Iklan

Iklan

Pertanyaan

Diketahui koordinat A ( − 1 , 3 ) dan B ( 3 , 6 ) . c. Tentukan vektor satuan AB .

Diketahui koordinat  dan .

c. Tentukan vektor satuan .

Iklan

D. Rajib

Master Teacher

Mahasiswa/Alumni Universitas Muhammadiyah Malang

Jawaban terverifikasi

Iklan

Pembahasan

Diketahui : Ditanya : Tentukanvektor satuan dari Jawab : Maka vektor satuan adalah :

Diketahui :

begin mathsize 14px style A left parenthesis negative 1 comma 3 right parenthesis end style 

begin mathsize 14px style B left parenthesis 3 comma 6 right parenthesis end style

Ditanya :

Tentukan vektor satuan dari begin mathsize 14px style stack A B with rightwards arrow on top end style 

Jawab :

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 stack A B with rightwards arrow on top equals open square brackets table row 3 row 6 end table close square brackets minus open square brackets table row cell negative 1 end cell row 3 end table close square brackets stack A B with rightwards arrow on top equals open square brackets table row cell 3 plus 1 end cell row cell 6 minus 3 end cell end table close square brackets stack A B with rightwards arrow on top equals open square brackets table row 4 row 3 end table close square brackets end style 

Maka vektor satuan begin mathsize 14px style stack A B with rightwards arrow on top end style adalah :

begin mathsize 14px style V e k t o r space s a t u a n space stack A B with rightwards arrow on top equals fraction numerator stack A B with rightwards arrow on top over denominator open vertical bar stack A B with rightwards arrow on top close vertical bar end fraction V e k t o r space s a t u a n space stack A B with rightwards arrow on top equals fraction numerator 1 over denominator square root of 4 squared plus 3 squared end root end fraction open square brackets table row 4 row 3 end table close square brackets V e k t o r space s a t u a n space stack A B with rightwards arrow on top equals fraction numerator 1 over denominator square root of 16 plus 9 end root end fraction open square brackets table row 4 row 3 end table close square brackets V e k t o r space s a t u a n space stack A B with rightwards arrow on top equals fraction numerator 1 over denominator square root of 25 end fraction open square brackets table row 4 row 3 end table close square brackets end style 

 

 

 

Perdalam pemahamanmu bersama Master Teacher
di sesi Live Teaching, GRATIS!

28

Alia Alodia

Bantu banget

Iklan

Iklan

Pertanyaan serupa

Misalkan P = 8 i + 15 j ​ . a. Carilah ∣ ∣ ​ P ∣ ∣ ​ b. Tunjukkan bahwa ∣ ∣ ​ P ∣ ∣ ​ P ​ adalah vektor satuan.

5

4.5

Jawaban terverifikasi

RUANGGURU HQ

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

Coba GRATIS Aplikasi Roboguru

Coba GRATIS Aplikasi Ruangguru

Download di Google PlayDownload di AppstoreDownload di App Gallery

Produk Ruangguru

Hubungi Kami

Ruangguru WhatsApp

+62 815-7441-0000

Email info@ruangguru.com

[email protected]

Contact 02140008000

02140008000

Ikuti Kami

©2024 Ruangguru. All Rights Reserved PT. Ruang Raya Indonesia