Components of Vectors — Question 3

Question 3

Inverse tangent needs a quadrant

A displacement sensor reports $A_x=-9$ m and $A_y=-12$ m in an east–north coordinate system. Values are exact.

1. Find the magnitude and a direction between $0^\circ$ and $360^\circ$.

2. Explain why the calculator value $\tan^{-1}(A_y/A_x)=53.13^\circ$ is insufficient.

3. Give the direction south of west, and specify the components of the reversed displacement.

Question 3 - Solution

Inverse tangent needs a quadrant

(a) Reconstruct the right triangle.

$$\boxed{A=\sqrt{(-9)^2+(-12)^2}=15\,\mathrm m},\quad\theta=180^\circ+\tan^{-1}(4/3)=233.130^\circ$$ $A=\sqrt{81+144}=15\ \mathrm m$. Both components are negative, so the arrow is in quadrant III. Thus $\theta=180^\circ+\tan^{-1}(12/9)\approx233.13^\circ$.

(b) A ratio loses two signs.

$$\tan(\theta+180^\circ)=\tan\theta,\quad\theta=\operatorname{atan2}(-12,-9)\bmod360^\circ$$ The quotient is positive for both quadrant I and quadrant III. Ordinary inverse tangent returns a principal value and cannot choose between opposite arrows. Inspect both signs, or use a two-argument angle function such as $\operatorname{atan2}(A_y,A_x)$ and convert negative angles to the chosen range.

(c) Reverse both projections.

$$-\vec A=(9,12)\,\mathrm m,\quad|-\vec A|=15\,\mathrm m,\quad\theta_{-A}=53.130^\circ$$ The original direction is $53.13^\circ$ south of west. The reverse has $(9,12)$ m, the same 15 m magnitude, and direction $53.13^\circ$ counterclockwise from east. Reversing only one sign would reflect the arrow across an axis rather than reverse it.