Question

Doing Vectors wrong

Answer 1

I have a vector:
\[\begin{bmatrix}
-4 \\
3
\end{bmatrix}\]

I want to turn that into polar co-ordinates. Firstly, I would use \(tan^{-1}\) to find that, right?
\[\theta=tan^{-1}\frac{y}{x}\]\[\theta=tan^{-1}\frac{3}{-4}\]\[\theta=323.40\]

The issue here is that the answer is: \(143.1\) degrees.\
What did I do wrong?

Answer 2

I remember my teacher saying that the equation doesn't know what quadrant the answer is in, so sometimes it might be off by \(90\) or \(180\) or something like that.

Answer 3

The matrix I showed you, @perl

Answer 4

ok that was lag

Answer 5

Oh, lol

Answer 6

Vector:
\[\begin{bmatrix}
-4 \\
3
\end{bmatrix}\]

Answer 7

So, how we do this,@perl ?

Answer 8

\(\theta\) has to be between 90 and 180

Answer 9

you need to notice that -4, 3 means you are in the 2nd quadrant, so subtract 180 degrees because tan inverse gives you stuff between -90 and 90

Answer 10

Oh, OK

Answer 11

SO you would need to find the quadrant, THEN choose if to use 90 or 180 as such?

Answer 12

I'm not sure how to read vectors in matrix form, but isn't vector notation written out as <x,y>? So in this case it would be <-4,3>?

Just clearing up my understanding of it :)

Answer 13

you need to find the quadrant, and note that tan inverse will always give answers in the 4th and 1st quadrant. So if you are not in that one, you need to add or subtract 180 degrees

Answer 14

well this is standard yes, but one would need to say that these vectors are in \(\mathbb{R}^2\)

Answer 15

So if the vector is from the 4rth or 1st quadrant, leave it?

Answer 16

And in second and third you would add or remove>

Answer 17

Right? @zzr0ck3r

Answer 18

correct

Answer 19

So in second quadrant you take away and third quadrant you add 180?

Answer 20

hey

Answer 21

sorry my computer is freezing

Answer 22

i have like a 20 minute lag

Answer 23

so what did you want to know

Answer 24

i lost the ability to draw

Answer 25

Oh. @perl Read my top question

Answer 26

it is because arctan has two solutions

Answer 27

arctan is negative in quadrant 2 and quadrant 4

Answer 28

?

Answer 29

the arctan is dividing a positive by a negative number

Answer 30

arctan doesnt 'know' which quadrants you have in mind

Answer 31

it loses that information, and gives you one answer

Answer 32

So what should I do then?

Answer 33

you have to use information from the problem

Answer 34

?

Answer 35

arctan can be in two different quadrants

Answer 36

ok, if you

Answer 37

do you realize that arctan (-3/4) = arctan (3/-4)

Answer 38

but (-3,4 ) and vector ( 3, -4) are different

Answer 39

Yes

Answer 40

arctan doesnt make a distinction in the input

-3/4 = 3/-4

Answer 41

, arctan is a function machine , it can only give one output for a given input

Answer 42

So what we do then?

Answer 43

you add or subtract pi to your answer, or 180 degrees

Answer 44

When?

Answer 45

In quadrant 2, do we add or subtract 180, @perl ?

Answer 46

you look at your vector

Answer 47

you can do either, you just want to make sure that if you add 180 and go over 360 degrees, that you then subtract 360 degrees so that you are in-between 0 and 360

Answer 48

your vector is (x,y) = (-4,3)

so you are quad. 2

Answer 49

similarly if you go below 0 when subtracting

Answer 50

So you can add or remove 180, and it won't make a difference?

Answer 51

As long it is in Quadrant 2 and 4?

Answer 52

:) rewind

Answer 53

Ok then :)

Answer 54

Just before than, what about quadrant 3?
Negative/Negative?

Answer 55

you should draw the vector on a piece of paper , and then draw the angle

Answer 56

if the angle doesn't make sense, then you know you have add 180 or not

Answer 57

draw the angle you get for your answer

Answer 58

But if it is Quadrant 3? When Negative/Negative?

Answer 59

that is the vector, but your 'solution angle' was what?