One of the first things you learn when starting college-level Physics is vectors; right in the very first chapter. Vectors are also taught at some level in some Algebra and Trigonometry textbooks. That’s where I’ve reached to in order to get a more solid understanding of them.
Let’s try this problem:
Find the component form of the vector that represents the velocity of an airplane descending at a speed of 100 mph at an angle of 30 degrees below the horizontal.
Let’s look at a graph of the situation:
To start, we know the magnitude of the vector (100) and its angle (180° + 30° = 210°):
\begin{align*}
\text{The components of vector } \overrightarrow{\mathbf{v}} \text{ are obtained from:} \\ \text{The product of its magnitude and angle of the unit vectors: } \overrightarrow{\mathbf{v}} &= \|\overrightarrow{\mathbf{v}}\| \cos\left(\theta\right)\,\mathbf{i} + \|\overrightarrow{\mathbf{v}}\| \sin\left(\theta\right)\,\mathbf{j} \\
\text{This angle is in the 3rd quadrant: } \; &= 100 \left(\cos 210°\right)\mathbf{i} + 100 \left(\sin 210°\right)\mathbf{j} \\
\text{The values of Trig functions of common angles should be memorized: } &= 100 \left(-\frac{\sqrt{3}}{2}\right)\mathbf{i} + 100 \left(-\frac{1}{2}\right)\mathbf{j} \\
&=\, – 50 \sqrt{3} \; \mathbf{i} \; – 50 \; \mathbf{j} \\
&=\, < – 50 \sqrt{3},\, – 50 >
\end{align*}
We have arrived at the answer: \(\boldsymbol{< – 50 \sqrt{3},\, – 50 >}\), which happens to be correct. Now let’s do it the other way. Let’s assume we are given this vector; can we extract from it its magnitude and angle?
We get the magnitude of the vector using the Pythagorean theorem:
\[\|\overrightarrow{\mathbf{v}}\| = \sqrt{\left( – 50 \sqrt{3}\right)^{2} + \left( – 50 \right)^{2}} = \sqrt{7500 + 2500} = \sqrt{10000} = 100\]
From the sign of the vector components we know the angle is in the 3rd quadrant:
\[\tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{ – 50}{ – 50 \sqrt{3}} = \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3}\] \[\theta = 180° + \arctan \left(\frac{\sqrt{3}}{3}\right) = 180° + 30° = 210°\]