Answer: D. 5.8 cm, 31 degrees South of the West axis
Explanation:
We have two vectors:
Vector 1: (5,0) W
Vector 2: (3,0) S
We need to find the magnitude and direction of the resultant vector:
For the magnitude we will use the formula to calculate the distance [tex]d[/tex] between two points:
[tex]d=\sqrt{{(Y2-Y1)}^{2} +{(X2-X1)}^{2}}[/tex] (1)
[tex]d=\sqrt{{(-5 cm - 0 cm)}^{2} +{(-3 cm - 0 cm)}^{2}}[/tex] (2)
[tex]d=\sqrt{34 cm^{2}}[/tex] (3)
[tex]d=5.83 cm[/tex] (4) This is the magnitude of the vector
For the direction, we will use the following formula. Taking into account that if we draw a vector diagram with the resultant vector, we will have a right triangle (whose hypotenuse [tex]d[/tex] was already calculated) and we can use trigonometry in this case:
[tex]sin \theta=\frac{Opposite-side}{d}[/tex] (5)
[tex]sin \theta=\frac{3}{5.83}[/tex] (6)
[tex]\theta= sin^{-1} \frac{3}{5.83}[/tex] (7)
[tex]\theta= 30.96\° \approx 31\°[/tex] (8)
Hence, the direction of this resultant vector is [tex]31\°[/tex] South of the West axis.
