Vector a with arrow has a magnitude of 12.3 units and points due west. vector b with arrow points due north. (a) what is the magnitude of b with arrow if a with arrow + b with arrow has a magnitude of 14.5 units? 1 7.67 units (b) what is the direction of a with arrow + b with arrow relative to due west? 2 °3 (c) what is the magnitude of b with arrow if a with arrow − b with arrow has a magnitude of 14.5 units? 4 7.67 units (d) what is the direction of a with arrow − b with arrow relative to due west? 5 °6

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:

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):

And the angle is

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, , but this time the direction is south-west instead of north-west.

positive reinforcement? ?

v=ir = 3x15 =45 v