## NotTim Group Title Consider u=[2,3] and v=[-4,-6]. Find 2 unit vectors parallel to u. 2 years ago 2 years ago

1. NotTim Group Title

I THINKING OF treating it as a parallel thing, but that's not correct, is it?

2. TuringTest Group Title

${\vec u\over||\vec u||}$makes it a unit vector, did you know that?

3. NotTim Group Title

what's that.

4. TuringTest Group Title

dividing by the norm/magnitude of the vector

5. FoolForMath Group Title

Unit vector parallel to $$\vec{u}$$ is : $$\frac{1}{\sqrt{13}} <2,3>$$

6. NotTim Group Title

ok... I don't understand, but I don't think I really need further explaination @TuringTest

7. NotTim Group Title

Also, what did you type FFM? That looks...strange

8. FoolForMath Group Title

But there is one one unique vector (parallel) isn't ?

9. NotTim Group Title

uhhh...i dunno. Was just gunna treat this as collinear. NEver saw how to deal with parallels in my txtbook

10. TuringTest Group Title

FFM I think is pointing out that the other, $$\vec v$$, is antiparallel I suppose... I just see it as i.e. magnitude is the length of the vector$\vec u=<2,3>$ and the magnitude is$||\vec u||=\sqrt{2^2+3^2}$$$\vec v$$ is parallel to $$\vec u$$ so you should be able to do the same trick there

11. NotTim Group Title

I think the equation editor is messing around with us. I see squares everywhere. Also, maybe I should skip to the chase. How would I get [0.55,0,83],[-0.55,-0.83]?

12. TuringTest Group Title

what is the magnitude of u ?

13. NotTim Group Title

[2,3]

14. TuringTest Group Title

no, read what I wrote above magnitude is the $$length$$ of the vector, which we can find with pythagorus$||\vec u||=\sqrt{2^2+3^2}$ and the magnitude is

15. TuringTest Group Title

to amke a vector into a unit vector, divide by its magnitude

16. TuringTest Group Title

*make

17. NotTim Group Title

aww...what exactly is a uni vector. I though the point [2,3] was a vector itself...

18. TuringTest Group Title

A unit vector is a vector that has a length (i.e. "magnitude) of 1

19. NotTim Group Title

ok. um. gimme a moment to do the math here

20. TuringTest Group Title

so a vector $$\vec u=<x,y>$$ is a unit vector iff$||\vec u||=\sqrt{x^2+y^2}=1$after dividing a vector by its magnitude, it should make sense that it will become a unit vector (you are essentially dividing a vector by it's length, which will logically give 1)

21. NotTim Group Title

wait, how do i divide [2,3] by sqrt 13?

22. TuringTest Group Title

do each component individually: it is a scalar multiplication

23. NotTim Group Title

im really fuzzy on this. I divide 2 by sqrt 13, and then 3 by sqrt 13 right?

24. TuringTest Group Title

yep, and those will be you new components

25. NotTim Group Title

umm.srry. pencil is breaking on me here...

26. TuringTest Group Title

try a calculator ;)

27. NotTim Group Title

im going to get a new pencil. bak in a flash

28. TuringTest Group Title

${2\over\sqrt{13}}$just requires a calculator!!!!

29. NotTim Group Title

argh. srry. wat nxt now

30. NotTim Group Title

got [0.55,0.83]

31. TuringTest Group Title

hooray

32. TuringTest Group Title

now for the next one, do the exact same trick ...or use a shortcut (which takes a little explanation) and get the answer in 2 seconds whichever you prefer

33. NotTim Group Title

I guess you don't wanna explain both methods?

34. TuringTest Group Title

well they are really based on the same idea: multiplying a vector be a scalar will leave it parallel to the original...

35. NotTim Group Title

wat do yu mean "exact same trick"

36. TuringTest Group Title

actually that was a poor choice of phrasing; upi need to regognize what I'm about to say either way: let's look at $$\vec u$$ and $$\vec v$$|dw:1333605852213:dw|let's say that $\vec v=c\vec u$then what does $$\vec v$$ look like?

37. TuringTest Group Title

let $$c=2$$ and we double the length of $$\vec u$$:|dw:1333606018155:dw|notice that they remain parallel

38. TuringTest Group Title

here we are doing the same thing: $$\vec v$$ is a scalar multiple of $$\vec u$$ with $$c=-1$$ this will change the direction of the vector (hence the term :anti-parallel")

39. NotTim Group Title

so wait, v and u are actually related? im suppose to use them both? i was just thinking of using u and random other vectrs

40. NotTim Group Title

oohhh...i see. scalar multitplier of -2.

41. TuringTest Group Title

|dw:1333606190135:dw|yeah, sorry my drawing is wrong, it should be -2 but note: that won't change the $$unit$$ $$vector$$ because all unit vectors have length 1 so all we need to do is notice that $$\vec v$$ is a scalar multiple of $$\vec u$$, so it has the same unit vector, but the - sign changes the direction it is still parallel though, in the sense that they never intersect

42. NotTim Group Title

ok.

43. NotTim Group Title

so, v is parallel to u. actually. wait. we already got the answer. thats magical.

44. TuringTest Group Title

...another way to say it is that a unit vector has only one parallel vector to it: it's negative therefor no calculation is really needed. it's better to understand it the first way though, in my opinion

45. TuringTest Group Title

yeah you got it :)

46. NotTim Group Title

wait...what's the "first" way.

47. TuringTest Group Title

that $$\vec v$$ is a (negative) scalar multiple of $$\vec u$$, and therefor has the same unit vector in the opposite direction

48. NotTim Group Title

does that mean its parallel to u?

49. TuringTest Group Title

though sorry, now that I reread the question v isn't even important

50. TuringTest Group Title

but yes, as I said v is a scalar multiple of u when one vector is a scalar multiple of the other, they are parallel

51. NotTim Group Title

yeah. i kidna just included every thing related to the question. given the chance, i would include the background info by the text too

52. TuringTest Group Title

scalar multiples only change the length of the vector

53. TuringTest Group Title

hence they do not change whether or not they are parallel

54. TuringTest Group Title

but it only asks for two unit vectors perpendicular to $$\vec u$$, so $$\vec v$$ has nothing to do with it; it's coincidentally parallel but is not needed to find the answer

55. NotTim Group Title

so vector, wit ha scalar multiple of the original vector, is not actually parallel to it? man. we've taken up quite some time with this.

56. TuringTest Group Title

unit vectors parallel to*

57. TuringTest Group Title

I repeat: scalar multiples only change the length of the vector, hence they do not change whether or not they are parallel so in$c\vec u=\vec v$ where $$c$$ is a scalar, $$\vec u$$ and $$\vec v$$ are parallel

58. NotTim Group Title

is there a situation when the vectors are not parallel, but are only different due to a scalar multiplier. many question i give

59. TuringTest Group Title

no, unless there is some huge exception that is slipping my mind...

60. TuringTest Group Title

scalar multiplication of a vector changes only $$direction$$ and/or $$magnitude$$ (i.e. length)

61. NotTim Group Title

so i pick, say [3,2,6]. Then it is parallel to [6,4,12]?

62. TuringTest Group Title

yes. in higher math, like linear algebra, you will learn that vectors that can be written as scalar multiples of each other are called "linearly dependent" that means they are parallel, which is very important in setting up coordinate systems and such, which is what you do in linear algebra

63. TuringTest Group Title

look at it like slope

64. NotTim Group Title

ok. thank you very much. all my questions have been cleared up, im not done my review, and i wish to take a nappy.

65. TuringTest Group Title

ok then, glad I could help rest well, a weary mind is not terribly useful in mathematics ;)

66. NotTim Group Title

should i sleep now, and risk not studying later, or study, but not sleep ever?