Question

let ( |u| = 10 ) at an angle of ( 45^{circ} ) and ( |w| = 13 ) at an angle of ( 150^{circ} ), and ( w = u + v ). what is the magnitude and direction angle of ( w )?
( |w| = 9.4; \theta = 72.9^{circ} )
( |w| = 9.4; \theta = 107.1^{circ} )
( |w| = 14.2; \theta = 72.9^{circ} )
( |w| = 14.2; \theta = 107.1^{circ} )

Explanation:

Step1: Convert vectors to component form

For vector \(u\) with \(|u| = 10\) and \(\theta_{u}=45^{\circ}\), the components are \(u_x = |u|\cos\theta_{u}=10\cos45^{\circ}=10\times\frac{\sqrt{2}}{2}\approx7.07\), \(u_y = |u|\sin\theta_{u}=10\sin45^{\circ}=10\times\frac{\sqrt{2}}{2}\approx7.07\).
For vector \(v\) with \(|v| = 13\) and \(\theta_{v}=150^{\circ}\), the components are \(v_x = |v|\cos\theta_{v}=13\cos150^{\circ}=13\times(-\frac{\sqrt{3}}{2})\approx - 11.26\), \(v_y = |v|\sin\theta_{v}=13\sin150^{\circ}=13\times\frac{1}{2}=6.5\).

Step2: Find the components of \(w = u + v\)

\(w_x=u_x + v_x\approx7.07-11.26=-4.19\), \(w_y=u_y + v_y\approx7.07 + 6.5 = 13.57\).

Step3: Calculate the magnitude of \(w\)

Using the formula \(|w|=\sqrt{w_x^{2}+w_y^{2}}\), \(|w|=\sqrt{(-4.19)^{2}+13.57^{2}}=\sqrt{17.5561 + 184.1449}=\sqrt{201.701}\approx14.2\).

Step4: Calculate the direction angle \(\theta\) of \(w\)

Using the formula \(\tan\theta=\frac{w_y}{w_x}\), \(\tan\theta=\frac{13.57}{-4.19}\approx - 3.24\). Since \(w_x<0\) and \(w_y>0\), \(\theta = 180^{\circ}-\arctan(3.24)\). \(\arctan(3.24)\approx73^{\circ}\), so \(\theta\approx107.1^{\circ}\).

Answer:

\(|w| = 14.2;\theta = 107.1^{\circ}\) (corresponds to the option \(|w| = 14.2;\theta=107.1^{\circ}\))