Question

If position vector is given as \(\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}\) and if its signs are reversed, then which of the following physical quantity remains unaffected?

• A. Velocity
• B. Displacement
• C. Acceleration
• D. Torque

Hint:

Quantities involving a {cross product} of two vectors (such as torque or angular momentum) can remain unchanged when both vectors reverse direction.

Answer 1

The Correct Option is D

Concept:
Reversing the sign of the position vector: \[ \vec{r} \rightarrow -\vec{r} \] represents a change of origin or spatial inversion. The effect on a physical quantity depends on how it is defined in terms of \(\vec{r}\).
Step 1: Effect on Different Physical Quantities
Velocity:
\[ \vec{v} = \frac{d\vec{r}}{dt} \Rightarrow \vec{v} \rightarrow -\vec{v} \] Velocity changes sign, hence it is affected
. Displacement:
Displacement is directly related to position vector. \[ \vec{s} \rightarrow -\vec{s} \] Thus, displacement is affected
. Acceleration:
\[ \vec{a} = \frac{d^2\vec{r}}{dt^2} \Rightarrow \vec{a} \rightarrow -\vec{a} \] Acceleration also changes sign, hence it is affected
. Torque:
Torque is defined as: \[ \vec{\tau} = \vec{r} \times \vec{F} \] If \(\vec{r} \rightarrow -\vec{r}\), then: \[ \vec{\tau} = (-\vec{r}) \times \vec{F} = -(\vec{r} \times \vec{F}) \] However, under spatial inversion, force \(\vec{F}\) also reverses direction: \[ \vec{F} \rightarrow -\vec{F} \] Thus, \[ \vec{\tau} = (-\vec{r}) \times (-\vec{F}) = \vec{r} \times \vec{F} \] So, torque remains unchanged
.