anonymous
  • anonymous
u = -4i + 1j and v = 4i + 1j; Find .||u+V|| the answer is u+v=2j
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
my teacher siad i was wrong
ganeshie8
  • ganeshie8
\(\|u+v\|\) represents the length of vector \(u+v\)
ganeshie8
  • ganeshie8
what you had there is just the vector \(u+v\) but your teacher wants its length

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anonymous
  • anonymous
ohh ok
ganeshie8
  • ganeshie8
how does vector 2j look ?
anonymous
  • anonymous
You are already given where vector u and vector v are pointing at, the distance between these points can be found using distance formula
anonymous
  • anonymous
how do i do that
anonymous
  • anonymous
wait sorry nevermind I messed up, forget that figure ok
anonymous
  • anonymous
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 \[|\vec u + \vec v|=\sqrt{?^2+?^2+?^2}\]
anonymous
  • anonymous
your vector \[\vec u + \vec v =0\hat i+2\hat j+0\hat k\]
anonymous
  • anonymous
im so counfuzzled this is why i dont do math lol
anonymous
  • anonymous
ok I will try to explain
anonymous
  • anonymous
|dw:1439388825827:dw| \[\vec a=x \hat i+y \hat j\]
anonymous
  • anonymous
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)
anonymous
  • anonymous
Do you understand this much?
anonymous
  • anonymous
so far
anonymous
  • anonymous
Now consider the triangle |dw:1439389090592:dw| How can you solve for the question mark?
anonymous
  • anonymous
It's a right triangle
anonymous
  • anonymous
umm i needa find the ? how do i do that
anonymous
  • anonymous
Yep, remember the good ol pythagorus thm?
anonymous
  • anonymous
i do put i never could do it properly
anonymous
  • anonymous
|dw:1439389330955:dw|\[c^2=a^2+b^2\] \[c=\sqrt{a^2+b^2}\]
anonymous
  • anonymous
remember that?
anonymous
  • anonymous
yes i do
anonymous
  • anonymous
Then if your sides are of x and y length, then what will be the "c" in that case?
anonymous
  • anonymous
instead of a and b, we have x and y
anonymous
  • anonymous
you there bro?
anonymous
  • anonymous
im there im thinking
anonymous
  • anonymous
ok
anonymous
  • anonymous
idk lol why am i supose to be doing sorry i have had no sleep
anonymous
  • anonymous
|dw:1439389871725:dw|\[?=\sqrt{x^2+y^2}\] Does that make sense?
anonymous
  • anonymous
yess captin
anonymous
  • anonymous
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
anonymous
  • anonymous
we have to do all this to get to the answer lol i think imma pass out
anonymous
  • anonymous
That is to say for a vector \[\vec a =x \hat i+y \hat j\] It's magnitude is given by \[|\vec a|=a=\sqrt{x^2+y^2}\]
anonymous
  • anonymous
No the question is really short, I'm doing all this so u can understand it !!
anonymous
  • anonymous
ok lol
anonymous
  • anonymous
|dw:1439390177628:dw|
anonymous
  • anonymous
what is all that lol
anonymous
  • anonymous
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 \[|\vec u +\vec v|=\sqrt{x^2+y^2}\]
sohailiftikhar
  • sohailiftikhar
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
anonymous
  • anonymous
i just wanna be done with this question i have three more questions lol
anonymous
  • anonymous
umm me and my tutor got that answer
anonymous
  • anonymous
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
sohailiftikhar
  • sohailiftikhar
oh! lol it was answer of v+u i thought it was the answer of \[\left| v+u \right|\]
anonymous
  • anonymous
yeah
anonymous
  • anonymous
so are we done here or what

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