## anonymous one year ago u = -4i + 1j and v = 4i + 1j; Find .||u+V|| the answer is u+v=2j

1. anonymous

my teacher siad i was wrong

2. ganeshie8

$$\|u+v\|$$ represents the length of vector $$u+v$$

3. ganeshie8

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

4. anonymous

ohh ok

5. ganeshie8

how does vector 2j look ?

6. 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

7. anonymous

how do i do that

8. anonymous

wait sorry nevermind I messed up, forget that figure ok

9. 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 $|\vec u + \vec v|=\sqrt{?^2+?^2+?^2}$

10. anonymous

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

11. anonymous

im so counfuzzled this is why i dont do math lol

12. anonymous

ok I will try to explain

13. anonymous

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

14. 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)

15. anonymous

Do you understand this much?

16. anonymous

so far

17. anonymous

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

18. anonymous

It's a right triangle

19. anonymous

umm i needa find the ? how do i do that

20. anonymous

Yep, remember the good ol pythagorus thm?

21. anonymous

i do put i never could do it properly

22. anonymous

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

23. anonymous

remember that?

24. anonymous

yes i do

25. anonymous

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

26. anonymous

instead of a and b, we have x and y

27. anonymous

you there bro?

28. anonymous

im there im thinking

29. anonymous

ok

30. anonymous

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

31. anonymous

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

32. anonymous

yess captin

33. 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

34. anonymous

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

35. anonymous

That is to say for a vector $\vec a =x \hat i+y \hat j$ It's magnitude is given by $|\vec a|=a=\sqrt{x^2+y^2}$

36. anonymous

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

37. anonymous

ok lol

38. anonymous

|dw:1439390177628:dw|

39. anonymous

what is all that lol

40. 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 $|\vec u +\vec v|=\sqrt{x^2+y^2}$

41. anonymous

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

42. anonymous

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

43. anonymous

umm me and my tutor got that answer

44. 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

45. anonymous

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

46. anonymous

yeah

47. anonymous

so are we done here or what