 8 , 4 right parenthesis end cell row blank equals cell left parenthesis negative 6 , 1 right parenthesis end cell row blank blank blank row cell stack D C with rightwards arrow on top end cell equals cell C minus D end cell row blank equals cell open parentheses 1 , 5 close parentheses minus left parenthesis 8 , 4 right parenthesis end cell row blank equals cell open parentheses negative 7 , 1 close parentheses end cell end table

Menentukan nilai n

table attributes columnalign right center left columnspacing 0px end attributes row cell 3 stack A B with rightwards arrow on top plus 2 stack D B with rightwards arrow on top minus stack D C with rightwards arrow on top end cell equals cell open parentheses table row cell negative n end cell row cell n minus 1 end cell end table close parentheses end cell row cell 3 left parenthesis negative 2 , 3 right parenthesis plus 2 left parenthesis negative 6 , 1 right parenthesis minus open parentheses negative 7 , 1 close parentheses end cell equals cell open parentheses table row cell negative n end cell row cell n minus 1 end cell end table close parentheses end cell row cell 3 open parentheses table row cell negative 2 end cell row 3 end table close parentheses plus 2 open parentheses table row cell negative 6 end cell row 1 end table close parentheses minus open parentheses table row cell negative 7 end cell row 1 end table close parentheses end cell equals cell open parentheses table row cell negative n end cell row cell n minus 1 end cell end table close parentheses end cell row cell open parentheses table row cell negative 6 end cell row 9 end table close parentheses plus open parentheses table row cell negative 12 end cell row 2 end table close parentheses minus open parentheses table row cell negative 7 end cell row 1 end table close parentheses end cell equals cell open parentheses table row cell negative n end cell row cell n minus 1 end cell end table close parentheses end cell row cell open parentheses table row cell negative 18 end cell row 11 end table close parentheses minus open parentheses table row cell negative 7 end cell row 1 end table close parentheses end cell equals cell open parentheses table row cell negative n end cell row cell n minus 1 end cell end table close parentheses end cell row cell open parentheses table row cell negative 11 end cell row 10 end table close parentheses end cell equals cell open parentheses table row cell negative n end cell row cell n minus 1 end cell end table close parentheses end cell end table

dari hasil di atas didapatkan 

table attributes columnalign right center left columnspacing 0px end attributes row cell negative n end cell equals cell negative 11 end cell row n equals 11 end table

Diperoleh n equals 11.

Jadi, dapat disimpulkan bahwa nilai dari n adalah 11.

Oleh karena itu, jawaban yang benar adalah C.

0

Roboguru

Vektor satuan dari vektor  adalah ....

Pembahasan Soal:

Panjang vektor:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell open vertical bar PQ with rightwards arrow on top close vertical bar end cell equals cell square root of 6 squared plus 8 squared end root end cell row blank equals cell square root of 36 plus 64 end root end cell row blank equals cell square root of 100 end cell row blank equals 10 end table end style 

Vektor satuan:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell e with rightwards arrow on top end cell equals cell fraction numerator PQ with rightwards arrow on top over denominator open vertical bar PQ with rightwards arrow on top close vertical bar end fraction end cell row blank equals cell fraction numerator 6 i with rightwards arrow on top plus 8 j with rightwards arrow on top over denominator 10 end fraction end cell row blank equals cell 3 over 5 i with rightwards arrow on top plus 4 over 5 j with rightwards arrow on top end cell end table end style 

0

Roboguru

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

Tapi Roboguru masih mau belajar. Menurut kamu pembahasan kali ini sudah membantu, belum?

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