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Course: Precalculus (Eureka Math/EngageNY)>Unit 2

Lesson 6: Topic D: Lessons 23-24: Why are vectors useful?

Vectors word problem: tug of war

Sal solves a word problem where three vector forces act upon a ring in different magnitudes & directions. Created by Sal Khan.

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• It seems to me that while Sal's answer is correct in that Team B won, his logic is not. The winning team was not supposed to be just who pulled the hardest (the magnitude of their vector), but whose area the ring ended up in. If Team B had pulled with the same magnitude, but in a direction closer to the X axis, they might have lost to team C. Am I missing something?
• You are right, just from the magnitudes it's not possible to state which team would have won. After Sal sums the 3 vectors, we can see that the resulting vector `-i + j` is the resulting force on the ring, and from that we can say for certain that the ring will move into the area of team B.

In fact, calculating the individual magnitudes of each force is completely useless for this problem.
• At when Sal does the final calculation, why doesn't he switch to radian mode?
• bellmanwells's comment is correct. The choice between degree mode and radian mode only determines how the trigonometric functions interpret their argument, or, in then case of the inverse trigonometric functions, how they return their result. Sal is just performing ordinary arithmetic operations, not trigonometric functions. The fact that the operations are on something whose units are in radians doesn't matter.
• When deciding which team would win wouldn't it be simpler to just add the i and j components and find what quadrant the point ends up in? I know it works in the example video but I can't think of why it wouldn't apply in other examples.
• I find the second question confusing. I interpreted it as the sum of all the vectors' magnitudes, so my math would've looked like: 4 + √20 + √18. If you are adding magnitudes, there should be no negative numbers, so why does he add it like that?
• Sal took the sum of all three vectors. The sum of vectors is not the sum of their magnitudes! This is easy to see because first and foremost, the sum of vectors must be another vector, but if you are just adding magnitudes (scalar quantities) together you end up with a scalar!

So Sal basically took the vector components of each vector and add them up. In particular, he had:
𝐚 = 4ℹ + 0𝐣
𝐛 = -2ℹ + 4𝐣
𝐜 = -3ℹ – 3𝐣
So to add all three vectors, he simply added all three of these equations:
𝐚 + 𝐛 + 𝐜 = 4ℹ – 2ℹ – 3ℹ + 0𝐣 + 4𝐣 – 3𝐣
The key here is to recognize ℹ and 𝐣 as vectors, namely the standard basis unit vectors in ℝ². So on the LHS, we are adding vectors, and on the RHS, we are adding vectors that make up the summands on the LHS so they are both equal! This is like saying:
2 + 5 + 7 = (1 + 1) + (2 + 3) + (6 + 1)
We just split each number on the LHS into parts all of which we add on the RHS. With vectors, this is the easiest way we can add them because we can only add horizontal components with themselves and vertical components with themselves (you cannot add ℹ terms to 𝐣 terms).
Comment if you have questions!
• When would the magnitude of a vector be negative?
• The magnitude of a vector is always positive. Remember, magnitude of the vector a (a1,a2,a3...an) is calculated using sqrt((a1)^2+(a2)^2+(a3)^2+...+(an)^2). Since there is a square root and a bunch of squares, the magnitude will always be positive.
• How can the magnitude of combined forces be smaller than the magnitude of each force?
• Think of it this way. (This example does not correlate to the video, but it will help explain vector addition and subtraction) Let's say each vector is equal to the distance and direction that you walked in a field. One was 3 feet north, the second was 3 feet south, and the other was 2 feet east. The idea behind adding vectors is that you get the displacement, not the total distance. So if you were to walk those directions and distances, your end point would be 2 feet east because 3 feet north and 3 feet south cancel each other out. What Sal is doing is basically the same thing I just did but just more complicated. The different magnitudes are cancelling each other out because they are pulling in opposite directions. The magnitude at the end is smaller because that is what is not cancelled out. I hope my explanation helps!
• In what direction is the ring getting pulled?Isn't the ring getting pulled in the direction of vector b?Because the force exerted by vector b is the strongest,the ring should be getting pulled in it's direction which can be obtained by taking the tan of y and x components of vector b.
Instead why is Sal calculating the direction angle from the combined force?
• We have to include all forces when calculating the movement of the metal ring. The ring will move in the direction of vector b only if neither team A nor team C are pulling in a different direction. Each team exerts its own force, and the ring moves according to the combined force.
• Why didn't Sal have to change "degrees" to "radians" on the calculator when he was trying to find out the direction of the vector in radians?
• That's a good question.
When you do a calculation involving π there is generally no reason to specify radians.
He is not using a trig function like sin, cos, tan etc in this case.
π is always just 3.14.... and so 2π is 6.28... and π/2 = 1.57...

It's when using the trig functions sin, cos, tan etc. that you really need to be careful because you will get completely different answers depending on your degree/radians setting.

For example
sin π = 0
In degrees:
sin π = sin 3.14 = 0.055
• Wouldn't the combined magnitude of the teams' forces be equal to 4+√20+√18?
:/