The sum of two vectors given magnitudes and included angle

AI Thread Summary
To find the sum of two vectors given their magnitudes and the angle between them, the law of cosines is applied. For vectors u and v with magnitudes of 15 and an included angle of 116 degrees, the resultant magnitude is calculated as approximately 15.9, with a direction angle of 58 degrees. In a second example with vectors of magnitudes 6 and 11 and an angle of 42 degrees, the calculated resultant is about 16, but there is confusion regarding the angle, as the answer key suggests 27 degrees instead of the calculated 21 degrees. The discussion highlights the importance of correctly applying trigonometric principles, particularly when the magnitudes of the vectors differ. Accurate vector addition requires careful consideration of angles and the law of cosines to avoid discrepancies in results.
Aaron H.
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Homework Statement



Given the magnitudes of vectors u and v and the angle θ between them, find sum of u + v. Give the magnitude to the nearest tenth when necessary and give the direction by specifying the angle that the resultant makes with u to the nearest degree.

Homework Equations



|u| = 15, |v| = 15, θ = 116°

The Attempt at a Solution



Knowing only the answer (15.9, 58°) and some trig ideas:

I draw an angle of 116 degrees in the starting point of the trig plane. I drop a line from the angle end side, forming a triangle with a 64 degree angle in quadrant II. The other angles of the triangle are both 58 degrees (116 deg / 2). Both opposite (U) and adjacent (V) sides are 15.

SAS - law of cosines

c^2 = (15)^2 + (15)^2 - 2 (15)(15) cos (64 deg)

c = 15.9

angle = 58 deg

(15.9, 58 deg)


What is the correct method?
 
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I think that is the correct method. Well done! Draw the quadrilateral that the two vectors and the sum make and use trig.
 
Can you verify this?

|u| = 6, |v| = 11, θ = 42

I drew the quadrilateral. From this, 138 deg captures (u + v). Law of cosines using 6, 11, and 138 deg. c = u + v = 15.97. Round to 16. 42 bisected = 21. Calculated answer (16, 21 deg), however, the answer key presents (16, 27 deg). I have no idea how that angle can be 27 so that must be a mistake.
 
Aaron H. said:
Can you verify this?

|u| = 6, |v| = 11, θ = 42

I drew the quadrilateral. From this, 138 deg captures (u + v). Law of cosines using 6, 11, and 138 deg. c = u + v = 15.97. Round to 16. 42 bisected = 21. Calculated answer (16, 21 deg), however, the answer key presents (16, 27 deg). I have no idea how that angle can be 27 so that must be a mistake.

In your last example you had |u|=|v| so you could just bisect the angle. Here |u| and |v| are different. So the angle of u+v won't bisect the angle. You'll have to do a little more trig to find the right angle. You'll want to find the angle between the sides that are 6 and 16 in the triangle whose sides are 6, 11 and 16.
 
Got it, thanks. The angles of the 6-11-16 triangle are 138 deg, 15 deg, and 27 deg. I used the law of cosines. 27 deg is the angle between u and u + v.
 
As written there are an infinite number of answers since u + v is a vector sum. Are you sure it isn't |u + v|?
 
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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