Components of Vectors — Question 1

Question 1

A component is a signed projection

A displacement has magnitude 12 m and direction $150^\circ$ counterclockwise from the positive $x$ axis. The axes are perpendicular, with east positive $x$ and north positive $y$.

1. Find its signed Cartesian components and explain the sign of each.

2. Recover the magnitude from the components. Is $|A_x|+|A_y|$ the magnitude?

3. An observer describes the direction as $30^\circ$ north of west. Resolve it using that angle and reconcile the results.

Question 1 - Solution

A component is a signed projection

(a) Project onto the specified axes.

$$A_x=12\cos150^\circ=-6\sqrt3,\quad A_y=12\sin150^\circ=6\quad\mathrm m$$ $A_x=A\cos\theta=-6\sqrt3\ \mathrm m\approx-10.39\ \mathrm m$ and $A_y=A\sin\theta=6.00\ \mathrm m$. The westward projection is negative; the northward projection is positive. A negative component does not make the magnitude negative.

(b) Use the right triangle.

$$\boxed{A=\sqrt{108+36}=12\,\mathrm m},\quad|A_x|+|A_y|=6(\sqrt3+1)\,\mathrm m>A$$ $\sqrt{A_x^2+A_y^2}=\sqrt{108+36}=12\ \mathrm m$. The sum of absolute components is $6(\sqrt3+1)\ \mathrm m\approx16.39\ \mathrm m$; that is the length of an axis-aligned two-leg route, not the straight arrow.

(c) Change the reference direction, not the vector.

$$A_x=-12\cos30^\circ=-6\sqrt3,\quad A_y=12\sin30^\circ=6\quad\mathrm m$$ From west, the horizontal leg is adjacent: $A_x=-12\cos30^\circ=-6\sqrt3$ m and $A_y=12\sin30^\circ=6$ m. The signs come from the quadrant. Substituting $30^\circ$ blindly into formulas referenced to positive $x$ would give the wrong direction.