What is the direction angle of vector <-2,-5>?

AI Thread Summary
The direction angle of the vector <-2,-5> is calculated using the magnitude, which is √29. The angle is initially found to be 111.891 degrees, but the correct direction angle is determined by subtracting this from 360 degrees, resulting in 248.109 degrees. This adjustment accounts for the vector's position in the third quadrant of the coordinate plane. Using the arctan function, one can derive the angle with respect to the negative x-axis and negative y-axis, confirming the need to add 180 degrees when the x-component is negative. Understanding these calculations and adjustments is essential for accurately determining direction angles in vector analysis.
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Homework Statement



Find the direction angle of vector <-2,-5>

Homework Equations



The components of a vector is v=<magnitude*cos ø, magnitude*sin ø>
Magnitude of a vector: √(a^2+b^2)

The Attempt at a Solution



I found the magnitude of the vector, which is √29. Then I set -2= √29 cos ø and solved for ø. It was 111.891 degrees but the answer to the problem was 360 MINUS 111.891 degrees. I thought it would be 180 PLUS 111.891.
 
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Ok you could draw the point in the plane. Make a right triangle with legs 2,5. You could then arcTan of that. That angle is the angle it makes with negative x-axis and negative y-axis. Since you are measuring from the positive x you add 180 to that. That is the angle.
 
360-111.891 is logical because of where on the coordinate plane you measure from.

By the way, welcome to the forums!
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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