Quick & easy vector question

[Andiio]

My teacher's been away for a few periods so we've had to work on our own for a few lessons. I just have a simple question.

After I drew the vector of '-8i - 6j' or '-5i + 12j', I was told to find both the magnitude and direction.

Once I got to the direction, I realised that I had to subtract the angle obtained from tan(theta) = y/x by 180. But why is this so? I know it has something to do with the fact that the vector is on the negative side of the x axis.. but i'm just blanking out

Thanks in advance!

[98.40_for_sure]

Direction is the angle from the positive direction of the x axis
or simply; this way -->
so use common logic... -8i - 6j is bottom left corner, so it would be 180 + angle
-5i + 12j is in top left corner, thus 180 minus angle

[TrueTears]

dont just do tan(theta) = y/x (this goes for complex numbers too)

sketch the vector, work it out geometrically.

eg, -8i-6j is in the 3rd quadrant.

sketch the triangle and see which angle you need to work out visually.

get into the habit of doing this, it is a bullet proof of way of checking if the angle you got is reasonable.

[Andiio]

dont just do tan(theta) = y/x (this goes for complex numbers too)

sketch the vector, work it out geometrically.

eg, -8i-6j is in the 3rd quadrant.

sketch the triangle and see which angle you need to work out visually.

get into the habit of doing this, it is a bullet proof of way of checking if the angle you got is reasonable.

Is the angle of direction for the vector denoted as the angle between the vector and the x-axis?

[TrueTears]

dont just do tan(theta) = y/x (this goes for complex numbers too)

sketch the vector, work it out geometrically.

eg, -8i-6j is in the 3rd quadrant.

sketch the triangle and see which angle you need to work out visually.

get into the habit of doing this, it is a bullet proof of way of checking if the angle you got is reasonable.

Is the angle of direction for the vector denoted as the angle between the vector and the x-axis?

good question, the convention is unlike that of argument and Argument for complex numbers, there is not set convention (as far as i'm concerned from a mathematical perspective), the question would often say "the direction made with the positive x axis", if it does not say, I would put the direction as the angle made by the vector with the positive x axis.

[Andiio]

Mmm.. i'm still a tad confused about the 180 + angle and 180 - angle concept..

I understand that the x-axis is a direction of 180 degrees, so for -8i - 6j, if you subtracted the angle then.. is it essentially just moving the left side of the angle down?
as in from ----- to  /----?

[98.40_for_sure]

Ok umm... you take your angle anti clockwise from ------> on the x-axis positive direction right?

for -8i - 6j, you know that -8i is left 8 units, and -6j is down 6 units, thus you would have a triangle in the 3rd quadrant (bottom left) that looks something like this:

        8
|----------/|-----------------------
|          o / |
|          /    |
|6      /      |
|      /        |
|    /          |
|  /            |
|/

my theta is "o"

so you're finding o = tan^-1(6/8)
but since you're trying to find the angle from --------------------->
the angle is 2 quadrants + o
= 180 + tan^-1(6/8)

[TrueTears]

the attached graph should unconfuse you.

dont think about this algebraically or mechanically.

visualise it.

the triangle is created from the vector.

the smaller angle is what u need to find.

find it using trig.

then you can see if u need to find the angle with the positive x axis, you'd use 180-smaller angle.


note when you use trig here to find the smaller angle you'd use the magnitude of the sides of the triangle not -8 and -6.

so in fact all questions which ask you to find the direction of vectors can be turned into a geometry question.

[98.40_for_sure]

Sorry truetears... but my drawing just destroys yours

[TrueTears]

lol yeh, its hot

[Andiio]

LOOOL 99.95 i'm surprised at how much effort you put into typing that
thanks guys!

so for -5i + 12j you would move from the positive x axis anticlockwise across quadrants 1 + 2 to get an angle of 180 degrees, then you'd minus the angle obtained from the vector?

[TrueTears]

correct.

[Andiio]

awesome so it's essentially the supplementary angles from the positive x axis which you want to find, yes?

[TrueTears]

if they are in 2nd or 3rd quadrants, then the answer to your question is yes.

if theyre in 1st or 4th, then no. (visualise this, you'll see why)

[98.40_for_sure]

Umm... you're getting caught up in the nitty gritty (i dont know what supplementary means), just visualize it in your head or something, and the angle should make sense you can't have 40 degrees for an obtuse angle or something