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Question

A particle P is moving with a constant speed of 6m/s in a direction 2^i^j2^k. When t=0,P is at a point whose position vector is 3^i+4^j7^k. Find the position vector of the particle P after 4 seconds.

A
18^i14^j23^k
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B
19^i4^j23^k
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C
19^i+4^j23^k
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D
19^i4^j+23^k
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Solution

The correct option is A 19^i4^j23^k
We have
Unit vector in the direction of velocity=2^i^j2^k4+1+4
So velocity=6×2^i^j2^k3=4^i2^j2^j
Hence displacement in 4sec=(4^i2^j4^k)×4=16^i8^j16^k
Displacement hence = final position Initial position
Final position=(16^i8^j16^k)+(3^i+4^j7^k)=(19^i4^j23^k)

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