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Exercise · Q10

Q.Define the position vector and the displacement vector of a particle. If a particle's position vectors at two instants are r⃗1\vec r_1 and r⃗2\vec r_2, derive an expression for its displacement in terms of r⃗1\vec r_1 and r⃗2\vec r_2, and explain why displacement does not depend on the choice of origin even though the position vector does.

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The position vector r⃗\vec r of a particle at any instant is the vector drawn from a chosen fixed origin OO to the particle's location at that instant. The displacement of the particle between an earlier instant t1t_1 (position vector r⃗1\vec r_1) and a later instant t2t_2 (position vector r⃗2\vec r_2) is the vector drawn directly from the earlier location to the later location, and is obtained by vector subtraction: Δr⃗=r⃗2−r⃗1\Delta\vec r = \vec r_2 - \vec r_1. If the origin is now shifted to a new point O′O', every position vector changes by the same constant vector c⃗\vec c (the vector from O′O' to OO): the new position vectors become r⃗1′=r⃗1+c⃗\vec r_1' = \vec r_1 + \vec c and r⃗2′=r⃗2+c⃗\vec r_2' = \vec r_2 + \vec c. The new displacement is r⃗2′−r⃗1′=(r⃗2+c⃗)−(r⃗1+c⃗)=r⃗2−r⃗1\vec r_2' - \vec r_1' = (\vec r_2+\vec c)-(\vec r_1+\vec c) = \vec r_2 - \vec r_1 — identical to before, because the constant c⃗\vec c cancels exactly in the subtraction. So while the position vector of a single point depends on where the origin is placed, the displacement between two points does not. [!ANSWER] Δr⃗=r⃗2−r⃗1\Delta\vec r = \vec r_2 - \vec r_1; it is origin-independent because any shift of origin adds the same constant to both r⃗1\vec r_1 and r⃗2\vec r_2, which cancels in the subtraction.

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