# Vector operations.

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?

• 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. – francescop21 Nov 23 '18 at 18:38
• Next time, I'll do that. Thanks for the suggestion. – MaverickEyedea Nov 23 '18 at 20:57

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: $$[3-2,2+4,4-6]=[1,6,-2]$$
Hint: Use the definition of velocity in a constant velocity movement$$\vec v={\vec x_2-\vec x_1\over t_2-t_1}$$