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## Precalculus

### Course: Precalculus>Unit 6

Lesson 6: Direction of vectors

# Direction of vectors from components: 3rd & 4th quadrants

Sal finds the direction angle of a vector in the third quadrant and a vector in the fourth quadrant.

## Want to join the conversation?

• How do we get tan to the power -1? • So the basic rule of this and the previous video is:
?? • Is there any way to find out the inverse tangent, sine, and cosine by hand?
Are there any methods? • Yes, but the math is too advanced for this level of study.
For example, here is the formula for the inverse sine of x (using radians, not degrees):
sin⁻¹ x = − i * ln [i x+√(1-x²)]
You will not be expected to do this kind of math, but you will be expected to memorize the inverse functions of the special angles.
• In Quadrant 3, is it possible to find the angle inside the triangle, and then subtract it from 270? Will that method also work? • No, you can't... when dealing with angle operations along the y-axis (90,270) you convert the sign to its complementary: sin <|> cos, tan <|> cot, but when you perform operations along the x-axis (180,360) you just change the sign, preserve the function type... I recommend you watching Trigonometry videos for further explanation... it all comes out of similarity... I hope this helps if you haven't figured it out by now :)
• At , what is the point of writing the vector as (-2i - 4j)? Because writing it as (-2, -4) is the same thing, except without the useless letters...? • Why write a number such as 345 as 3.45 x 10²? The latter is scientific notation - it has its place.
Why write a vector, such as (2, 4) as 2i + 4j? The latter is engineering notation - it has its place.

Some conventions may seem pointless to you now, but if you ever get into the areas they are used, they will make total sense.
• Why in 2nd & 3rd quadrant, we add 180 degrees to the angle? and why in 4th quadrant, we add 360 degrees?

How do we know that when we should add 180 and 360 degrees to get the correct angle of the vector?

Somebody pls clarify it:(
(1 vote) • Taking the inverse tangent of the ratio of sides of a right triangle will only give results from -90 to 90, so you need to know how to manipulate the answer, because we want the answer to be anywhere from 0 to 360.

if both coordinates are positive, you are fine, you will get the right answer.

if both are negative, so in quadrant 3, you are taking the inverse tangent of a fraction with a negative numerator and denominator so it would be positive. If you try a vector like 2i + 3j and then -2i - 3j, you'll get the same answer. So you need to realize the tangent and angle is the same as the tangent of 180 plus that angle.

Now, if one is positive and one is negative that puts it in either quadrant 2 or 4. In both cases you are taking the inverse tangent of of a negative number, which gives you some value between -90 and 0 degrees. So for all positive ratios you take the inverse tangent of the result is between 0 and 90.

Now, if you have a positive x value and negative y value, so quadrant 4, the answer is technicallyc correct. Most answers want the value between 0 and 360, so you need one more full revolution to get it there.

Quadrant 2 meanwhile has the same logic as quadrant 3 from before.

If it helps lets use the coordinates 2i + 3j again. This makes a triangle in quadrant 1. if you used -2i + 3j it makes the same triangle in quadrant 2. The relevant angle is obviously 180 minus that angle, I will call x. Using tangent you get -x so you add 180, which is the same as 180 - x

-2i - 3j makes the same triangle in quadrant 3 where the relevant angle is 180 + x

So that means if you take the tangent of a vector in quadrant 2 or 3 you add 180 to that.

If you have -2i - 3j then you have the same triangle in quadrant 4. You could look at the relevant angle as -x or 360 - x, the 360 - x is more useful. Taking the inverse tangent gets you -x again, so adding 360 to it puts it at the appropriate range of numbers.

I really really hope that helped, if not though let me know.
• In the 3rd qudrant, I did tan(270-theta) = 4/2

=> theta = 206.56 degrees. This answer isn't the same as Sal who calculates it as 243.4 degrees. Can somebody help me here? • In the 'Direction of vectors' videos we are only dealing in two dimensions, so it is easy to visualise. However, with three dimensions or higher we might not be able to determine whether the tan result is correct by visual inspection. Will the rules of adding 180 and 360 still hold at these higher dimensions? Do we apply the same thinking at higher dimensions or rely on something else entirely?   