Roboguru

Diketahui koordinat titik dan . Jika , koordinat titik P adalah ....

Pertanyaan

Diketahui koordinat titik M left parenthesis 3 comma 2 comma minus sign 4 right parenthesis dan N left parenthesis 1 comma minus sign 1 comma 2 right parenthesis. Jika MN with rightwards arrow on top colon N with P yields on top equals 1 colon 2, koordinat titik
P adalah ....

  1. open parentheses 5 over 3 comma 0 comma 0 close parentheses

  2. open parentheses 7 over 3 comma minus sign 1 comma 2 close parentheses

  3. open parentheses 7 comma 8 comma minus sign 16 close parentheses

  4. open parentheses 3 comma 7 comma minus sign 14 close parentheses

  5. open parentheses negative sign 3 comma minus sign 7 comma 14 close parentheses

Pembahasan Soal:

Diketahui koordinat titik M left parenthesis 3 comma 2 comma minus sign 4 right parenthesis dan N left parenthesis 1 comma minus sign 1 comma 2 right parenthesis. Jika MN with rightwards arrow on top colon N with P yields on top equals 1 colon 2, koordinat titik
P adalah

table attributes columnalign right center left columnspacing 0px end attributes row cell MN with rightwards arrow on top colon N with P yields on top end cell equals cell 1 colon 2 end cell row cell fraction numerator MN with rightwards arrow on top over denominator N with P yields on top end fraction end cell equals cell 1 half end cell row cell fraction numerator N bond M over denominator P bond N end fraction end cell equals cell 1 half end cell row cell 2 open parentheses N bond M close parentheses end cell equals cell P bond N end cell row cell 2 N minus sign 2 M end cell equals cell P bond N end cell row P equals cell 2 N minus sign 2 M and N end cell row P equals cell 3 N minus sign 2 M end cell row P equals cell 3 open parentheses 1 comma minus sign 1 comma 2 close parentheses minus sign 2 open parentheses 3 comma 2 comma minus sign 4 close parentheses end cell row P equals cell open parentheses 3 comma minus sign 3 comma 6 close parentheses minus sign open parentheses 6 comma 4 comma minus sign 8 close parentheses end cell row P equals cell open parentheses 3 minus sign 6 comma minus sign 3 minus sign 4 comma 6 plus 8 close parentheses end cell row P equals cell open parentheses negative sign 3 comma minus sign 7 comma 14 close parentheses end cell end table

Jadi, jawaban yang tepat adalah E.

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

Y. Umi

Mahasiswa/Alumni Universitas Gadjah Mada

Terakhir diupdate 06 Juni 2021

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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

Membantu

Kurang Membantu

Apakah pembahasan ini membantu?

Belum menemukan yang kamu cari?

Post pertanyaanmu ke Tanya Jawab, yuk

Mau Bertanya

Pertanyaan yang serupa

Titik-titik sudut segitiga adalah . Titik membagi di dalam dengan perbandingan . Titik adalah titik tengah dan titik membagi   di luar dengan perbandingan . a. Carilah koordinat b. Tunjukkan segaris ...

Pembahasan Soal:

Ingat kembali,

Rumus perbandingan pada vektor

Titik straight B membagi vektor top enclose AC dengan perbandingan m colon n

AB with bar on top space colon space BC with bar on top space equals space m space colon space n

OB with bar on top equals fraction numerator n OA with bar on top space plus space m OC with bar on top over denominator n plus m end fraction

Titik straight B membagi di luar vektor top enclose AC dengan perbandingan m colon n

AB with bar on top space colon space BC with bar on top space equals space minus m space colon space n space equals space m space colon space minus n

Titik segaris

AB with bar on top space equals space k BC with bar on top

Berdasarkan penjelasan tersebut, diperoleh sebagai berikut

a. Titik straight D membagi AB di dalam dengan perbandingan 1 colon 2

stack A D with bar on top space colon space stack D B with bar on top space equals space 1 space colon space 2 stack O D with bar on top space equals fraction numerator 2 open parentheses 3 comma 1 comma 6 close parentheses plus 1 open parentheses 0 comma 4 comma negative 3 close parentheses over denominator 2 plus 1 end fraction equals fraction numerator open parentheses 6 comma 2 comma 12 close parentheses plus open parentheses 0 comma 4 comma negative 3 close parentheses over denominator 3 end fraction equals fraction numerator open parentheses 6 comma 6 comma 9 close parentheses over denominator 3 end fraction equals open parentheses 2 comma 2 comma 3 close parentheses

Sehingga koordinat straight D left parenthesis 2 comma 2 comma 3 right parenthesis

Titik straight E adalah titik tengah top enclose AC

stack O E with bar on top equals fraction numerator open parentheses 3 comma 1 comma 6 close parentheses plus open parentheses 1 comma 1 comma negative 4 close parentheses over denominator 2 end fraction equals fraction numerator open parentheses 4 comma 2 comma 2 close parentheses over denominator 2 end fraction equals open parentheses 2 comma 1 comma 1 close parentheses

Sehingga koordinat straight E left parenthesis 2 comma 1 comma 1 right parenthesis

Titik straight F membagi top enclose BC di luar dengan perbandingan 2 colon 1 

stack B F with bar on top space colon space stack C F with bar on top space equals space 2 space colon space 1 stack B F with bar on top space colon space stack F C with bar on top space equals space 2 space colon space minus 1 stack O F with bar on top space equals fraction numerator negative 1 open parentheses 0 comma 4 comma negative 3 close parentheses plus left parenthesis 2 right parenthesis open parentheses 1 comma 1 comma negative 4 close parentheses over denominator negative 1 plus 2 end fraction equals fraction numerator open parentheses 0 comma negative 4 comma 3 close parentheses plus open parentheses 2 comma 2 comma negative 8 close parentheses over denominator 1 end fraction equals open parentheses 2 comma negative 2 comma negative 5 close parentheses

Sehingga koordinat F left parenthesis 2 comma negative 2 comma negative 5 right parenthesis

b. Tunjukkan straight D comma space straight E comma space dan space straight F segaris

Apabila straight D comma space straight E comma space dan space straight F segaris, maka stack D E with bar on top space equals k space stack E F with bar on top

stack D E with bar on top equals stack O E with bar on top minus stack O D with bar on top equals left parenthesis 2 comma 1 comma 1 right parenthesis minus open parentheses 2 comma 2 comma 3 close parentheses equals open parentheses 0 comma negative 1 comma negative 2 close parentheses stack E F with bar on top equals stack O F with bar on top minus stack O E with bar on top equals open parentheses 2 comma negative 2 comma negative 5 close parentheses minus left parenthesis 2 comma 1 comma 1 right parenthesis equals open parentheses 0 comma negative 3 comma negative 6 close parentheses

table attributes columnalign right center left columnspacing 0px end attributes row cell stack D E with bar on top space end cell equals cell k space stack E F with bar on top end cell row cell open parentheses 0 comma negative 1 comma negative 2 close parentheses end cell equals cell 1 third space open parentheses 0 comma negative 3 comma negative 6 close parentheses end cell end table

Jadi, terbukti bahwa straight D comma space straight E comma space dan space straight F segaris

c. Tentukan top enclose DE space end enclose colon space top enclose EF

stack D E with bar on top equals stack O E with bar on top minus stack O D with bar on top equals left parenthesis 2 comma 1 comma 1 right parenthesis minus open parentheses 2 comma 2 comma 3 close parentheses equals open parentheses 0 comma negative 1 comma negative 2 close parentheses stack E F with bar on top equals stack O F with bar on top minus stack O E with bar on top equals open parentheses 2 comma negative 2 comma negative 5 close parentheses minus left parenthesis 2 comma 1 comma 1 right parenthesis equals open parentheses 0 comma negative 3 comma negative 6 close parentheses

stack D E with bar on top space colon space stack E F with bar on top space equals space 1 space colon space 3space space 

Roboguru

Diketahui titik P dan Q. Jika titik T pada garis PQ dan . Maka koordinat titik T adalah ....

Pembahasan Soal:

Perhatikan gambar berikut!

begin mathsize 14px style stack text PT end text with rightwards arrow on top colon stack text TQ end text with rightwards arrow on top equals text m:n end text end style 

Koordinat titik R

begin mathsize 14px style text T end text equals fraction numerator m text Q end text plus n text P end text over denominator m plus n end fraction end style 

Diketahui titik titik Pbegin mathsize 14px style open parentheses negative 1 comma 5 comma 2 close parentheses end style dan Qbegin mathsize 14px style open parentheses 5 comma negative 4 comma 17 close parentheses end style dan begin mathsize 14px style text PT:QT end text equals text 2:1 end text end style dan begin mathsize 14px style text PT:TQ end text equals text 2:-1 end text end style. Diperoleh

begin mathsize 14px style text T end text equals fraction numerator text 2 end text left parenthesis 5 comma negative 4 comma 17 right parenthesis text-end text 1 left parenthesis negative 1 comma 5 comma 2 right parenthesis over denominator 2 minus 1 end fraction text T end text equals fraction numerator left parenthesis 10 comma negative 8 comma 34 right parenthesis plus left parenthesis 1 comma negative 5 comma negative 2 right parenthesis over denominator 1 end fraction text T end text equals fraction numerator left parenthesis 10 plus 1 comma negative 8 minus 5 comma 34 minus 2 right parenthesis over denominator 1 end fraction text T end text equals left parenthesis 11 comma negative 13 comma 32 right parenthesis end style 

Jadi, koordinat titik T adalah begin mathsize 14px style text T end text open parentheses 11 comma negative 13 comma 32 close parentheses end style.

Jadi, tidak ada jawaban yang tepat.

Roboguru

Diketahui titik , , dan . Titik D terletak pada AB sehingga . Panjang CD adalah ....

Pembahasan Soal:

Diketahui: 

Titik begin mathsize 14px style straight A left parenthesis 1 comma space minus 1 comma space 2 right parenthesis end stylebegin mathsize 14px style straight B left parenthesis 4 comma space 5 comma space 2 right parenthesis end style, dan begin mathsize 14px style straight C left parenthesis 1 comma space 0 comma space 4 right parenthesis end style. Titik D terletak pada AB sehingga begin mathsize 14px style AD space colon space DB space equals space 2 space colon space 1 end style.

Ditanya:

Panjang CD.

Jawaban:

Untuk menentukan panjang CD terlebih dahulu kita cari titik D, Yaitu:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell AD space colon space DB space end cell equals cell space 2 space colon space 1 end cell row cell AD over DB end cell equals cell 2 over 1 end cell row AD equals cell 2 DB end cell row cell left parenthesis straight d minus straight a right parenthesis end cell equals cell 2 left parenthesis straight b minus straight d right parenthesis end cell row cell straight d minus straight a end cell equals cell 2 straight b minus 2 straight d end cell row cell straight d plus 2 straight d end cell equals cell straight a plus 2 straight b end cell row cell 3 straight d end cell equals cell straight a plus 2 straight b end cell row straight d equals cell 1 third left parenthesis straight a plus 2 straight b right parenthesis end cell row straight D equals cell 1 third times open square brackets open parentheses table row 1 row cell negative 1 end cell row 2 end table close parentheses plus 2 open parentheses table row 4 row 5 row 2 end table close parentheses close square brackets end cell row blank equals cell 1 third times open square brackets open parentheses table row 1 row cell negative 1 end cell row 2 end table close parentheses plus open parentheses table row 8 row 10 row 4 end table close parentheses close square brackets end cell row blank equals cell 1 third times open square brackets open parentheses table row 9 row 9 row 6 end table close parentheses close square brackets end cell row blank equals cell open parentheses table row 3 row 3 row 2 end table close parentheses end cell end table end style 

Dengan menerapkan rumus panjang vektor yaitu:

Jika begin mathsize 14px style straight A left parenthesis straight a subscript 1 comma space straight a subscript 2 comma space straight a subscript 3 right parenthesis end style dan begin mathsize 14px style straight B left parenthesis straight b subscript 1 comma space straight b subscript 2 comma space straight b subscript 3 right parenthesis end style maka:

 begin mathsize 14px style open vertical bar AB close vertical bar equals square root of left parenthesis straight b subscript 1 minus straight a subscript 1 right parenthesis squared plus left parenthesis straight b subscript 2 minus straight a subscript 2 right parenthesis squared plus left parenthesis straight b subscript 2 minus straight a subscript 2 right parenthesis squared end root end style.

Koordinat titik begin mathsize 14px style straight C left parenthesis 1 comma space 0 comma space 4 right parenthesis end style dan begin mathsize 14px style straight D left parenthesis 3 comma space 3 comma space 2 right parenthesis end style Sehingga kita peroleh:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell open vertical bar CD close vertical bar end cell equals cell square root of left parenthesis 3 minus 1 right parenthesis squared plus left parenthesis 3 minus 0 right parenthesis squared plus left parenthesis 2 minus 4 right parenthesis squared end root end cell row blank equals cell square root of 2 squared plus 3 squared plus left parenthesis negative 2 right parenthesis squared end root end cell row blank equals cell square root of 4 plus 9 plus 4 end root end cell row blank equals cell square root of 17 end cell end table end style 

Jadi, Jika diketahui titik begin mathsize 14px style straight A left parenthesis 1 comma space minus 1 comma space 2 right parenthesis end stylebegin mathsize 14px style straight B left parenthesis 4 comma space 5 comma space 2 right parenthesis end style, dan begin mathsize 14px style straight C left parenthesis 1 comma space 0 comma space 4 right parenthesis end style. Titik D terletak pada AB sehingga begin mathsize 14px style AD space colon space DB space equals space 2 space colon space 1 end style. Panjang begin mathsize 14px style open vertical bar CD close vertical bar equals square root of 17 end style.

Oleh karena itu, Jawaban yang benar adalah A

Roboguru

Diketahui . Titik  pada  sehingga . Titik  pada pertengahan . Jika  adalah titik potong  dan , tentukan  dan .

Pembahasan Soal:

Untuk menyelesaikan permasalahan ini dengan menggunakan vektor posisi.

Misalkan basisnya adalah begin mathsize 14px style CA with rightwards arrow on top end style dan undefined (vektor begin mathsize 14px style CA with rightwards arrow on top equals a end style dan begin mathsize 14px style CB with rightwards arrow on top equals b end style). Berdasarkan hal tersebut diperoleh:

begin mathsize 14px style table attributes columnalign right center left columnspacing 2px end attributes row cell AE with rightwards arrow on top end cell equals cell 1 half AB with rightwards arrow on top end cell row blank equals cell 1 half open parentheses AC with rightwards arrow on top plus CB with rightwards arrow on top close parentheses end cell row cell AE with rightwards arrow on top end cell equals cell 1 half open parentheses negative a plus b close parentheses space space space... open parentheses 1 close parentheses end cell end table end style 

 begin mathsize 14px style table attributes columnalign right center left columnspacing 2px end attributes row cell CE with rightwards arrow on top end cell equals cell CA with rightwards arrow on top plus AE with rightwards arrow on top end cell row blank equals cell a plus 1 half open parentheses negative a plus b close parentheses end cell row blank equals cell a minus 1 half a plus 1 half b end cell row blank equals cell 1 half a plus 1 half b end cell row cell CE with rightwards arrow on top end cell equals cell 1 half open parentheses a plus b close parentheses space space... open parentheses 2 close parentheses end cell end table end style   

 begin mathsize 14px style table attributes columnalign right center left columnspacing 2px end attributes row cell AD with rightwards arrow on top end cell equals cell AC with rightwards arrow on top plus CD with rightwards arrow on top end cell row blank equals cell AC with rightwards arrow on top plus 1 third CB with rightwards arrow on top end cell row cell AD with rightwards arrow on top end cell equals cell negative a plus 1 third b space space... left parenthesis 3 right parenthesis end cell end table end style   

Karena begin mathsize 14px style AZ with rightwards arrow on top end style searah dengan begin mathsize 14px style AD with rightwards arrow on top end style, (begin mathsize 14px style straight lambda end style adalah skalar) maka diperoleh:

begin mathsize 14px style table attributes columnalign right center left columnspacing 2px end attributes row cell AZ with rightwards arrow on top end cell equals cell lambda AD with rightwards arrow on top end cell row blank equals cell lambda space open parentheses negative a plus 1 third b close parentheses end cell row cell AZ with rightwards arrow on top end cell equals cell negative lambda space a plus 1 third lambda space b space space... open parentheses 4 close parentheses end cell end table end style 

Di lain pihak begin mathsize 14px style ZE with rightwards arrow on top end style  adalah kelipatan dari begin mathsize 14px style CE with rightwards arrow on top end style, (begin mathsize 14px style straight mu end style adalah skalar) sehingga begin mathsize 14px style ZE with rightwards arrow on top equals straight mu space CE with rightwards arrow on top end style, maka nilai lain dari undefined diperoleh:

begin mathsize 14px style table attributes columnalign right center left columnspacing 2px end attributes row cell AZ with rightwards arrow on top end cell equals cell AE with rightwards arrow on top plus EZ with rightwards arrow on top end cell row blank equals cell AE with rightwards arrow on top minus ZE with rightwards arrow on top end cell row blank equals cell AE with rightwards arrow on top minus straight mu space CE with rightwards arrow on top end cell row blank equals cell 1 half open parentheses negative a plus b close parentheses minus straight mu space 1 half open parentheses a plus b close parentheses end cell row blank equals cell negative 1 half a plus 1 half b minus 1 half straight mu space a minus 1 half straight mu space b end cell row cell ZE with rightwards arrow on top end cell equals cell open parentheses fraction numerator negative 1 minus straight mu over denominator 2 end fraction close parentheses a plus open parentheses fraction numerator 1 minus straight mu over denominator 2 end fraction close parentheses b space space... open parentheses 5 close parentheses end cell end table end style  

Dengan menyamakan koefisien begin mathsize 14px style a end style dan begin mathsize 14px style b end style pada persamaan (4) dan (5), diperoleh:
Koefisien begin mathsize 14px style a end stylebegin mathsize 14px style negative straight lambda equals fraction numerator negative 1 minus straight mu over denominator 2 end fraction end style  
Koefisien begin mathsize 14px style b end stylebegin mathsize 14px style 1 third straight lambda equals fraction numerator 1 minus straight mu over denominator 2 end fraction end style 

Sederhanakan koefisien begin mathsize 14px style a end style, diperoleh sebagai berikut:

begin mathsize 14px style table attributes columnalign right center left columnspacing 2px end attributes row cell negative lambda end cell equals cell fraction numerator negative 1 minus mu over denominator 2 end fraction end cell row lambda equals cell fraction numerator 1 plus mu over denominator 2 end fraction end cell end table end style 

Substitusikan hasil tersebut pada koefisien begin mathsize 14px style b end style, diperoleh:

begin mathsize 14px style table attributes columnalign right center left columnspacing 2px end attributes row cell 1 third straight lambda end cell equals cell fraction numerator 1 minus straight mu over denominator 2 end fraction end cell row cell 1 third open parentheses fraction numerator 1 plus straight mu over denominator 2 end fraction close parentheses end cell equals cell fraction numerator 1 minus straight mu over denominator 2 end fraction end cell row cell fraction numerator 1 plus straight mu over denominator 6 end fraction end cell equals cell fraction numerator 1 minus straight mu over denominator 2 end fraction end cell row cell 1 plus straight mu end cell equals cell 3 open parentheses 1 minus straight mu close parentheses end cell row cell 1 plus straight mu end cell equals cell 3 minus 3 straight mu end cell row cell straight mu plus 3 straight mu end cell equals cell 3 minus 1 end cell row cell 4 straight mu end cell equals 2 row straight mu equals cell 1 half end cell end table end style 

Substitusikan begin mathsize 14px style straight mu equals 1 half end style ke koefisien begin mathsize 14px style a end style, diperoleh:

begin mathsize 14px style table attributes columnalign right center left columnspacing 2px end attributes row straight lambda equals cell fraction numerator 1 plus straight mu over denominator 2 end fraction end cell row straight lambda equals cell fraction numerator 1 plus begin display style 1 half end style over denominator 2 end fraction end cell row straight lambda equals cell fraction numerator begin display style 3 over 2 end style over denominator 2 end fraction end cell row straight lambda equals cell 3 over 4 end cell end table end style 

Sehingga diperoleh begin mathsize 14px style straight mu equals 1 half end style dan begin mathsize 14px style straight lambda equals 3 over 4 end style.

Karena begin mathsize 14px style AZ with rightwards arrow on top equals straight lambda stack space AD with rightwards arrow on top end style dan begin mathsize 14px style straight lambda equals 3 over 4 end style, maka begin mathsize 14px style AZ with rightwards arrow on top equals 3 over 4 stack space AD with rightwards arrow on top end style.

Dengan menggunakan konsep panjang vektor, perhatikan perhitungan berikut:

 begin mathsize 14px style table attributes columnalign right center left columnspacing 2px end attributes row cell open vertical bar AZ with rightwards arrow on top close vertical bar end cell equals cell 3 over 4 open vertical bar AD with rightwards arrow on top close vertical bar end cell row AZ equals cell 3 over 4 AD end cell row cell AZ over AD end cell equals cell 3 over 4 space space open parentheses ingat space ZD equals AD minus AZ close parentheses end cell row cell AZ over ZD end cell equals cell fraction numerator 3 over denominator 4 minus 3 end fraction end cell row cell AZ over ZD end cell equals cell 3 over 1 end cell row cell AZ space colon space ZD end cell equals cell 3 space colon space 1 end cell end table end style 

Karena begin mathsize 14px style ZE with rightwards arrow on top equals straight mu space CE with rightwards arrow on top end style dan begin mathsize 14px style straight mu equals 1 half end style, maka begin mathsize 14px style ZE with rightwards arrow on top equals 1 half CE with rightwards arrow on top end style.

Dengan menggunakan konsep panjang vektor, perhatikan perhitungan berikut:

begin mathsize 14px style table attributes columnalign right center left columnspacing 2px end attributes row cell open vertical bar ZE with rightwards arrow on top close vertical bar end cell equals cell 1 half open vertical bar CE with rightwards arrow on top close vertical bar end cell row ZE equals cell 1 half CE end cell row cell ZE over CE end cell equals cell 1 half space space space open parentheses ingat space CZ equals CE minus ZE close parentheses end cell row cell CZ over ZE end cell equals cell fraction numerator 2 minus 1 over denominator 1 end fraction end cell row cell CZ over ZE end cell equals cell 1 over 1 end cell row cell CZ space colon space ZE end cell equals cell 1 space colon space 1 end cell end table end style 

Jadi, diperoleh begin mathsize 14px style AZ space colon space ZD equals 3 space colon space 1 end style dan begin mathsize 14px style CZ space colon space ZE equals 1 space colon 1 end style.

Roboguru

Diketahui titik . Jika titik P membagi KL sehingga , vektor yang diwakili oleh  adalah …

Pembahasan Soal:

Diketahui:

begin mathsize 14px style straight K left parenthesis 3 comma space 1 comma space minus 4 right parenthesis comma space straight L left parenthesis 2 comma space minus 6 comma space 8 right parenthesis space dan space straight M left parenthesis 2 comma space 8 comma space 4 right parenthesis end style
begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell KP with rightwards arrow on top space colon space PL with rightwards arrow on top end cell equals cell 2 space colon space 3 end cell row cell straight m space colon space straight n end cell equals cell 2 space colon space 3 end cell end table end style 

Mencari koordinat P:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row straight P equals cell open parentheses fraction numerator straight m times Xl plus straight n times Xk over denominator straight m plus straight n end fraction comma blank fraction numerator straight m times Yl plus straight n times Yk over denominator straight m plus straight n end fraction comma blank fraction numerator straight m times Zl plus straight n times Zk over denominator straight m plus straight n end fraction close parentheses end cell row blank equals cell open parentheses fraction numerator 2 times 2 plus 3 times 3 over denominator 2 plus 3 end fraction comma blank fraction numerator 2 times left parenthesis negative 6 right parenthesis plus 3 times 1 over denominator 2 plus 3 end fraction comma fraction numerator 2 times 8 plus 3 times left parenthesis negative 4 right parenthesis over denominator 2 plus 3 end fraction close parentheses blank end cell row blank equals cell open parentheses fraction numerator 4 plus 9 over denominator 5 end fraction comma blank fraction numerator negative 12 plus 3 over denominator 5 end fraction comma blank fraction numerator 16 plus left parenthesis negative 12 right parenthesis over denominator 5 end fraction close parentheses end cell row blank equals cell open parentheses 13 over 5 comma blank minus 9 over 5 comma blank 4 over 5 close parentheses end cell end table end style 

Mencari vektor begin mathsize 14px style PM with rightwards arrow on top end style:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell PM with rightwards arrow on top end cell equals cell straight m with rightwards arrow on top minus straight p with rightwards arrow on top end cell row blank equals cell open parentheses 2 comma space 8 comma space 4 close parentheses minus open parentheses 13 over 5 comma blank minus 9 over 5 comma blank 4 over 5 close parentheses end cell row blank equals cell open parentheses 2 minus 13 over 5 comma blank 8 minus open parentheses negative 9 over 5 close parentheses comma blank 4 minus 4 over 5 close parentheses end cell row blank equals cell open parentheses negative 3 over 5 comma blank 49 over 5 comma blank 16 over 5 close parentheses end cell end table end style 

Jadi, vektor yang diwakili oleh begin mathsize 14px style PM with rightwards arrow on top end style adalah begin mathsize 14px style open parentheses table row cell negative 3 over 5 end cell row cell 49 over 5 end cell row cell 16 over 5 end cell end table close parentheses end style.

Dengan demikian, jawaban yang tepat adalah D.

Roboguru

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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

Membantu

Kurang Membantu

Apakah pembahasan ini membantu?

Belum menemukan yang kamu cari?

Post pertanyaanmu ke Tanya Jawab, yuk

Mau Bertanya

RUANGGURU HQ

Jl. Dr. Saharjo No.161, Manggarai Selatan, Tebet, Kota Jakarta Selatan, Daerah Khusus Ibukota Jakarta 12860

Coba GRATIS Aplikasi Ruangguru

Produk Ruangguru

Produk Lainnya

Hubungi Kami

Ikuti Kami

©2021 Ruangguru. All Rights Reserved