Question

let ( mathbf{a}=(5,-9) ) and ( mathbf{b}=(-3,1) ), and ( mathbf{c}=mathbf{b}-mathbf{a} ). what is the magnitude and direction angle of ( mathbf{c} )?
( |mathbf{c}|=12.8, \theta=126.7^{circ} )
( |mathbf{c}|=18.0, \theta=128.7^{circ} )
( |mathbf{c}|=12.8, \theta=309.7^{circ} )
( |mathbf{c}|=18.0, \theta=308.7^{circ} )

Explanation:

Step1: Calculate the vector \(c\)

Given \(a=(5,-9)\) and \(b = (-3,1)\), then \(c=b - a=(-3 - 5,1-(-9))=(-8,10)\)

Step2: Calculate the magnitude of \(c\)

The magnitude of a vector \((x,y)\) is \(|c|=\sqrt{x^{2}+y^{2}}\). For \(c=(-8,10)\), \(|c|=\sqrt{(-8)^{2}+10^{2}}=\sqrt{64 + 100}=\sqrt{164}\approx12.8\)

Step3: Calculate the direction angle \(\theta\)

The formula for the direction angle of a vector \((x,y)\) is \(\tan\theta=\frac{y}{x}\). Here \(x=-8,y = 10\), so \(\tan\theta=\frac{10}{-8}=-\frac{5}{4}\). Since the vector \((-8,10)\) is in the second quadrant (\(x<0,y>0\)), \(\theta=\arctan(-\frac{5}{4})+180^{\circ}\). \(\arctan(\frac{5}{4})\approx51.3^{\circ}\), so \(\theta\approx180 - 51.3=128.7^{\circ}\)

Answer:

\(|c| = 12.8,\theta=128.7^{\circ}\) (First option)