Representing quantities with vectors | Vectors | Precalculus | Khan Academy
We're told a powerful magnet is attracting a metal ball on a flat surface. The magnet is pulling the ball at a force of 15 newtons, and the magnet is 20 degrees to the south from the eastward direction relative to the ball. Here are a few vectors where the magnitude of vector A is equal to the magnitude of vector C, which is equal to 50 newtons, and the magnitude of vector B is equal to the magnitude of vector D, which is equal to 20 newtons.
Which vectors can represent the force of the team's pull? All right, pause this video and see if you can think about that on your own before we do it together.
All right, now let's do it together. So before I even look at this, I'm just going to look at the description. It has a magnitude of 15 newtons. If we're talking about a force, you could view it as a strength of 50 newtons, and the magnet which is pulling on the ball is 20 degrees to the south from the eastward direction relative to the ball.
So if this is the ball right over here, and if this is the eastward direction, it says the magnet is 20 degrees to the south from the eastward direction relative to the ball. So the magnet would be in this direction, and this angle right over here is 20 degrees. The magnet is pulling on the ball, so the vector would go in that direction towards the magnet, and we know it has a force of 15 newtons. That's the magnitude, so it needs to be a 15 newton magnitude right over here.
So when we look at the choices, choice A is interesting; at least the direction looks right. Looks like it's 20 degrees south of due east, and they also tell us that the magnitude of A is 50 newtons, so I am liking A.
Now let's look at B. Well, B looks 15 degrees south of due east, not 20 degrees south, so I will rule that out. Also, B's magnitude is wrong; it's 20 newtons.
C, the magnitude is right; it's 15 newtons, but the direction looks like 20 degrees north of due east, so I'll rule that one out. And last but not least, D, the direction is clearly wrong; it looks like 15 degrees north of due east, and its magnitude is 20 newtons, not 15 newtons, so I'd rule that one out.
Now, to be clear, a vector is only defined by its magnitude and its direction, not by its starting point. So if I had some other vector that looked like this right over here, that had the same magnitude and direction—if this was right over here, a 20 degree angle, and it had a magnitude of 50 newtons—then I would have selected this one as well.
You can shift a vector around like this as long as it has the same magnitude and it has the same direction, it is an equivalent vector.