Vectors and Parametric Equations
PhD. in Mathematics
Norm was 4th at the 2004 USA Weightlifting Nationals! He still trains and competes occasionally, despite his busy schedule.
Let’s try a harder example. Here we have a plane travelling 350 per hour at a heading of 210 degrees. The wind is 70 miles per hour from 270 degrees. Find the course and ground speed. Remember all these directions are given as clockwise from the north. So for example the air plane at a heading of 210 degrees, we have to measure that angle from north.
Now 210 degrees is 30 degrees past 180. So that’s going to look something like this, I’m just going to approximate it. So this is my heading 210 degrees and my area speed is 250. That’s the magnitude of this vector. So let me label this, 350. Now what about the wind? The wind is 70 miles per hours from 270 degrees so it’s coming from 270 degrees, that’s this angle over here.
So it's blowing into this direction. I usually over-exaggerate the length and the width vector just because it will make the picture easier to understand. This is going to be a magnitude of 70, and again this is 270 degrees.
Now I need to find the course in ground speed. The course in ground speed are going to be the resultant of these two vectors. So let me first copy this vector down here to show you what the result looks like. So I’m just going to eyeball this. It’s about that long, 70. Let me draw the resultant.
So this is the resultant and the magnitude of the resultant is going to be the ground speed and the direction of the resultant, this direction angle is going to be the course. So we'll have to find that eventually. But let’s start by trying to find the magnitude of this vector, the ground speed. I’ll call that G. Now I have almost enough information to solve this triangle. I’ve got two sides but I need at least an angle. And I can get one pretty easily if I notice that I can actually find this angle in here.
Remember, the wind direction is 270 degrees, so this is pointing in the 270 degree direction. And this vector is pointing in the 210 degree direction and that means that this angle here is 60 degrees. Now if that’s 60 degrees then this is also 60 degrees, why?
These are alternate interior angles, parallel lines cut by a transversal these angles have to be congruent. So I have 350, 70 and 60 degrees that’s enough to find G using the law of cosines so let’s do that now.
G² is 350² plus 70² minus twice the product 2 times 70 time 350, times cosine of the angle between which is 60. So let me use my calculator for that 350², plus 70² minus 2 times 70 times 350 times cosine 60. I get 102,900. And so G is going to be approximately square root, answer 320.8 miles per hour. So you could see that I actually loses a lot of a speed because of the severe angle here. The wind is kind of blowing it back a little bit, it’s a little bit of a head wind.
So we have a magnitude of 320.8 and that’s going to be our ground speed, now all we have to do is find the course. Now to find the course, I’m going to find this angle here. Once I find that angle, I can add that angle to or subtract that angle rather from the 210 degrees that will give me my course. So how do I find that angle? Law of sines.
I now know this length and the angle opposite. So it gives me the law of sines. Sine theta over the length of the side opposite 70, equals sine of 60 over G. Remember I have G stored in my calculator so I can still use that. That means sine that equals 70 sine 60 degrees over G, which means theta is inverse sine of all this. Inverse sine of 70 sine 60 over G. So let’s calculate that.
Inverse sine 70 sine 60 divided by G. And that gives me 10.9 approximately degrees. 10.9 degrees that’s what this little angle is in here. So to get the final answer for my course, I subtract that from 210. So of course it's approximately 210 degrees minus 10.9 degrees. And that’s going to equal 199.1 degrees.
So the final answer is that the plane is heading at the speed of 320.8 miles per hour, and its course is 199.1 degrees from north.