u = -4i + 1j and v = 4i + 1j; Find .||u+V|| the answer is u+v=2j

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u = -4i + 1j and v = 4i + 1j; Find .||u+V|| the answer is u+v=2j

Mathematics
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my teacher siad i was wrong
\(\|u+v\|\) represents the length of vector \(u+v\)
what you had there is just the vector \(u+v\) but your teacher wants its length

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Other answers:

ohh ok
how does vector 2j look ?
You are already given where vector u and vector v are pointing at, the distance between these points can be found using distance formula
how do i do that
wait sorry nevermind I messed up, forget that figure ok
for a vector \[\vec a=x \hat i+y \hat j+z \hat k\] It's magnitude is given by \[|\vec a|=a=\sqrt{x^2+y^2+z^2}\] so for \[|\vec u + \vec v|=\sqrt{?^2+?^2+?^2}\]
your vector \[\vec u + \vec v =0\hat i+2\hat j+0\hat k\]
im so counfuzzled this is why i dont do math lol
ok I will try to explain
|dw:1439388825827:dw| \[\vec a=x \hat i+y \hat j\]
You can think of it has moving x amount along x-axis(in the direction of i hat) and moving y units along y-axis (in the direction of j hat)
Do you understand this much?
so far
Now consider the triangle |dw:1439389090592:dw| How can you solve for the question mark?
It's a right triangle
umm i needa find the ? how do i do that
Yep, remember the good ol pythagorus thm?
i do put i never could do it properly
|dw:1439389330955:dw|\[c^2=a^2+b^2\] \[c=\sqrt{a^2+b^2}\]
remember that?
yes i do
Then if your sides are of x and y length, then what will be the "c" in that case?
instead of a and b, we have x and y
you there bro?
im there im thinking
ok
idk lol why am i supose to be doing sorry i have had no sleep
|dw:1439389871725:dw|\[?=\sqrt{x^2+y^2}\] Does that make sense?
yess captin
Ok so that question mark is the hypotenuse of the triangle so if we consider... |dw:1439389973299:dw| Then that question mark indeed tells us the length of the vector
we have to do all this to get to the answer lol i think imma pass out
That is to say for a vector \[\vec a =x \hat i+y \hat j\] It's magnitude is given by \[|\vec a|=a=\sqrt{x^2+y^2}\]
No the question is really short, I'm doing all this so u can understand it !!
ok lol
|dw:1439390177628:dw|
what is all that lol
your vector \[\vec u +\vec v=0 \hat i +2 \hat j\] Compare it to the form \[\vec a = x \hat i +y \hat j\] You find that \[x=0\]\[y=2\] then the magnitude is given by \[|\vec u +\vec v|=\sqrt{x^2+y^2}\]
you just have to take the squares of all the components of vector and add them then take their square root to find the lengh of a vector
i just wanna be done with this question i have three more questions lol
umm me and my tutor got that answer
I'm sorry but there's nothing I can do about that. I've really tried to dumb it down as much as possible, sorry if you can't understand, everyone learns in a different way. If you're not sure about pythagorus theorem, you should clear your concept about that before attempting vectors
oh! lol it was answer of v+u i thought it was the answer of \[\left| v+u \right|\]
yeah
so are we done here or what

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