Question

find the magnitude and direction angle (to the nearest tenth) for each vector. give the measure of the direction angle as an angle in 0,360°.
(-2√2, -2√2)

a. 4√2; 135°
b. 8; 45°
c. 4; 45°
d. 4; 225°

Explanation:

Step1: Calculate the magnitude of the vector

For a vector \((x,y)\), the magnitude \(r=\sqrt{x^{2}+y^{2}}\). Here \(x = - 2\sqrt{2}\), \(y=-2\sqrt{2}\).

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Step2: Calculate the direction angle \(\theta\)

We know that \(\tan\theta=\frac{y}{x}\). Here \(\tan\theta=\frac{-2\sqrt{2}}{-2\sqrt{2}} = 1\). Since \(x<0\) and \(y<0\), the vector lies in the third - quadrant.
If \(\tan\theta = 1\), the reference angle \(\theta_{ref}=45^{\circ}\). Then \(\theta=180^{\circ}+45^{\circ}=225^{\circ}\)

Answer:

D. \(4;225^{\circ}\)