Basis vectors

For describing position of a particle we need vector coordinates r=(x,y,z). But when you go to higher physics like lagrangian formalism also basis vectors are introduced. The position vector then becomes r=xi+yj+zk, where i,j,k are basis vectors. Why do we have do do that? Why is not enough just to use the r=x+y+z?
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 One doesn't need to lagrangian mech. to enounter, xi+yj+zk, newtonian would do. x+y+z is just a number, which is different from (x,y,z), an ordered triplet. The i,j,k specify the directions.
 It's merely a notational preference.

Basis vectors

I did not mean it so simply. It has to do with equations of motion, that they are independent of basis. So any elaboration in this regard would be very much appreciated. Why do the equations of motion become independent of basis when we introduce the i,j,k vectors?
 erg... xi+yj+zk is not independent of basis, the standard basis is being used here. i,j,k are e1,e2,e3. xi+yi+zk=x(1,0,0)+y(0,1,0)+z(0,0,1)=(x,y,z) or maybe im wrong, though this is how this things were shown in some books i have at home(30 years old or so, from uni of california)
 It's because i,j,k are Cartesian rectangular co-ordiantes: the preferred basis for Newtonian mechanics. Newton's eqns. look ugly in anything other than Cartesian coordinates. So when doing Newton's physics in a Cartesian basis, the vectors themselves don't feature in the equations, so it appears as if they're not really there. When doing physics in other co-ordinate systems, the basis vectors also appear in the equations. If you tried doing non-Lagrangian (i.e. simpler Newtonian mechanics) in a non-Cartesian basis the basis vectors would still feature. e.g. try finding the equations of motion in planar polar co-ordinates using just Newton and differentiating vectors: you'll (obviously) get the same result that Lagrangian mechanics would give you. note: i've made the simple association with basis vectors and co-ordinate systems using the notion of the co-ordinate basis.
 Thanks Masudr!

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