Question

let |u| = 4 at an angle of 210° and |v| = 9 at an angle of 315°, and w = u − v. what is the magnitude and direction angle of w?
|w| = 5.5; θ = 156.1°
|w| = 5.5; θ = 203.9°
|w| = 10.8; θ = 156.1°
|w| = 10.8; θ = 203.9°

Explanation:

Step1: Convert vectors to component form

For vector \(u\) with \(|u| = 4\) and \(\theta_{u}=210^{\circ}\):
The \(x\) - component \(u_{x}=|u|\cos\theta_{u}=4\cos210^{\circ}=4\times(-\frac{\sqrt{3}}{2})=- 2\sqrt{3}\)
The \(y\) - component \(u_{y}=|u|\sin\theta_{u}=4\sin210^{\circ}=4\times(-\frac{1}{2})=-2\)

For vector \(v\) with \(|v| = 9\) and \(\theta_{v}=315^{\circ}\):
The \(x\) - component \(v_{x}=|v|\cos\theta_{v}=9\cos315^{\circ}=9\times\frac{\sqrt{2}}{2}=\frac{9\sqrt{2}}{2}\)
The \(y\) - component \(v_{y}=|v|\sin\theta_{v}=9\sin315^{\circ}=9\times(-\frac{\sqrt{2}}{2})=-\frac{9\sqrt{2}}{2}\)

Since \(w = u - v\), then \(w_{x}=u_{x}-v_{x}=-2\sqrt{3}-\frac{9\sqrt{2}}{2}\approx - 3.464 - 6.364=-9.828\)
\(w_{y}=u_{y}-v_{y}=-2+\frac{9\sqrt{2}}{2}\approx - 2 + 6.364 = 4.364\)

Step2: Calculate the magnitude of \(w\)

The magnitude \(|w|=\sqrt{w_{x}^{2}+w_{y}^{2}}\)
\(w_{x}^{2}+w_{y}^{2}=(-9.828)^{2}+(4.364)^{2}=96.59 + 19.04=115.63\)
\(|w|=\sqrt{115.63}\approx10.8\)

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

\(\tan\theta=\frac{w_{y}}{w_{x}}=\frac{4.364}{-9.828}\)
Since \(w_{x}<0\) and \(w_{y}>0\), \(\theta\) is in the second quadrant.
\(\theta = 180^{\circ}+\arctan(\frac{4.364}{-9.828})\)
\(\arctan(\frac{4.364}{9.828})\approx23.9^{\circ}\)
\(\theta\approx180^{\circ}-23.9^{\circ}=156.1^{\circ}\)

Answer:

\(|w| = 10.8;\theta = 156.1^{\circ}\) (the third option)