Find the position vector of P

[kay]

A particle P is moving with a const. speed of 6m/s in a direction 2î - j - 2k. When t=0, P is at a point with position vector 3i + 4 j -7k. Find the position vector of P after (1) t seconds
(2) 4 seconds.
The solutionnstates that ' change the particle speed is constant and is a long text direction and its velocity is also constant velocity can be written as v vector = 6( v cap)
Where v cap is the unit vector in the direction 2i -j -2k. Thus,
v vector= 6 (2i -j -2k)/(√(2 sq + 1 sq + 2 sq)) = 4i -2j -4k.'
Please explain what the solution stated. :(
(It is not the complete solution, there's more; but still)

 

[Nathanael]

[Edited]

This post was not helpful, let me try again in post #4

[Edited]

 

[kay]

See, i do know what a unit vector is, but i don't know how to find a unit vector. I had thought that unit vector could be calculated by ' vector/ magnitude of vector ' but here i can't understand how it has been applied. :/

 

[Nathanael]

Right, a unit vector is "vector / magnitude of vector" which means that (if you think about it,) all it represents is a direction
(just like "i hat" or "j hat" or "k hat")

In this problem you want to find the change in the (coordinates of the) position of the particle.

Now it's clear that the object has gone 6 meters, right? If they asked you, "how far does an object going 6 m/s travel in one second" you would say, "wow, that was an easy question," right?

The tricky part about the problem is that this 6 meters is not parallel to any axis, so finding the new coordinates is not as simple as "add 6 meters"

But we can use this idea of a unit vector to find the new coordinates. We multiply the distance (of 6 meters) by the direction, ("v hat") and add it to our original position (just like we would if it was in the direction of "i hat" or "j hat" instead)

The formula "vector / magnitude of vector" just gives us a convenient way to convert the "v hat" direction into the directions of the axes ("i hat" and "j hat" and "k hat")
(That formula tells you "v hat" in terms of the other directions, so you can just multiply by "v hat" and add to the initial position like normal)


Did this help?


Edit:
I kept calling it "hat" whereas you called it "cap"

 

[vela]

See, i do know what a unit vector is, but i don't know how to find a unit vector. I had thought that unit vector could be calculated by ' vector/ magnitude of vector ' but here i can't understand how it has been applied. :/

Where v cap is the unit vector in the direction 2i -j -2k. Thus,
v vector= 6 (2i -j -2k)/(√(2 sq + 1 sq + 2 sq)) = 4i -2j -4k.'

Isn't that exactly what you said above?

 

[kay]

Right, a unit vector is "vector / magnitude of vector" which means that (if you think about it,) all it represents is a direction
(just like "i hat" or "j hat" or "k hat")

In this problem you want to find the change in the (coordinates of the) position of the particle.

Now it's clear that the object has gone 6 meters, right? If they asked you, "how far does an object going 6 m/s travel in one second" you would say, "wow, that was an easy question," right?

The tricky part about the problem is that this 6 meters is not parallel to any axis, so finding the new coordinates is not as simple as "add 6 meters"

But we can use this idea of a unit vector to find the new coordinates. We multiply the distance (of 6 meters) by the direction, ("v hat") and add it to our original position (just like we would if it was in the direction of "i hat" or "j hat" instead)

The formula "vector / magnitude of vector" just gives us a convenient way to convert the "v hat" direction into the directions of the axes ("i hat" and "j hat" and "k hat")
(That formula tells you "v hat" in terms of the other directions, so you can just multiply by "v hat" and add to the initial position like normal)Did this help?Edit:
I kept calling it "hat" whereas you called it "cap"

Yes Nathaniel, it helped a lot, thank you so much. :D
And another thing, you said:
'We multiply the distance of 6m by the direction v cap'
So one can find out the vector by multiplying the magnitude and the (unit vector of) direction as given in the question?

 

[Nathanael]

So one can find out the vector by multiplying the magnitude and the (unit vector of) direction as given in the question?

Yes, exactly. One can find the 'displacement vector' by multiplying the magnitude of the displacement and the (unit vector of the) direction.

(But of course the question didn't ask for the 'displacement vector', it asked for the new 'position vector'; so you must add the 'displacement vector' with the 'initial position vector' to get the answer)

 

[kay]

Yes. Thank you so much. :)