Answer:
magnitude=34.45 m
direction=[tex]55.52\°[/tex]
Explanation:
Assuming the initial point P1 of this vector is at the origin:
P1=(X1,Y1)=(0,0)
And knowing the other point is P2=(X2,Y2)=(19.5,28.4)
We can find the magnitude and direction of this vector, taking into account a vector has a initial and a final point, with an x-component and a y-component.
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{{(28.4 m - 0 m)}^{2} +{(19.5 m - 0m)}^{2}}[/tex] (2)
[tex]d=\sqrt{1186.81 m^{2}}[/tex] (3)
[tex]d=34.45 m[/tex] (4) This is the magnitude of the vector
For the direction, which is the measure of the angle the vector makes with a horizontal line, we will use the following formula:
[tex]tan \theta=\frac{Y2-Y1}{X2-X1}[/tex] (5)
[tex]tan \theta=\frac{24.8 m - 0m}{19.5 m - 0m}[/tex] (6)
[tex]tan \theta=\frac{24.8}{19.5}[/tex] (7)
Finding [tex]\theta[/tex]:
[tex]\theta= tan^{-1}(\frac{24.8}{19.5})[/tex] (8)
[tex]\theta= 55.52\°[/tex] (9) This is the direction of the vector
