INSTRUCTOR: "An airplane flies due north at an air speed of 550 miles per hour. The wind is blowing from the northwest at 50 miles per hour. What is the ground speed of the airplane?"

Let's begin by noticing what all the information is that we have. We have an airplane flying due north at 550 miles per hour. We can represent it as a vector. I'm going to draw just an initial point and going straight north.

That's 90 degrees with the x-axis. Or we could just say that's traveling along the positive y-axis with a magnitude of 550 miles per hour. That's just how our plane is traveling.

Now, the wind blowing from the northwest at 50 miles per hour means I'm going to have-- or the way I'm going to draw this and represent this is that I will begin at the origin, be going down and to the right. That is into the fourth quadrant. That's where the northwest comes in. And we have a magnitude of 50 miles per hour.

Now, in component form, in standard unit vector notation, I should say-- let's use the standard unit vectors. I'm going to have that our plane-- I'm going to represent that as the vector p-- is 550j because there is no x component to that. It's just going due north. And the wind, I'm going to represent that as the vector w.

And we could write this as, well, let's see, our magnitude is 50. So we're going to write 50 multiplied by cosine of, let's see, what angle would that be? That would be an angle of 315 degrees.

OK, and, again, for reference, that would be a 45-degree angle with the positive x-axis going down, OK? So as a reference angle, we're going to use 45 degrees when we go to calculate this. But it would be 315 degrees from the positive x-axis. Or I guess I should have said or minus 45 degrees, rotating the opposite way.

So 50 times cosine 315 degrees i. OK, and then we have minus 50. And then that would be a sine of 315 degrees. OK, actually, I should probably write that just as plus 50 sine 315 degrees j.

But the sine of 315 is going to be negative. So I want to be careful with that. All right, so the vector w for our wind is then equal to 25 square root of 2 i minus 25 square root of 2 j.

Now, since we are concerned with the effect of the wind on the plane, we are going to find next is the vector p plus the vector w. If we add those two vectors together, noting that we do not have an x component or i component to the vector p, this will be 550j. So we'll combine it with the last term so that we are going to have the vector 25 square root of 2 i plus, in parentheses, 550 minus 25 root 2, close parentheses, j.

So that we can find the ground speed, we want to find the magnitude of this vector, so the magnitude of vector p plus vector w. We'll square each of those components and then take the square root of that sum. So this will be the square root of 25 root 2 squared plus, in parentheses, 550 minus 25 root 2, close parentheses, squared.

And when we calculate that, we will get something that is approximately 515.86 miles per hour. So we can sum this up by saying that as a result of the wind, the plane is traveling at approximately 516 miles per hour relative to the ground. And that answers our question.