Components of Vectors — Question 6

Question 6

Three projections describe a spatial arrow

A spatial displacement has perpendicular-axis components $(A_x,A_y,A_z)=(2,-3,6)$ m. Let $\alpha,\beta,\gamma$ denote its angles with the positive $x,y,z$ axes, each in $[0^\circ,180^\circ]$.

1. Find the displacement magnitude and the length of its projection into the $xy$ plane.

2. Find the three axis angles using $A_x=A\cos\alpha$ and analogous relations.

3. Check the relation among the squared direction cosines. Explain why the angle with positive $y$ is obtuse.

Question 6 - Solution

Three projections describe a spatial arrow

(a) Separate a planar projection from the full arrow.

$$A_{xy}=\sqrt{2^2+(-3)^2}=\sqrt{13}\,\mathrm m,\quad\boxed{A=\sqrt{13+36}=7\,\mathrm m}$$ The $xy$ projection has length $\sqrt{2^2+(-3)^2}=\sqrt{13}$ m. The full length is $A=\sqrt{13+6^2}=7$ m. A projection generally discards a component and is shorter than the original.

(b) Normalize each projection to obtain its cosine.

$$(\cos\alpha,\cos\beta,\cos\gamma)=\tfrac17(2,-3,6),\quad(\alpha,\beta,\gamma)=(73.398^\circ,115.377^\circ,31.003^\circ)$$ $\alpha=\cos^{-1}(2/7)\approx73.40^\circ$, $\beta=\cos^{-1}(-3/7)\approx115.38^\circ$, and $\gamma=\cos^{-1}(6/7)\approx31.00^\circ$. These are three axis angles, not three independent planar bearings.

(c) Check a necessary geometric identity.

$$\cos^2\alpha+\cos^2\beta+\cos^2\gamma=\frac{4+9+36}{49}=1,\quad\cos\beta<0\Rightarrow\beta>90^\circ$$ $\cos^2\alpha+\cos^2\beta+\cos^2\gamma=(4+9+36)/49=1$. The negative $y$ projection requires $\cos\beta<0$, so $\beta>90^\circ$. Using $|A_y|$ in the cosine formula would lose the direction toward negative $y$.