Question

let |r| = 6 at an angle of 30° and |s| = 11 at an angle of 225°. which expression represents |r − s|?
○ √(6² + 11² − 2(6)(11) cos(15°))
○ √(6² + 11² − 2(6)(11) cos(45°))
○ √(6² + 11² − 2(6)(11) cos(165°))
○ √(6² + 11² − 2(6)(11) cos(255°))

Explanation:

Step1: Find the angle between vectors \(r\) and \(s\)

The angle of vector \(r\) is \(\theta_{r}=30^{\circ}\), and the angle of vector \(s\) is \(\theta_{s} = 225^{\circ}\). The angle \(\alpha\) between the two vectors is \(\alpha=\theta_{s}-\theta_{r}\) (when \(\theta_{s}>\theta_{r}\)). So \(\alpha=225^{\circ}-30^{\circ}=195^{\circ}\). But in the law of cosines for \(|\vec{r}-\vec{s}|\), the angle used is \(360^{\circ}-\alpha\) (because the law of cosines formula \(c^{2}=a^{2}+b^{2}-2ab\cos C\) where \(C\) is the angle between the two - sides \(a\) and \(b\) in the triangle formed by \(\vec{r}\), \(\vec{s}\), and \(\vec{r}-\vec{s}\)). Another way: The formula for \(|\vec{A}-\vec{B}|=\sqrt{|\vec{A}|^{2}+|\vec{B}|^{2}-2|\vec{A}||\vec{B}|\cos\theta}\), where \(\theta\) is the angle between \(\vec{A}\) and \(\vec{B}\). If we consider the geometric interpretation of vectors, the angle between \(\vec{r}\) and \(\vec{s}\) for the law of cosines (to find \(|\vec{r}-\vec{s}|\)) is \(360^{\circ}-(225^{\circ} - 30^{\circ})=165^{\circ}\).

Step2: Apply the law of cosines

The law of cosines formula for the magnitude of the difference of two vectors \(\vec{r}\) and \(\vec{s}\) is \(|\vec{r}-\vec{s}|=\sqrt{|\vec{r}|^{2}+|\vec{s}|^{2}-2|\vec{r}||\vec{s}|\cos\theta}\), where \(|\vec{r}| = 6\), \(|\vec{s}|=11\), and \(\theta\) is the angle between \(\vec{r}\) and \(\vec{s}\). Substituting \(|\vec{r}| = 6\), \(|\vec{s}| = 11\), and \(\theta = 165^{\circ}\) into the formula, we get \(|\vec{r}-\vec{s}|=\sqrt{6^{2}+11^{2}-2(6)(11)\cos(165^{\circ})}\).

Answer:

\(\sqrt{6^{2}+11^{2}-2(6)(11)\cos(165^{\circ})}\) (the third option)