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

- anonymous

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

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- anonymous

my teacher siad i was wrong

- ganeshie8

\(\|u+v\|\) represents the length of vector \(u+v\)

- ganeshie8

what you had there is just the vector \(u+v\)
but your teacher wants its length

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## More answers

- anonymous

ohh ok

- ganeshie8

how does vector 2j look ?

- 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

how do i do that

- anonymous

wait sorry nevermind I messed up, forget that figure ok

- 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

your vector
\[\vec u + \vec v =0\hat i+2\hat j+0\hat k\]

- anonymous

im so counfuzzled this is why i dont do math lol

- anonymous

ok I will try to explain

- anonymous

|dw:1439388825827:dw|
\[\vec a=x \hat i+y \hat j\]

- 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

Do you understand this much?

- anonymous

so far

- anonymous

Now consider
the triangle
|dw:1439389090592:dw|
How can you solve for the question mark?

- anonymous

It's a right triangle

- anonymous

umm i needa find the ? how do i do that

- anonymous

Yep, remember the good ol pythagorus thm?

- anonymous

i do put i never could do it properly

- anonymous

|dw:1439389330955:dw|\[c^2=a^2+b^2\]
\[c=\sqrt{a^2+b^2}\]

- anonymous

remember that?

- anonymous

yes i do

- anonymous

Then if your sides are of x and y length, then what will be the "c" in that case?

- anonymous

instead of a and b, we have x and y

- anonymous

you there bro?

- anonymous

im there im thinking

- anonymous

ok

- anonymous

idk lol why am i supose to be doing sorry i have had no sleep

- anonymous

|dw:1439389871725:dw|\[?=\sqrt{x^2+y^2}\]
Does that make sense?

- anonymous

yess captin

- 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

we have to do all this to get to the answer lol i think imma pass out

- 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

No the question is really short, I'm doing all this so u can understand it !!

- anonymous

ok lol

- anonymous

|dw:1439390177628:dw|

- anonymous

what is all that lol

- 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

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

i just wanna be done with this question i have three more questions lol

- anonymous

umm me and my tutor got that answer

- 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

oh! lol it was answer of v+u i thought it was the answer of \[\left| v+u \right|\]

- anonymous

yeah

- anonymous

so are we done here or what

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