Vectors and the Laws of Physics — Question 10

Question 10

Audit two records claimed to describe the same arrow

Lab A measures a displacement $(8,6)$ m. Lab B uses axes turned $30^\circ$ counterclockwise from A. Three candidate B records are I: $(4\sqrt3+3,3\sqrt3-4)$ m; II: $(4\sqrt3-3,4+3\sqrt3)$ m; III: $(8,8)$ m. For perpendicular axes rotated counterclockwise by $\phi$ about the same origin, a fixed vector transforms as $A_{x\prime}=A_x\cos\phi+A_y\sin\phi$ and $A_{y\prime}=-A_x\sin\phi+A_y\cos\phi$.

1. Use magnitude alone to reject any candidates that cannot describe the same arrow.

2. Apply the known transformation to select the correct record.

3. Explain what candidate II represents and why an invariant check is necessary but not sufficient.

Question 10 - Solution

Audit two records claimed to describe the same arrow

Here $Q_\phi$ transforms components when axes turn counterclockwise by $\phi$; $R_\phi=Q_{-\phi}$ rotates the arrow in fixed axes. An unsubscripted $Q$ denotes the same component transformation, and $I$ is the identity.

(a) Test the common physical length.

$$A_I^2=A_{II}^2=100\,\mathrm{m^2},\quad A_{III}^2=128\,\mathrm{m^2}\ne A^2$$ The original length is 10 m. Candidate I has squared length 100; candidate II also has squared length 100. Candidate III has length $8\sqrt2\approx11.314$ m and is immediately incompatible with a mere rotation.

(b) Use the specified orientation, not just the norm.

$$Q_{30}(8,6)=\boxed{(4\sqrt3+3,3\sqrt3-4)\,\mathrm m}\quad\text{(I)}$$ The transformation gives $A_{x\prime}=8\cos30^\circ+6\sin30^\circ=4\sqrt3+3$ and $A_{y\prime}=-8\sin30^\circ+6\cos30^\circ=3\sqrt3-4$. Therefore I is correct.

(c) Diagnose the surviving false candidate.

$$R_{30}(8,6)=(4\sqrt3-3,4+3\sqrt3)\,\mathrm m\quad\text{(II)},\qquad R_{30}=Q_{-30}$$ Candidate II is the result of physically rotating the original arrow counterclockwise by $30^\circ$ in the original fixed axes, or of rotating the axes clockwise. It preserves length but uses the wrong relative orientation for Lab B. A norm test rules out some errors, but equal magnitudes alone do not establish equality of vectors.