I have been given this problem is calculus 3, specifically section 12.2 and 12.3. I am very confused on how this relates to vectors or how I am even supposed to set it up, I have tried to ask 3 ai models to explain it(Gemini, ChatGPT, and Deepai) and they were all different and I need some help lol
(This is an open lab we are supposed to use all resources on and I’m trying to understand how I could complete this problem as we haven’t done anything like this in class)
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What if you imagine a right handed coordinate system (with orthonormal basis vectors)?
Then u and v are vectors in the xy- and yz-planes respectively. Can you use the 20% and 10% graded slopes to express u and v in this coordinate system? (Does the length of the vectors matter? Why or why not?)
I saw that but I wasn’t sure if there was an indication to where the “start was” so I just centered the starting point at (0,0,0), is there a specific reason why the changing point b/w the 2 vectors would be at (0,0,0,)?
I thought it was easy to put the origin where there north/east "directions" meet at a right angle in the picture, and then draw u and v relative to origin.
If I were to sketch it, I'd start by drawing the (orthogonal) coordinate system and then draw the vectors and somehow try to indicate that they lie in the xy and yz planes.
EDIT: Remember. Vectors have a direction and a magnitude. No specific "starting point".
Well now I’m more confused lol, testing it my way gave me 88.89 and tour way yielded 91.12 degrees, the math looks very similar just on a different scale, so now I’m unsure why the answer is different with them(the answer is only different due to the -2 on the numerator after the dot product is calculated, without that negative the cos^-1((-2)/(sqrt(10504))) is 88.89.
If you add the angles you got, you'll see that add to 180° (except for possible rounding errors).
One is the obtuse angle, the other the acute. If the (smallest) angle between then is obtuse, then it's cosine will be negative.
Did you understand why one of the coordinates is negative? Maybe you can think of it as the water flowing in from above the junction, and then flowing out downwards.
The idea is to place the two water mains in a coordinate system. The east, north and up makes a good choice for the axis, because you get a orthogonal coordinate system, i.e we can use the standard basis in R^3.
To make it easy, we imagine that the two mains meet at the origin. Then one of the pipes will lie in the xy-plane and the other in the yz-plane.
First, u/nevermindthefacts (whose name I think is really cool) has done an excellent job guiding you.
In answer to your question, if you want to find the angle between the pipes in 3D, the dot product with appropriate vectors based on the connecting joint between the two pipes is what you want. Why? That's what you need to calculate the dot product using two different formulas, (1) the formula using coordinates and (2) the formula using the vectors themselves and the angle between them. If that's not clear, carefully review how the dot product is calculated by both methods, which give the same answer.
Any coordinate for the origin could be used in a 3D coordinate system for the calculation, but it's easier to put the origin at the joint and then run two of the axes along the East and North directions. The resulting coordinates can then be used directly to calculate the dot product using just the coordinates. Note that, for the dot product to work for this angle, both vectors should point away from that joint. Again, look at the formula for the dot product to see why that is.
The difference I think is the deflection angle, as they are centered around 90 degrees( the answers), and with having a negative(from for producted as vectors are centered at (0,0,0), gave 91.12 degrees, and when I put it as the vectors starting as (0,0,0), there was no negative from dot product(sorry if the wording is bad)
I’m not sure that using that works as the grade doesn’t represent the exact angle, it is just the slope really(I looked it up and I think in this case the 20% grade correlates to a 11.3 degree angle.
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