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a positron undergoes a displacement delta'r'vector=2.i^-3.0j^+6.0k^.ending with position vector'r'=3.0j^-4.0k^ in meter . what was the posistron's initial position vector?

a positron undergoes a displacement delta'r'vector=2.i^-3.0j^+6.0k^.ending with position vector'r'=3.0j^-4.0k^ in meter . what was the posistron's initial position vector?

Grade:11

4 Answers

Chetan Mandayam Nayakar
312 Points
10 years ago

initial position vector=3j-4k-(2i-3j+6k)= -2.0i+6.0j-10.0k meter

vikas askiitian expert
509 Points
10 years ago

final position vector(rf)-initial position vector(ri)=displacement vector(r)

     ri=rf-r

       =3j - 4k - (2i-3j+6k)

initial position vector =-2i+6j-10k

Tarun Saxena
29 Points
10 years ago

it will be  = -2.i^+6.j^-10.k^

Sudheesh Singanamalla
114 Points
10 years ago

2i + 6j - 10k is the required initial vector.

final vector - initial vector = delta'r'

initial vector = final vector - delta 'r'.

just subtract and get the answer :)

 

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