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Alok gupta Grade: 11
        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?
6 years ago

Answers : (4)

Chetan Mandayam Nayakar
312 Points
										

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

6 years ago
vikas askiitian expert
510 Points
										

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

6 years ago
Tarun Saxena
29 Points
										

it will be  = -2.i^+6.j^-10.k^

6 years ago
Sudheesh Singanamalla
114 Points
										

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 :)


 


Do approve if this helps you

6 years ago
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