part a)
Vector a has magnitude 12.3 and its direction is west, while Vector b has unknown magnitude and its direction is north. This means that the two vectors form a right-angle triangle, so a and b are two sides, while a+b is the hypothenuse.
We know the magnitude of a+b, which is 14.5, so we can use the Pythagorean theorem to calculate the magnitude of b:
[tex] |b|=\sqrt{(a+b)^2-a^2}=\sqrt{(14.5)^2-(12.3)^2}=7.68 [/tex]
part b) The direction of the vector a+b relative to west can be found by calculating the tangent of the angle of the right-angle triangle described in the previous part; the tangent of the angle is equal to the ratio between the opposite side (b) and the adjacent side (a):
[tex] tan x=\frac{b}{a}=\frac{7.68}{12.3}=0.62 [/tex]
And the angle is
[tex] x=tan^{-1} (0.62)=31.8^{\circ} [/tex]
with direction north-west.
part c)
This is exactly the same problem as the one we solved in part a): the only difference here is that the hypothenuse of the triangle is now given by a-b rather than a+b. In order to find a-b, we have to reverse the direction of b, which now points south. However, the calculations to get the magnitude of b are exactly the same as before, since the magnitude of (a-b) is the same as (a+b) (14.5 units), therefore the magnitude of b is still 7.68 units.
part d)
Again, this part is equivalent to part b); the only difference is that b points now south instead of north, so the vector (a-b) has direction south-west instead of north-west as before. Since the magnitude of the vectors involved are the same as part b), we still get the same angle, [tex] 31.8^{\circ} [/tex], but this time the direction is south-west instead of north-west.
