Vectors and the Laws of Physics — Question 9

Question 9

A spatial rotation leaves the axial component unchanged

A spatial vector is $\vec A=(2,-3,6)$ m. Rotate the x and y coordinate axes counterclockwise by $90^\circ$ as viewed from positive z toward the origin, while keeping the z axis fixed. Use a right-handed orthonormal frame.

1. Find all three new components.

2. Check the full magnitude and the magnitude of the projection into the xy plane.

3. Explain why changing the z component to $-6$ m would not describe the specified rotation.

Question 9 - Solution

A spatial rotation leaves the axial component unchanged

(a) Rotate only the perpendicular pair of axes.

$$\boxed{\vec A'=(A_y,-A_x,A_z)=(-3,-2,6)\,\mathrm m}$$ The positive new x direction is old positive y, and positive new y is old negative x. Thus $(A_{x\prime},A_{y\prime},A_{z\prime})=(-3,-2,6)$ m. The z projection remains 6 m because its axis has not moved.

(b) Verify both preserved lengths.

$$A'^2=9+4+36=49\,\mathrm{m^2},\quad A'_{xy}=\sqrt{9+4}=\sqrt{13}\,\mathrm m$$ The full squared norm is $4+9+36=49\ \mathrm{m^2}$ before and $9+4+36=49\ \mathrm{m^2}$ after, giving length 7 m. The planar projection length remains $\sqrt{13}$ m. This is a rigid change of orientation, not a stretch.

(c) Respect the stated axis and handedness.

$$\hat i'=\hat j,\quad\hat j'=-\hat i,\quad\hat i'\times\hat j'=\hat k=\hat k'$$ Reversing z is an additional transformation absent from the problem. Keeping the stated new x and y directions while reversing z would also make the ordered frame left-handed. A sign change must follow from the actual basis directions, not from an expectation that all components should change.