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 formula is \(|\vec{r}-\vec{s}|=\sqrt{|\vec{r}|^{2}+|\vec{s}|^{2}-2|\vec{r}||\vec{s}|\cos\beta}\), where \(\beta = 180^{\circ}-\alpha\) (the angle used in the law - of - cosines formula for the magnitude of the difference of two vectors). Substituting \(\alpha = 195^{\circ}\), we get \(\beta=180^{\circ}-(225^{\circ} - 30^{\circ})=165^{\circ}\)

Step2: Apply the law of cosines

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

Answer:

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