How do I answer this vector question?

Question

A jogger travels a route that has two parts. The first is a displacement A of 2.80 km due south, and the second involves a displacement B that points due east.
(a) The resultant displacement A + B has a magnitude of 4.56 km. What is the magnitude of B? __________ km
What is the direction of A + B relative to due south? _______degrees (east of south)
(b) Suppose that A - B had a magnitude of 4.56 km What then would be the magnitude of B and what is the direction of A - B relative to due south?
_________ km
direction of A - B relative to due south
__________ degrees (west of south)

Answer 1

draw a diagram that is relatively to scale, then use pathagoras on the first one (a^2+b^2=c^2) u can also use soh cah toa for the other angle ones. also use the sin law (sin a/a = sinb/b) etc

Answer 2

well draw it out and you'll see that if you head due south then head due east ... you'll form a ...hmm triangle of vectors if you pardon my lazy terminology; a right triangle at that..... *cough*
[draw your A vector straight down if you'd like, your B vector straight to the right with it's initial point where A left off .... then draw a resultant or displacement vector from your start point to where vector B terminated .... and you are told the magnitude ...]

since you'll have a triangle of sorts you can determine other resultant angles.

anyhow, gl & visualization or sketching can be your best friend sometimes :)

Answer 3

Use the pythagorean theorem, A^2 + B^2 = C^2, to get B:

2.8^2 + B^2 = 4.56^2
B^2 = 4.56^2 - 2.8^2
B = 3.599

To get the angle t,

sin t = B/4.56 = 3.599/4.56
t = arcsin (3.599/4.56) = 52.118

Not completely sure I understand the question in part (b) but it's trivial if my interpretation is correct. You try and figure it out.

Answer 4

Just draw the picture. You have a right triangle and are given two of the sides. You can easily find the 3rd side. To find degrees just use the arcsin or arccos to find the angle.