Question

Find a vector that bisects the smaller of the two angles formed by <4,3> and <9,-40>

I will fan&medal if you can explain how you got the answer, I'm so confused.

Answer 1

Let's name the 2 vectors A<4,3> and B<9,40>, the scalar product would be :
(A,B) = |A||B|cos(x)
So you can find the angle x between A and B from that equation knowing that the modulus is |A|=sqrt(4²+3²)

Answer 2

Then you can find cos(x/2). To find C<a,b> the unit vector bisecting x, you can solve the system sqrt(a²+b²)=1 et (A,C)=|A|*1*cos(x/2).
Good luck !

Answer 3

okay so would it be (4*9)+(3*40)=|<4,3>|*|<9,40>|*cos(x) to solve for angle x?

Answer 4

Please hold on for 5 min I need to check something. I'll walk you through the entire process don't worry.

Answer 5

oky doky, thanks.

Answer 6

Hi again !
Forget what I said before, it's for 3D vectors. For 2D vectors which is the case here, it's much more easier. You find the unit vectors for A and B, and you add them something called the parallelogram rule.
Now for unit vectors here's the formula u=A/|A|

Answer 7

how do you find A? I know vector A=<4,3> but that looks weird <4,3>/|A| unless its <4/|A|,3/|A|>

Answer 8

That's exactly what you said @xlovely_Lizardx
So the unit vector for A would be u=<4/5,3/5>.

Answer 9

okay, makes sense. so how do i use that to find the angle and essentially the bisecting vector?

Answer 10

Well like I said before you find unit vector "v" for B too and you add them like this C=u+v
C would be the vector that bisects the smaller of the two angles formed by A and B.

Answer 11

so we really don't care about the value of the angle? we just add the unit vectors?

Answer 12

Yep !

Answer 13

so \[<\frac{ 4 }{ 5 },\frac{ 3 }{ 5 }>+<\frac{ 9 }{ 41 },\frac{ 40 }{ 41 }>=answer?\]

Answer 14

As easy as that !

Answer 15

i got \[<\frac{ 209 }{ 205 },\frac{ 323 }{ 205}>\] and it said it was wrong

Answer 16

Give me a sec to check your answer

Answer 17

Ah sorry forgot you need to divide by the magnitude :p
Calculate C/|C| that's your answer

Answer 18

I got <0.54784,0.4473> and it's still wrong

Answer 19

Well we used the right method, I don't know what's wrong. Let me call in someone to check @hartnn

Answer 20

I'm not sure the method is right, my professor has never talked about doing it that way before.

Answer 21

Any ideas @hartnn ?

Answer 22

it was also too easy, he likes to challenge us a lot. like an 8 problem homework should take us about 3 hours to finish

Answer 23

what method your prof had taught ?
(by that time i will see the method used above...)

Answer 24

were on dot product stuff right now, there's a lot of different methods he expects us to do, but i've just never seen the way math&ing001 explained.

Answer 25

http://www.physicsforums.com/showthread.php?t=467380 <---this looks more familiar

Answer 26

i'm just not sure how to convert what they did into my problem

Answer 27

did u see this there ?
================================
Find a vector W that bisects the smaller of the two angles formed by 12i+5j and 15i - 8j.

Step 1: rewrite as vector A = <12, 5> and Vector B = <15, -8> and find the unit vector for both A and B.

Unit Vector = A/|A| = (12, 5)/Sqrt(12^2+5^2) = (12/13, 5/13)
Unit Vector B = B/|B| = (15, -8)/Sqrt(15^2+-8^2) = (15/17, -8/17)

Step 2: Plug into the formula (Unit Vector A + Unit Vector B)/2 = W

((12/13)+(15/17), (5/13)-(8/17))/2 = W = <399/442, -19/442>

Read more: http://www.physicsforums.com

Answer 28

you were half way there,
you found the sum, just divide it by 2 :P

Answer 29

so that means i would take the:\[<\frac{ 209 }{ 205 },\frac{ 323 }{ 205 }>\] and divide by 2? or do i still need to find the unit vector of w and then divide by 2?

Answer 30

Nevermind, they're both wrong -_- ughhh

Answer 31

do you guys have any other suggestions? i'm kind of on a time crunch, I have to leave for work in about an hour and i have to have all my homework done before i leave or i wont get credit for it

Answer 32

divide that by 2
\(<\frac{ 209 }{2 \times 205 },\frac{ 323 }{2 \times 205 }>\)

Answer 33

yeah, i did that its not right

Answer 34

strange, that should have been correct...

Answer 35

@ganeshie8 do you have any ideas? we're all stumped

Answer 36

must be a grader problem.. .

Answer 37

is this the question :
Find a vector that bisects the smaller of the two angles formed by <4,3> and <9,40>

?

Answer 38

yeah

Answer 39

wait no its -40

Answer 40

AHHHH that's probs why it was wrong -_- i'm an idiot

Answer 41

lol

Answer 42

lol okie work it again :)

Answer 43

i first thought only sign would change, but no
work it all again

Answer 44

just kidding its still wrong...ughhhh

Answer 45

-77/410 is still wrong ?

Answer 46

i got \[<\frac{ 209 }{ 410},\frac{ -77 }{ 410 }>\] and it was wrong

Answer 47

it says both coordinates are wrong, even the 209/410

Answer 48

if you're trying, also try
209/205, -77/205

Answer 49

well when i did the 209/205, 323/205 it said they were both wrong then too, so the 209/205 is not correct

Answer 50

actually it just accepted it. :0!

Answer 51

lol, thank god!

Answer 52

I hate this homework software, its so touchy and tells you things are wrong when they aren't. it told me 209/205 was wrong before but now its saying its right

Answer 53

forget it, now atleast now we won't forget the method
just add 2 units vectors, thats it!

Answer 54

Lol yeah I'm definitely not going to forget it now, I've redone this problem like 10 times xD I wish my professor would have gone over this method before assigning it.

Answer 55

has he covered rectangular<->polar conversion yet ?

Answer 56

nope

Answer 57

maybe thats how your prof wants you to learn it...the hard way!

Answer 58

while i have you here, another one of the problems i was having a hard time with i have to find the distance of 2 vectors using points. Do you know the distance formula for vectors? or is there such a thing? I can't find it in m notes

Answer 59

it is same as distance between points

Answer 60

does it matter the order of the points due to vectors having a direction?

Answer 61

distance is a scalar quantity
\((a-b)^2 \)
and
\((b-a)^2\)
will give same result

Answer 62

so, no, it doesn't matter.

Answer 63

mkay :D thanks so much for all your help, you too @math&ing001 and @ganeshie8

Answer 64

welcome ^_^