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Question

A particle is projected with initial velocity v=3^i+4^j+12^k & a = 8^i+6^j4^k then path of particle is

A
parabolic
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B
curved but not parabolic
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C
straight line
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D
not defined as vector in 3D
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Solution

The correct option is A parabolic
a.v=(8^i+6^j4^k).(3^i+4^j+12^k)

=24+24-48=0

Hence a is to v and not parallel. So path will not be straight line. At any other angle between constant acceleration and velocity only parabolic path is possible.



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