Re: vector components

On 2007-09-21 02:35:49 -0400, conrad <conrad@xxxxxxxxxx> said:

Two airplanes leave an airport at the
same time. The first flies 175 km/h in a direction of
305.6 deg. The second flies 220 km/h in a direction of
195.5 deg. After 2 hr, how far apart are the planes?

My first method: use the law of cosines
This forms a triangle where one angle
is 110.1 deg and two of the adjacent
sides are 350 km and 440 km.

Answer: 649.57 km

My second method: represent each
vector in terms of its components
then add the component parts to
find the magnitude of the resultant
vector.

The first forms an acute angle with
the North axis that is 54.4 deg.

w = sin(54.4)*350

Ok. Assuming West is positive, then you don't need a minus sign here.

x = cos(54.4)*350

Assuming North is positive, this is correct.


The second forms an acute
angle with the south axis
that is 15.5 deg.

y = sin(15.5)*440

Again, assuming West is positive, then you don't need a minus sign here.

z = cos(15.5)*440

This is incorrect. Since you have implicitly assumed north is positive, this value should have a minus sign. So, it should be
z = -cos(15.5)*440.


<w, x> - <y, z> = <x - z, w - y>

|<w - y, x - z>| = ([w - y]^2 + [x - z]^2)^1/2

Which comes out to 256.48 km

I believe this should give you the same answer as the first method. Assuming your arithmetics were correct there.


This conflicts with my answer from
method 1, which is the correct solution.


Where did I error in my thinking?

--

-kira

.