I need help on all of these problems, sometimes using this website: http://phet.colorado.edu/sims/vector-addition/vector-addition_en.html. The questions are attached below. @zepdrix

- Summersnow8

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- Kash_TheSmartGuy

For the first few questions.

- Kash_TheSmartGuy

https://www.mathsisfun.com/algebra/vectors.html

- Kash_TheSmartGuy

https://www.khanacademy.org/math/precalculus/vectors-precalc

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## More answers

- Kash_TheSmartGuy

http://www.physicsclassroom.com/class/vectors/Lesson-1/Vectors-and-Direction

- Kash_TheSmartGuy

Maybe, those will help^^

- Summersnow8

it didn't

- Summersnow8

##### 1 Attachment

- Summersnow8

attempt at #7

##### 1 Attachment

- Kash_TheSmartGuy

sry, :(

- zepdrix

link is broken :(

- Summersnow8

@zepdrix http://phet.colorado.edu/sims/vector-addition/vector-addition_en.html

- zepdrix

so 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

@zepdrix I did #1, that's as far as I got. Everything else I attempted but didn't work

- zepdrix

Ok 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

does that solve it?

##### 1 Attachment

- zepdrix

Mmm 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

So 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

Do you see where this is going? :)

- Summersnow8

yes, I understand what you are implying

- zepdrix

11 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

okay, haha, yes I understand, now on to the next one :)

- zepdrix

mmmm number 3 should be pretty straight forward.

- Summersnow8

but what would it look like?

- Summersnow8

like this?

##### 1 Attachment

- zepdrix

if 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

Yah I think yours will work also.
But it'll be easier to setup if both tails start from the origin.

- zepdrix

i mean, if both tails are touching.

- zepdrix

i dont really understand how to use the `show sum` tool.
trying to figure that out...

- zepdrix

oh oh ok i get it now

- zepdrix

actually, your example was fine, ya just go with that :)
neeeeext

- zepdrix

On 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

so my example works or no?

- zepdrix

ya

- Summersnow8

okay cool. next

- zepdrix

4 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

when I try to press the show sum button it doesnt come up

- Summersnow8

no, I dont really get 4 either

- zepdrix

|dw:1436388637712:dw|So for problem 3, we did something like this.
Two vectors added together, gave us nothing, zero.

- zepdrix

|dw:1436388727454:dw|

- zepdrix

Now for this next part, we want to add 3 vectors together.
A+B+C and end up with only A.

- zepdrix

So we could draw A ..... anywhere else.|dw:1436388787394:dw|

- zepdrix

If 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.

- Summersnow8

okay.....

- zepdrix

\(\large\rm \color{orangered}{\vec C=-\vec B}\)
Therefore adding some vectors A, B and C:
\(\large\rm \vec A+\vec B+\color{orangered}{\vec C}=?\)

- zepdrix

Can you fill in the blank? :)

- Summersnow8

A?

- Summersnow8

That is how I would describe it?

- zepdrix

Well, 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

oh okay, that makes sense

- zepdrix

how bout 5, figure that one out? :d
comeonnnn that one is eassyyyy :D

- Summersnow8

no......?

- zepdrix

you `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

i gotta go :3 i be back later maybe

- Summersnow8

okay, thanks for the help so far

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