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At 12:00 pm, a spaceship is at position [3,2,4]km away from the origin with respect to some 3 dimensional co ordinate system. The ship is travelling with velocity [-1,2,-3]km/h What is the location of the spaceship after 2 hours have passed?

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  • $\begingroup$ Hi and welcome to math.SE. You're more likely to be well received if you show us your efforts and attempts, instead of merely stating your question. $\endgroup$ – francescop21 Nov 23 '18 at 18:38
  • $\begingroup$ Next time, I'll do that. Thanks for the suggestion. $\endgroup$ – MaverickEyedea Nov 23 '18 at 20:57
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Initial position: $$[3,2,4]$$ Velocity: $$[-1,2,-3]$$ So, in $2$ hours, it travels $-2,4$ and $-6$ kms resp. in $3-D$ co-ordinate system.

It's final position becomes: $$[3-2,2+4,4-6]=[1,6,-2]$$

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  • $\begingroup$ Thank you so much for your answer. i was thinking too much and using vector projections to do it. You've been helpful. Thanks, again. $\endgroup$ – MaverickEyedea Nov 23 '18 at 20:56
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Hint: Use the definition of velocity in a constant velocity movement$$\vec v={\vec x_2-\vec x_1\over t_2-t_1}$$

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