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Summersnow8
 one year ago
I need help on all of these problems, sometimes using this website:
http://phet.colorado.edu/sims/vectoraddition/vectoraddition_en.html.
The questions are attached below. @zepdrix
Summersnow8
 one year ago
I need help on all of these problems, sometimes using this website: http://phet.colorado.edu/sims/vectoraddition/vectoraddition_en.html. The questions are attached below. @zepdrix

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Kash_TheSmartGuy
 one year ago
Best ResponseYou've already chosen the best response.0For the first few questions.

Kash_TheSmartGuy
 one year ago
Best ResponseYou've already chosen the best response.0https://www.khanacademy.org/math/precalculus/vectorsprecalc

Kash_TheSmartGuy
 one year ago
Best ResponseYou've already chosen the best response.0http://www.physicsclassroom.com/class/vectors/Lesson1/VectorsandDirection

Kash_TheSmartGuy
 one year ago
Best ResponseYou've already chosen the best response.0Maybe, those will help^^

Summersnow8
 one year ago
Best ResponseYou've already chosen the best response.0@zepdrix http://phet.colorado.edu/sims/vectoraddition/vectoraddition_en.html

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.0so uhhhh which one you up to right now? 0_o by the way, i would recommend using `style 2`, that one makes the most sense.

Summersnow8
 one year ago
Best ResponseYou've already chosen the best response.0@zepdrix I did #1, that's as far as I got. Everything else I attempted but didn't work

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.0Ok let's use trig to explain problem 2, as they have requested. If the `x component` is `twice the size` of the `y component` we can write that relationship like this: \(\large\rm x=2y\)

Summersnow8
 one year ago
Best ResponseYou've already chosen the best response.0does that solve it?

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.0Mmm that's the right idea, yes! But they want this generalized for ANY vector. You showed that it works for a specific set of lengths, 11 and 22.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.0So what we would rather do is call our \(\large\rm \color{orangered}{x=2y}\) and our \(\large\rm y=y\) x is twice the length of y, that's how we got that relationship. So again plug this into your inverse tangent.\[\large\rm \tan^{1}\left(\frac{y}{\color{orangered}{x}}\right)=\tan^{1}\left(\frac{y}{\color{orangered}{2y}}\right)\]

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.0Do you see where this is going? :)

Summersnow8
 one year ago
Best ResponseYou've already chosen the best response.0yes, I understand what you are implying

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.011 divided by 22 can be simplified to 1/2. The same thing will happen with our setup here! We can cancel the y's and see that the argument simplifies to 1/2. Again, you'll get 26.6 degrees, but now we've generalized it. yay team

Summersnow8
 one year ago
Best ResponseYou've already chosen the best response.0okay, haha, yes I understand, now on to the next one :)

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.0mmmm number 3 should be pretty straight forward.

Summersnow8
 one year ago
Best ResponseYou've already chosen the best response.0but what would it look like?

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.0if i have some vector V and then I add to that some vector V, they'll give me 0, ya? So draw a vector going off in some direction. And draw another vector going in the opposite direction. Illustrated here as an example:dw:1436388162556:dw

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.0Yah I think yours will work also. But it'll be easier to setup if both tails start from the origin.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.0i mean, if both tails are touching.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.0i dont really understand how to use the `show sum` tool. trying to figure that out...

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.0actually, your example was fine, ya just go with that :) neeeeext

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.0On your picture that you posted, if you click the "show sum" button, you should notice that nothing happens. That's because the sum is zero, so there is nothing to display

Summersnow8
 one year ago
Best ResponseYou've already chosen the best response.0so my example works or no?

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.04 is the same idea... this should be pretty straight forward. We just showed that we can add two vectors to get zero. So now we want to add three vectors and end up with only the first one...... meaning that `the other two vectors added to zero`. ya? ;o

Summersnow8
 one year ago
Best ResponseYou've already chosen the best response.0when I try to press the show sum button it doesnt come up

Summersnow8
 one year ago
Best ResponseYou've already chosen the best response.0no, I dont really get 4 either

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.0dw:1436388637712:dwSo for problem 3, we did something like this. Two vectors added together, gave us nothing, zero.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.0Now for this next part, we want to add 3 vectors together. A+B+C and end up with only A.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.0So we could draw A ..... anywhere else.dw:1436388787394:dw

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.0If they want us to "describe it" though, instead of graphing it... let's umm.... Let's just say that vector C is the negative of vector B.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.0\(\large\rm \color{orangered}{\vec C=\vec B}\) Therefore adding some vectors A, B and C: \(\large\rm \vec A+\vec B+\color{orangered}{\vec C}=?\)

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.0Can you fill in the blank? :)

Summersnow8
 one year ago
Best ResponseYou've already chosen the best response.0That is how I would describe it?

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.0Well, don't forget the middle step though.\[\large\rm \vec A+\vec B+\color{orangered}{\vec C}=\vec A+\vec B+\color{orangered}{\vec B}\]Which yes, you can go a little further and say,\[\large\rm =\vec A+\vec 0=\vec A\]Because B and B cancel out.

Summersnow8
 one year ago
Best ResponseYou've already chosen the best response.0oh okay, that makes sense

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.0how bout 5, figure that one out? :d comeonnnn that one is eassyyyy :D

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.0you `could` get all fancy and try to make the y's add up to zero. That seems tedious though. let's just make it really easy on ourselves and draw horizontal lines. Example: Let's say I want three vectors to add up to `20 in the x direction` `0 in the y direction`. Well I could do something like this:dw:1436389431399:dw\[\large\rm \vec A+\color{royalblue}{\vec B}+\vec C=5x+0y+\color{royalblue}{5x+0y}+10x+0y\]\[\large\rm =20x+0y\]

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.0i gotta go :3 i be back later maybe

Summersnow8
 one year ago
Best ResponseYou've already chosen the best response.0okay, thanks for the help so far
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